ELECTRICITY

Pre-requisite : Before going through this chapter, you
should be thorough with the basic concepts of the
chapter explained in X NCERT.
 = Permittivity of the medium
and  = 0r
r = relative Permittivity or dielectric constant of the medium.
Coulomb’s law is based on physical observation and
it is not logically derived from any other concept.
CONSERVATION OF ELECTRIC CHARGE
Whenever two bodies are charged by rubbing, one gets
positively charged and the other gets negatively
charged. The net charge on the two bodies, however,
remains zero–the same as that before rubbing. In other
words, charge is conserved. It can neither be created
nor be destroyed. The only thing that happens on
rubbing is that charged particles (electrons) get
transferred from one body to the other.
In some phenomena, charged particles are created.
But even then the conservation of charge holds. For
example, a free neutron converts itself into an electron
and the proton taken together is also zero. So, there is
no change in the conversion of a neutron to an electron
and a proton.
COULOMB’S LAW
Charles Augustine de Coulomb studied the interaction
forces of charged particles in detail in 1784. He used a
torsion balance. On the basis of his experiments he
established Coulomb’s law. According to this law the
magnitude of the electric force between two point
charges is directly proportional to the product of the
magnitude of the two charges and inversely
proportional to the square of the distance between
them and acts along the straight line joining the two
charges.
In mathematical terms, the force that each of the two
point charges q1 and q2 at a distance r apart exerts on
the other can be expressed as–
F= k
r2
This force is repulsive for like charges and attractive
for unlike charges.
1
4  0 ,
here 0 is absolute permittivity of free space.
The force is directed along the line joining the centres
of the two charged particles.
For any other medium except air, free space or vacuum
coulomb’s law reduces to
F
1 q1q2
=
4 r 2
1.
Find out the electrostatics force between two point
charges placed in air (each of +1 C) if they are
separated by 1m .
Sol. Fe =
kq1q2
r
=
2
9  10 9  1 1
= 9×10 9 N
12
Note : From the above result we can say that 1 C
charge is too large to realize. In nature, charge is
usually of the order of C
2.
A particle of mass m carrying charge q 1 is revolving
around a fixed charge –q 2 in a circular path of radius
r. Calculate the period of revolution and its speed
also.
4 2mr
1 q1q2
2
Sol.
= mr =
'
4 0 r 2
T2
T2 =
or
( 4 0 )r 2 ( 4 2mr )
q1q2
 0mr
q1q2
T = 4r
and also we can say that
q1q 2
4 0r 2
=
mv 2
r
V=

q1q2
4 0mr
PROPERTIES OF ELECTRIC FIELD INTENSITY
q1q2
Where k is a constant of proportionality. k =
ILLUSTRATIONS
(i) It is a vector quantity. Its direction is the same
as the force experienced by positive charge.
(ii) Electric field due to positive charge is always
away from it while due to negative charge always
towards it.
(iii) Its S.. unit is Newton/Coulomb.
(iv) Electric force on a charge q placed in a region
of electric field at a point where the electric field



intensity is E is given by F  qE .
Electric force on point charge is in the same
direction of electric field on positive charge
and in opposite direction on a negative charge.
PAGE # 1 1
(v) It obeys the superposition principle, that is, the
field intensity at a point due to a system of charges is
vector sum of the field intensities due to individual
point charges.
4.

Sol. As force on a charge q in an electric field E is
 


E  E1  E 2  E 3 + .....


F q = qE
A
Five point charges, each of value q are placed on five
vertices of a regular hexagon of side L. What is the
magnitude of the force on a point charge of value – q
coulomb placed at the centre of the hexagon?


| Fq |  | W |
i.e., E =
L
E
q
|q|E = mg
W
mg
= 10 N/C., in downward direction.
|q|
The position where the resultant force on a charged
particle becomes zero is called equilibrium position.
O
q
F
i.e.,
E
q
ELECTROSTATIC EQUILIBRIUM
D
q
Fe
So according to given problem
(vi) It is produced by source charges. The electric
field will be a fixed value at a point unless we change
the distribution of source charges.
3.
Calculate the electric field intensity which would be
just sufficient to balance the weight of a particle of
charge –10 c and mass 10 mg.
q
(a) Stable Equilibrium :
C
-q
A charge is initially in equilibrium position and is
displaced by a small distance. If the charge tries to
return back to the same equilibrium position then this
equilibrium is called position of stable equilibrium.
q
A
B
(b) Unstable Equilibrium :
If charge is displaced by a small distance from its
equilibrium position and the charge has no tendency
to return to the same equilibrium position. Instead it
goes away from the equilibrium position.
Sol. If there had been a sixth charge +q at the remaining
vertex of hexagon force due to all the six charges on –
q at O would be zero (as the forces due to individual
charges will balance each other), i.e.,
FR  0

(c) Neutral Equilibrium :
If charge is displaced by a small distance and it is still
in equilibrium condition then it is called neutral
equilibrium.

Now if f is the force due to sixth charge and F due to
remaining five charges.
 
F + f =0
or, |F| = |f| =
i.e.


F =–f
1 q2
1 q q
=
4 0 L2
4 0 L2
1 q2


=
=
FNet
FCO 4  2 along CO
 L
5.
Two equal positive point charges 'Q' are fixed at points
B(a, 0) and A(–a, 0). Another test charge q0 is also
placed at O(0, 0). Show that the equilibrium at 'O' is
(i) stable for displacement along X-axis.
(ii) unstable for displacement along Y-axis.
Sol. (i)




Initially FAO + FBO = 0  | FAO | = | FBO | =
KQq0
a2
When charge is slightly shifted towards + x axis by a
small distance x, then.


| FAO | < | FBO |
PAGE # 2 2
Therefore the particle will move towards origin (its
original position) hence the equilibrium is stable.
(ii) When charge is shifted along y axis
(ii) The electric intensity at a point is the number
of lines of force streaming through per unit area
normal to the direction of the intensity at that point.
The intensity will be more where the density of
lines is more.
After resolving components net force will be along y
axis so the particle will not return to its original
position so it is unstable equilibrium. Finally the
charge will move to infinity.
ELECTRIC LINES OF FORCE (ELOF)
The line of force in an electric field is an imaginary
line, the tangent to which at any point on it
represents the direction of electric field at the given
point.
(iii) Number of lines originating (terminating) from
(on) is directly proportional to the magnitude of
the charge.
(a) Properties :
(i) Line of force originates out from a positive
charge and terminates on a negative charge. If
there is only one positive charge then lines start
from positive charge and terminate at . If there
is only one negative charge then lines start from
 and terminates at negative charge.
(iv) ELOF of resultant electric field can never intersect
with each other.
(v) Electric lines of force produced by static
charges do not form close loop.
(vi ) E lec tr ic lin es of fo rc e en d o r s ta rt
perpendicularly on the surface of a conductor.
(vii) Electric lines of force never enter into
conductors.
6.
If number of electric lines of force from charge q
are 10 then find out number of electric lines of
force from 2q charge.
Sol. No. of ELOF  charge
10  q
 20  2q
So number of ELOF will be 20.
7.
A charge + Q is fixed at a distance of d in front of
an infinite metal plate. Draw the lines of force
indicating the directions clearly.
PAGE # 3 3
Sol. There will be induced charge on two surfaces of
conducting plate, so ELOF will start from +Q charge
and terminate at conductor and then will again start
from other surface of conductor.
ELECTRIC POTENTIAL ENERGY
Consider a charge Q placed at a point P as shown in
figure. If another charge q of the same sign is now
brought from a very far away distance (infinity) to point
O near P, then charge q will experience a force of
repulsion due to charge Q. If charge q is still pushed
towards P, work is done. This work done is the potential
energy of the system of these two charges.
ELECTRIC FLUX
Q

Consider some surface in an electric field E . Let us
select a small area element dS on this surface.
The electric flux of the field over the
 
area element is given by dE = E . ds
Direction of dS is normal to the surface. It is along
n̂
or
or
or
dS
E
dE = EdS cos 
dE = (E cos ) dS
dE = En dS
r
P
O
direction of dS .
q
From infinity
Thus, the electric potential energy of a system of
charges is defined as the amount of work done in
bringing the various charges from infinite separation
to their present positions to form the required
system. It is denoted by U. For the system of two
charges separated by distance r as shown in figure,
the electric potential energy is given by :
U=
where En is the component of electric field in the
q
kQq
r
Electric potential energy is the from of energy, therefore
it is measured in joule (J).
SUPER CONDUCTOR AND ITS APPLICATIONS
If the electric field is uniform over that area then
 
E = E  S
(a) Physical Meaning :
The electric flux through a surface inside an electric
field represents the total number of electric lines of
force crossing the surface in a direction normal the
surface. It is a property of electric field
(b) Unit :
(i) The SI unit of electric flux is Nm2 C–1 (gauss) or J
m2 C–1.
(ii) Electric flux is a scalar quantity. (It can be positive,
negative or zero)
8.
Prof. K. Onnes in 1911 discovered that certain metals
and alloys at very low temperature lose their
resistance considerably. This phenomenon is known
as super-conductivity. As the temperature decreases,
the resistance of the material also decreases, but when
the temperature reaches a certain critical value (called
critical temperature or transition temperature), the
resistance
of
the
material
completely
disappears i.e. it becomes zero. Then the material
behaves as if it is a super-conductor and there will be
flow of electrons without any resistance whatsoever.
The critical temperature is different for different material.
It has been found that mercury at critical temperature
4.2 K, lead at 7.25 K and niobium at critical temperature
9.2 K become super-conductor.
The electric field in a region is given by,
Applications of super conductors :
 3  4 
3
E  E 0 i  E 0 j with E0 = 2.0 × 10 N/C. Find the
5
5
flux of this field through a rectangular surface of area
0.2m2 parallel to the Y–Z plane.
Sol.
 
E = E  S
2
 3  4 
E 0 i  E 0 j  . 0.2 î = 240 N  m
5
5

C
=
 
(i) Super conductors are used for making very strong
electromagnets.
(ii) Super conductivity is playing an important role in
material science research and high energy particle
physics.
(iii) Super conductivity is used to produce very high
speed computers.
(iv) Super conductors are used for the transmission of
electric power.
PAGE # 4 4
CELL
It converts chemical energy into electrical energy.
Electrochemical cells are of three types :
(a) Primary cell
(b) Secondary cell
(c) Fuel cell
(a) Primary Cell :
also so that current flows through resistor R. If r is the
internal resistance of the cell and V is the reading
shown by voltmeter, then
I=
E
Rr
 E = IR + Ir
Here, IR = V the potential difference
So,
E = V + r
It is an electrochemical cell, which cannot be recharged,
but the chemicals have to be replaced after a long use.
The reactions taking place in the cell are irreversible.
Eg. : Daniel cell, Lechlanche cell, Dry cell etc.
(b) Secondary Cell :
 E = I (R + r)
r=
 V = IR or I =
E–V
I
.........(i)
V
R
So for equation (i)
Electrical energy can be converted into chemical energy
and chemical energy can be converted into electrical
energy in these cells, i.e. secondary cells can be
recharged after use. Chemical reaction taking place in
these cells are reversible.
r=
(E – V)R
V
........(ii)
GROUPING OF CELLS
Example : Lead accumulator, Edison cell (alkali cell)
and iron nickel cell.
(a) Cells in Series :
(c) Electro Motive Force of a Cell (E.M.F.) :
It is the maximum potential difference between the two
electrodes of the cell when no current is drawn from
the cell or cell is in the open circuit.
(d) Potential Difference of a Cell :
It is the difference of potential between two terminals
of the cell when current is drawn from it or the cell is in
closed circuit.
(e) Internal Resistance of a Cell :
It is the resistance offered to the flow of current inside
the cell i.e. internal resistance is the resistance offered
to the flow of current by electrolyte. Internal resistance
decreases with the increase of the area of plates and
also with the decrease of the distance between plates.
Enrn
E1,r1 E2,r2 E3,r3
B
Eeq,req
A
B
Equivalent EMF
Eeq = E1 + E2 + ......... + En (write EMF’s with polarity)
Equivalent internal resistance req = r1 + r2 + r3 + r4 + ........ rn
If n cells each of emf E, are arranged in series and if r is
internal resistance of each cell, then total emf = nE
E,r E,r E,r
E,r
A
B
Upto n
I
R
Determination of internal resistance of a cell :
So current in the circuit,
I=
nE
R  nr
There may by two cases :
(i) If nr << R, then I =
nE
= n × current due to one cell.
R
So, series combination is advantageous.
(ii) If nr >> R, then I =
Connect a voltmeter to a cell through key K1. Also
connect a resistor R to cell through K2. First put in key
K1. The reading shown by voltmeter gives us the e.m.f.
of the cell since negligible current flows through cell
due to high resistance of the voltmeter. Insert key K2
E
= current due to one cell.
r
So, Series combination is not advantageous.
Note : If polarity of m cells is reversed, then equivalent
e.m.f. = (n–2m) E while the equivalent resistance is
still nr + R, so current in R will be
i=
(n  2m)E
nr  R
PAGE # 5 5
(b) Cells in Parallel :
The combination of cells is equivalent to single cell of
If m cells each of emf E and internal resistance r be
connected in parallel and if this combination is
connected to an external resistance then the emf of
the circuit is E.
emf = mE and internal resistance =
Internal resistance of the circuit =
Current I =
r
.
m
mr
n
mE
mr
R
n
For maximum current, nR = mr
mr
= internal resistance of the equivalent
n
or R =
battery.
Imax =
E
and I =
r
R
m

mE
mR  r
using mn = N in above equation we get number of
rows n =
There may by two cases :
(i) If mR << r, then I =
mE
= m × current due to one cell.
r
So, Parallel combination is advantageous.
(ii) If mR >> r, then I =
nE mE

.
2r
2R
E
= current due to one cell.
R
So, parallel combination is not advantageous.
9.
9 cells, each having the same emf and 3 ohm internal
resistance, are used to draw maximum current through
an external resistance of 3 ohm. find the combination
of cells.
Sol. For the condition of maximum current number of rows
n=
Nr
R
If emf and internal resistances of each cell are different,
then,
Eeq =
E1 / r1  E2 / r2  .....En / rn
1/ r1  1/ r2  .....1/ rn
for two cells E =
E1r2  E2r1
(Use emf with polarity)
r1  r2
E1
E2
E3
r1
En
rn
r2
r3
Nr
R
so n =
93
=3
3
so combination will be like 3 rows and 3 cells in each
row.
BATTERY
Battery is an arrangement that creates a constant
potential difference between its terminals. It is a
combination of a number of cells in series.
The impact of battery :
(c) Cells in Multiple Arc :
n = number of rows
m = number of cells in each row.
mn = N (total number of identical cells) :
With the discovery of voltaic cell, it was soon realised
that if one constructs a number of cells and joins the
negative terminal of one with the positive terminal of
the other and so on, then the emf (which is the potential
difference between the electrodes in an open circuit)
of the combination of cells will be the sum of the emf’s
of the individual cells. This observation led to a burst of
scientific activity in 1802. Humphrey Davy, an English
chemist, made a battery of 60 pairs of zinc and copper
plates. The large emf thus produced, was used to get
high current, which could melt iron and platinum wires.
By 1807, he had a battery of almost 300 plates with
which he was able to decompose chemical salts. This
led to the discovery of new elements.
PAGE # 6 6
By 1808, Davy had assembled 2,000 pairs of plates.
W ith this battery, he created electric arcs and
succeeded in extracting the elements like barium,
calcium and magnesium from their compounds. Thus,
electricity took a front seat in exploring the nature of
matter.
or
R  r 4
or
R
r
 4
R
r
 4  0.15 % = 0.60%
EFFECT OF STRETCHING OF A WIRE ON RESISTANCE
ELECTRICAL RESISTANCE
The property of a substance by virtue of which it
opposes the flow of electric current through it is
termed as electrical resistance. Electrical resistance
depends on the size, geometry, temperature and
internal structure of the conductor.
In stretching, the density of wire usually does not
change. Therefore
Volume before stretching = Volume after stretching
 1A 1   2 A 2
We known that, vd = eE 
m
=
and
eV
m
If information of lengths before and after stretching
I = Anevd
R2   2 
 
R1   1 
V
m

I
Ane 2 
2
A
 1
1 A 2
 =
R 2  A1 


R1  A 2 
m
Ane 2 

A
2
If information of radius r1 and r2 is given then use
V
m

R=
I
Ane 2 
 R =
A1  2

A 2 1
is given, then use
eV
= Ane
m
Ane2 V 
I=
m
R=
R 2  2 A1


R1  1 A 2
  =
 is called resistivity (it is also called specific
m
4
(a) Reciprocal of resistivity of a conductor is called
its conductivity. It is generally represented by  .
ne 2 
ne 2 
r 
  1 
 r2 
CONDUCTIVITY :
RA

m
resis tance), and  =
2
=
1
,  is called

conductivity. Therefore current in conductors is
proportional to potential difference applied across
(b)  
1

(c) Unit : ohm 1. metre 1
EFFECT OF TEMPERATURE ON RESISTANCE
AND RESISTIVITY
its ends. This is Ohm's Law. Units:    1m 1
also called siemens m –1.
10. If a copper wire is stretched to make its radius
decrease by 0.15%, Find the percentage increase
in resistance (approximately).
Sol. Due to stretching resistance changes are in the ratio
R 2  r1 
 
R1  r2 
4
The resistance of a conductor depends upon the
temperature. As the temperature increases, the
random motion of free electrons also increases. If
the number density of charge carrier electrons
remains constant as in the case of a conductor, then
the increase of random motion increases the
resistivity. The variation of resistance with temperature
is given by the following relation
R t  R 0 1  t   t 2


PAGE # 7 7
where Rt and R0 are the resistance at t0C and 00C
respectively and  and  are constants.
The
constant  is very small so its may be assumed
negligible.

R t  R 0 1  t 
or

R t  R0
R0  t
This constant  is called as temperature coefficient
of resistance of the substance.
If R0 = 1 ohm, t = 10C, then
  R t  R 0 
Thus, the temperature coefficient of resistance is
equal to the increase in resistance of a conductor
having a resistance of one ohm on raising its
temperature by 10C. The temperature coefficient of
resistance may be positive or negative.
From calculations it is found that for most of the
metals the value of  is nearly
1 0
/ C . Hence
273
substituting  in the above equation
t 

R t  R 0 1 

273 

T
 273  t 
 R0 
  R0
273
273


The resistivity of alloys increases with the rise of
temperature but less than that of metals.
On applying pressure on pure metals, its resistivity
decreases but on applying tension, the resistivity
increases.
The resistance of alloys such as eureka, manganin
etc., increases in smaller amount with the rise in
temperature. Their temperature coefficient of
resistance is negligible. On account of their high
resistivity and negligible temperature coefficient of
resistance these alloys are used to make wires for
resistance boxes, potentiometer, meter bridge etc.,
The resistance of semiconductors, insulators,
electrolytes etc., decreases with the rise in
temperature. Their temperature coefficients of
resistance are negative.
On increasing the temperature of semi conductors a
large number of electrons get free after breaking their
bonds. These electrons reach the conduction band
from valence band. Thus conductivity increases or
resistivity decreases with the increase of free electron
density.
11. A wire has a resistance of 2 ohm at 273 K and a
resistance of 2.5 ohm at 373 K. What is
the temperature coefficient of resistance of the
material?
Sol.  
R  R0
2.5  2

R 0 T  T0  2  373  273 
where T is the absolute temperature of the conductor.

Rt  T

Thus, the resistance of a pure metallic wire is directly
proportional to its absolute temperature.
The graph drawn between the resistance R t and
temperature t is found to be a straight line
0 .5
 2.5  10  3 / 0K
200
WHEATSTONE BRIDGE
Wheatstone bridge is an arrangement of four resistors
in the shape of a quadrilateral which can be used to
measure unknown resistance in terms of the
remaining three resistances.
Rt
R0
tºC
The resistivity or specific resistance varies with
temperature. This variation is due to change in
resistance of a conductor with temperature. The
dependence of the resistivity with temperature is
represented by the following equation.
 t   0 1  t 
With the rise of temperature the specific resistance
or resistivity of pure metals increases and that of
semi-conductors and insulators decrease.
The arrangement of Wheatstone bridge is shown in
figure below. Out of four resistors, two resistances R1,
R2 and R3, R4 are connected in series and are joined in
parallel across two points a and c. A battery of emf E is
connected across junctions a and c and a galvanometer
(G) between junction b and d. The keys K1 and K2 are
used for the flow of current in the various branches of
bridge.
Principle of Wheatstone Bridge :
When key K1 is closed, current i from the battery is
divided at junction a in two parts. A part i1 goes through
R1 and the rest i2 goes through R3. When key K2 is
closed, galvanometer shows a deflection.
PAGE # 8 8
(a) Ammeter :
Ammeter is an electrical instrument which measures
the strength of current in ‘ampere’ in a circuit. Ammeter
is a pivoted coil galvanometer which is always
connected in series in circuit so that total current (to be
measured) may pass through it. For an ammeter of
good quality, the resistance of its coil should be very
low so that it may measure the strength of current
accurately (without affecting the current passing
through the circuit). The resistance of an ideal ammeter
is zero (practically it should be minimum). So, to
minimize the effective resistance of an ammeter, a low
value resistance (shunt) as per requirement is
connected in parallel to the galvanometer to convert it
to ammeter of desired range.
In electric circuit, the positive terminal of an ammeter
is connected to positive plate and negative terminal is
connected to negative plate of battery.
The direction of deflection depends on the value of
potential difference between b and d. When the value
of potential at b and d is same, then no current will flow
through galvanometer. This condition is known as the
condition of balanced bridge or null deflection
condition. This situation can be obtained by choosing
suitable values of the resistances. Thus, in null
deflection state, we have :
Va – Vb = Va – Vd
or
i1 R1 = i2 R3
...(i)
Similarly :
Vb – Vc = Vd – Vc
or
i1 R2 = i2 R4
...(ii)
On dividing equation (i) by (ii), we get
i2 R 3
i1 R1
i1 R 2 = i2 R 4
or
Desired value of shunt depends on the range
(measurable maximum current) of ammeter converted
from galvanometer.
If pivoted galvanometer of resistance G is to measure
current i (as an ammeter) then from figure.
i
 R2 

 R1 
R4 = (R3) 
For better accuracy of the bridge one should choose
resistances R1, R2, R3 and R4 of same order.
GALVANOMETER
Galvanometer is a simple device, used to detect the
current, to find direction of current and also to compare
the currents.
With the help of galvanometer we make two
important devises known as Ammeter and voltmeter
as discussed below.
G
is
ig G = (i – ig) S
ig
S
or
i
ig G
S=
(i  ig )
Where ig is an amount of current required for full
deflection in galvanometer. By using a low value of
resistance S (shunt) in parallel to the galvanometer
(resistance G), the effective resistance of
R1 R 3

R 2 R 4 ...(iii)
Equation (iii) states the condition of balanced bridge.
Thus, in null deflection condition the ratio of
resistances of adjacent arms of the bridge are same.
The resistor of unknown resistance is usually
connected in one of the arm of the bridge. The
resistance of one of the remaining three arms is
adjusted such that the galvanometer shows zero
deflection. If resistance of unknown resistor is R4. Then
ig
converted ammeter RA =

GS
becomes very low..
(G  S)
NOTE :
Shunt : If anyhow, the flowing current through
galvanometer becomes more than its capacity, the coil
has possibility of burning due to heat produced by
flowing current. Secondly, its pointer may break up due
to impact with ‘stop pin’ as its proportional deflection
as per amount of flowing current.
In order to minimize these possibilities a low resistance
wire (or strip) is connected in parallel with galvanometer,
which is known as shunt.
(b) Voltmeter :
It is an electrical instrument which measures the
potential difference in ‘volt’ between two points of
electric circuit. It’s construction is similar as that of
ammeter. The only difference between ammeter and
voltmeter is that ammeter has its negligible
(approximately zero) resistance so that it may measure
current of circuit passing through it more accurately
giving the deflection accordingly, while the voltmeter
passes negligible current through itself so that potential
difference developed due to maximum current passing
PAGE # 9 9
through circuit may be measured. Therefore, an
appropriate value of high resistance is required to be
connected in series of galvanometer to convert it into a
voltmeter of desired range.
Voltmeter is connected in parallel to the electric circuit.
 V1  V2 + V3  V4 = 0. Boxes may contain resistor
or battery or any other element (linear or nonlinear).
It is also known as KVL
12. Figure shows, current in a part of electrical circuit, what
will be the value of current (i) ?
If a galvanometer of resistance G is to be converted
into a voltmeter of range V, then required value of high
resistance RH will be
1A
2A
V = ig (RH + G)
V
 
or RH =  I  – G
 g
P
2A
Q
3
1.
R
S
3A
A
i
V
i
ig
RH
Connecting this value of high resistance in the series
of galvanometer, it will be converted to a voltmeter of
range V. After connecting high resistance RH in series
of galvanometer of resistance G, the effective resistance
of voltmeter becomes RV = (RH + G) very high (high in
comparison to G).
Ideal voltmeter has infinite resistance of its own. When
ideal voltmeter is connected parallel to a part of an
electric circuit, it passes zero amount of current through
itself from the circuit so that measurement of potential
difference across the points of connection may be
perfectly accurate.
1A
2A
G
Sol.
P
2A
i2
i1
R
Q
3A
3
1.
i3
A
S
i
From KCL, current at junction P, i1 = 2 + 3 = 5 A
From KCL, current at junction Q, i2 = i1 + 1 = 5 + 1 = 6 A
From KCL, current at junction R, i3 = i2 – 2 = 6 – 2 = 4 A
From KCL, current at junction S, i = i3 – 1.3 = 4 – 1.3
= 2.7 A
13. In the circuit shown, calculate the value of R in ohm
that will result in no current through the 30 V battery.
KIRCHHOFF'S LAWS
(a) Kirchhoff’s Current Law (Junction law) :
This law is based on law of conservation of charge.
It states that "The algebraic sum of the currents
meeting at a point of the circuit is zero" or total
current entering a junction is equal total current
leaving the junction.
  in =  out.
It is also known as KCL (Kirchhoff's current law).
Sol. Applying KVL in loop CEFDC
50 – iR – 20 i = 0
i =
(b) Kirchhof f ’s Voltage Law (Loop law) :
50
20  R
i
C
A
“The algebraic sum of all the potential differences
along a closed loop is zero. IR + EMF =0”. The
closed loop can be traversed in any direction. While
traversing a loop if potential increases, put a positive
sign in expression and if potential decreases put
a negative sign.
E
50V
i
R
20
B
10
D
i
F
Potential drop across R = Potential drop across AB
iR = 30

50
.R = 30
20  R
 R = 30 
PAGE # 1010
8.
EXERCISE
Charge and coulomb’s Law :
1.
Conductivity of superconductor is :
(A) infinite
(B) very large
(C) very small
(D) zero
12 positive charges of magnitude q are placed on a
circle of radius R in a manner that they are equally
spaced. A charge +Q is placed at the centre. If one of
the charges q is removed, then the force on Q is :
(KVPY/2010)
(A) zero
qQ
(B)
2.
3.
4.
Two charges of +1 C & + 5 C are placed 4 cm apart,
the ratio of the force exerted by both charges on each
other will be (A) 1 : 1
(B) 1 : 5
(C) 5 : 1
(D) 25 : 1
A body has 80 microcoulomb
additional electrons on it will
(A) 8 x 10–5
(C) 5 x 1014
5.
W
q
away from the position of the removed
qQ
(D)
40R2
towards the position of the removed
charge.
In a neon discharge tube 2.8 × 1018 Ne+ ions move to
the right per second while 1.2 ×1018 electrons move to
the left per second. Therefore , the current in the
discharge tube is :
(IJSO/Stage-I/2011)
(A) 0.64 A towards right
(B) 0.256 A towards right
(C) 0.64 A towards left
(D) 0.256 A towards left
Electric filed and Potential :
F
2
(B) 2F
F
(D)
4
(C) 4F
7.
40R2
charge.
9.
Two particles having charges q1 and q2 when kept at a
certain distance, exert force F on each other. If distance
is reduced to half, force between them becomes :
(A)
6.
11qQ
(C)
Which of the following relation is wrong ?
(A) Q = It
(D) V =
away from the position of the removed
charge.
of charge. Number of
be :
(B) 80 x 1015
(D) 1.28 x 10–17
1Coulomb
(B) 1 ampere = 1Second
(C) V = Wq
40R2
10. If Q = 2 coloumb and force on it is F = 100 newton,
then the value of field intensity will be:
(A) 100 N/C
(B) 50 N/C
(C) 200 N/C
(D) 10 N/C
4
Coulomb of charge contains.............................
25
electrons :
(A) 1015
(B) 1018
20
(C) 10
(D) none of these
5 charges each of magnitude 10–5 C and mass 1 kg
are placed (fixed) symmetrically about a movable
central charges of magnitude 5 × 10–5C and mass 0.5
kg as shown. The charges at P1 is removed. The
acceleration of the central charge is : (KVPY/2009)
11. In the electric field of charge Q, another charge is
carried from A to B. A to C, A to D and A to E, then work
done will be -
A
B
Q +
centre
P1
P2
P5
C D
O
P3
P4
1
[Given OP1 = OP2 = OP3 = OP4 = OP5 1 m ;
4 0 = 9 ×
109 in SI units]
(A) 9 m s–2 upwards
(C) 4.5 m s–2 upwards
(B) 9 m s–2 downwards
(D) 4.5 m s–2 downwards
(A)
(B)
(C)
(D)
E
minimum along path AB.
minimum along path AD.
minimum along path AE.
zero along all the paths.
PAGE # 1111
12. A negatively charged particle initially at rest is placed in
an electric field that varies from point to point. There
are no other fields. Then :
(KVPY/2008)
16. In the given circuit, the equivalent resistance between
points A and B will be.
(A) the particle moves along the electric line of force
passing through it.
(B) the particle moves opposite to the electric line of
force passing through it.
(C) the direction of acceleration of the particle is
tangential to the electric line of force at every instant.
(D) the direction of acceleration of the particle is normal
to the electric line of force at every instant.
13. Two charges +q and –q are placed at a distance b
apart as shown in the figure below.
(KVPY/2009)
B
P
A
C
b/2
+q
–q
b
The electric field at a point P on the perpendicular
bisector as shown as :
(A) along vector 
A
(B) along vector 
B
(C) along vector 
C
(D) Zero
14. Two charges +Q and _2Q are located at points A and B
on a horizontal line as shown below :
The electric field is zero at a point which is located at a
finite distance :
(KVPY/2011)
(A) On the perpendicular bisector of AB
(B) left of A on the line
(C) between A and B on the line
(D) right of B on the line
(A)
8
R
3
(C) 6R
(B) 4R
(D) 10R
17. Resistance of a conductor of length 75 cm is 3.25 .
What will be the length of a similar conductor whose
resistance is 13.25 ?
(A) 305.76 cm
(B) 503.76 cm
(C) 200 cm
(D) 610 cm
18. A piece of wire of resistance 4 is bent through 1800
at its mid point and the two halves are twisted together,
then resistance is :
(A) 1 
(B) 2 
(C) 5 
(D) 8 
19. In how many parts (equal) a wire of 100  be cut so
that a resistance of 1  is obtained by connecting
them in parallel ?
(A) 10
(B) 5
(C) 100
(D) 50
20. If a wire of resistance 1  is stretched to double its
length, then the resistance will become :
(A)
1

2
(B) 2 
(C)
1

4
(D) 4 
21. Two copper wires, one of length 1 m and the other of
length 9 m, are found to have the same resistance.
Their diameters are in the ratio :
(A) 3 : 1
(B) 1 : 9
(C) 9 : 1
(D) 1 : 3
22. Reading of ammeter in ampere for the following circuit
is :
Resistance :
15. There are three resistance 5, 6and 8connected in
parallel to a battery of 15 V and of negligible resistance.
The potential drop across 6resistance is :
(A) 10 V
(B) 15 V
(C) 20 V
(D) 8 V
(A) 0.8
(C) 0.4
(B) 1
(D) 2
PAGE # 1212
23. Two resistors are joined in series, then their equivalent
resistance is 90  . When the same resistors are
joined in parallel, the equivalent resistance is 20  .
The resistances of the two resistors will be :
(A) 70 , 20 
(B) 80 , 10 
(C) 60 , 30 
(D) 50 , 40 
24. In the ladder network shown, current through the
resistor 3  is 0.25 A. The input voltage ‘V’ is equal
to
29. The resistance of a wire of cross-section ‘a’ and length
‘  ’ is R ohm. The resistance of another wire of the
same material and of the same length but cross-section ‘4a’ will be
(A) 4R
(C)
(B)
R
16
R
4
(D) 16 R
30. In the following circuit the value of total resistance between X and Y in ohm is :
X
r
r
r
Y
(A) 10 V
(B) 20 V
(C) 5 V
15
(D)
V
2
25. The reading of voltmeter is
(A) 50V
(C) 40V
10
(B) 60 V
(D) 80 V
r
3 )R
(C) 
r to 
r
(B) ( 3 – 1)R
(D) 50 r
32. In case of the circuit arrangement shown below, the
equivalent resistance between A and B is :
(IAO/Jr./Stage-I/2009)
2
26. 1.4A
(A) 0.4
(C) 0.6
r
31. Wires A and B are made from the same material. Wire
A has length 12m and weight 50 g, while wire B is 18 m
long and weighs 40 g. Then the ratio (RA / RB) of their
resistances will be :
(IAO/Jr./Stage-I/2008)
(A) 16 / 45
(B) 4 / 5
(C) 8 / 15
(D) 4 / 9
A
25
(A) (1 +
r
r
A
B
1.4A
5
(B) 1
(D) 1.2
27. Three identical bulbs are connected in parallel with a
battery. The current drawn from the battery is 6 A.
If one of the bulbs gets fused, what will be the total
current drawn from the battery ?
(A) 6A
(B) 2A
(C) 4A
(D)
28. A uniform wire of resistance R is uniformly compressed
along its length, until its radius becomes n times the
original radius. Now, the resistance of the wire
becomes :
(A) R/n
(B) R/n4
2
(C) R/n
(D) n R
(A) 10
(C)
40
W
3
(B) 2.5 W
(D) None of the above
33. The net resistance between points P and Q in the
circuit shown in fig. is
(A) R/2
(B) 2R/5
(C) 3R/5
(D) R/3
PAGE # 1313
34. A wire of resistance 10.0 ohm is stretched so as to
increase its length by 20%. Its resistance then would
be :
(IAO/Sr/Stage-I/2008)
(A) 10.0 ohm
(C) 14.4 ohm
38. In the circuit arrangement shown, if the point A and B are
joined by a wire the current in this wire will be :
(IJSO/Stage-I/2011)
(B) 12.0 ohm
(D) 10.2 ohm
A
35. In the circuit shown below, all the resistances are equal,
each equal to R. The equivalent resistance between
points A and C is :
B
(IAO/Sr./Stage-I/2009)
B
24Volt
C
R
R
R
R
(A) 1A.
(C) 4A.
R
R
R
39. In the following circuit, each resistor has a resistance
of 15  and the battery has an e.m.f. of 12 V with
R
A
D
(A) R
(C) R /2
(B) 2A.
(D) zero.
(B) 4R
(D) none of the above
negligible internal resistance. (IJSO/Stage-II/2011)
36. A battery or 10 V and negligible internal resistance is
connected across the diagonally opposite corner of a
cubical network consisting of 12 resistors each of
resistance 1 . The total current 1 in the circuit external
to the network is :
(KVPY/2007)
A
10V
(A) 0.83 A
(B) 12 A
(C) 1 A
(D) 4 A
37. Figure (a) below shows a Wheatstone bridge in which
P, Q, R, S are fixed resistances, G is a galvanometer
and B is a battery. For this particular case the
galvanometer shows zero deflection. Now, only the
positions of B and G are interchanged,. as shown in
figure (b). The new deflection of the galvanometer.
(KVPY/2010)
(A) is to the left.
(B) is to the right.
(C) is zero.
(D) depends on the values of P, Q, R, S
When a resistor of resistance R is connected between
D & F, no current flows through the galvanometer (not
shown in the figure) connected between C & F.
Calculate the value of R.
(A) 10 
(B) 15 
(C) 5 
(D) 30 
40. The circuit given below is for the operation of an
industrial fan. The resistance of the fan is 3 ohms. The
regulator provided with the fan is a fixed resistor and a
variable resistor in parallel.
(IJSO/Stage-II/2011)
Under what value of the variable resistance given
below, Power transferred to the fans will be maximum?
The power source of the fan is a dc source with internal
resistnace of 6 ohm.
(A) 3 
(B) 0
(C) 
(D) 6 
PAGE # 1414
41. When all the resistances in the circuit are 1 each,
then the equivalent resistance across points A & B will
be :
(IJSO/Stage-II/2011)
 46. In the circuit shown below,
(IAO/Sr./Stage-I/2007)
10V
X
Y
(A) current flowing in the circuit is 200 mA
(B) power supplied by the battery is 2 watt
(C) current from X to Y is zero
(D) potential difference across 10 is equal to zero
(A) 5/6 
(B) 1/2 
(C) 2/3 
(D) 1/3 
47. We are given n resistors, each of resistance R. The
ratio of the maximum to minimum resistance that can
be obtained by combining them is :
(KVPY/2008)
(A) nn
(B) n
(C) n2
(D) logn
Cell :
42. A cylindrical copper rod has length L and resistance R.
If it is melted and formed into another rod of length 2L.
the resistance will be :
(KVPY/2011)
(A) R
(B) 2R
(C) 4R
(D) 8R
 43. There are four resistors of 12 ohm each. Which of the
following values is/are possible by their combinations
(series and / or parallel) ?
(IJSO/Stage-I/2008)
(A) 9 ohm
(B) 16 ohm
(C) 12 ohm
(D) 30 ohm
 44. In case of the circuit shown below, which of the following
statements is/are true ?
(IJSO/Stage-I/2009)
+ –
1
•
A
R1
B
R2
•
(A) R(E – V)
(C)
(E  V )
E
R
EV
R
(E  V )
R
(D)
V
(B)
49. 24 cells, each having the same e.m.f. and 2 ohm
internal resistance, are used to draw maximum current
through an external resistance of 3 ohm. The cells
should be connected
(A) in series
(B) in parallel
(C) in 4 rows, each row having 6 cells
(D) in 6 rows, each row having 4 cells
4
R3
2
3
(A) R1, R2 and R3 are in series.
(B) R2 and R3 are in series.
(C) R2 and R3 are in parallel.
(D) The equivalent resistance of the circuit is
R1+
48. A cell of emf E is connected across a resistance R.
The potential difference between the terminals of the
cell is found to be V. The internal resistance of the cell
is given as :
R 2R3
R 2  R3
 45. A current i reaching at a point in a circuit gets branched
and flows through two resistors R1 and R2. Then, the
current through R1 varies as : (IAO/Jr./Stage-I/2007)
(A) R1
(B) R2
(C) (R1+ R2)
(D) 1l (R1 + R2)
50. The cells are joined in parallel to get the maximum
current when
(A) external resistance is very large as compared to
the total internal resistance
(B) internal resistance is very large as compared to
the external resistance
(C) internal resistance and external resistance are
equal
(D) emf of each cell is very large
51. In secondary cells :
(A) Chemical changes can be reversed by heating
electrodes
(B) Chemical changes can be reversed by passing
electric current
(C) Current is produced by photo chemical reactions
(D) None of these
PAGE # 1515
52. Three types of electric cells which provide current are :
(A) Button cell, solar cell & secondary cell
(B) Solar cell, electrolytic cell, electro chemical cell
(C) (A) and (B) both are correct
(D) Neither (A) nor (B) is correct
Electric Energy and Power :
53. In which of the following cells, the potential difference
between the terminals of a cell exceeds its emf.
58. Two identical heater wires are first connected in series
and then in parallel with a source of electricity. The
ratio of heat produced in the two cases is :
(A) 2 : 1
(B) 1 : 2
(C) 4 : 1
(D) 1 : 4
57. An electric iron of heating element of resistance 88 
is used at 220 volt for 2 hours. The electric energy
spent, in unit, will be :
(A) 0.8
(B) 1.1
(C) 2.2
(D) 8.8
59. You are given three bulbs 25 W, 40 W and 60 W . Which
of them has the lowest resistance?
(A) 25 watt bulb
(B) 40 watt bulb
(C) 60 watt bulb
(D) insufficient data
(A) a
(C) c
(B) b
(D) d
54. A cell, an ammeter and a voltmeter are all connected in
series. The ammeter reads a current I and the
voltmeter a potential difference V. If a torch bulb is
connected across the voltmeter, then.
(IJSO/Stage-I/2009)
(A) both I and V will increase
(B) both I and V will decrease
(C) I will increase but V will decrease
(D) I will decrease but V will increase
55. In the process of electrostatic induction.
(IJSO/Stage-II/2011)
(A) a conductor is rubbed with an insulator.
(B) a charge is produced by friction.
(C) negative and positive charges are separated.
(D) electrons are ‘sprayed’ on the object.
56. Consider the circuit below. The bulb will light up if :
(KVPY/2009)
S1
S2
60. If R1 and R2 are the filament resistances of a 200 W
bulb and a 100 W bulb respectively designed to operate
on the same voltage, then :
(A) R1 = 2 R2
(B) R2 = 2 R1
(C) R2 = 4 R1
(D) R1 = 4 R2
61. If two bulbs, whose resistance are in the ratio of 1 : 2,
are connected in series. The power dissipated in them
has the ratio of :
(A) 1 : 1
(B) 1 : 2
(C) 2 : 1
(D) 1 : 4
62. When a voltage of 20 volt is applied between the two
ends of a coil, 800 cal/s heat is produced. The value of
resistance of the coil is :(1 calorie = 4.2 joule) :
(A) 1.2 
(B) 1.4 
(C) 0.12 
(D) 0.14 
63. You are given two fuse wires A and B with current rating
2.5 A and 6 A respectively. Which of the two wires would
you select for use with a 1100 W, 220 V room heater ?
(A) A
(B) B
(C) A and B
(D) none of these
64. An electric current of 2.0 A passes through a wire of
resistance 25  . How much heat (in joule) will be
developed in 1 minute ?
(A) 6
(B) 6000
(C) 50
(D) 10
65 . Two bulbs, one of 200W and the other of 100W, are
connected in series with a 100 V battery which has no
internal resistance. Then,
(KVPY/2009)
100V
~
S3
200W
(A) S1 S2 and S3 are all closed.
(B) S1 is closed but S2 and S3 are open.
(C) S1 and S3 are closed but S2 is open.
(D) none of these
100W
(A) the current passing through the 200W bulb is more
than that through the 100W bulb.
(B) the power dissipation in the 200W bulb is more
than that In the 100 W bulb.
(C) the voltage drop across the 200W bulb is more
than that across the 100W bulb.
(D) the power dissipation In the 100W bulb is more
than that in the 200W bulb.
PAGE # 1616
66. An electric heater consists of a nichrome coil and runs
under 220 V, consuming 1 kW power. Part of its coil
burned out and it was reconnected after cutting off the
burnt portion. The power it will consume now is :
(KVPY/2010)
(A) more than 1 kW.
(B) less that 1 kW, but not zero.
(C) 1 kW.
(D) 0 kW.
71. The circuit shown has 3 identical light bulbs A, B, C
and 2 identical batteries E1, E2 . When the switch is
open, A and B glow with equal brightness. When the
switch is closed:
(KVPY/2007)
67. In the following circuit, the 1 resistor dissipates
power P. If the resistor is replaced by 9. the power
dissipated in it is :
(KVPY/2011)
(A) A and B will maintain their brightness and C will be
dimmer than A and B.
(B) A and B will become dimmer and C will be brighter
than A and B.
(C) A and B will maintain their brightness and C will not
glow.
(D) A, B and C will be equally bright.
C
E1
A
S
E2
B
72. A student connects two lamps in the circuit shown.
The emf of the two batteries is different.
(IJSO/Stage-II/2011)
(A) P
(C) 9P
(B) 3P
(D) P/3
68. A neon lamp is connected to a voltage a.c. source. The
voltage is gradually increased from zero volt. It is
observed that the neon flashes at 50 V. The a.c. source
is now replaced by a variable dc source and the
experiment is repeated. The neon bulb will flash at :
(IAO/Sr./Stage-I/2008)
(A) 50V
(B) 70V
(C) 100V
(D) 35V
69. A certain network consists of two ideal and indentical
voltage sources in series and a large number of ideal
resistor. The power consumed in one of the resistor is
4W when either of the two sources is active and other
is replaced by a short circuit. The power consumed by
same resistor when both sources are simultaneously
active would be :
(IJSO/Stage-II/2011)
(A) 0 or 16W
(B) 4W or 8W
(C) 0 or 8W
(D) 8W or 16W
Which of the following statements are correct?
i. When keys 1, 2, 3 and 4 are closed, bulbs
A and B will both glow
ii. When key 2 and 4 are closed bulb A will glow
iii.When 1 and 4 are closed, bulb A will glow
iv.When 2, 3 and 4 are closed, both A and B will
glow
(A) only ii
(B) only iv
(C) i, ii and iv
(D) ii and iii
73. Figure below shows a portion of an electric circuit with
the currents in ampere and their directions. The
magnitude and direction of the current in the portion
PQ is :
(KVPY/2011)
Circuit and Other :
70. A galvanometer can be converted into a voltmeter by
connecting
(A) A high resistance in series with the galvanometer
(B) A high resistance in parallel with the galvanometer
(C) a low resistance in series with the galvanometer
(D) a low resistance in parallel with the galvanometer
(A) 0A
(C) 4A from Q to P
(B) 3A from P to Q
(D) 6A from Q to P
PAGE # 1717
MOLE CONCEPT
ATOMS
The symbol of Copper is Cu (from its Latin name
Cuprum)
All the matter is made up of atoms. An atom is the
smallest particle of an element that can take part in a
chemical reaction. Atoms of most of the elements
are very reactive and do not exist in the free state (as
single atom).They exist in combination with the atoms
of the same element or another element.
Atoms are very, very small in size. The size of an atom
is indicated by its radius which is called "atomic
radius" (radius of an atom). Atomic radius is
measured in "nanometres"(nm).
1 metre = 109 nanometre or 1nm = 10-9 m.
Atoms are so small that we cannot see them under
the most powerful optical microscope.

It should be noted that in a "two letter" symbol, the
first letter is the "capital letter" but the second letter is
the small letter
Symbol Derived from English Names
Note :
Hydrogen atom is the smallest atom of all , having an
atomic radius 0.037nm.
(a) Symbols of Elements :
A symbol is a short hand notation of an element which
can be represented by a sketch or letter etc.
Dalton was the first to use symbols to represent
elements in a short way but Dalton's symbols for
element were difficult to draw and inconvenient to
use, so Dalton's symbols are only of historical
importance. They are not used at all.

English Name of
the Element
Hydrogen
Helium
Lithium
Boron
Carbon
Nitrogen
Oxygen
Fluorine
Neon
Magnesium
Aluminium
Silicon
Phosphorous
Sulphur
Chlorine
Argon
Calcium
Symbol
H
He
Li
B
C
N
O
F
Ne
Mg
Al
Si
P
S
Cl
Ar
Ca
Symbols Derived from Latin Names
It was J.J. Berzelius who proposed the modern
system of representing en element.
The symbol of an element is the "first letter" or the
"first letter and another letter" of the English name or
the Latin name of the element.
e.g. The symbol of Hydrogen is H.
The symbol of Oxygen is O.
There are some elements whose names begin with
the same letter. For example, the names of elements
Carbon, Calcium, Chlorine and Copper all begin with
the letter C. In such cases, one of the elements is
given a "one letter "symbol but all other elements are
given a "first letter and another letter" symbol of the
English or Latin name of the element. This is to be
noted that "another letter" may or may not be the
"second letter" of the name. Thus,
The symbol of Carbon is C.
The symbol of Calcium is Ca.
The symbol of Chlorine is Cl.
English Name of
the Element
Symbol
Latin Name of
the Element
Sodium
Na
Natrium
Potassium
K
Kalium
(b) Significance of The Symbol of an
Element :
(i) Symbol represents name of the element.
(ii) Symbol represents one atom of the element.
(iii) Symbol also represents one mole of the element.
That is, symbol also represent 6.023 × 1023 atoms of
the element.
(iv) Symbol represent a definite mass of the element
i.e. atomic mass of the element.
Example :
(i) Symbol H represents hydrogen element.
(ii) Symbol H also represents one atom of hydrogen
element.
(iii) Symbol H also represents one mole of hydrogen
atom.
(iv) Symbol H also represents one gram hydrogen
atom.
PAGE # 18
An anion bears that much units of negative charge as
there are the number of electrons gained by the
neutral atom to form that anion.
e.g. A nitrogen atom gains 3 electrons to form nitride
ion, so nitride ion bears 3 units of negative charge
and it is represented as N3-.
Note :
Size of a cation is always smaller and anion is always
greater than that of the corresponding neutral atom.
IONS
An ion is a positively or negatively charged atom or
group of atoms.
Every atom contains equal number of electrons
(negatively charged) and protons (positively charged).
Both charges balance each other, hence atom is
electrically neutral.

(a) Cation :
(c) Monoatomic ions and polyatomic ions :
(i) Monoatomic ions : Those ions which are formed
from single atoms are called monoatomic ions or
simple ions.
e.g. Na+, Mg2+ etc.
If an atom has less electrons than a neutral atom,
then it gets positively charged and a positively
charged ion is known as cation.
e.g. Sodium ion (Na+), Magnesium ion (Mg2+) etc.
A cation bears that much units of positive charge as
there are the number of electrons lost by the neutral
atom to form that cation.
(ii) Polyatomic ions : Those ions which are formed
from group of atoms joined together are called
polyatomic ions or compound ions.
e.g. Ammonium ion (NH4+) , hydroxide ion (OH–) etc.
which are formed by the joining of two types of atoms,
nitrogen and hydrogen in the first case and oxygen and
hydrogen in the second.
e.g. An aluminium atom loses 3 electrons to form
aluminium ion, so aluminium ion bears 3 units of
positive charge and it is represented as Al3+.
(d) Valency of ions :
The valency of an ion is same as the charge present
on the ion.
If an ion has 1 unit of positive charge, its valency is 1
and it is known as a monovalent cation. If an ion has
2 units of negative charge, its valency is 2 and it is
known as a divalent anion.
(b) Anion :
If an atom has more number of electrons than that of
neutral atom, then it gets negatively charged and a
negatively charged ion is known as anion.
e.g. Chloride ion (Cl¯), oxide ion (O2-) etc.
LIST OF COMMON ELECTROVALENT POSITIVE RADICALS
LIST OF COMMON ELECTROVALENT NEGATIVE RADICALS
Bivalent
Electronegative
Monovalent Electronegative
1. Fluoride
F–
1. Sulphate
2. Sulphite
SO 3
3. Sulphide
S
4. Thiosulphate
S 2O3
5. Zincate
ZnO2
6. Oxide
O
7. Peroxide
O2
8. Dichromate
Cr 2O7
–
9. Carbonate
–
10. Silicate
2CO 3
2SiO 3
Br
I
6. Hydroxide
7. Nitrite
8.Nitrate
H
–
–
OH
–
NO2
–
NO3
9. Bicarbonate or Hydrogen carbonate HCO3
10. Bisulphite or Hydrogen sulphite
11. Bisulphide or Hydrogen sulphide
12. Bisulphate or Hydrogen sulphate
13. Acetate

2-
–
Cl
3. Bromide
5. Hydride
2-
SO 4
–
2. Chloride
4. Iodide
Trivalent
Electronegative
HSO3
–
HS
22-
1. Nitride
N
3-
Tetravalent
Electronegative
1. Carbide
C
4-
3-
2. Phosphide
P
3. Phosphite
PO 3
4. Phosphate
PO 4
33-
2-
222-
–
HSO4
–
CH3COO
Note :
Cation contains less no. of electrons and anion contains more no. of electrons than the no. of protons present in
them.
PAGE # 19
Example :
LAWS OF CHEMICAL COMBINATION
Water is a compound of hydrogen and oxygen. It can
be obtained from various sources (like river, sea, well
etc.) or even synthesized in the laboratory. From
whatever source we may get it, 9 parts by weight of
water is always found to contain 1 part by weight of
hydrogen and 8 parts by weight of oxygen. Thus, in
water, this proportion of hydrogen and oxygen always
remains constant.
The laws of chemical combination are the
experimental laws which led to the idea of
atoms being the smallest unit of matter. The laws of
chemical combination played a significant role in the
development of Dalton’s atomic theory of matter.
There are two important laws of chemical combination.
These are:

(i) Law of conservation of mass
(ii) Law of constant proportions
(a) Law of Conservation of Mass or Matter :
(c) Law of Multiple Proportions :
This law was given by Lavoisier in 1774 . According to
the law of conservation of mass, matter can neither
be created nor be destroyed in a chemical reaction.
Or
The law of conservation of mass means that in a
chemical reaction, the total mass of products is equal
to the total mass of the reactants. There is no change
in mass during a chemical reaction.
Suppose we carry out a chemical reaction between A
and B and if the products formed are C and D then,
A + B  C + D
Suppose 'a' g of A and 'b' g of B react to produce 'c' g of
C and 'd' g of D. Then, according to the law of
conservation of mass, we have,
a+b = c+d
Example :
When Calcium Carbonate (CaCO 3) is heated, a
chemical reaction takes place to form Calcium Oxide
(CaO) and Carbon dioxide (CO2). It has been found
by experiments that if 100 grams of calcium carbonate
is decomposed completely, then 56 grams of Calcium
Oxide and 44 grams of Carbon dioxide are formed.
Note :
The converse of Law of definite proportions that when
same elements combine in the same proportion, the
same compound will be formed, is not always true.
According to it, when one element combines with the
other element to form two or more different compounds,
the mass of one element, which combines with a
constant mass of the other, bears a simple ratio to
one another.
Simple ratio means the ratio between small natural
numbers, such as 1 : 1, 1 : 2, 1 : 3
e.g.
Carbon and oxygen when combine, can form two
oxides that are CO (carbon monoxide), CO2 (carbon
dioxide).
In CO,12 g carbon combined with 16 g of oxygen.
In CO2,12 g carbon combined with 32 g of oxygen.
Thus, we can see the mass of oxygen which combine
with a constant mass of carbon (12 g) bear simple
ratio of 16 : 32 or 1 : 2

Note :
The law of multiple proportion was given by Dalton in
1808.
Sample Problem :
1.
Since the total mass of products (100g ) is equal to
the total mass of the reactants (100g), there is no
change in the mass during this chemical reaction.
The mass remains same or conserved.
(b) Law of Constant Proportions / Law of
Definite Proportions :
Proust, in 1779, analysed the chemical composition
(types of elements present and percentage of
elements present ) of a large number of compounds
and came to the conclusion that the proportion of
each element in a compound is constant (or fixed).
According to the law of constant proportions: A
chemical compound always consists of the same
elements combined together in the same proportion
by mass.

Note :
The chemical composition of a pure substance is
not dependent on the source from which it is obtained.
Carbon is found to form two oxides, which contain
42.8% and 27.27% of carbon respectively. Show that
these figures illustrate the law of multiple proportions.
Sol. % of carbon in first oxide = 42.8
% of oxygen in first oxide = 100 - 42.8 = 57.2
% of carbon in second oxide = 27.27
 % of oxygen in second oxide = 100 - 27.27 = 72.73
For the first oxide Mass of oxygen in grams that combines with 42.8 g
of carbon = 57.2
 Mass of oxygen that combines with 1 g of carbon =
57.2
 1.34 g
42.8
For the second oxide Mass of oxygen in grams that combines with 27.27 g
of carbon = 72.73
 Mass of oxygen that combines with 1 g of carbon =
72.73
 2.68 g
27.27
Ratio between the masses of oxygen that combine
with a fixed mass (1 g) of carbon in the two oxides
= 1.34 : 2.68 or 1 : 2 which is a simple ratio. Hence,
this illustrates the law of multiple proportion.
PAGE # 20
(d) Law of Reciprocal Proportions :
(e) Gay Lussac’s Law of Gaseous Volumes :
Gay Lussac found that there exists a definite
relationship among the volumes of the gaseous
reactants and their products. In 1808, he put forward
a generalization known as the Gay Lussac’s Law of
combining volumes. This may be stated as follows :
According to this law the ratio of the weights of two
element A and B which combine separately with a
fixed weight of the third element C is either the same
or some simple multiple of the ratio of the weights in
which A and B combine directly with each other.
e.g.
When gases react together, they always do so in
volumes which bear a simple ratio to one another
and to the volumes of the product, if these are also
gases, provided all measurements of volumes are
done under similar conditions of temperature and
pressure.
e.g.
Combination between hydrogen and chlorine to form
hydrogen chloride gas. One volume of hydrogen and
one volume of chlorine always combine to form two
volumes of hydrochloric acid gas.
The elements C and O combine separately with the
third element H to form CH4 and H2O and they combine
directly with each other to form CO2.
4 H2
CH4
12
C
12
H2O
CO2
16
O
32
In CH4, 12 parts by weight of carbon combine with 4
parts by weight of hydrogen. In H2O, 2 parts by weight
of hydrogen combine with 16 parts by weight of
oxygen. Thus the weight of C and O which combine
with fixed weight of hydrogen (say 4 parts by weight)
are 12 and 32 i.e. they are in the ratio 12 : 32 or 3 : 8.
Now in CO2, 12 parts by weight of carbon combine
directly with 32 parts by weight of oxygen i.e. they
combine directly in the ratio 12 : 32 or 3 : 8 which is
the same as the first ratio.

H2 (g) + Cl2 (g) 2HCl (g)
1vol. 1 vol.
2 vol.
The ratio between the volume of the reactants and
the product in this reaction is simple, i.e., 1 : 1 : 2.
Hence it illustrates the Law of combining volumes.
(f) Avogadro’s Hypothesis :
This states that equal volumes of all gases under
similar conditions of temperature and pressure
contain equal number of molecules.
This hypothesis has been found to explain elegantly
all the gaseous reactions and is now widely
recognized as a law or a principle known as Avogadro’s
Law or Avogadro’s principle.
The reaction between hydrogen and chlorine can be
explained on the basis of Avogadro’s Law as follows :
Note :
The law of reciprocal proportion was put forward by
Ritcher in 1794.
2.
Sample Problem :
Ammonia contains 82.35% of nitrogen and 17.65%
of hydrogen. Water contains 88.90% of oxygen and
11.10% of hydrogen. Nitrogen trioxide contains
63.15% of oxygen and 36.85% of nitrogen. Show that
these data illustrate the law of reciprocal proportions.
Hydrogen + Chlorine
Sol
In NH3, 17.65 g of H combine with N = 82.35 g
 1g of H combine with N =
82.35
17.65
88.90
11.10
1 vol.
2 vol.
n molecules.
n molecules.
2n molecules.(By Avogadro's Law)
1 molecules.
2
1 molecules.
2
1 molecules. (By dividing throughout by 2n)
1 Atom
1 Atom
1 Molecule
(By experiment)
(Applying Avogadro's hypothesis)
g = 4.67 g
In H2O, 11.10 g of H combine with O = 88.90 g
 1 g H combine with O =
Hydrogen chloride gas
1 vol.
g = 8.01 g
 Ratio of the weights of N and O which combine
It implies that one molecule of hydrogen chloride gas
is made up of 1 atom of hydrogen and 1 atom of
chlorine.
(i) Applications of Avogadro’s hypothesis :
with fixed weight (=1g) of H
= 4.67 : 8.01 = 1 : 1.72
(A) In the calculation of atomicity of elementary
gases.
In N2O3, ratio of weights of N and O which combine
with each other = 36.85 : 63.15 = 1 : 1.71
e.g.
Thus the two ratio are the same. Hence it illustrates
the law of reciprocal proportions.
2 volumes of hydrogen combine with 1 volume of
oxygen to form two volumes of water vapours.
Hydrogen + Oxygen  Water vapours
2 vol.
1 vol.
2 vol.
PAGE # 21
Applying Avogadro’s hypothesis
GRAM-ATOMIC MASS
Hydrogen
+ Oxygen  Water vapours
2 n molecules n molecules 2 n molecules
1
or 1 molecule
molecule
2
1 molecule
1
Thus1 molecule of water contains
molecule of
2
The atomic mass of an element expressed in grams
is called the Gram Atomic Mass of the element.
The number of gram -atoms
=
oxygen. But 1 molecule of water contains 1 atom of
oxygen. Hence.
1
molecule of oxygen = 1 atom of
2
e.g.
Calculate the gram atoms present in (i) 16g of oxygen
oxygen or 1 molecules of oxygen = 2 atoms of oxygen
i.e. atomicity of oxygen = 2.
and (ii) 64g of sulphur.
(B) To find the relationship between molecular mass
and vapour density of a gas.
(i) The atomic mass of oxygen = 16.
Density of gas
Vapour density (V.D.) = Density of hydrogen
=
Mass of a certain volume of the gas
Mass of the same volume of hydrogen at
the same temp. and pressure
 Gram-Atomic Mass of oxygen (O) = 16 g.
No. of Gram-Atoms =
V.D. =
64
64
= Gram Atomic Mass of sulphur =
=2
32
Atomic
Number
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Mass of n molecules of the gas
Mass of n molecules of hydrogen
=
Mass of 1 molecule of the gas
Mass of 1 molecule of hydrogen
=
Molecular mass of the gas
Molecular mass of hydrogen
Molecular mass
2
(since molecular mass of hydrogen is 2)
Hence, Molecular mass = 2 × Vapour density
=
ATOMIC MASS UNIT
The atomic mass unit (amu) is equal to one-twelfth
(1/12) of the mass of an atom of carbon-12.The mass
of an atom of carbon-12 isotope was given the atomic
mass of 12 units, i.e. 12 amu or 12 u.
The atomic masses of all other elements are now
expressed in atomic mass units.
RELATIVE ATOMIC MASS
The atomic mass of an element is a relative quantity
and it is the mass of one atom of the element relative
to one -twelfth (1/12) of the mass of one carbon-12
atom. Thus, Relative atomic mass
=
Mass of one atom of the element
1
12
 mass of one C  12 atom
[1/12 the mass of one C-12 atom = 1 amu, 1 amu =
1.66 × 10–24 g = 1.66 × 10–27 kg.]
Note :
One amu is also called one dalton (Da).
16
=1
16
(ii) The gram-atoms present in 64 grams of sulphur.
If n molecules are present in the given volume of a gas
and hydrogen under similar conditions of temperature
and pressure.

Mass of the element in grams
Gram Atomic mass of the element
Element
Symbol
Hydrogen
H
Helium
He
Lithium
Li
Beryllium
Be
Boron
B
Carbon
C
Nitrogen
N
Oxygen
O
Fluorine
F
Neon
Ne
Sodium
Na
Magnesium
Mg
Aluminium
Al
Silicon
Si
Phosphorus
P
Sulphur
S
Chlorine
Cl
Argon
Ar
Potassium
K
Calcium
Ca
Atomic
mass
1
4
7
9
11
12
14
16
19
20
23
24
27
28
31
32
35.5
40
39
40
RELATIVE MOLECULAR MASS
The relative molecular mass of a substance is the
mass of a molecule of the substance as compared
to one-twelfth of the mass of one carbon -12 atom
i.e.,
Relative molecular mass
=
Mass of one molecule of the substance
1
12
 mass of one C  12 atom
The molecular mass of a molecule, thus, represents
the number of times it is heavier than 1/12 of the
mass of an atom of carbon-12 isotope.
PAGE # 22
GRAM MOLECULAR MASS
The molecular mass of a substance expressed in
grams is called the Gram Molecular Mass of the
substance . The number of gram molecules
Mass of the subs tance in grams
=
Gram molecular mass of the subs t ance
e.g.
(i) Molecular mass of hydrogen (H2) = 2u.
 Gram Molecular Mass of hydrogen (H2) = 2 g .
(ii) Molecular mass of methane (CH4) = 16u
 Gram Molecular Mass of methane (CH4) = 16 g.
EQUIVALENT MASS
(a) Definition :
Equivalent mass of an element is the mass of the
element which combine with or displaces 1.008 parts
by mass of hydrogen or 8 parts by mass of oxygen or
35.5 parts by mass of chlorine.
(b) Formulae of Equivalent Masses of different
substances :
(i) Equivalent mass of an element =
Atomic wt. of the element
Valency of the element
(ii) Eq. mass an acid =
Mol. wt. of the acid
Basicity of the acid
e.g. the number of gram molecules present in 64 g of
methane (CH4).
Basicity is the number of replaceable H+ ions from
one molecule of the acid.
64
64
= Gram molecular mass of CH =
= 4.
4
16
(iii) Eq. Mass of a base =
(a) Calculation of Molecular Mass :
The molecular mass of a substance is the sum of
the atomic masses of its constituent atoms present
in a molecule.
Ex.1 Calculate the molecular mass of water.
(Atomic masses : H = 1u, O = 16u).
Sol. The molecular formula of water is H2O.
Molecular mass of water = ( 2 × atomic mass of H)
+ (1 × atomic mass of O)
= 2 × 1 + 1 × 16 = 18
i.e., molecular mass of water = 18 amu.
Ex.2 Find out the molecular mass of sulphuric acid.
(Atomic mass : H = 1u, O = 16u, S = 32u).
Sol. The molecular formula of sulphuric acid is H2SO4.
 Molecular mass of H2SO4
= (2 × atomic mass of H) + ( 1 × atomic mass of S)
+ ( 4 × atomic mass of O)
= (2 × 1) + (1× 32) + (4×16) = 2 + 32 + 64 = 98
i.e., Molecular mass of H2SO4= 98 amu.
FORMULA MASS
The term ‘formula mass’ is used for ionic compounds
and others where discrete molecules do not exist,
e.g., sodium chloride, which is best represented as
(Na+Cl–)n, but for reasons of simplicity as NaCl or
Na+Cl–. Here, formula mass means the sum of the
masses of all the species in the formula.
Thus, the formula mass of sodium chloride = (atomic
mass of sodium) + (atomic mass of chlorine)
= 23 + 35.5
= 58.5 amu
Mol. wt. of the base
Acidity of the base
Acidity is the number of replaceable OH– ions from
one molecule of the base
(iv) Eq. mass of a salt
=
Mol. wt. of the salt
Number of metal atoms  valency of metal
(v) Eq. mass of an ion =
Formula wt. of the ion
Charge on the ion
(vi) Eq. mass of an oxidizing/reducing agent
=
Mol wt. or At. wt
No. of electrons lost or gained by one
molecule of the substance
Equivalent weight of some compounds are given in
the table :
S.No.
Compound
Equivalent
weight
1
HCl
36.5
2
H2SO4
49
3
HNO3
63
4
45
COOH
5
COOH
.2H2O
63
6
NaOH
40
7
KOH
56
8
CaCO3
50
9
NaCl
58.5
10
Na2CO3
53
PAGE # 23
SOME IMPORTANT RELATIONS AND FORMULAE
In Latin, mole means heap or collection or pile. A
mole of atoms is a collection of atoms whose total
mass is the number of grams equal to the atomic
mass in magnitude. Since an equal number of moles
of different elements contain an equal number of
atoms, it becomes convenient to express the
amounts of the elements in terms of moles. A mole
represents a definite number of particles, viz, atoms,
molecules, ions or electrons. This definite number is
called the Avogadro Number (now called the Avogadro
constant) which is equal to 6.023 × 1023.
(i) 1 mole of atoms = Gram Atomic mass = mass of
6.023 × 1023 atoms
(ii) 1 mole of molecules = Gram Molecular Mass
= 6.023 x 1023 molecules
(iii) Number of moles of atoms
=
(iv) Number of moles of molecules
A mole is defined as the amount of a substance that
contains as many atoms, molecules, ions, electrons
or other elementary particles as there are atoms in
exactly 12 g of carbon -12 (12C).
=
(a) Moles of Atoms :
=
(i) 1 mole atoms of any element occupy a mass which
is equal to the Gram Atomic Mass of that element.
e.g. 1 Mole of oxygen atoms weigh equal to Gram
Atomic Mass of oxygen, i.e. 16 grams.
(ii) The symbol of an element represents 6.023 x 1023
atoms (1 mole of atoms) of that element.
e.g : Symbol N represents 1 mole of nitrogen atoms
and 2N represents 2 moles of nitrogen atoms.
(b) Moles of Molecules :
(i) 1 mole molecules of any substance occupy a mass
which is equal to the Gram Molecular Mass of that
substance.
Mass of element in grams
Gram Atomic Mass of element
Mass of substance in grams
Gram Molecular Mass of substance
(v) Number of moles of molecules
No. of molecules of element
N

Avogadro number
NA
Ex.3 To calculate the number of moles in 16 grams of
Sulphur (Atomic mass of Sulphur = 32 u).
Sol. 1 mole of atoms = Gram Atomic Mass.
So, 1 mole of Sulphur atoms = Gram Atomic Mass of
Sulphur = 32 grams.
Now, 32 grams of Sulphur = 1 mole of Sulphur
So, 16 grams of Sulphur
= (1/32) x 16 = 0.5 moles
Thus, 16 grams of Sulphur constitute 0.5 mole of
Sulphur.
22.4 litre
e.g. : 1 mole of water (H2O) molecules weigh equal to
Gram Molecular Mass of water (H2O), i.e. 18 grams.
In term of
volume
23
(ii) The symbol of a compound represents 6.023 x
10 23 molecules (1 mole of molecules) of that
compound.
6.023 × 10
(NA) Atoms
23
1 Mole
e.g. : Symbol H 2O represents 1 mole of water
molecules and 2 H2O represents 2 moles of water
molecules.

Note :
The symbol H2O does not represent 1 mole of H2
molecules and 1 mole of O atoms. Instead, it
represents 2 moles of hydrogen atoms and 1 mole
of oxygen atoms.

Note :
The SI unit of the amount of a substance is Mole.
(c) Mole in Terms of Volume :
Volume occupied by 1 Gram Molecular Mass or 1
mole of a gas under standard conditions of
temperature and pressure (273 K and 1atm.
pressure) is called Gram Molecular Volume. Its value
is 22.4 litres for each gas.
Volume of 1 mole
= 22.4 litre (at STP)

Note :
The term mole was introduced by Ostwald in 1896.
6.023 × 10
(NA) molecules
In terms of
particles
In terms of
mass
1 gram atom
of element
1 gram molecule
of substance
1 gram formula
mass of substance
PROBLEMS BASED ON THE MOLE CONCEPT
Ex.4 Calculate the number of moles in 5.75 g of
sodium. (Atomic mass of sodium = 23 u)
Sol. Number of moles
Mass of the element in grams
=
Gram Atomic Mass of element
=
5.75
23
= 0.25 mole
or,
1 mole of sodium atoms = Gram Atomic mass of
sodium = 23g.
23 g of sodium = 1 mole of sodium.
5.75 g of sodium =
5.75
23
mole of sodium = 0.25 mole
PAGE # 24
Ex.5 What is the mass in grams of a single atom of
chlorine ? (Atomic mass of chlorine = 35.5u)
Sol. Mass of 6.022 × 1023 atoms of Cl = Gram Atomic
Mass of Cl = 35.5 g.
 Mass of 1 atom of Cl =
35.5 g
6.022  10 23
= 5.9 × 10–23 g.
Ex.6 The density of mercury is 13.6 g cm . How many
moles of mercury are there in 1 litre of the metal ?
(Atomic mass of Hg = 200 u).
–3
Sol. Mass of mercury (Hg) in grams = Density
(g cm–3)× Volume (cm3)
= 13.6 g cm–3 × 1000 cm3 = 13600 g.
 Number of moles of mercury
Mass of mercury in grams
13600
= Gram Atomic Mass of mercury =
= 68
200
Ex.7 The mass of a single atom of an element M is
3.15× 10–23 g . What is its atomic mass ? What
could the element be ?
Sol. Gram Atomic Mass = mass of 6.022 × 1023 atoms
= mass of 1 atom × 6.022 × 1023
= (3.15 × 10–23g) × 6.022 × 1023
= 3.15 × 6.022 g = 18.97 g.
 Atomic Mass of the element = 18.97u
Thus, the element is most likely to be fluorine.
Ex.8 An atom of neon has a mass of 3.35 × 10–23 g.
How many atoms of neon are there in 20 g of the
gas ?
Sol. Number of atoms
=
20
Total mass
=
= 5.97 × 1023
Mass of 1 atom
3.35  10 – 23
Ex.9 How many grams of sodium will have the same
number of atoms as atoms present in 6 g of
magnesium ?
(Atomic masses : Na = 23u ; Mg =24u)
Sol. Number of gram -atom of Mg
Mass of Mg in grams
1
6
= Gram Atomic Mass =
=
24
4
 Gram Atoms of sodium should be =
1
4
 1 Gram Atom of sodium = 23 g
1
4
gram atoms of sodium = 23 ×
1
= 5.75 g
4
Ex.10 How many moles of Cr are there in 85g of Cr2S3 ?
(Atomic masses : Cr = 52 u , S =32 u)
Sol. Molecular mass of Cr2S3 = 2 × 52 + 3 × 32 = 104
+ 96 = 200 u.
200g of Cr2S3 contains = 104 g of Cr.
 85 g of Cr2S3 contains =
104  85
200
Thus, number of moles of Cr =
g of Cr = 44.2g
44.2
52
Ex.11 What mass in grams is represented by
(a) 0.40 mol of CO2,
(b) 3.00 mol of NH3,
(c) 5.14 mol of H5IO6
(Atomic masses : C=12 u, O=16 u, N=14 u,
H=1 u and I = 127 u)
Sol. Weight in grams = number of moles × molecular
mass.
Hence,
(a) mass of CO2 = 0.40 × 44 = 17.6 g
(b) mass of NH3 = 3.00 × 17 = 51.0 g
(c) mass of H5IO6 = 5.14 × 228 = 1171.92g
Ex.12 Calculate the volume in litres of 20 g of hydrogen
gas at STP.
Sol. Number of moles of hydrogen
Mass of hydrogen in grams
=
Gram Molecular Mass of hydrogen
=
20
= 10
2
 Volume of hydrogen = number of moles × Gram
Molecular Volume.
= 10 ×22.4 = 224 litres.
Ex.13 The molecular mass of H 2 SO 4 is 98 amu.
Calculate the number of moles of each element
in 294 g of H2SO4.
Sol. Number of moles of H2SO4 =
294
98
=3.
The formula H 2SO 4 indicates that 1 molecule of
H2SO4 contains 2 atoms of H, 1 atom of S and 4
atoms of O. Thus, 1 mole of H2SO4 will contain 2
moles of H,1 mole of S and 4 moles of O atoms
Therefore, in 3 moles of H2SO4 :
Number of moles of H = 2 × 3 = 6
Number of moles of S = 1 × 3 = 3
Number of moles of O = 4 × 3 = 12
Ex.14 Find the mass of oxygen contained in 1 kg of
potassium nitrate (KNO3).
Sol. Since 1 molecule of KNO 3 contains 3 atoms of
oxygen, 1 mol of KNO 3 contains 3 moles of
oxygen atoms.
 Moles of oxygen atoms = 3 × moles of KNO3
=3×
1000
= 29.7
101
(Gram Molecular Mass of KNO3 = 101 g)
 Mass of oxygen = Number of moles × Atomic
mass
= 29.7 × 16 = 475.2 g
Ex.15 You are asked by your teacher to buy 10 moles of
distilled water from a shop where small bottles
each containing 20 g of such water are available.
How many bottles will you buy ?
Sol. Gram Molecular Mass of water (H2O) = 18 g
 10 mol of distilled water = 18 × 10 = 180 g.
Because 20 g distilled water is contained in 1
bottle,
180 g
of distilled water is contained in =
= 0.85 .
bottles = 9 bottles.
 Number of bottles to be bought = 9
PAGE # 25
180
20
Ex.16 6.022 × 1023 molecules of oxygen (O2) is equal to
how many moles ?
Sol. No. of moles =
No. of molecules of oxygen
Avogadro' s no. of molecules
N

N
=
A
6.023  10 23
6.023  10 23
(iii) If the simplest ratio is fractional, then values of
simplest ratio of each element is multiplied by
smallest integer to get the simplest whole number
for each of the element.
=1
(iv) To get the empirical formula, symbols of various
elements present are written side by side with their
respective whole number ratio as a subscript to the
lower right hand corner of the symbol.
PERCENTAGE COMPOSITION
The percentage composition of elements in a
compound is calculated from the molecular formula
of the compound.
The molecular mass of the compound is calculated
from the atomic masses of the various elements
present in the compound. The percentage by mass
of each element is then computed with the help of the
following relations.
Percentage mass of the element in the compound
(v) The molecular formula of a substance may be
determined from the empirical formula if the molecular
mass of the substance is known. The molecular
formula is always a simple multiple of empirical
formula and the value of simple multiple (n) is
obtained by dividing molecular mass with empirical
formula mass.
n
=
Total mass of the element
=
× 100
Molecular mass
Ex.17 What is the percentage of calcium in calcium
carbonate (CaCO3) ?
Sol. Molecular mass of CaCO 3 = 40 + 12 + 3 × 16
= 100 amu.
Mass of calcium in 1 mol of CaCO3 = 40g.
Percentage of calcium =
40  100
Ex-20 A compound of carbon, hydrogen and nitrogen
contains these elements in the ratio of 9:1:3.5 respectively.
Calculate the empirical formula. If its molecular mass is
108, what is the molecular formula ?
Sol.
32  100
98
Sol. Molecular mass of water, H2O = 2 + 16 = 18 amu.
H2O has two atoms of hydrogen.
So, total mass of hydrogen in H2O = 2 amu.
2  100
18
= 11.11 %
Similarly,
percentage of oxygen =
16  100
18
Mass Atomic Relative Number
Ratio Mass
of Atoms
Carbon
9
12
Hydrogen
1
1
Nitrogen
3.5
14
= 88.88 %
The following steps are involved in determining the
empirical formula of a compound :
(i) The percentage composition of each element is
divided by its atomic mass. It gives atomic ratio of the
elements present in the compound.
(ii) The atomic ratio of each element is divided by the
minimum value of atomic ratio as to get the simplest
ratio of the atoms of elements present in the
compound.
9
 0.75
12
Simplest
Ratio
0.75×4 = 3
1
 1
1
1×4=4
3.5
 0.25
14
0.25 ×4 = 1
Empirical ratio = C3H4N
Empirical formula mass = (3 × 12) + (4× 1) + 14 = 54
= 32.65 %
Ex.19 W hat are the percentage compositions of
hydrogen and oxygen in water (H2O) ?
(Atomic masses : H = 1 u, O = 16 u)
Percentage of H =
Element
= 40 %
100
Ex.18 What is the percentage of sulphur in sulphuric
acid (H2SO4) ?
Sol. Molecular mass of H2SO4 = 1 × 2 + 32 + 16 × 4 = 98
amu.
 Percentage of sulphur =
Molecular Mass
Empirical Formula Mass
n=
Molecular Mass
108
2
Empirical Formula Mass = 54
Thus, molecular formula of the compound
= (Empirical formula)2
= (C3H4N)2
= C6H8N2
Ex.21 A compound on analysis, was found to have the
following composition :
(i) Sodium = 14.31%, (ii) Sulphur = 9.97%, (iii) Oxygen
= 69.50%, (iv) Hydrogen = 6.22%. Calculate the
molecular formula of the compound assuming that
whole hydrogen in the compound is present as water
of crystallisation. Molecular mass of the compound
is 322.
Sol.
Element Percentage
Atomic
mass
Relative Number
of atoms
Simplest ratio
Sodium
14.31
23
14 .31

23
0.622
0.622
2
0.311
Sulphur
9.97
32
9 .97

32
0.311
0.311
1
0.311
Hydrogen
6.22
1
6.22
6.22
 20
0.311
Oxygen
69.50
16
4.34
4.34
 14
0.311
6 . 22
1
69. 50

16

PAGE # 26
The empirical formula = Na2SH20O14
Empirical formula mass
= (2 × 23) + 32 + (20 × 1) + (14 × 16)
= 322
Molecular mass = 322
Molecular formula = Na2SH20O14
Whole of the hydrogen is present in the form of water
of crystallisation. Thus, 10 water molecules are
present in the molecule.
So, molecular formula = Na2SO4. 10H2O
Normality of a solution is defined as the number of
gram equivalents of the solute dissolved per litre (dm3)
of given solution. It is denoted by ‘N’.
Mathematically,
Number of gram equivalent s of solute
N =
N=
Weight of solute in gram / equivalent weight of solute
Volume of the solution in litre
Strength in grams per litre
(a) Strength in g/L :
N=
The strength of a solution is defined as the amount of
the solute in grams present in one litre (or dm3) of the
solution, and hence is expressed in g/litre or g/dm3.
Equivalent mass of solute
=
S
E
If ‘w’ gram of the solute is present in V cm3 of a given
solution.
1000
w
Weight of solute in gram
N = Equivalent mass of the solute ×
Volume of solution in litre
(b) Molarity :
Molarity of a solution is defined as the number of
moles of the solute dissolved per litre (or dm3) of
solution. It is denoted by ‘M’. Mathematically,
w
N = Equivalent mass of the solute ×
N=
Volume of the solution in litre
V
e.g. A solution of sulphuric acid having 0.49 gram of
it dissolved in 250 cm 3 of solution will have its
normality,
Number of moles of solute
M=
Volume of the solution in litre
N can be calculated from the strength as given below :
CONCENTRATION OF SOLUTIONS
Strength in g/L =
(d) Normality :
0.49
49
1000
×
250
1000
V
= 0.04
(Eq. mass of sulphuric acid = 49).
Mass of solute in gram/Gram Molecular Mass of solute
Volume of solution in litre
M can be calculated from the strength as given below :
M=
1000
w
Molecular
×
mass
V
e.g. a solution of sulphuric acid having 4.9 grams of it
dissolved in 500 cm3 of solution will have its molarity,
w
M=
Normality
Deci
normal
1
2
1000
Molecular mass
4.9
98
1000
×
500
×
Centi
normal
1
10
1
100
Some Important Formulae :
Molecular mass of solute
If ‘w’ gram of the solute is present in V cm of a given
solution , then
M=
Semi
normal
Strength in grams per litre
3
M=
Solution
V
= 0.1
(c) Formality :
In case of ionic compounds like NaCl, Na2CO3 etc.,
formality is used in place of molarity. The formality of
a solution is defined as the number of gram formula
masses of the solute dissolved per litre of the
solution. It is represented by the symbol ‘F’. The term
formula mass is used in place of molecular mass
because ionic compounds exist as ions and not as
molecules. Formula mass is the sum of the atomic
masses of the atoms in the formula of the compound.
Mass of solute in gram/Formula Mass of solute
Volume of solution in litre
(i) Milli equivalent of substance = N × V
where , N  normality of solution
V  Volume of solution in mL
(ii) If weight of substance is given,
milli equivalent (NV) =
w  1000
E
Where, W  Weight of substance in gram
E  Equivalent weight of substance
(iii) S = N × E
S  Strength in g/L
N  Normality of solution
E  Equivalent weight
(iv) Calculation of normality of mixture :
N
N
HCl is mixed with 50 ml of
H SO .
10
5 2 4
Find out the normality of the mixture.
Sol. Milli equivalent of HCl + milli equivalent H 2SO 4
= milli equivalent of mixture
{ where, V3 =V1 + V2 )
N1 V1 + N2 V2 = N3 V3
Ex.22 100 ml of
 1
 1

 100     50   N3 × 150

10
5

 

N3 =
20
150
=
2
15
= 0.133
PAGE # 27
Ex.23 100 ml of
N
HCl is mixed with 25 ml of
10
Find out the normality of the mixture.
N
5
NaOH.
Sol. Milli equivalent of HCl – milli equivalent of NaOH
= milli equivalent of mixture
N1 V1 – N2 V2 = N3 V3
{ where, V3 =V1 + V2 )
 1
 1

 100  –   25  = N3 × 125

10
5

 

N3 =

1
25
Note :
1 milli equivalent of an acid neutralizes 1 milli
equivalent of a base.
(e) Molality :
Molality of a solution is defined as the number of
moles of the solute dissolved in 1000 grams of the
solvent. It is denoted by ‘m’.
Mathematically,
Number of moles of the solute
× 1000 ‘m’ can
Weight of the solvent in gram
be calculated from the strength as given below :
m=
Strength per 1000 gram of solvent
m=
Molecular mass of solute
If ‘w’ gram of the solute is dissolved in ‘W’ gram of the
solvent then
m=
w
1000
×
Mol. mass of the solute
W
e.g. A solution of anhydrous sodium carbonate
(molecular mass = 106) having 1.325 grams of it,
dissolved in 250 gram of water will have its molality m=

1.325 1000

= 0.05
106 250
Ex.25 Find the molarity and molality of a 15% solution
of H2SO4 (density of H2SO4 solution = 1.02 g/cm3)
(Atomic mass : H = 1u, O = 16u , S = 32 u)
Sol. 15% solution of H2SO4 means 15g of H2SO4 are
present in 100g of the solution i.e.
Wt. of H2SO4 dissolved = 15 g
Weight of the solution = 100 g
Density of the solution = 1.02 g/cm3 (Given)
Calculation of molality :
Weight of solution = 100 g
Weight of H2SO4 = 15 g
Wt. of water (solvent) = 100 – 15 = 85 g
Molecular weight of H2SO4 = 98
 15 g H2SO4 =
15
98
= 0.153 moles
Thus ,85 g of the solvent contain 0.153 moles .
1000 g of the solvent contain=
0.153
85
× 1000 = 1.8 mole
Hence ,the molality of H2SO4 solution = 1.8 m
Calculation of molarity :
15 g of H2SO4 = 0.153 moles
Wt. of solution
Vol. of solution =
=
100
1.02
Density of solution
= 98.04 cm3
This 98.04 cm3 of solution contain H2SO4 = 0.153
moles
1000 cm3 of solution contain H2SO4
0.153
=
98.04
× 1000 = 1.56 moles
Note :
Hence the molarity of H2SO4 solution = 1.56 M
Relationship Between Normality and Molarity of a
Solution :
Normality of an acid = Molarity × Basicity
Normality of base = Molarity × Acidity
(f) Mole Fraction :
Ex.24 Calculate the molarity and normality of a solution
containing 0.5 g of NaOH dissolved in 500 cm 3
of solvent.
Sol. Weight of NaOH dissolved = 0.5 g
Volume of the solution = 500 cm3
(i) Calculation of molarity :
Molecular weight of NaOH = 23 + 16 + 1 = 40
Weight of solute/ molecular weight of solute
Volume of solution in litre
0.5/40
Molarity =
=
500/1000
= 0.025
(ii) Calculation of normality :
Normality
Weight of solute/ equivalent weight of solute
=
Volume of solution in litre
0.5/40
=
= 0.025
500/1000
The ratio between the moles of solute or solvent to
the total moles of solution is called mole fraction.
Moles of solute
mole fraction of solute =
=
Moles of solution

n
nN
w/m
w/m  W/M
Moles of solvent
Mole fraction of solvent =
Moles of solution

N
nN =
W/M
w/m  W/M
where,
n  number of moles of solute
N  number of moles of solvent
m  molecular weight of solute
M  molecular weight of solvent
w  weight of solute
W  weight of solvent
PAGE # 28
Ex.26 Find out the mole fraction of solute in 10% (by weight)
urea solution.
weight of solute (urea) = 10 g
weight of solution = 100 g
weight of solvent (water) = 100 – 10 = 90g
Moles of solution
10 / 60
w/m
w/m  W/M

=
10 / 60  90 / 18
2KClO3(s)  2KCl(s) + 3O2(g)
2 mole KClO3  3 mole O2
 3 mole O2 formed by 2 mole KClO3
2
Moles of solute
mole fraction of solute =
Sol. Decomposition of KClO3 takes place as,
=
= 0.032
Note :
Sum of mole fraction of solute and solvent is always
equal to one.

 2.4 mole O 2 will be formed by  3  2.4 


mole KClO3 = 1.6 mole KClO3
Mass of KClO3 = Number of moles × molar mass
= 1.6 × 122.5 = 196 g
(ii) Calculations based on mass-mass relationship:
In making necessary calculation, following steps are
followed (a) Write down the balanced chemical equation.
STOICHIOMETRY
(a) Quantitative Relations in Chemical
Reactions :
Stoichiometry is the calculation of the quantities of
reactants and products involved in a chemical
reaction.
It is based on the chemical equation and on the
relationship between mass and moles.
N2(g) + 3H2(g)  2NH3(g)
(b) Write down theoretical amount of reactants and
products involved in the reaction.
(c) The unknown amount of substance is calculated
using unitary method.
Ex.28 Calculate the mass of CaO that can be prepared
by heating 200 kg of limestone CaCO3 which is
95% pure.
Sol. Amount of pure CaCO 3 =
95
 200 = 190 kg
100
A chemical equation can be interpreted as follows -
= 190000 g
1 molecule N2 + 3 molecules H2  2 molecules
NH 3 (Molecular interpretation)
CaCO3(s)  CaO(s) + CO2(g)
1 mol N2 + 3 mol H2  2 mol NH3
(Molar interpretation)
100 g CaCO3  56 g CaO
28 g N2 + 6 g H2  34 g NH3
(Mass interpretation)
1 volume N2 + 3 volume H2  2 volume NH3
(Volume interpretation)
Thus, calculations based on chemical equations are
divided into four types (i) Calculations based on mole-mole relationship.
(ii) Calculations based on mass-mass relationship.
(iii) Calculations based on mass-volume relationship.
(iv) Calculations based on volume -volume
relationship.
(i) Calculations based on mole-mole relationship :
In such calculations, number of moles of reactants
are given and those of products are required.
Conversely, if number of moles of products are given,
then number of moles of reactants are required.
Ex.27 Oxygen is prepared by catalytic decomposition
of potassium chlorate (KClO 3). Decomposition
of potassium chlorate gives potassium chloride
(KCl) and oxygen (O2). How many moles and how
many grams of KClO 3 are required to produce
2.4 mole O2.
1 mole CaCO3  1 mole CaO
100 g CaCO3 give 56 g CaO
56
 190000 g CaCO3 will give=
× 190000 g CaO
100
= 106400 g = 106.4 kg
Ex.29 Chlorine is prepared in the laboratory by treating
manganese dioxide (MnO 2 ) with aqueous
hydrochloric acid according to the reaction MnO2 + 4HCl  MnCl2 + Cl2 + 2H2O
How many grams of HCl will react with 5 g MnO2 ?
Sol. 1 mole MnO2 reacts with 4 mole HCl
or 87 g MnO2 react with 146 g HCl
 5 g MnO2 will react with =
146
× 5 g HCl = 8.39 g HCl
87
Ex.30 How many grams of oxygen are required to burn
completely 570 g of octane ?
Sol. Balanced equation
2C8H18 + 25O2
2 mole
2 × 114
16CO2 + 18H2O
25 mole
25 × 32
First method : For burning 2 × 114 g of the octane,
oxygen required = 25 × 32 g
For burning 1 g of octane, oxygen required =
25  32
g
2  114
Thus, for burning 570 g of octane, oxygen required
=
25  32
× 570 g = 2000 g
2  114
PAGE # 29
Mole Method : Number of moles of octane in 570
grams
Ex-33 What quantity of copper (II) oxide will react with
2.80 litre of hydrogen at STP ?
570
= 5.0
114
Sol. CuO +
1 mol
79.5 g
For burning 2.0 moles of octane, oxygen required
= 25 mol = 25 × 32 g
For burning 5 moles of octane, oxygen required
=
25  32
× 5.0 g = 2000 g
2 .0
Proportion Method : Let x g of oxygen be required for
burning 570 g of octane. It is known that 2 × 114 g of
the octane requires 25 × 32 g of oxygen; then, the
proportion.
x
25  32 g oxygen
=
570 g oc tane
2  114 g oc tan e
2Fe2O3 + 8H2SO4
4 mole
4 × 120 g
8 mole
8 × 98 g
4 × 120 g of FeS2 yield H2SO4 = 8 × 98 g
1000 g of FeS2 will yield H2SO4 =
8  98
× 1000
4  120
= 1633.3 g
(iii) Calculations involving mass-volume relationship :
In such calculations masses of reactants are given
and volume of the product is required and vice-versa.
1 mole of a gas occupies 22.4 litre volume at STP.
Mass of a gas can be related to volume according to
the following gas equation PV = nRT
PV =
w
RT
m
NH4Cl(s)
=
79.5
× 2.80 g = 9.93 g
22.4
Ex-34 Calculate the volume of carbon dioxide at STP
evolved by strong heating of 20 g calcium carbonate.
Sol. The balanced equation is -
CaCO3
NH3(g) + HCl(g)
1 mol
 53.5 g NH4Cl give 1 mole NH3
1
 26.75 g NH4Cl will give
× 26.75 mole NH3
53.5
= 0.5 mole
PV = nRT
1 ×V = 0.5 × 0.0821 × 300
V = 12.315 litre
CaO +
1 mol
100 g
CO2
1 mol
= 22.4 litre at STP
22.4
× 20 = 4.48 litre
100
Ex.35 Calculate the volume of hydrogen liberated at 27ºC
and 760 mm pressure by heating 1.2 g of magnesium
with excess of hydrochloric acid.
Sol. The balanced equation is
Mg + 2HCl
MgCl2
+
24 g
H2
1 mol
24 g of Mg liberate hydrogen = 1 mole
1.2 g Mg will liberate hydrogen = 0.05 mole
PV = nRT
1 × V = 0.05 × 0.0821 × 300
V = 1.2315 litre
(iv) Calculations based on volume volume
relationship :
These calculations are based on two laws :
(i) Avogadro’s law
(ii) Gay-Lussac’s Law
e.g.
N 2(g)
+
1 mol
1 × 22.4 L
3H 2(g)
3 mol
3 × 22.4 L
2NH 3(g) (Avogadro's law)
2 mol
2 × 22.4 L
(under similar conditions of temperature and
pressure, equal moles of gases occupy equal
volumes)
N2(g)
+
1 vol
Ex-32. What volume of NH3 can be obtained from 26.75 g
of NH4Cl at 27ºC and 1 atmosphere pressure.
Sol. The balanced equation is 1 mol
53.5 g
22.4 litre of hydrogen at STP reduce CuO = 79.5 g
2.80 litre of hydrogen at STP will reduce CuO
=
Ex.31 How many kilograms of pure H 2SO 4 could be
obtained from 1 kg of iron pyrites (FeS2) according to
the following reactions ?
4FeS2 + 11O2  2Fe2O3 + 8SO2
2SO2 + O2  2SO3
SO3 + H2O  H2SO4
Sol. Final balanced equation,
4FeS2 + 15O2 + 8H2O
+ H2O
100 g of CaCO3 evolve carbon dioxide = 22.4 litre
20 g CaCO3 will evolve carbon dioxide
25  32  570
= 2000 g
2  114
x=
Cu
H2
1 mol
22.4 litre at NTP
3H2(g)
3 vol
2NH3(g) (Gay- Lussac's Law)
2 vol
(under similar conditions of temperature and
pressure, ratio of coefficients by mole is equal to ratio
of coefficient by volume).
Ex-36 One litre mixture of CO and CO2 is taken. This is
passed through a tube containing red hot charcoal.
The volume now becomes 1.6 litre. The volume are
measured under the same conditions. Find the
composition of mixture by volume.
Sol. Let there be x mL CO in the mixture , hence, there will
be (1000 – x) mL CO2. The reaction of CO2 with red
hot charcoal may be given as -
CO2(g) + C(s)
1 vol.
(1000 –x)
2CO(g)
2 vol.
2(1000 – x)
Total volume of the gas becomes = x + 2(1000 – x)
x + 2000 – 2x = 1600
x = 400 mL
 volume of CO = 400 mL and volume of CO2 = 600 mL
PAGE # 30
Ex-37 What volume of air containing 21% oxygen by volume
is required to completely burn 1kg of carbon containing
100% combustible substance ?
Sol. Combustion of carbon may be given as,
C(s) + O2(g)
1 mol
12 g
CO2(g)
1 mol
32 g
USEFUL FORMULAE FOR
VOLUMETRIC CALCULATIONS
(i) milliequivalents = normality × volume in millilitres.
(ii) At the end point of titration, the two titrants, say 1
and 2, have the same number of milliequivalents,
i.e., N1V1 = N2V2, volume being in mL.
 12 g carbon requires 1 mole O2 for complete
combustion
1
 1000 mole O2 for
12
combustion, i.e. , 83.33 mole O2
 1000 g carbon will require
m.e.
.
1000
(iv) No. of equivalents for a gas =
(iii) No. of equivalents =
Volumeat STP
equivalent volume( vol. of 1eq. at STP)
Volume of O2 at STP = 83.33 × 22.4 litre
(v) Strength in grams per litre = normality × equivalent
= 1866.66 litre
weight.
 21 litre O2 is present in 100 litre air
100
1866.66 litre O2 will be present in
× 1866.66 litre air
21
= 8888.88 litre or 8.89 × 103 litre
Ex-38 An impure sample of calcium carbonate contains
80% pure calcium carbonate 25 g of the impure
sample reacted with excess of hydrochloric acid.
Calculate the volume of carbon dioxide at STP
obtained from this sample.
Sol. 100 g of impure calcium carbonate contains = 80 g
pure calcium carbonate
25 g of impure calcium carbonate sample will contain
=
80
× 25 = 20 g pure calcium carbonate
100
The desired equation is CaCO3 + 2HCl
(vi) (a) Normality = molarity × factor relating mol. wt.
and eq. wt.
(b) No. of equivalents = no. of moles × factor relat
ing mol. wt. and eq. wt.
Ex.39 Calculate the number of milli equivalent of H2SO4
present in 10 mL of N/2 H2SO4 solution.
1
× 10 = 5.
2
Ex.40 Calculate the number of m.e. and equivalents of
Sol. Number of m.e. = normality × volume in mL =
NaOH present in 1 litre of N/10 NaOH solution.
Sol. Number of m.e. = normality × volume in mL
=
1
× 1000 = 100
10
Number of equivalents =
CaCl2 + CO2 + H2O
22.4 litre
at STP
1 mol
100 g
100 g pure CaCO3 liberate = 22.4 litre CO2.
20 g pure CaCO3 liberate =
22.4
 20
100
= 4.48 litre CO2
VOLUMETRIC CALCULATIONS
no. of m.e.
100
=
= 0.10
1000
1000
Ex.41 Calculate number of m.e. of the acids present in
(i) 100 mL of 0.5 M oxalic acid solution.
(ii) 50 mL of 0.1 M sulphuric acid solution.
Sol. Normality = molarity × basicity of acid
(i) Normality of oxalic acid = 0.5 × 2 = 1 N
m.e. of oxalic acid = normality × vol. in mL = 1 × 100
= 100.
(ii) Normality of sulphuric acid = 0.1 × 2 = 0.2 N
m.e. of sulphuric acid
= 0.2 × 50 = 10
The quantitative analysis in chemistry is primarily
carried out by two methods, viz, volumetric analysis
and gravimetric analysis.In the first method the mass
of a chemical species is measured by measurement
of volume, whereas in the second method it is determined by taking the weight.
Ex.42 A 100 mL solution of KOH contains 10
The strength of a solution in volumetric analysis is
generally expressed in terms of normality, i.e., number of equivalents per litre but since the volume in the
volumetric analysis is generally taken in millilitres
(mL), the normality is expressed by milliequivalents
per millilitre.
Again, strength in grams/litre = normality × eq. wt.
milliequivalents of KOH. Calculate its strength in normality and grams/litre.
no. of m.e.
Sol. Normality = volume in mL
=
10
 0 .1
100
 strength of the solution = N/10
=
1
 56 = 5.6 gram/litre.
10


molecular wt. 56
 eq. wt. of KOH 

 56 
acidity
1


PAGE # 31
Ex.43 What is strength in gram/litre of a solution of H2SO4,
N
12 cc of which neutralises 15 cc of
NaOH
10
solution ?
Sol. m.e. of NaOH solution =
1
× 15 = 1.5
10
m.e. of 12 cc of H2SO4 = 1.5
1.5
12
Strength in grams/litre = normality × eq. wt.
 normality of H2SO4 =
1.5
× 49 grams/litre
12
= 6.125 grams/litre.
=


molecular wt. 98
 eq. wt. of H2 SO 4 

 49 
basicity
2


Ex.44 What weight of KMnO4 will be required to prepare
N
solution if eq. wt. of KMnO4 is 31.6 ?
10
Sol. Equivalent weight of KMnO4 = 31.6
250 mL of its
1
10
Volume of solution (V) = 250 ml
Normality of solution (N) =
W
NEV ; W = 1  31 .6  250 31 .6
 0.79 g
10
1000
1000
40
Ex.45 100 mL of 0.6 N H2SO4 and 200 mL of 0.3 N HCl
were mixed together. What will be the normality of the
resulting solution ?
Sol. m.e. of H2SO4 solution = 0.6 × 100 = 60
m.e. of HCl solution = 0.3 × 200 = 60
 m.e. of 300 mL (100 + 200) of acidic mixture
= 60 + 60 = 120.
m.e.
total vol.
120
2
=
=
N.
300
5
Normality of the resulting solution =
Ex.46 A sample of Na2CO3. H2O weighing 0.62 g is added
to 100 mL of 0.1 N H2SO4. Will the resulting solution
be acidic, basic or neutral ?
0.62
Sol. Equivalents of Na2CO3. H2O =
= 0.01
62
124


 62 
 eq. wt. of Na 2 CO 3 .H2 O 
2


m.e. of Na2CO3. H2O = 0.01 × 1000 = 10
m.e. of H2SO4 = 0.1 × 100 = 10
Since the m.e. of Na2CO3. H2O is equal to that of H2SO4,
the resulting solution will be neutral.
(a) Introduction :
Volumetric analysis is a method of quantitative
analysis. It involves the measurement of the volume
of a known solution required to bring about the
completion of the reaction with a measured volume
of the unknown solution whose concentration or
strength is to be determined. By knowing the volume
of the known solution, the concentration of the solution
under investigation can be calculated. Volumetric
analysis is also termed as titrimetric analysis.
(b) Important terms used in volumetric
analysis :
(i) Titration : The process of addition of the known
solution from the burette to the measured volume of
solution of the substance to be estimated until the
reaction between the two is just complete, is termed
as titration. Thus, a titration involves two solutions:
(a) Unknown solution and (b) Known solution or standard solution.
(ii) Titrant : The reagent or substance whose solution is employed to estimate the concentration of unknown solution is termed titrant. There are two types
of reagents or titrants:
(A) Primary titrants : These reagents can be
accurately weighed and their solutions are not to be
standardised before use. Oxalic acid, potassium
dichromate, silver nitrate, copper sulphate, ferrous
ammonium sulphate, sodium thiosulphates etc., are
the examples of primary titrants.
(B) Secondary titrants : These reagents cannot
accurately weighed and their solutions are to be
standardised before use. Sodium hydroxide,
potassium hydroxide, hydrochloric acid, sulphuric
acid, iodine, potassium permanganate etc. are the
examples of secondary titrants.
(iii) Standard solution : The solution of exactly known
concentration of the titrant is called the standard
solution.
(iv) Titrate : The solution consisting the substance to
be estimated is termed unknown solution. The
substance is termed titrate.
(v) Equivalence point : The point at which the reagent
(titrant) and the substance (titrate) under investigation
are chemically equivalent is termed equivalence point
or end point.
(vi) Indicator : It is the auxiliary substance used for
physical (visual) detection of the completion of titration
or detection of end point is termed as indicator.
Indicators show change in colour or turbidity at the
stage of completion of titration.
(c) Concentraion representation of solution
(A) Strength of solution : Grams of solute dissolved
per litre of solution is called strength of solution'
(B) Parts Per Million (ppm) : Grams of solute
dissolved per 10 6 grams of solvent is called
concentration of solution in the unit of Parts Per Million
(ppm). This unit is used to represent hardness of
water and concentration of very dilute solutions.
(C) Percentage by mass : Grams of solute dissolved
per 100 grams of solution is called percentage by
mass.
(D) Percentage by volume : Millilitres of solute per
100 mL of solution is called percentage by volume.
For example, if 25 mL ethyl alcohol is diluted with
water to make 100 mL solution then the solution thus
obtained is 25% ethyl alcohol by volume.
(E) Mass by volume percentage :Grams of solute
present per 100 mL of solution is called percentage
mass by volume.
For example, let 25 g glucose is dissolved in water to
make 100 mL solution then the solution is 25% mass
by volume glucose.
PAGE # 32
(d) Classification of reactions involved in
volumetric analysis
(A) Neutralisation reactions
The reaction in which acids and bases react to form
salt called neutralisation.
e.g., HCI + NaOH  NaCI + H2O
H+(acid) + OH–(base)  H2O (feebly ionised)
The titration based on neutralisation is called
acidimetry or alkalimetry.
(B) Oxidation-reduction reactions
The reactions involving simultaneous loss and gain
of electrons among the reacting species are called
oxidation reduction or redox reactions, e.g., let us
consider oxidation of ferrous sulphate (Fe2+ ion) by
potassium permanganate (MnO 4– ion) in acidic
medium.
MnO4– + 8H+ + 5e–  Mn2+ + 4H2O
(Gain of electrons or reduction)
5  [Fe2+  Fe3+ + e–]
(Loss of electrons or oxidation)
MnO4– + 5Fe2+ + 8H+  Mn2+ + 5Fe3+ + 4H2O
________________________________________________________________
In the above reaction, MnO4– acts as oxidising agent
and Fe2+ acts as reducing agent.
The titrations involving redox reactions are called redox
titrations. These titrations are also called according
to the reagent used in the titration, e.g., iodometric,
cerimetric, permanganometric and dichromometric
titrations
(C) Precipitation reactions :
A chemical reaction in which cations and anions
combine to form a compound of very low solubility (in
the form of residue or precipitate), is called
precipitation.
BaCl2 + Na2SO4  BaSO4 + 2NaCl
(white precipitate)
The titrations involving precipitation reactions are
called precipitation titrations.
(D) Complex formation reactions :
These are ion combination reactions in which a
soluble slightly dissociated complex ion or
compound is formed.
Complex compounds retain their identity in the
solution and have the properties of the constituent
ions and molecules.
e.g. CuSO4 + 4NH3  [Cu(NH3)4]SO4
(complex compound)
AgNO3 + 2KCN  K[Ag(CN)2] + KNO3
(complex compound)
2CuSO4 + K4[Fe(CN)6]  Cu2[Fe(CN)6] + 2K2SO4
(complex compound)
The titrations involving complex formation reactions
are called complexometric titrations.
The determination of concentration of bases by
titration with a standard acid is called acidimetry and
the determination of concentration of acid by titration
with a standard base is called alkalimetry.
The substances which give different colours with
acids and base are called acid base indicators. These
indicators are used in the visual detection of the
equivalence point in acid-base titrations.
The acid-base indicators are also called pH
indicators because their colour change according to
the pH of the solution.
In the selection of indicator for a titration, following
two informations are taken into consideration :
(i) pH range of indicator
(ii) pH change near the equivalence point in the
titration.
The indicator whose pH range is included in the pH
change of the solution near the equivalence point, is
taken as suitable indicator for the titration.
(i) Strong acid-strong base titration : In the titration
of HCl with NaOH, the equivalence point lies in the
pH change of 4–10. Thus, methyl orange, methyl red
and phenolphthalein will be suitable indicators.
(ii) Weak acid-strong base titration : In the titration
of CH3COOH with NaOH the equivalence point lies
between 7.5 and 10. Hence, phenolphthalein (8.3–
10) will be the suitable indicator.
(iii) Weak base-strong acid titration : In the titration
of NH4OH (weak base) against HCl (strong acid) the
pH at equivalence point is about 6.5 and 4. Thus,
methyl orange (3.1–4.4) or methyl red (4.2–6.3) will
be suitable indicators.
(iv) Weak acid-weak base titration : In the titration of
a weak acid (CH3COOH) with weak base (NH4OH)
the pH at the equivalence point is about 7, i.e., lies
between 6.5 and 7.5 but no sharp change in pH is
observed in these titrations. Thus, no simple indicator can be employed for the detection of the equivalence point.
(v) Titration of a salt of a weak acid and a strong
base with strong acid:
H2CO3
+ 2NaOH  Na2CO3 + 2H2O
Weak acid Strong base
Na 2CO3 when titrated with HCl, the following two
stages are involved :
Na2CO3 + HCl  NaHCO3 + NaCl (First stage)
pH = 8.3, near equivalence point
NaHCO3 + HCl  NaCl + H2CO3 (Second stage)
pH = 4, near equivalence point
For first stage, phenolphthalein and for second stage,
methyl orange will be the suitable indicator.
PAGE # 33
TITRATION OF MIXTURE OF NaOH, Na2CO3 AND
NaHCO3 BY STRONG ACID LIKE HCl
In this titration the following indicators are mainly used :
(i) Phenolphthalein (weak organic acid) : It shows
colour change in the pH range (8 – 10)
(ii) Methyl orange (weak organic base) : It shows
colour change in the pH range (3.1 – 4.4). Due to
lower pH range, it indicates complete neutralisation
of whole of the base.
Let for complete neutralisation of Na2CO3, NaHCO3 and NaOH, x,y and z mL of standard HCl are required. The
titration of the mixture may be carried by two methods as summarised below :
Volume of HCl
used with
Volume of HCl used
Mixture
Phenlphthalein
from
beginning
Methyl orange
from
beginning
Phenolphthalein
from
beginning
Methyl orange
after first end
point
1. NaOH
+ Na2CO3
z + (x/2)
(x + z)
z + x/2
x/2 (for remaining 50%
Na2CO3)
2. NaOH
+ NaHCO3
z+0
(z + y)
z+0
y (for remaining 100%
NaHCO3
3. Na2CO3
+ NaHCO3
(x/2) + 0
(x + y)
(x/2) + 0
x/2 + y (for remaining 50%
of Na2CO3 and 100%
NaHCO3 are indicated)
An indicator is a substance which is used to determine the end point in a titration. In acid-base titrations organic substances (weak acids or weak bases)
are generally used as indicators. They change their
colour within a certain pH range. The colour change
and the pH range of some common indicators are
tabulated below:
________________________________________
Indicator
pH range
Colour
change
________________________________________
Methyl orange
3.2 – 4.5
Orange to red
Methyl red
4.4 – 6.5
Red to yellow
Litmus
5.5 – 7.5
Red to blue
Phenol red
6.8 – 8.4
Yellow to red
Phenolphthalein
8.3 – 10.5 Colourless to pink
________________________________________
Theory of acid-base indicators : Two theories have
been proposed to explain the change of colour of
acid-base indicators with change in pH.
1. Ostwald's theory:
According to this theory
(a) The colour change is due to ionisation of the acidbase indicator. The unionised form has different
colour than the ionised form.
(b) The ionisation of the indicator is largely affected in
acids and bases as it is either a weak acid or a weak
base. In case, the indicator is a weak acid, its
ionisation is very much low in acids due to common
ions while it is fairly ionised in alkalies. Similarly if the
indicator is a weak base, its ionisation is large in
acids and low in alkalies due to common ions.
Considering two important indicators phenolphthalein (a weak acid) and methyl orange (a weak base),
Ostwald's theory can be illustrated as follows:
PAGE # 34
Phenolphthalein: It can be represented as HPh. It
ionises in solution to a small extent as:
This theory also explains the reason why phenolphthalein is not a suitable indicator for titrating a weak
–
base against strong acid. The OH ions furnished by
a weak base are not sufficient to shift the equilibrium
towards right hand side considerably, i.e., pH is not
reached to 8.3. Thus, the solution does not attain
pink colour. Similarly, it can be explained why methyl
orange is not a suitable indicator for the titration of
weak acid with strong base.
H+ + Ph–
HPh
Colourless
Pink
Applying law of mass action,
K=
[H ][Ph  ]
[HPh]
The undissociated molecules of phenolphthalein are
colourless while ph– ions are pink in colour. In presence of an acid, the ionisation of HPh is practically
negligible as the equilibrium shifts to left hand side
due to high concentration of H+ ions. Thus, the solution would remain colourless. On addition of alkali,
hydrogen ions are removed by OH– ions in the form of
water molecules and the equilibrium shifts to right
hand side. Thus, the concentration of ph– ions increases in solution and they impart pink colour to the
solution.
SOLUBILITY
The solubility of a solute in a solution is always
expressed with respect to the saturated solution.
(a) Definition :
The maximum amount of the solute which can be
dissolved in 100g (0.1kg) of the solvent to form a
saturated solution at a given temperature.
Suppose w gram of a solute is dissolved in W gram
of a solvent to make a saturated solution at a fixed
temperature and pressure. The solubility of the solute
will be given by -
Let us derive Hendetson's equation for an indicator
HIn + H2O
H3+O + In–
'Acid form'
'Base form'
Mass of the solute
w
Conjugate acid-base pair
KIn =
W
[In  ][H3 O]
[HIn]
[HIn]
[In  ]
+
pH = –log10 [H3 O] = –log10 [KIn] – log10
[In  ]
The solubility of a substance in liquids generally
increases with rise in temperature but hardly changes
with the change in pressure. The effect of temperature
depends upon the heat energy changes which
accompany the process.
[In ]
(Henderson's equation for
[HIn]
indicator)
At equivalence point ;
[In–] = [HIn] and pH = pKIn
Methyl orange : It is a very weak base and can be
represented as MeOH. It is ionised in solution to give
Me+ and OH– ions.
MeOH
Orange
Me+ +OH–
Red
Applying law of mass action,
K=
× 100
(b) Effect of Temperature and Pressure on
Solubility of a Solids :
[HIn]

pH = pKIn + log10
Mass of the solvent
For example, the solubility of potassium chloride in
water at 20ºC and 1 atm. is 34.7 g per 100g of water.
This means that under normal conditions 100 g of
water at 20ºC and 1 atm. cannot dissolve more than
34.7g of KCl.
KIn = Ionization constant of indicator
[H3+O] = KIn 
× 100 =
[Me  ][OH  ]
[MeOH]
In presence of an acid, OH– ions are removed in the
form of water molecules and the above equilibrium
shifts to right hand side. Thus, sufficient Me+ ions are
produced which impart red colour to the solution. On
addition of alkali, the concentration of OH– ions increases in the solution and the equilibrium shifts to
left hand side, i.e., the ionisation of MeOH is practically negligible. Thus, the solution acquires the colour
of unionised methyl orange molecules, i.e. orange.

Note :
If heat energy is needed or absorbed in the process,
it is of endothermic nature. If heat energy is evolved
or released in the process, it is of exothermic nature.
(i) Effect of temperature on endothermic dissolution
process : Most of the salts like sodium chloride,
potassium chloride, sodium nitrate, ammonium
chloride etc. dissolve in water with the absorption of
heat. In all these salts the solubility increases with
rise in temperature. This means that sodium chloride
becomes more soluble in water upon heating.
(ii) Effect of temperature on exothermic dissolution
process : Few salts like lithium carbonate, sodium
carbonate monohydrate, cerium sulphate etc.
dissolve in water with the evolution of heat. This
means that the process is of exothermic nature. In
these salts the solubility in water decreases with rise
in temperature.
PAGE # 35

Note :
1. While expressing the solubility, the solution must
be saturated but for expressing concentration (mass
percent or volume percent), the solution need not to
be saturated in nature.
2. While expressing solubility, mass of solvent is
considered but for expressing concentration the
mass or volume of the solution may be taken into
consideration.
EXERCISE
1.
minimum amount of water required to dissolve 4 g
K2SO4 is (A) 10 g
(C) 50 g
2.
(c) Ef fect of Temperature on the Solubility
of a Gas
3.
(ii) The solubility of gases in liquids increases on
increasing the pressure and decreases on decreasing the pressure.
mass of solute
 100
mass of solvent
12
×100 = 16 g
4.
Ex.48 4 g of a solute are dissolved in 40 g of water to
form a saturated solution at 25ºC. Calculate the
solubility of the solute.
Mass of solvent
5.
6.
=
46
× (50g) = 23 g
100
(b) When the solution is cooled, the solubility of
salt in water will decrease. This means, that upon
cooling, it will start separating from the solution
in crystalline form.
(B) 1.10%
(D) 110%
Two elements A (at. wt. 75) and B (at. wt. 16) combine
to yield a compound. The % by weight of A in the
(A) A2B
(C) AB
7.
(B) A2B3
(D) AB2
No. of oxalic acid molecules in 100 mL of 0.02 N
oxalic acid are (A) 6.023 × 1020
(C) 6.023 × 1022
8.
(b) What will happen if this solution is cooled ?
Sol. (a) Mass of potassium chloride in 100 g of water
in saturated solution = 46 g
Mass of potassium chloride in 50 g of water in
saturated solution.
The percentage of sodium in a breakfast cereal
labelled as 110 mg of sodium per 100 g of cereal is -
compound was found to be 75.08. The empirical
formula of the compound is -
Mass of solute = 4 g
Mass of solvent = 40 g
Ex.49 (a) What mass of potassium chloride would be
needed to form a saturated solution in 50 g of
water at 298 K ? Given that solubility of the salt is
46g per 100g at this temperature.
In a compound AxBy -
(A) 11%
(C) 0.110%
× 100
4
× 100 = 10 g
40
(D) 0.25 mole
(C) X × mole of A = y × mole of B = (x + y) × mole of AxBy
(D) X × mole of A = y × mole of B
Mass of solute
Solubility =
8g of sulphur are burnt to form SO2, which is oxidised
by Cl 2 water. The solution is treated with BaCl 2
(A) Mole of A = Mole of B = mole of AxBy
(B) Eq. of A = Eq. of B = Eq. of AxBy
Thus, the solubility of potassium sulphate in
water is 16 g at 60ºC.
Sol. Solubility =
(B) 200
(D) 18
(C) 0.75 mole
Ex.47 12 grams of potassium sulphate dissolves in
75 grams of water at 60ºC. What is the solubility
of potassium sulphate in water at that temperature ?
75
Molarity of H2SO4 (density 1.8g/mL) is 18M. The
molality of this solution is -
solution. The amount of BaSO4 precipitated is (A) 1.0 mole
(B) 0.5 mole
SAMPLE PROBLEMS
=
(B) 25 g
(D) 75 g
(A)36
(C) 500
(i) The solubility of a gas in a liquid decreases with
the rise in temperature.
Sol. Solubility =
The solubility of K2SO4 in water is 16 g at 50ºC. The
(B) 6.023 × 1021
(D) 6.023 × 1023
Which of the following sample contains the maximum
number of atoms (A) 1 mg of C4H10
(C) 1 mg of Na
9.
(B) 1 mg of N2
(D) 1 mL of water
The total number of protons, electrons and neutrons
in 12 g of
12
6 C is
-
25
(A) 1.084 × 10
(C) 6.022 × 1022
(B) 6.022 × 1023
(D) 18
10. 4.4 g of CO2 and 2.24 litre of H2 at STP are mixed in a
container. The total number of molecules present in
the container will be (A) 6.022 × 1023
(B) 1.2044 × 1023
(C) 2 mole
(D) 6.023 × 1024
PAGE # 36
11. Which is not a molecular formula ?
21. How many atoms are contained in a mole of Ca(OH)2 -
(A) C6H12O6
(B) Ca(NO3)2
(A) 30 × 6.02 × 1023 atoms/mol
(C) C2H4O2
(D) N2O
(B) 5 × 6.02 × 1023 atoms/mol
12. The hydrated salt, Na2SO4. nH2O undergoes 55.9%
loss in weight on heating and becomes anhydrous.
(C) 6 × 6.02 × 1023 atoms/mol
(D) None of these
22. Insulin contains 3.4% sulphur. The minimum
The value of n will be (A) 5
(B) 3
molecular weight of insulin is -
(C) 7
(D) 10
(A) 941.176 u
(B) 944 u
(C) 945.27 u
(D) None of these
13. W hich of the following mode of expressing
concentration is independent of temperature -
23. Number of moles present in 1 m3 of a gas at NTP are -
(A) Molarity
(B) Molality
(A) 44.6
(B) 40.6
(C) Formality
(D) Normality
(C) 42.6
(D) 48.6
14. The haemoglobin of the red blood corpuscles of most
24. Weight of oxygen in Fe2O3 and FeO is in the simple
of the mammals contains approximately 0.33% of
ratio of -
iron by weight. The molecular weight of haemoglobin
(A) 3 : 2
(B) 1 : 2
is 67,200. The number of iron atoms in each molecule
(C) 2 : 1
(D) 3 : 1
of haemoglobin is (Atomic weight of iron = 56) (A) 2
(B) 3
(C) 4
(D) 5
15. An oxide of metal have 20% oxygen, the eq. wt. of
metal oxide is -
25. 2.76 g of silver carbonate on being strongly heated
yield a residue weighing (A) 2.16g
(B) 2.48 g
(C) 2.32 g
(D) 2.64 g
26. How many gram of KCl would have to be dissolved in
(A) 32
(B) 40
(C) 48
(D) 52
16. How much water is to be added to dilute 10 mL of 10
N HCl to make it decinormal ?
60 g of H2O to give 20% by weight of solution (A) 15 g
(B) 1.5 g
(C) 11.5 g
(D) 31.5 g
27. When the same amount of zinc is treated separately
(A) 990 mL
(B) 1010 mL
(C) 100 mL
(D) 1000 mL
17. The pair of compounds which cannot exist in solution is (A) NaHCO3 and NaOH
(B) Na2SO3 and NaHCO3
with excess of H2SO4 and excess of NaOH, the ratio
of volumes of H2 evolved is (A) 1 : 1
(B) 1 : 2
(C) 2 : 1
(D) 9 : 4
28. Amount of oxygen required for combustion of 1 kg of a
(C) Na2CO3 and NaOH
mixture of butane and isobutane is -
(D) NaHCO3 and NaCl
(A) 1.8 kg
(B) 2.7 kg
(C) 4.5 kg
(D) 3.58 kg
18. If 250 mL of a solution contains 24.5 g H2SO4 the
molarity and normality respectively are (A) 1 M, 2 N
(B) 1M,0.5 N
(C) 0.5 M, 1N
(D) 2M, 1N
19. The mole fraction of NaCl, in a solution containing 1
mole of NaCl in 1000 g of water is (A) 0.0177
(B) 0.001
(C) 0.5
(D) 0.244
20. 3.0 molal NaOH solution has a density of 1.110 g/
mL. The molarity of the solution is (A) 2.9732
(B) 3.05
(C) 3.64
(D) 3.0504
29. Rakesh needs 1.71 g of sugar (C12H22O11) to sweeten
his tea. What would be the number of carbon atoms
present in his tea ?
(A) 3.6 × 1022
(B) 7.2 × 1021
(C) 0.05 × 1023
(D) 6.6 × 1022
30. The total number of AlF3 molecule in a sample of AlF3
containing 3.01 × 1023 ions of F– is (A) 9.0 × 1024
(C) 7.5 × 10
(B) 3.0 × 1024
23
(D)1023
31. The volume occupied by one molecule of water
(density 1 g/cm3) is (A) 18 cm3
(C) 6.023 × 10
(B) 22400 cm3
–23
(D) 3.0 × 10–23 cm3
PAGE # 37
32. 224 mL of a triatomic gas weigh 1 g at 273 K and 1
atm. The mass of one atom of this gas is (A) 8.30 × 10–23 g
(B) 2.08 × 10–23 g
(C) 5.53 × 10–23 g
(D) 6.24 × 10–23 g
33. The percentage of P2O5 in diammonium hydrogen
phosphate is (A) 77.58
(B) 46.96
(C) 53.78
(D) 23.48
43. The number of molecules present in 11.2 litre CO2 at
STP is (A) 6.023 × 1032
(C) 3.011 × 10
23
(B) 6.023 × 1023
(D) None of these
44. 250 ml of 0.1 N solution of AgNO3 are added to 250
ml of a 0.1 N solution of NaCl. The concentration of
nitrate ion in the resulting solution will be (A) 0.1N
(B) 1.2 N
(C) 0.01 N
(D) 0.05 N
34. The mole fraction of water in 20% (wt. /wt.) aqueous
45. Amount of BaSO4 formed on mixing the aqueous
solution of H2O2 is (A)
(C)
77
68
20
80
(B)
(D)
68
77
solution of 2.08 g BaCl2 and excess of dilute H2SO4 is (A) 2.33 g
(B) 2.08 g
(C) 1.04 g
(D) 1.165 g
80
20
35. Which of the following has the maximum mass ?
(A) 25 g of Hg
(B) 2 moles of H2O
(C) 2 moles of CO2
46. 2g of NaOH and 4.9 g of H2SO4 were mixed and
volume is made 1 litre. The normality of the resulting
solution will be (A) 1N
(B) 0.05 N
(C) 0.5 N
(D) 0.1N
47. 1g of a metal carbonate neutralises completely 200
(D) 4 g atom of oxygen
36. Total mass of neutrons in 7mg of 14C is (A) 3 × 1020 kg
(B) 4 × 10–6 kg
(C) 5 × 10–7 kg
(D) 4 × 10–7 kg
37. Vapour density of a metal chloride is 66. Its oxide
contains 53% metal. The atomic weight of metal is (A) 21
(B) 54
(C) 26.74
(D) 2.086
38. The number of atoms in 4.25 g NH3 is approximately (A) 1 × 1023
(B) 1.5 × 1023
(C) 2 × 1023
(D) 6 × 1023
39. The modern atomic weight scale is based on (A) C12
(B) O16
(C) H1
(D) C13
40. Amount of oxygen in 32.2g of Na2SO4. 10H2O is (A) 20.8 g
(B) 22.4 g
(C) 2.24 g
(D) 2.08 g
mL of 0.1N HCl. The equivalent weight of metal
carbonate is (A) 25
(B) 50
(C) 100
(D) 75
48. 100 mL of 0.5 N NaOH were added to 20 ml of 1N
HCl and 10 mL of 3 N H2SO4. The solution is (A) acidic
(B) basic
(C) neutral
(D) none of these
49. 1M solution of H2SO4 is diluted from 1 litre to 5 litres ,
the normality of the resulting solution will be (A) 0.2 N
(B) 0.1 N
(C) 0.4 N
(D) 0.5 N
50. The volume of 7g of N2 at S.T.P. is (A) 11.2 L
(B) 22.4 L
(C) 5.6 L
(D) 6.5 L
51. One mole of calcium phosphide on reaction with
excess of water gives -
41. Which of the followings does not change on dilution ?
(A) Molarity of solution
(A) three moles of phosphine
(B) one mole phosphoric acid
(B) Molality of solution
(C) two moles of phosphine
(C) Millimoles and milli equivalent of solute
(D) Mole fraction of solute
42. Equal masses of O2, H2 and CH4 are taken in a
container. The respective mole ratio of these gases
in container is (A) 1 : 16 : 2
(B) 16 : 1 : 2
(C) 1 : 2 : 16
(D) 16 : 2 : 1
(D) one mole of P2O5
52. Mg (OH)2 in the form of milk of magnesia is used to
neutralize excess stomach acid. How many moles of
stomach acid can be neutralized by 1 g of Mg(OH)2 ?
(Molar mass of Mg(OH)2 = 58.33)
(A) 0.0171
(B) 0.0343
(C) 0.686
(D) 1.25
PAGE # 38
53. Calcium carbonate decomposes on heating
60. The volume of CO2 (in litres) liberated at STP when
according to the following equation -
10 g of 90% pure limestone is heated completely, is-
CaCO3(s)
(A) 22.4 L
(B) 2.24 L
(C) 20.16 L
(D) 2.016 L
How
CaO(s) + CO2(g)
many moles of CO 2 will be obtained by
decomposition of 50 g CaCO3 ?
3
(A)
2
1
(C)
2
61. A metal oxide has the formula Z2O3. It can be reduced
5
(B)
2
by hydrogen to give free metal and water. 0.1596 g of
(D) 1
reduction. The atomic mass of the metal is -
the metal requires 6 mg of hydrogen for complete
54. Sulphur trioxide is prepared by the following two
(A) 27.9
(B) 159.6
(C) 79.8
(D) 55.8
reactions -
Question number 62, 63, 64 and 65 are based on the
S8(s) + 8O2(g)  8SO2(g)
following information :
2SO2(g) + O2(g)  2SO3(g)
How many grams of SO3 are produced from 1 mole
Q.
redox reaction. Following equations describe the
of S8 ?
(A) 1280
(B) 640
(C) 960
(D) 320
Dissolved oxygen in water is determined by using a
procedure -
I
2Mn2+(aq) + 4OH–(aq) + O2 (g)
I
MnO2(s)+2I–(aq)+4H+(aq)  Mn2+(aq)+I2(aq) + 2H2O(  )
phosphorous and hydrogen. The change in volume
III
2S 2O32 – (aq) + I2(aq)  2S 4 O 26 – (aq) + 2I–(aq)
when 100 mL of such gas decomposed is -
62. How many moles of S 2O 23 – are equivalent to each
55. PH 3 (g) decomposes on heating to produce
(A) + 50 mL
(B) + 500 mL
(C) – 50 mL
(D) – 500 mL
56. What amount of BaSO4 can be obtained on mixing
0.5 mole BaCl2 with 1 mole of H2SO4 ?
(A) 0.5 mol
(B) 0.15 mol
(C) 0.1 mol
(D) 0.2 mol
57. In the reaction , CrO5 + H2SO4  Cr2(SO4)3 + H2O + O2
one mole of CrO5 will liberate how many moles of O2 ?
 2MnO2(s) + 2H2O(  )
mole of O2 ?
(A) 0.5
B) 1
(C) 2
(D) 4
63. What amount of I2 will be liberated from 8 g dissolved
oxygen ?
(A) 127 g
(B) 254 g
(C) 504 g
(D) 1008 g
64. 3 × 10 moles O2 is dissolved per litre of water, then
–3
(A) 5/2
(B) 5/4
what will be molarity of I– produced in the given
(C) 9/2
(D) None
reaction ?
(A) 3 × 10–3 M
58. Calcium carbonate decomposes on heating
1
 3  10 – 3 M
2
65. 8 mg dissolved oxygen will consume -
(C) 2 × 3 × 10–3 M
according to the equation CaCO3(s)  CaO(s) + CO2(g)
At STP the volume of CO 2 obtained by thermal
decomposition of 50 g of CaCO3 will be -
(B) 2.5 × 10–4 mol Mn2+
(B) 44 litre
(C) 10 mol Mn2+
(C) 11.2 litre
(D) 1 litre
(D) 2 mol Mn2+
oxygen, the following reaction takes place4FeCl3(s) + 3O2(g)  2Fe2O3(s) + 6Cl2(g)
If 3 moles of FeCl3 are ignited in the presence of 2
moles of O2 gas, how much of which reagent is
(D)
(A) 5 × 10–4 mol Mn+2
(A) 22.4 litre
59. When FeCl 3 is ignited in an atmosphere of pure
(B) 4 × 3 × 10–3 M
66. 2 g of a base whose eq. wt. is 40 reacts with 3 g of an
acid. The eq. wt. of the acid is(A) 40
(B) 60
(C) 10
(D) 80
67. Normality of 1% H2SO4 solution is nearly -
present in excess and therefore, remains unreacted ?
(A) 2.5
(B) 0.1
(A) 0.33 mole FeCl3 remains unreacted
(C) 0.2
(D) 1
(B) 0.67 mole FeCl3 remains unreacted
68. What volume of 0.1 N HNO3 solution can be prepared
(C) 0.25 mole O2 remains unreacted
from 6.3 g of HNO3 ?
(D) 0.50 mole O2 remains unreacted
(A) 1 litre
(B) 2 litres
(C) 0.5 litre
(D) 4 litres
PAGE # 39
seminormal HCl solution to make it decinormal is (A) 200 mL
(B) 400 mL
(C) 600 mL
(D) 800 mL
70. 0.2 g of a sample of H2O2 required 10 mL of 1N KMnO4
in a titration in the presence of H2SO4. Purity of H2O2 is(A) 25%
(B) 85%
(C) 65%
(D) 95%
71. Which of the following has the highest normality ?
(A) 1 M H2SO4
(C) 1 M H3PO4
(B) 1 M H3PO3
(D) 1 M HNO3
72. The molarity of 98% H2SO4(d = 1.8g/mL) by wt. is (A) 6 M
(B) 18.74 M
(C) 10 M
(D) 4 M
80. A 3 N solution of H2SO4 in water is prepared from
Conc. H2SO4 (36 N) by diluting [KVPY-PartII-2007]
(A) 20 ml of the conc. H2SO4 to 240 ml
(B) 10 ml of the conc. H2SO4 to 240 ml
(C) 1 ml of the conc. H2SO4 to 36 ml
(D) 20 ml of the conc. H2SO4 to 36 ml
81. The solubility curve of KNO3 as a function of temperature is given below
[KVPY-Part-II-2007]
Solubility (g/100 ml water)
69. The volume of water to be added to 200 mL of
250
200
150
100
50
0
73. 0.7 g of Na2CO3 . xH2O is dissolved in 100 mL. 20 mL
of which required to neutralize 19.8 mL of 0.1 N HCl.
The value of x is (A) 4
(C) 2
(B) 3
(D) 1
74. 0.45 g of an acid of molecular weight 90 was
neutralised by 20 mL of 0.5 N caustic potash. The
basicity of the acid is (A) 1
(B) 2
(C) 3
(D) 4
75. 1 litre of 18 molar H2SO4 has been diluted to 100
litres. The normality of the resulting solution is (A) 0.09 N
(B) 0.18
(C) 1800 N
(D) 0.36
N
HCl is required to react completely with
10
1.0 g of a sample of limestone. The percentage purity
76. 150 ml of
of calcium carbonate is (A) 75%
(C) 80%
(B) 50%
(D) 90%
N
N
77. 50 ml of
HCl is treated with 70 ml
NaOH.
10
10
Resultant solution is neutralized by 100 ml of
sulphuric acid. The normality of H2SO4 (A) N/50
(C) N/30
0
N
HCl were added to 1 g calcium car10
bonate, what would remain after the reaction ?
(A) CaCO3
(B) HCl
(C) Neither of the two
(D) Part of both
40
60
80
100
Temperature (°C)
The amount of KNO3 that will crystallize when a saturated solution of KNO3 in 100 ml of water is cooled
from 90°C to 30 °C, is
(A) 16 g
(B) 100 g
(C) 56 g
(D)160 g
82. The volume of 0.5 M aqueous NaOH solution required
to neutralize 10 ml of 2 M aqueous HCl solution is :
[KVPY-Part-I-2008]
(A) 20ml
(B) 40ml
(C) 80ml
(D) 120ml
83. 3.01×1023 molecules of elemental Sulphur will react
with 0.5 mole of oxygen gas completely to produce
[KVPY-Part-I-2008]
(A) 6.02 × 1023 molecules of SO3
(B) 6.02 × 1023 molecules of SO2
(C) 3.01 × 1023 molecules of SO3
(D) 3.01 x 1023 molecules of SO2
84. The solubility of a gas in a solution is measured in
three cases as shown in the figure given below where
w is the weight of a solid slab placed on the top of the
cylinder lid. The solubility will follow the order :
[KVPY-Part-I-2008]
w
(B) N/25
(D) N/10
78. 200 mL of
20
gas
solution
w
gas
solution
w
w
w
w
gas
solution
79. Equivalent mass of KMnO4, when it is converted to
MnSO4 is (A) M/5
(B) M/3
(C) M/6
(D) M/2
(A) a > b > c
(C) a = b = c
(B) a < b < c
(D) a >b < c
PAGE # 40
85. The density of a salt solution is1.13 g cm–3 and it
contains 18% of NaCI by weight. The volume of the
solution containing 36.0 g of the salt will be :
[KVPY-Part-II-2008]
(A) 200 cm3
(B) 217 cm3
3
(C) 177 cm
(D) 157cm3
86. One mole of nitrogen gas on reaction with 3.01 x 1023
molecules of hydrogen gas produces [KVPY-Part-I-2009]
(A) one mole of ammonia
(B) 2.0 x 1023 molecules of ammonia
(C) 2 moles of ammonia
(D) 3.01 × 1023 molecules of ammonia
87.
[KVPY-Part-I-2009]
Solubility
g/I
250
KNO3
200
150
KCl
100
50
20
40
60 80 100
Temperature (ºC)
Given the solubility curves of KNO3 and KCl, which of
the following statements is not true ?
(A) At room temperature the solubility of KNO3 and
KCI are not equal
(B) The solubilities of both KNO3 and KCI increase
with temperature
(C) The solubility of KCI decreases with temperature
(D) The solubility of KNO3 increases much more compared to that of KCl with increase in temperature
88. 10 ml of an aqueous solution containing 222 mg of
calcium chloride (mol. wt. = 111) is diluted to 100 ml.
The concentration of chloride ion in the resulting solution is [KVPY-Part-II-2009]
(A) 0.02 mol/lit.
(B) 0.01 mol/lit.
(C) 0.04 mol/lit
(D) 2.0 mol/lit.
89. Aluminium reduces manganese dioxide to manganese at high temperature. The amount of aluminium
required to reduce one gram mole of manganese
dioxide is [KVPY-Part-II-2009]
(A) 1/2 gram mole
(B) 1 gram mole
(C) 3/4 gram mole
(D) 4/3 gram mole
90. One mole of oxalic acid is equivalent to
[IJSO-2009]
(A) 0.5 mole of NaOH
(B) 1 mole of NaOH
(C) 1.5 mole of NaOH
(D) 2 mole of NaOH

PAGE # 41
NUMBER
SYSTEM
(vi) Real numbers : Numbers which can represent
(i) Natural numbers :
Counting numbers are known as natural numbers.
N = { 1, 2, 3, 4, ... }.
actual physical quantities in a meaningful way are
known as real numbers. These can be represented
on the number line. Number line is geometrical straight
line with arbitrarily defined zero (origin).
(ii) Whole numbers :
(vii) Prime numbers : All natural numbers that have
All natural numbers together with 0 form the collection
of all whole numbers.
W = { 0, 1, 2, 3, 4, ... }.
one and itself only as their factors are called prime
numbers i.e. prime numbers are exactly divisible by
1 and themselves. e.g. 2, 3, 5, 7, 11, 13, 17, 19, 23,...etc.
If P is the set of prime number then P = {2, 3, 5, 7,...}.
(iii) Integers :
All natural numbers, 0 and negative of natural numbers
form the collection of all integers.
I or Z = { ..., – 3, – 2, – 1, 0, 1, 2, 3, ... }.
(iv) Rational numbers :
These are real numbers which can be expressed in the
(viii) Composite numbers : All natural numbers, which
are not prime are composite numbers. If C is the set
of composite number then C = {4, 6, 8, 9, 10, 12,...}.

(ix) Co-prime Numbers : If the H.C.F. of the given
p
, where p and q are integers and q  0 .
q
e.g. 2/3, 37/15, -17/19.
form of


All natural numbers, whole numbers and integers are
rational.
Rational numbers include all Integers (without any
decimal part to it), terminating fractions ( fractions in
which the decimal parts are terminating e.g. 0.75,
– 0.02 etc.) and also non-terminating but recurring
decimals e.g. 0.666....., – 2.333...., etc.
Fractions :
(a) Common fraction : Fractions whose denominator
is not 10.
(b) Decimal fraction : Fractions whose denominator is
10 or any power of 10.
(c) Proper fraction : Numerator < Denominator i.e.
3
5
.
5
(d) Improper fraction : Numerator > Denominator i.e. .
3
(e) Mixed fraction : Consists of integral as well as
2
.
7
(f) Compound fraction : Fraction whose numerator and
fractional part i.e. 3

2/3
denominator themselves are fractions. i.e.
.
5/7
Improper fraction can be written in the form of mixed
fraction.
All real number which are not rational are irrational
numbers. These are non-recurring as well as
non-terminating type of decimal numbers.
2,
3
4 , 2 3 ,
numbers (not necessarily prime) is 1 then they are
known as co-prime numbers. e.g. 4, 9 are co-prime
as H.C.F. of (4, 9) = 1.

Any two consecutive numbers will always be co-prime.
(x) Even Numbers : All integers which are divisible by 2
are called even numbers. Even numbers are denoted
by the expression 2n, where n is any integer. So, if E is
a set of even numbers, then E = { ..., – 4, –2, 0, 2, 4,...}.
(xi) Odd Numbers : All integers which are not
divisible by 2 are c alled odd numbers. Odd
numbers are denoted by the general expression
2n –1 where n is any integer. If O is a set of odd
numbers, then O = {..., –5, –3, –1, 1, 3, 5,...}.
(xii) Imaginary Numbers : All the numbers whose
square is negative are called imaginary numbers.
e.g. 3i, -4i, i, ... ; where i =
- 1.
(xiii) Complex Numbers : The combined form of real
and imaginary numbers is known as complex
numbers. It is denoted by Z = A + iB where A is real part
and B is imaginary part of Z and A, B  R.
 The set of complex number is the super set of all the
sets of numbers.
Squares : When a number is multiplied by itself then
the product is called the square of that number.
(v) Irrational Numbers :
For Ex. :
1 is neither prime nor composite number.
2 3 ,
47
3 etc.
Perfect Square : A natural number is called a perfect
square if it is the square of any other natural number
e.g. 1, 4, 9,... are the squares of 1, 2, 3,... respectively.
PAGE # 4242
Ex.1 Find the smallest number by which 300 must be
multiplied so that the product is a perfect square.
Sol. Given number is 300, first we resolve it into prime
factors.
Ex.3 Find the square root of 3 +
Sol. Let
3+
2 300
2 150
25

2 pq =
5
5

4pq = 2
Clearly, 3 has no pair. Thus if we multiply it by 3 then
product will be a perfect square.
 Required smallest number is 3 .
Ex.2 Find the smallest number by which 1575 must be
divided so that the quotient becomes a perfect square.
Sol. Given number is 1575, first we write it as the product of
prime factors
3 1575
3 525
7
7
1


(p – q)2 = 7

p–q=
...(iv)
7
[By eqn (i)]
 p+q=3

1
3 7
2
1
3 7
q=
2
p=


3 2 = 


[On adding (i) & (iv)]
[On subtracting (i) & (iv)]
1
 3 7  3 7 



2 
then the result is called the cube of that number.
Perfect cube : A natural number is said to be a perfect
cube if it is the cube of any other natural number.
Ex.4 What is the smallest number by which 675 must be
multiplied so that the product is a perfect cube.
Sol. Resolving 675 into prime factor, we get
3 675
225 = 15 ...etc.
We can calculate the square root of positive numbers
only. However the square root of a positive number
may be a positive or a negative number.
e.g.
[By squaring both sides ]
Cube : If any number is multiplied by itself three times
Square roots : The square root of a number x is that
number which when multiplied by itself gives x as the
product. As we say square of 3 is 9, then we can also
say that square root of 9 is 3.
The symbol use to indicate the square root of a number
81 = 9,
...(iii)
(p – q)2 = 9 – 2

 1575 = 3 × 3 × 5 × 5 × 7.
Clearly, 7 has no pair, so if we divide it by 7 then quotient
become a perfect square.
’ , i.e.
...(ii)
2


35
[By equating the parts]
(p – q)2 = (p + q)2 – 4 pq
300 = 2 × 2 × 3 × 5 × 5
175
[By squaring both sides]
...(i)
5
5
q
pq
p+q=3
75
5
p +
2 =p+q+2
3
1
is ‘
3 2 =
2.
25 = + 5 or – 5.
3 225
3 75
5
25
5
5
1
Properties of Square Roots :
(i) If the unit digit of a number is 2, 3, 7 or 8, then it does
not have a square root in N.
675 = (3 × 3 × 3) × 5 × 5
Grouping the factor in triplets of equal factors, we get
(ii) If a number ends in an odd number of zeros, then it
does not have a square root in N.
675 = (3 × 3 × 3) × 5 × 5
We find that 3 occurs as a prime factor of 675 thrice but
(iii) The square root of an even number is even and
5 occurs as a prime factor only twice. Thus, if we multiply
675 by 5, 5 will also occur as a prime factor thrice and
square root of an odd number is odd.e.g. 81 = 9,
= 16,
256
324 = 18 ...etc.
(iv) Negative numbers have no square root in set of
real numbers.
the product will be 3 × 3 × 3 × 5 × 5 × 5, which is a
perfect cube.
Hence, we must multiply 675 by 5 so that the product
becomes a perfect cube.
PAGE # 4343
Ex.5 What is the smallest number by which 18522 must be
divided so that the quotient is a perfect cube ?
Sol. Resolving 18522 into prime factors, we get
Laws of Surds :
2 18522
3 9261
n
n
(i)
 a
(ii)
n
(iii)
a  n b  n ab
n
3087
3
1029
(iv)
nm
7
7
343
49
(v)
n
7
7
18522 = 2 × 3 × 3 × 3 × 7 × 7 × 7
Grouping the factors in triplets of equal factors, we get
18522 = 2 × (3 × 3 × 3) × (7 × 7 × 7)
Clearly, if we divide 18522 by 2, the quotient would be
3 × 3 × 3 × 7 × 7 × 7 = 33 × 73 which is a perfect cube.
Therefore, we must divide 18522 by 2 so that the
quotient ‘9261’ is a perfect cube.
a  nm a  m n a
a
np
a is given a special
name Surd. Where ‘a’ is called radicand, rational. Also
Sol.
the symbol
Ex.8 Divide :
n
n
is called the radical sign and the index
6
3
8a 5 b  4a 2 b 2
3
8 3 a15 b 3  6 4 2 a 4b 4 = 4a b6 2ab .
n is called order of the surd.
n
a is read as nth root of ‘a’ and can also be written
1
an
as
Sol.
3
6
24
3

200
3
24 by
6
(24) 3
(200)
200
6
2
216
.
625
Comparison of Surds :
.
It is clear that if x > y > 0 and n > 1 is a (+ve) integer
Identification of Surds :
(i)
np
a p or, n a m 
amp
[Important for changing order of surds]
Ex.6 If x = 1 + 21/3 + 22/3, then find the value of x3 – 3x2 – 3x – 1.
Sol. x = 1 + 21/3 + 22/3
x – 1 = (21/3 + 22/3)
(x – 1)3 = ( 21/3 + 22/3)3
x3 – 3x2 + 3x – 1 = (21/3)3 + (22/3)3 + 3. 21/3. 22/3 (21/3 + 22/3)
x3 – 3x2 + 3x – 1 = 2 + 22 + 3.21 (x – 1)
x3 – 3x2 + 3x – 1 = 6 + 6 (x – 1)
x3 – 3x2 + 3x – 1 = 6x
x3 – 3x2 – 3x – 1 = 0.
Ex.7 Simplify :
Any irrational number of the form
[Here order should be same]
a
b
a n b  n
3
1
 n an = a
then
4 is a surd as radicand is a rational number..
Similar examples :
3
5 , 4 12 , 5 7 , 12 , ...
x >
n
n
y . e.g.
3
16 >
3
12 ,
5
36 >
5
25 and so
on.
,
Ex.9 Which is greater in each of the following :
(ii) 2 + 3 is a surd (as surd + rational number will
give a surd)
Similar examples : 3 – 2 , 3  1, 3 3  1,...
(iii)
3
(i)
Sol. (i)

of 2 – 3 .
Similar examples : 7  4 3 , 9 – 4 5 , 9  4 5 ,...
1
(iv)
3
3 is a surd as
Similar examples :
3 3
3
1
 1 3
3  3 2   3 6  6 3
 
 
5
8
1
and
2
(ii)
3
1
3
L.C.M. of 3 and 5 is 15.
7 – 4 3 is a surd as 7 – 4 3 is a perfect square

6 and
3
6  35 6 5  15 7776
5
8  35 8 3  15 512
15 7776  15 512

3
5
6  8
(ii) L.C.M. of 2 and 3 is 6.

3
6
5 , 4 5 6 , ...
 1
  and
2
6
 1
 
3
2
(v) These are not a surds :
(A)
3
8 , because
3
3
8  23
which is a rational
=
6
number.
(B)
2  3 , because 2 + 3 is not a perfect square.
(C) 3 1 3 , because radicand is an irrational number..
1
and
8
as 8 < 9 
So, 6
1
8
6
1
9
1 1

8 9
6
1
9
or,
1
>
2
3
1
.
3
PAGE # 4444
Ex.10 Arrange 2, 3 3 and
Sol. L.C.M. of 2, 3, 4 is 12.
2

and
26
4
2 6  12 64 ,
3
3  34 3 4  12 81 ,
4
5
4 3
5 in ascending order..
5 3  12 125
As, 64 < 81 < 125.

12
64  12 81  12 125
2 33 45
Conjugate Surds :

R.F. of a  b and
conjugate surds .
a  b type surds are called
Ex.11 (i) 2 – 3 is conjugate of 2  3 .

(ii) 5  1 is conjugate of 5 – 1.
Sometimes conjugate surd and reciprocals are same.
Ex.12 (i) 2  3 , it’s conjugate is 2 – 3 , its reciprocal is
2 – 3 & vice versa.
(ii) 5 – 2 6 , it’s conjugate is 5  2 6 , its reciprocal is
5  2 6 & vice versa.
Factors : ‘a’ is a factor of ‘b’ if there exists a relation
such that a × n = b, where ‘n’ is any natural number.

1 is a factor of all numbers as 1 × b = b.

Factor of a number cannot be greater than the number
(infact the largest factor will be the number itself). Thus
factors of any number will lie between 1 and the number
itself (both inclusive) and they are limited.
Multiples : ‘a’ is a multiple of ‘b’ if there exists a relation
of the type b × n = a. Thus the multiples of 6 are
6 × 1 = 6, 6 × 2 = 12, 6 × 3 = 18, 6 × 4 = 24, and so on.

The smallest multiple will be the number itself and the
number of multiples would be infinite.

NOTE :
To understand what multiples are, let’s just take an
example of multiples of 3. The multiples are 3, 6, 9,
12,.... so on. We find that every successive multiples
appears as the third number after the previous.
So if one wishes to find the number of multiples of 6
less than 255, we could arrive at the number through
255
= 42 (and the remainder 3). The remainder is of
6
no consequence to us. So in all there are 42 multiples.
255
=7
36
(and the remainder is 3). Hence, there are 7 multiples
of 36.
If one wishes to find the multiples of 36, find
Factorisation : It is the process of splitting any number
into form where it is expressed only in terms of the
most basic prime factors.
For example, 36 = 22 × 3 2. 36 is expressed in the
factorised form in terms of its basic prime factors.
Number of factors : For any composite number C,
which can be expressed as C = ap × bq × cr ×....., where
a, b, c ..... are all prime factors and p, q, r are positive
integers, then the number of factors is equal to
(p + 1) × (q + 1) × (r + 1).... e.g. 36 = 22 × 32. So the
factors of 36 = (2 +1) × (2 + 1) = 3 × 3 = 9.
Ex.13 If N = 123 × 34 ×52, find the total number of even
factors of N.
Sol. The factorised form of N is
(22 × 31)3 × 34 × 52  26 × 37 × 52.
Hence, the total number of factors of N is
(6 + 1) (7 + 1) (2 + 1) = 7 × 8 × 3 = 168.
Some of these are odd multiples and some are even.
The odd multiples are formed only with the combination
of 3s and 5s.
So, the total number of odd factors is (7 + 1)(2 + 1) = 24.
Therefore, the number of even factors
= 168 – 24 = 144.
Ex.14 A number N when factorised can be written
N = a 4 × b 3 × c 7. Find the number of perfect squares
which are factors of N (The three prime numbers
a, b, c > 2).
Sol. In order that the perfect square divides N, the powers
of ‘a’ can be 0, 2 or 4, i.e. 3.
Powers of ‘b’ can be 0, 2, i.e. 2. Power of ‘c’ can be 0, 2,
4 or 6, i.e. 4.
Hence, a combination of these powers given 3 × 2 × 4
i.e. 24 numbers.
So, there are 24 perfect squares that divides N.
Ex.15 Directions : (i to iv) Answer the questions based on
the given information.
There are one thousand lockers and one thousand
students in a school. The principal asks the first student
to go to each locker and open it. Then he asks the
second student go to every second locker and close it.
The third student goes to every third locker, and if it is
closed, he opens it, and it is open, he closes it. The
fourth student does it to every fourth locker and so on.
The process is completed with all the thousand
students.
(i) How many lockers are closed at the end of the
process ?
(ii) How many students can go to only one locker ?
(iii) How many lockers are open after 970 students
have done their job ?
(iv) How many student go to locker no. 840 ?
PAGE # 4545
Sol. (i to iv) : Whether the locker is open or not depends on
the number of times it is accessed. If it is accessed
odd number of times, then it is open while if it is
accessed even number of times then it is closed.
How many times a locker will be accessed depends
on the locker no. If it contains odd number of factors,,
then it will be open and if it contains even number of
factors. Then it will be closed. We know that a perfect
square contains odd number of factors while a nonperfect square contains even number of factors. Thus
the lockers with perfect square number will be open
and the number of these perfect squares from 1 to
1000 determines the no. of open lockers.
(i) No. of closed lockers = No. of non-perfect square
numbers from 1 to 1000 = 1000 – 31 = 969.
(ii) Upto 500 students they can go to two or more than
two lockers, while the rest 500 can go to only one locker.
(iii) The 31 perfect squares ( the last being 312 = 961)
will be open while the lockers from 971 to 1000 is yet
to be accessed last time so they all are open. The total
being = 31 + 30 = 61
(iv) The no. of students that have gone to locker no.
840 is same as the no. of factors of 840.
840 = 23 × 3 × 5 × 7.
So, the no. of factors = (3 + 1) (1 + 1) (1 + 1) (1 + 1) = 32.
LCM (least Common Multiple) : The LCM of given
numbers, as the name suggests is the smallest
positive number which is a multiple of each of the given
numbers.
HCF (Highest Common factor) : The HCF of given
numbers, as the name suggests is the largest factor
of the given set of numbers.
Consider the numbers 12, 20 and 30. The factors and
the multiples are :
Factors
1, 2, 3, 4, 6, 12
1, 2, 4, 5, 10, 20
1, 2, 3, 5, 6, 10, 15, 30
Given
numbers
12
20
30
Multiples
12, 24, 36, 48, 60, 72, 84, 96, 108, 120....
20, 40, 60, 80, 100, 120.....
30, 60, 90, 120....
The common factors are 1 and 2 and the common
multiples are 60, 120...
Thus the highest common factor is 2 and the least
common multiple is 60. Meaning of HCF is that the
HCF is the largest number that divides all the given
numbers.
Also since a number divides its multiple, the meaning
of LCM is that it is smallest number which can be
divided by the given numbers.
 HCF will be lesser than or equal to the least of the
numbers and LCM will be greater than or equal to the
greatest of the numbers.
Ex.16 Find a number greater then 3 which when divided by
4, 5, and 6 always leaves the same remainder 3.
Sol. The smallest number which, when divided by 4, 5 and
6, leaves the remainder 3 in each case is
LCM (4, 5 and 6) + 3 = 60 + 3 = 63.
Ex.17 In a school 437 boys and 342 girls have been divided
into classes, so that each class has the same number
of students and no class has boys and girls mixed.
What is the least number of classes needed?
Sol. We should have the maximum number of students in
a class. So we have to find HCF (437, 342) = 19.
HCF is also the factor of difference of the number.


437
342
+
= 23 + 18
19
19
= 41 classes.
Number of classes =
For any two numbers x and y :
x × y = HCF (x, y) × LCM (x, y).
HCF and LCM of fractions :
LCM of numerators
LCM of fractions = HCF of deno min ators
HCF of numerators
HCF of fractions = LCM of deno min ators
Make sure the fractions are in the most reducible form.
Ex.18 Find the least number which when divided by 6, 7, 8,
9 and 10 leaves remainder 1.
Sol. As the remainder is same
Required number = LCM of divisors + Remainder
= LCM (6, 7, 8, 9, 10) +1
= 2520 + 1 = 2521.
Ex.19 Six bells start tolling together and they toll at intervals
of 2, 4, 6, 8, 10, 12 sec. respectively, find.
(i) after how much time will all six of them toll together ?
(ii) how many times will they toll together in 30 min ?
Sol. The time after which all six bells will toll together must
be multiple of 2, 4, 6, 8, 10, 12.
Therefore, required time = LCM of time intervals.
= LCM (2, 4, 6, 8, 10, 12) = 120 sec.
Therefore after 120 s all six bells will toll together.
After each 120 s, i.e. 2 min, all bell are tolling together.
 30

 1
Therefore in 30 min they will toll together 
 2

= 16 times
1 is added as all the bells are tolling together at the
start also, i.e. 0th second.
Ex.20 LCM of two distinct natural numbers is 211. What is
their HCF ?
Sol. 211 is a prime number. So there is only one pair of
distinct numbers possible whose LCM is 211,
i.e. 1 and 211. HCF of 1 and 211 is 1.
Ex.21 An orchard has 48 apple trees, 60 mango trees and
96 banana trees. These have to be arranged in rows
such that each row has the same number of trees and
all are of the same type. Find the minimum number of
such rows that can be formed.
Sol. Total number of trees are 204 and each of the trees
are exactly divisible by 12. HCF of (48, 60, 96).

204
= 17 such rows are possible.
12
PAGE # 4646
Division Algorithm : General representation of result is,
Dividend
Re mainder
 Quotient 
Divisor
Divisor
Dividend = (Divisor × Quotient ) + Remainder
NOTE :
(i) (xn – an) is divisible by (x – a) for all the values of n.
(ii) (xn – an) is divisible by (x + a) and (x – a) for all the
even values of n.
(iii) (xn + an) is divisible by (x + a) for all the odd values of n.
Test of Divisibility :
No.
Divisiblity Test
2
Unit digit should be 0 or even
3
The sum of digits of no. should be divisible by 3
4
The no formed by last 2 digits of given no. should be divisible by 4.
5
Unit digit should be 0 or 5.
6
No should be divisible by 2 & 3 both
8
The number formed by last 3 digits of given no. should be divisible by 8.
9
Sum of digits of given no. should be divisible by 9
The difference between sums of the digits at even & at odd places
should be zero or multiple of 11.
25 Last 2 digits of the number should be 00, 25, 50 or 75.
11
Rule for 7 : Double the last digit of given number and
subtract from remaining number the result should be
zero or divisible by 7.
Ex.22 Check whether 413 is divisible by 7 or not.
Sol. Last digit = 3, remaining number = 41, 41 – (3 x 2) = 35
(divisible by 7). i.e. 413 is divisible by 7.
This rule can also be used for number having more
than 3 digits.
Ex.23 Check whether 6545 is divisible by 7 or not.
Sol. Last digit = 5, remaining number 654, 654 – (5 x 2)
= 644; 64 – (4 x 2) = 56 divisible by 7. i.e. 6545 is
divisible by 7.
Rule for 13 : Four times the last digit and add to
remaining number the result should be divisible by
13.
Ex.24 Check whether 234 is divisible by 13 or not .
Sol. 234, (4 x 4) + 23 = 39 (divisible by 13), i.e. 234 is divisible
by 13.
Rule for 17 : Five times the last digit of the number and
subtract from previous number the result obtained
should be either 0 or divisible by 17.
Ex.25 Check whether 357 is divisible by 17 or not.
Sol. 357, (7 x 5) – 35 = 0, i.e. 357 is divisible by 17.
Ex.28 Find the largest four digit number which when
reduced by 54, is perfectly divisible by all even natural
numbers less than 20.
Sol. Even natural numbers less than 20 are 2, 4, 6, 8, 12,
14, 16, 18.
Their LCM = 2 × LCM of first 9 natural numbers
= 2 × 2520 = 5040
This happens to be the largest four-digit number
divisible by all even natural numbers less than 20. 54
was subtracted from our required number to get this
number.
Hence, (required number – 54) = 5040
 required number = 5094.
Ex.29 Ajay multiplied 484 by a certain number to get the
result 3823a. Find the value of ‘a’.
Sol. 3823a is divisible by 484, and 484 is a factor of 3823a.
4 is a factor of 484 and 11 is also a factor of 484.
Hence, 3823a is divisible by both 4 and 11.
To be divisible by 4, the last two digits have to be
divisible by 4.
‘a’ can take two values 2 and 6.
38232 is not divisible by 11, but 38236 is divisible by
11.
Hence, 6 is the correct choice.
Ex.30 Which digits should come in place of  and $ if the
number 62684$ is divisible by both 8 and 5 ?
Sol. Since the given number is divisible by 5, so 0 or 5 must
come in place of $. But, a number ending with 5 in
never divisible by 8. So, 0 will replace $.
Now, the number formed by the last three digits is 40,
which becomes divisible by 8, if  is replaced by 4 or 8.
Hence, digits in place of  and $ are 4 or 8 and 0
respectively.
Ex.31 On dividing 15968 by a certain number, the quotient
is 89 and the remainder is 37. Find the divisor.
Dividend  Re mainder 15968  37

Sol. Divisor =
= 179.
Quotient
89
Ex.32 How many numbers between 200 and 600 are
divisible by 4, 5, and 6 ?
Sol. Every such number must be divisible by L.C.M. of 4, 5,
6, i.e.60.
Such numbers are 240, 300, 360, 420, 480, 540
Clearly, there are 6 such numbers.
Rule for 19 : Double the last digit of given number and
add to remaining number The result obtained should
be divisible by 19.
The method of finding the remainder without actually
performing the process of division is termed as
remainder theorem.
Ex.26 Check whether 589 is divisible by 19 or not.
Sol. 589, (9 x 2) + 58 = 76 (divisible by 19), i.e. the number
is divisible by 19.
Ex27. Find the smallest number of six digits which is exactly
divisible by 111.
Sol. Smallest number of 6 digits is which is 100000.
On dividing 100000 by 111, we get 100 as remainder.
 Number to be added = (111 - 100) = 11.
Hence, required number = 100011.

Remainder should always be positive. For example if
we divide –22 by 7, generally we get –3 as quotient and
–1 as remainder. But this is wrong because remainder
is never be negative hence the quotient should be –4
and remainder is +6. We can also get remainder 6 by
adding –1 to divisor 7 ( 7–1 = 6).
PAGE # 4747
Ex.33 Two numbers, x and y, are such that when divided by
6, they leave remainders 4 and 5 respectively. Find the
remainder when (x2 + y2) is divided by 6.
Sol. Suppose x = 6k1 + 4 and y = 6k2 + 5
x2 + y2 = (6k1 + 4)2 + (6k2 + 5)2
= 36k12 + 48k1 + 16 + 36k22 + 60k2 + 25
= 36k12 + 48k1 + 36k22 + 60k2 + 41
Obviously when this is divided by 6, the remainder will
be 5.
Ex.34 A number when divided by 259 leaves a remainder
139. What will be the remainder when the same
number is divided by 37 ?
Sol. Let the number be P.
So, P – 139 is divisible by 259.
P  139
Let Q be the quotient then,
=Q
259
 P = 259Q + 139

Sol.
5 x
9 y 4
13 z 8
1 12
Now, 1169 when divided by 585 gives remainder
= 584.
To find the remainder of big number
 NOTE :
(i) Binomial Expansion :
(a + b)n = an +
(a – b)n = an –
P
259 Q  139
=
37
37
3 x
5 y 1
8 z 4
1 7
 z = (8 × 1 + 7) = 15 ; y (5z + 4) = (5 × 15 + 4) = 79 ;
x = (3y + 1) = (3 × 79 + 1) = 238.
8 238
5 29 6
Now, 3 5
4
1
2

Respective remainders are 6, 4, 2.
Ex.36 A number was divided successively in order by 4, 5
and 6. The remainders were respectively 2, 3 and 4.
Then find out the number.
Sol.
n
1!
n(n  1)
an–1b +
an–1b +
2!
n(n  1)
2!
an – 2b2 + .... + bn, or
an– 2b2 – ... + (– 1)nbn.
Hence, first term is pure of a i.e an and last digit is pure
of b, i.e. bn.
 259 is divisible by 37,
 When 139 divided by 37, leaves a remainder of 28.
Ex.35 A number being successively divided by 3, 5 and 8
leaves remainders 1, 4 and 7 respectively. Find the
respective remainders if the order of divisors be
reversed.
Sol.
n
1!
(ii) Total number of terms in the expansion of (a + b)n is
(n + 1).
Ex.38 What is the remainder when 738 is divided by 48.
19
Sol.
19
19
49 
48  1
7 38
72
=
=
=
so by using
48
48
48
48
binomial expansion, we can say that 18 terms are
completely divisible by 48 but the last term which is
 
 119
48
is not divisible. So, 119 = 1 is the remainder..
Ex.39 What is the remainder if 725 is divided by 4?
Sol. 725 can be written (8–1)25. There are 26 terms in all and
first 25 terms are divisible by 8, hence also by 4. The
last term is (–1)25. Hence, (8 –1)25 can be written 8X – 1
or 4Y –1 ( where Y = 2X). So, 4Y – 1 divided by 4 leaves
the remainder 3.
Ex.40 What is the remainder if 345 is divided by 8 ?
Sol. 345 can be written 922 × 3. 9 can be written as (8 + 1).
Hence, any power of 9 can be written 8N + 1. In other
words, any power of 9 is 1 more than a multiple of 8.
Hence, (8N + 1) × 3 leaves remainder 3 when divided
by 8.
16
4 x
5 y 2
6 z 3
1 4
 z = (6 × 1 + 4) = 10
 y = (5 × z + 3) = (5 × 10 + 53) = 53
 x = (4 × y + 2) = (4 × 53 + 2) = 214
Hence, the required number is 214.
Ex.37 In dividing a number by 585, a student employed the
method of short division. He divided the number
successively by 5, 9 and 13 (factors of 585) and got the
remainders 4, 8 and 12. If he had divided number by
585, then find out the remainder.
Ex.41 What is the remainder when 1415
is divided by 5 ?
1516
Sol. 14
= (15 –1)odd = 15n + (–1)odd, i.e. a (multiple of 5)
–1. Thus when divided by 5 the remainder will be (–1),
i.e. 4.
Ex.42 What is the remainder when 357 + 27 is divided by
28?
Sol. 357 = (33)19
 357 + 27 = (27)19 + 27
= (28 – 1)19 + 27
= 28M + (–1)19 + 27
[Expand by binomial theorem]
= 28M – 1 + 27
= 28M + 26
When 28M + 26 divided by 28, the remainder is 26.
Hence, the required remainder is 26.
PAGE # 4848
Ex.43 What is the remainder when 82361 + 83361 + 84361
+ 85361 + 86361 is divided by 7?
Sol. 82 361 + 83 361 + 84 361 + 85 361 + 86 361 = [(84 – 2) 361
+ (84 – 1)361 + + 84361 + (84 + 1)361 + (84 + 2)361]
Since, 84 is a multiple of 7, then the remainder will be
when, (– 2)361 + (–1)361 + 1361 + 2361 is divided by 7.
is (– 2)361 + (–1)361 + 1361 + 2361 = 0.
So the remainder is zero.
We are having 10 digits in our number systems and
some of them shows special characterstics like they,
repeat their unit digit after a cycle, for example 1 repeat
its unit digit after every consecutive power. So, its
cyclicity is 1 on the other hand digit 2 repeat its unit
digit after every four power, hence the cyclicity of 2 is
four. The cyclicity of digits are as follows :
Digit
Cyclicity
0, 1, 5 and 6
1
4 and 9
2
Ex.51 In (57)9 unit digit is 7
Ex.52 In (97)99 unit digit is 3
(vii) When there is 8 in unit’s place of any number.
Since, in 81 unit digit is 8, in 82 unit digit is 4, in 83 unit
digit is 2, in 84 unit digit is 6, after that unit’s digit repeats
after a group of 4.
(viii) When there is 9 in unit’s place of any number.
Since, in 91 unit’s digit is 9, in 92 unit’s digit is 1, after
that unit’s digit repeats after a group of 2.
(ix) When there is zero in unit’s place of any number.
There will always be zero in unit’s place.
Ex.53 Find the last digit of
(i) 357
(ii) 1359
57
gives the remainder
4
1. So, the last digit of 357 is same as the last digit of 31,
i.e. 3.
Sol. (i) The cyclicity of 3 is 4. Hence,
(ii) The number of digits in the base will not make a
difference to the last digit. It is last digit of the base
which decides the last digit of the number itself. For
59
which gives a remainder 3. So the
4
59
last digit of 13 is same as the last digit of 33, i.e. 7.
1359, we find
2, 3, 7 and 8
4
So, if we want to find the last digit of 245, divide 45 by 4.
The remainder is 1 so the last digit of 245 would be
same as the last digit of 21 which is 2.
To Find the Unit Digit in Exponential Expressions :
(i) When there is 2 in unit’s place of any number.
Since, in 21 unit digit is 2, in 22 unit digit is 4, in 23 unit
digit is 8, in 24 unit digit is 6, after that the unit’s digit
repeats. e.g. unit digit(12)12 is equal to the unit digit of,
24 i.e.6
Ex.44 In (32)33 unit digit is equal to the unit digit of 21 i.e. 2.
(ii) When there is 3 in unit’s place of any number.
Since, in 31 unit digit is 3, in 32 unit digit is 9, in 33 unit
digit is 7, in 34 unit digit is 1, after that the unit’s digit
repeats.
Ex.45 In (23)13 unit digit be 3
Ex.46 In (43)46 unit digit be 9
(iii) When there is 4 in unit’s place of any number.
Since, in 41 unit digit is 4, in 42 unit digit is 6, after that
the unit’s digit repeats.
Ex.47 In (34)14 unit digit is 6
Ex.54 Find the last digit of the product 723 x 813.
Sol. Both 7 and 8 exhibit a cyclicity of 4. The last digits are
71 = 7
81 = 8
2
7 =9
82 = 4
73 = 3
83 = 2
4
7 =1
84 = 6
5
7 =7
85 = 8
The cycle would repeat itself for higher powers.
723 ends with the same last digit as 73, i.e. 3.
813 ends with the same last digit as 81, i.e. 8. Hence,
the product of the two numbers would end with the
same last digit as that of 3 × 8, i.e. 4.
Ex.55Find unit’s digit in y = 717 + 734
Sol. 717 + 734 = 71 + 72 = 56, Hence the unit digit is 6
Ex.56What will be the last digit of (73 )75
6476
6476
76
Sol. Let (73 )75
= (73)x where x = 75 64 = (75)even power
 Cyclicity of 3 is 4
 To find the last digit we have to find the remainder
when x is divided by 4.
x = (75)even power = (76 – 1)even power , where n is divided by
4 so remainder will be 1.
Therefore, the last digit of (73 )75
6476
will be 31 = 3.
33
Ex.48 In (34) unit digit is 4
(iv) When there is 5 in unit’s place of any number.
Since, in 51 unit digit is 5, in 52 unit digit is 5 and so on.
Ex.49 In (25)15 unit digit is 5
(v) When there is 6 in unit’s place of any number.
Since, in 61 unit digit is 6, in 62 unit digit is 6 & so on.
Ex.50 In (46)13 unit digit is 6.
,
(vi) When there is 7 in unit’s place of any number.
Since, in 71 unit digit is 7, in 72 unit digit is 9, in 73 unit
digit is 3, in 74 unit digit is 1, after that the unit’s digit
repeats.
Ex.57What will be the unit digit of (87 )75
55
75 63
6355
.
55
Sol. Let (87 )
= (87)x where x = 75 63 = (75)odd
 Cyclicity of 7 is 4.
 To find the last digit we have to find the remainder
when x is divided by 4.
x = (75)odd power = (76 – 1)odd power
where x is divided by 4 so remainder will be –1 or 3, but
remainder should be positive always
therefore, the last digit of (87 )75
Hence, the last digit is of (87 )75
6355
6355
will be 73 = 343.
is 3.
PAGE # 4949
To Find the Last Tw o Digits in Exponential
Expressions :
 We know that the binomial theorem :
(a + b)n = an +
n
an–1b +
n(n  1)
an – 2b2 + .... + bn.
1!
2!
(i) Last two digits of numbers ending in 1 :
Let's start with some examples.
Ex.58What are the last two digits of 31 786 ?
Sol. 31 786 = (30 + 1) 786 = 30 786
+
786 (786  1)
+ 786 × 30 785 × 1
× 30 784 × 1 2 + .... + 1 786 .
2!
Note that all the terms excluding last two terms will
end in two or more zeroes. The last two terms are
786 × 30 ×1 785 and 1 786. Now, the second last term
will end with one zero and the tens digit of the second
last term will be the product of 786 and 3 i.e. 8.
Therefore, the last two digits of the second last term
will be 80. The last digit of the last term is 1. So the
last two digits of 31 786 are 81.
Ex.59Find the last two digits of 41 2789.
Sol. According to the previous example we can calculate
the answer 61 (4 × 9 = 36. Therefore, 6 will be the
tens digit and one will be the units digit).
Ex.60 Find the last two digits of 71 56747.
Sol. Last two digits will be 91 (7 × 7 gives 9 and 1 as
units digit)
Ex.61 Find the last two digits of 51 456 × 61 567.
Sol. The last two digits of 51 456 will be 01 and the last
two digits of 61 567 will be 21. Therefore, the last two
digits of 51 456 × 61 567 will be the last two digits of
01 × 21 = 21.
(ii) Last two digits of numbers ending in 3, 7 or 9 :
Ex.62 Find the last two digits of 19 266.
Sol. 19 266 = (19 2)133. Now, 19 2 ends in 61 (19 2 = 361)
therefore, we need to find the last two digits of (61)133.
Once the number is ending in 1 we can straight away
(iii) Last two digits of numbers ending in 2, 4, 6 or 8 :
There is only one even two-digit number which
always ends in itself (last two digits) - 76 i.e. 76
raised to any power gives the last two digits as 76.
Therefore, our purpose is to get 76 as last two digits
for even numbers. We know that 24 2 ends in 76 and
2 10 ends in 24. Also, 24 raised to an even power
always ends with 76 and 24 raised to an odd power
always ends with 24. Therefore, 24 34 will end in 76
and 24 53 will end in 24.
Ex.65Find the last two digits of 2 543.
Sol. 2 543 = (2 10)54 × 2 3 = (24) 54 (24 raised to an even
power). 2 3 = 76 × 8 = 08.
NOTE :
Here if you need to multiply 76 with 2 n, then you can
straightaway write the last two digits of 2 n because
when 76 is multiplied with 2 n the last two digits
remain the same as the last two digits of 2 n .
Therefore, the last two digits of 76 ×2 7 will be the
last two digits of 2 7 = 28.
Ex.66Find the last two digits of 64 236.
Sol. 64 236 = (2 6)236 = 2 1416 = (2 10)141 × 2 6 = 24 141 (24 raised
to odd power) × 64 = 24 × 64 = 36.
Now those numbers which are not in the form of 2 n
can be broken down into the form 2 n × odd number.
We can find the last two digits of both the parts
separately.
Ex.67Find the last two digits of 62 586.
Sol. 62 586 = (2 × 31)586 = 2 586 × 31 586 = (2 10)58 × 2 6 × 31 586
= 76 × 64 × 81 = 84.
Ex.68Find the last two digits of 54 380.
Sol. 54 380 = (2 × 3 3)380 = 2 380 × 3 1140 = (2 10)38 × (3 4)285 = 76
× 81 285 = 76 × 01 = 76.
Ex.69Find the last two digits of 56 283.
Sol. 56 283 = (23 × 7)283 = 2849 × 7283 = (210)84 × 29 × (74)70 × 73
= 76 × 12 × (01)70 × 43 = 16.
Ex.70 Find the last two digits of 78 379.
Sol. 78 379 = (2 × 39)379 = 2 379 × 39 379 = (2 10)37 × 2 9 × (39 2)189
× 39 = 24 × 12 × 81 × 39 = 92.
get the last two digits with the help of the previous
method. The last two digits are 81 (6 × 3 = 18, so
the tenth digit will be 8 and last digit will be 1).
Factorial n : Product of n consecutive natural numbers
Ex.63 Find the last two digits of 33 288.
is known as ‘factorial n’ it is denoted by ‘n!’.
Sol. 33 288 = (33 4)72. Now 33 4 ends in 21 (33 4 = 33 2 × 33 2
= 1089 × 1089 = xxxxx21) therefore, we need to find
the last two digits of 21 72. By the previous method,
the last two digits of 21 72 = 41 (tens digit = 2 × 2 = 4,
unit digit = 1)
472
factorial one. Hence 0! = 1 = 1!
The approach to finding the highest power of x dividing
Ex.64Find the last two digits of 87
474
So, n! = n(n – 1)(n – 2).............321.
e.g. 5! = 5 × 4 × 3 × 2 × 1 = 120.
 The value of factorial zero is equal to the value of
2
4 118
474
.
2
Sol. 87 = 87 × 87 = (87 ) × 87 = (69 × 69)
[The last two digits of 87 2 are 69]
= 61 118 × 69 = 81 × 69 = 89.
118
× 69
y  y   y 
y! is     2    3  ......., where [ ] represents just
x x  x 
the integral part of the answer and ignoring the fractional
part.
PAGE # 5050
Ex.71 What is the highest power of 2 that divides 20!
completely?
Sol. 20! = 1 × 2 × 3 × 4 ×....× 18 × 19 × 20
= 1 × (21) × 3 × (22) × 5 × (21 × 31) × 7 × (23) × ..... so on.
In order to find the highest power of 2 that divides the
above product, we need to find the sum of the powers
of all 2 in this expansion. All numbers that the divisible
by 21 will contribute 1 to the exponent of 2 in the product
20
= 10. Hence, 10 numbers contribute 2 1 to the
21
product. Similarly, all numbers that are divisible by 22
will contribute an extra 1 to the exponent of 2 in the
20
product, i.e
= 5. Hence, 5 numbers contribute an
22
extra 1 to exponents. Similarly, there are 2 numbers that
are divisible by 23 and 1 number that is divisible by 24.
Hence, the total 1s contributed to the exponent of 2 in
20! is the sum of ( 10 + 5 +2 +1) = 18. Hence, group of
all 2s in 20! gives 218 x (N), where N is not divisible by
2. If 20! is divided by 2x then maximum value of x is 18.
Ex.72 What is the highest power of 5 that divides of
x = 100! = 100 × 99 × 98 × ...... × 3 × 2 × 1.
Sol. Calculating contributions of the different powers of 5,
we have
100
= 20,
100
= 4.
5
52
Hence, the total contributions to the power of 5 is 24, or
the number 100! is divisible by 524.
1
Ex.73 How many zeros at the end of first 100 multiples of
10.
Sol. First 100 multiple of 10 are = 10 × 20 × 30 × ......× 1000
= 10100 (1 × 2 × 3 × .......× 100)
= 10100 × 1024 × N
= 10124 × N
Where N is not divisible by 10
So, there are 124 zero at the end of first 100 multiple of
10.
Ex.74 What is the highest power of 6 that divides 9!
9
9
= 1 and 2 = 0. Thus
6
6
answers we get is 1 which is wrong. True there is just
one multiple of 6 from 1 to 9 but the product 2 × 3 = 6
and also 4 × 9 = 36, can further be divided by 6. Thus,
when the divisor is a composite number find the
highest power of its prime factors and then proceed. In
this case, 9! can be divided by 27 and 34 and thus by 64
(In this case we need not have checked power of 2 as
it would definitely be greater then that of 3).
Sol. By the normal method.
Ex.75What is the largest power of 12 that would divide 49! ?
Sol. To check the highest power of 12 in 49!, we need to
check the highest powers of 4 and 3 in it.
Highest power of 3 in 49! = 22
Highest power of 2 in 49! = 46
46
= 23
2
 Highest power of 12 will be 22. (Since the common
power between 3 and 4 is 22).
 Highest power of 4 in 49! =
Ex.76 How many zeros will be there at the end of 36!36! ?
Sol. Highest power of 5 in 36! is 8.
So there will be 8 zeros at the end of 36!
So at the end of 36!36! , there will be 8 × 36! zeros.
The number system that we work in is called the
‘decimal system’. This is because there are 10 digits
in the system 0-9. There can be alternative system that
can be used for arithmetic operations. Some of the
most commonly used systems are : binary, octal and
hexadecimal.
These systems find applications in computing.
Binary system has 2 digits : 0, 1.
Octal system has 8 digits : 0, 1, 2..., 7.
Hexadecimal system has 16 digits : 0, 1, 2,..., 9, A , B,
C, D, E, F.
After 9, we use the letters to indicate digits. For instance,
A has a value 10, B has a value 11, C has a value 12,...
so on in all base systems.
The counting sequences in each of the systems would
be different though they follow the same principle.
Conversion : conversion of numbers from (i) decimal
system to other base system. (ii) other base system to
decimal system.
(i) Conversion from base 10 to any other base :
Ex.77 Convert (122)10 to base 8 system.
8 122
8
Sol. 8
15
1
0
2
7
1
The number in decimal is consecutively divided by the
number of the base to which we are converting the
decimal number. Then list down all the remainders in
the reverse sequence to get the number in that base.
So, here (122) 10 = (172)8.
Ex.78 Convert (169)10 in base 7
7 169
7 24
7 3
Sol. 0
1
3
3
Remainder
(169)10 =(331)7
Ex.79 Convert (0.3125)10 to binary equivalent.
Sol.
Integer
2  0.3125 = 0.625
0
2  0.625 = 1.25
1
2  0.25 = 0.50
0
2  0.50 = 1.00
1
Thus
(0.3125)10 = (0.1010)2
PAGE # 5151
Ex.80 Convert (1987.725)10  (........)8
Sol. First convert non-decimal part into base 8.
Ex.85 If a – b = 2,
8 1987
8 248 3
8
8
31
3
0
a a
0
7
3
and
b b
_____
then find the value of a, b and c.
cc 0
Sol. These problems involve basic number
(1987)10 = (3703)8
Now we have to convert (0.725)10 (........)8
(i) aa + bb = 11(a + b) (ii) aa, bb are two-digit numbers.
Multiply
Hence, their sum cannot exceed 198. So, c must be 1.
(iii) Hence, cc0 = 110. This implies a + b = 10 or a = 6
0.725 × 8 = [5.8]
0.8 × 8 = [6.4]
0.4 × 8 = [3.2]
0.2 × 8 = [1.6]
0.6 × 8 = [4.8]
...5
...6
...3
...1
...4
and b = 4.
Such problems are part of a category of problems called
alphanumerices.
Keep on accomplishing integral parts after
a 3b
multiplication with decimal part till decimal part is zero.
a c
Ex.86 If _____
a a 9

(0.725)10 = (0.56314...)8

(1987.725)10 = (3703.56314...)8
then find a, b and c if each of them is
distinctly different digit.
(ii) Conversion from any other base to decimal system
Ex.81 Convert (231)8 into decimal system.
Sol. (231)8 , the value of the position of each of the numbers
( as in decimal system) is :
Sol. (i) since the first digit of (a 3 b) is written as it is after
subtracting ac carry over from a to 3.
(ii) there must be a carry over from 3 to b, because if no
carry over is there, it means 3 – a = a.

1 = 80 × 1
2a = 3
3
2
which is not possible because a is a digit. For a carry

3 = 81 × 3
2 = 82 × 2
Hence, (231)8 = (80 × 1 + 81 × 3 + 82 × 2)10
(231)8 = (1 + 24 + 128)10
a=
over 1, 2 – a = a
a=1
(iii) it means b and c are consecutive digit (2, 3),
(231)8 = (153)10
(3, 4),.... (8, 9)
Ex.82 Convert (0.03125)10 to base 16.
Sol. 16  0.3125 = 0.5
0
16  0.5 = 8.0
8
So (0.03125)10 = (0.08)16
1a4
x3b
Ex.87
If
8c8
s 72
t 5d8
Ex.83 Convert (761.56)8  (......)16
Sol. In such conversion which are standard form
conversions, it is easier to
then, find the value of a, b, c, d, s and t, where all of
(761.56)8  (.....)2  (.....)16
them are different digits.
Converting every digit in base 8 to base 2,
(111110001.10110)2  (1F1.B8)16
Ex.84 Convert (3C8.08)16 to decimal
Sol. (3C8.08)16 = 3  162 + C  161
+ 8  16 + 0  16–1 + 8  16–2
Sol. Let us consider 1 a 4 × 3 = s72.
a × 3 results in a number ending in 6.
As 16 and 26 is ruled out, a is 2.
Thus, s = 3, t = 4
Now 1 a 4 × b = 8c8 ; b = 2 or 7
Again 2 is ruled out because in that case, product would
be much less than 800.
= 768 + 192 + 8 + 0 + 0.03125 = (968.03125)10
 b = 7.
So (3C8.08)16 = (968.03125)10
Hence, a = 2, b = 7, c = 6, d = 8, s = 3 and t = 4.
PAGE # 5252
1.
If the number 357y25x is divisible by both 3 and 5, then
find the missing digit in the unit’s place and the
thousand place respectively are :
(A) 0, 6
(B) 5, 6
(C) 5, 4
(D) None of these
2.
There are four prime numbers written in ascending
order. The product of the first three is 385 and that of
the last three is 1001. The last number is :
(A) 11
(B) 13
(C) 17
(D) 19
3.
Find the square root of 7 – 4 3 .
(A) 2 – 3
(B) 5 – 3
(C) 2 – 5
(D) None of these
4.
The number of prime factors of (3 × 5)12 (2 × 7)10 (10)25
is :
(A) 47
(B) 60
(C) 72
(D) 94
5.
How many three-digit numbers would you find, which
when divided by 3, 4, 5, 6, 7 leave the remainders 1, 2,
3, 4, and 5 respectively ?
(A) 4
(B) 3
(C) 2
(D) 1
6.
Six strings of violin start vibrating simultaneously and
they vibrate at 3, 4, 5, 6,10 and 12 times in a minute,
find :
i. After how much time will all six of them vibrate
together ?
ii. How many times will they vibrate together in 30 min ?
(A) 60 sec, 31 times
(B) 60 min, 31 times
(C) 120 sec, 15 times
(D) None of these
7.
The HCF of 2 numbers is 11 and their LCM is 693. If
their sum is 176, find the numbers.
(A) 99,77
(B) 110, 66
(C) 88,77
(D) 121, 44
8.
If P is a prime number, then the LCM of P and (P + 1) is :
(A) P(P +1)
(B) (P + 2)P
(C) (P + 1)(P – 1)
(D) None of these
9.
Find out (A + B + C + D) such that AB x CB = DDD, where
AB and CB are two-digit numbers and DDD is a threedigit number.
(A) 21
(B) 19
(C) 17
(D) 18
10. Three pieces of cakes of weights 4
1
3
Ibs, 6 Ibs and
2
4
1
Ibs respectively are to be divided into parts of equal
5
weights. Further, each must be as heavy as possible.
If one such part is served to each guest, then what is
the maximum number of guests that could be
entertained ?
(A) 54
(B) 72
(C) 20
(D) 41
7
11. How many natural numbers between 200 and 400 are
there which are divisible by
i. Both 4 and 5?
ii. 4 or 5 or 8 or 10 ?
(A) 9, 79
(B) 10, 80
(C) 10, 81
(D) None of these
12. 4 61  4 62  4 63  4 64 is divisible by :
(A) 3
(C) 11
(B) 10
(D) 13
13. If x is a whole number, then x2 (x2 – 1) is always divisible
by :
(A) 12
(B) 24
(C) 12 – x
(D) Multiple of 12
14. If 653 xy is exactly divisible by 80 then the find the value
of (x + y)
(A) 2
(B) 3
(C) 4
(D) 6
15. Find the unit digit of (795 – 358).
(A) 6
(B) 4
(C) 3
(D) None of these
16. When a number P is divided by 4 it leaves remainder
3. If the twice of the number P is divided by the same
divisor 4 than what will be the remainder ?
(A) 0
(B) 1
(C) 2
(D) 6
17. If (232 +1) is divisible by a certain number then which of
the following is also divisible by that number.
(A) (216 – 1)
(B) 216 + 1
96
(C) 2 + 1
(D) None of these
18. When 1! + 2! + 3! + ... + 25! is divided by 7, what will be
the remainder ?
(A) 0
(B) 5
(C) 1
(D) None of these
19. A number when divided by 342 gives a remainder 47.
When the same number is divided by 19, what would
be the remainder ?
(A) 3
(B) 5
(C) 9
(D) None of these
20. What is the remainder when 9875347 × 7435789
× 5789743 is divided by 4 ?
(A) 1
(B) 2
(C) 3
(D) None of these
21. P is a prime number greater than 5. What is the
remainder when P is divided by 6?
(A) 5
(B) 1
(C) 1 or 5
(D) None of these
22. What is the remainder when 587 is divided by 15?
(A) 0
(B) 5
(C) 10
(D) None of these
PAGE # 5353
23. What is the remainder when 763 is divided by 344?
(A) 1
(C) 6
(B) 343
(D) 338
24. What is the remainder when 3040 is divided by 17?
(A) 1
(B) 16
(C) 13
(D) 4
25. What is the remainder when 7413 – 4113 + 7513 – 4213 is
divided by 66?
(A) 2
(B) 64
(C) 1
(D) 0
26. A number when divided successively by 4 and 5 leaves
remainders 1 and 4 respectively. W hen it is
successively divided by 5 and 4, then the respective
remainders will be :
(A) 1, 2
(B) 2, 3
(C) 3, 2
(D) 4, 1
27. How many zeros will be there at the end of the product
2!2!  4!4!  6!6!  8!8!  10!10! ?
(A) 10! + 6!
(C) 10! + 8! + 6!
(B) 2 (10!)
(D) 6! + 8! + 2 (10!)
28. W hen Sholey screened on the TV there was a
commercial break of 5 min after every 15 min of the
movie. If from the start of the movie to the end of the
movie there was in all 60 min of commercials that was
screened what is the duration the movie ?
(A) 180 min
(B) 195 min
(C) 169 min
(D) 165 min
29. Which of the following numbers is exactly divisible by
all prime numbers between 1 and 17 ?
(A) 345345
(B) 440440
(C) 510510
(D) 515513
Directions : (32 to 36) Read the following information
carefully and answer the questions given below.
In a big hostel, there are 1,000 rooms. In that hostel
only even numbers are used for room numbers, i.e.
the room numbers are 2, 4, 6, ...., 1998, 2000. All the
rooms have one resident each. One fine morning, the
warden calls all the residents and tells them to go
back to their rooms as well as multiples of their room
numbers. When a guy visits a room and finds the door
open, he closes it, and if the door is closed, he opens
it, All 1,000 guys do this operation. All the doors were
open initially.
30. The last room that is closed is room number ?
(A) 1936
(B) 2000
(C) 1922
(D) None of these
31. The 38th room that is open is room number :
(A) 80
(B) 88
(C) 76
(D) None of these
32. If only 500 guys, i.e. residents of room number 2 to
1000 do the task, then the last room that is closed is
room number
(A) 2000
(B) 1936
(C) 1849
(D) None of these
33. In the case of previous question, how many rooms
will be closed in all ?
(A) 513
(B) 31
(C) 13
(D) 315
34. If you are a lazy person, you would like to stay in a room
whose number is :
(A) more than 500
(B) more than 1000
(C) 500
(D) 2000
35. A 4-digit number is formed by repeating a 2-digit
number such as 2525, 3232 etc. Any number of this
form is exactly divisible by :
(A) 7
(B) 11
(C) 13
(D) Smallest 3-digit prime number
36. If we write all the whole numbers from 200 to 400, then
how many of these contain the digit 7 only once ?
(A) 32
(B) 34
(C) 35
(D) 36
37. If (12 + 22 + 32 + .....+ 102) = 385, then the value of
(22 + 42 + 62 +...... + 202).
(A) 770
(B) 1155
(C) 1540
(D) (385 × 385)
38. How many four-digit numbers are there such that the
digits are repeated at least once, i.e. all the digits do
not occur only once ?
(A) 9000
(B) 4536
(C) 4464
(D) None of these
39. Find the total number of prime factors in the expression
(4)11 × (7)5 × (11)2.
(A) 37
(B) 33
(C) 26
(D) 29
40. The largest number which exactly divides the product
of any four consecutive natural numbers is :
(A) 6
(B) 12
(C) 24
(D) 120
41. The largest natural number by which the product of
three consecutive even natural numbers is always
divisible, is :
(A) 6
(B) 24
(C) 48
(D) 96
42. The sum of three consecutive odd numbers is always
divisible by :
I. 2
II. 3
III. 5
IV. 6
(A) Only I
(B) Only II
(C) Only I and III
(D) Only II and IV
43. A number is multiplied by 11 and 11 is added to the
product. If the resulting number is divisible by 13, the
smallest original number is :
(A) 12
(B) 22
(C) 26
(D) 53
PAGE # 5454
44. A 3-digit number 4a3 is added to another 3-digit
number 984 to give the four-digit number 13b7, which
is divisible by 11. Then ,(a + b) is :
(A) 10
(B) 11
(C) 12
(D) 15
45. A number when divided by the sum of 555 and 445
gives two times their difference as quotient and 30 as
the remainder. The number is :
(A) 1220
(B) 1250
(C) 22030
(D) 220030
46. In a 4-digit number, the sum of the first two digits is
equal to that of the last two digits. The sum of the first
and last digits is equal to the third digit. Finally, the sum
of the second and fourth digits is twice the sum of the
other two digits. What is the third digit of the number ?
(A) 5
(B) 8
(C) 1
(D) 4
47. Anita had to do a multiplication. Instead of taking 35 as
one of the multipliers, she took 53. As a result, the
product went up by 540. What is the new product ?
(A) 1050
(B) 540
(C) 1440
(D) 1590
48. Three friends, returning from a movie, stopped to eat
at a restaurant. After dinner, they paid their bill and
noticed a bowl of mints at the front counter. Sita took 1/
3 of the mints, but returned four because she had a
monetary pang of guilt. Fatima then took 1/4 of what
was left but returned three for similar reasons. Eswari
then took half of the remainder but threw two back into
the bowl. The bowl had only 17 mints left when the raid
was over. How many mints were originally in the bowl?
(A) 38
(B) 31
(C) 41
(D) 48
49. In a number system, the product of 44 and 11 is 3414.
The number 3111 of this system, when converted to
the decimal number system, becomes :
(A) 406
(B) 1086
(C) 213
(D) 691
50. A set of consecutive positive integers beginning with 1
is written on the blackboard. A student came and erased
one number. The average of the remaining numbers
is 35
(A) 7
(C) 9
7
. What was the number erased?
17
(B) 8
(D) None of these
51. Let D be a recurring decimal of the form D = 0. a1 a2 a1
a2 a1 a2 ....., where digits a1 and a2 lie between 0 and 9.
Further, at most one of them is zero. Which of the
following numbers necessarily produces an integer,
when multiplied by D?
(A) 18
(B) 108
(C) 198
(D) 288
52. W hat is the value of the following expression
 1   1   1 


1

 
 



 (2 2  1)    ( 4 2  1)    (6 2  1)   .....   (20 2  1)  ?

 
 



(A)
9
19
(B)
10
19
(C)
10
21
(D)
11
21
53. Consider a sequence of seven consecutive integers.
The average of the first five integers is n. The average
of all the seven integers is :
(A) n
(B) n + 1
(C) k × n, where k is a function of n
2
(D) n +  
7
54. Let N = 553 + 173 – 723 . N is divisible by :
(A) both 7 and 13
(B) both 3 and 13
(C) both 17 and 7
(D) both 3 and 17
55. Convert the number 1982 from base 10 to base 12.
The results is :
(A) 1182
(B) 1912
(C) 1192
(D) 1292
56. The LCM of two numbers is 864 and their HCF is 144.
If one of the numbers is 288, then the other
number is :
(A) 576
(B) 1296
(C) 432
(D) 144
57. The LCM of two numbers is 567 and their HCF is 9. If
the difference between the two numbers is 18, find the
two numbers :
(A) 36 and 18
(B) 78 and 60
(C) 63 and 81
(D) 52 and 34
58. If the product of n positive real numbers is unity, then
their sum is necessarily :
(A) a multiple of n
(C) never less than n
1
n
(D) a positive integer
(B) equal to n –
59. How many even integers n, where 100  n  200, are
divisible neither by seven nor by nine ?
(A) 40
(B) 37
(C) 39
(D) 38
60. In a certain examination paper, there are n question. For
j = 1, 2....n, there are 2n-j students who answered j or
more questions wrongly. If the total number of wrong
answers is 4095, then the value of n is :
(A) 12
(B) 11
(C) 10
(D) 9
61. The number of positive n in the range 12  n  40 such
that the product (n –1) (n – 2).... 3.2.1 is not divisible by
n is :
(A) 5
(B) 7
(C) 13
(D) 14
PAGE # 5555
62. The owner of a local jewellery store hired 3 watchmen
to guard his diamonds, but a thief still got in and stole
some diamonds. On the way out, the thief met each
1
watchman, one at a time. To each he gave
of the
2
diamonds he had then, and 2 more besides. He
escaped with one diamond. How many did he steal
originally ?
(A) 40
(B) 36
(C) 25
(D) none of these
63. A rich merchant had collected many gold coins. He did
not want any body to know about him. One day, his wife
asked, “How many gold coins do we hire?” After
pausing a moment he replied, “Well ! if divide the the
coins into two unequal numbers, then 48 times the
difference between the two numbers equals the
difference between the square of the two numbers.
“ The wife looked puzzled. Can you help the merchant’s
wife by finding out how many gold coins the merchant
has ?
(A) 96
(B) 53
(C) 43
(D) 48
64. A child was asked to add first few natural numbers
(that is 1 + 2 + 3.....) so long his patience permitted. As
he stopped, he gave the sum as 575. When the teacher
declared the result wrong, the child discovered he had
missed one number in the sequence during addition.
The number he missed was :
(A) less than 10
(B) 10
(C) 15
(D) more than 15
65. 76n – 66n, where n is an integer > 0, is divisible by :
(A) 13
(B) 127
(C) 559
(D) All of these
66. The value of
10  25  108  154  225
(A) 4
(C) 8
is :
(B) 6
(D) 10
67 When (55) 10 is represented in base 25 then the
expression is :
(A) (25)25
(B) (35)25
(C) (55)25
(D) none of these
68. Arrange the following numbers in ascending order
3 4 7 1
, , , .
7 5 9 2
(A)
4 7 3 1
, , ,
5 5 9 2
(B)
3 1 7 4
, , ,
7 2 9 5
(C)
4 7 1 3
, , ,
5 9 2 7
(D)
1 3 7 4
, , ,
2 7 9 5
69. Which of the following surds is greatest in magnitude :
6
17 , 2,12 25 , 3 4
(A) 6 17
(B)
(C)
(D)
3
4
12
25
2
70. What is the decimal equivalent of the hexadecimal 25
digit number (100.....001)16 ?
(A) 223 + 1
(B) 224 + 1
(C) 292 + 1
(D) 296 + 1
71. The square root of a perfect square containing n digits
has ............ digits.
n 1
(A)
(B) n/2
2
(C) A or B
(D) None
72. If the decimal number 2111 is written in the octal system,
then what is its unit place digit ?
(A) 0
(B) 1
(C) 2
(D) 3
73. The 288th term of the series a, b, b, c, c, c, d, d, d, d, e,
e, e, e, e, f, f, f, f, f, f,....is :
(A) u
(B) v
(C) x
(D) w
74. If n = 67 then find the unit digit of [3n + 2n ].
(A) 1
(B) 10
(C) 5
(D) None
75. The number of 2-digit numbers n such that 3
divides n – 2 and 5 divides n – 3 is : [KVPY 2007]
(A) 5
(B) 6
(C) 7
(D) 10
76. How many 4-digit numbers are there with the property
that it is a square and the number obtained by
increasing all its digits by 1 is also a square ?
[KVPY 2007]
(A) 0
(B) 1
(C) 2
(D) 4
77. The number of positive fractions m/n such that
1/3 < m/n < 1 and having the property that the fraction
remains the same by adding some positive integer to
the numerator and multiplying the denominator by the
same positive integer is :
[KVPY 2007]
(A) 1
(B) 3
(C) 6
(D) infinite
78. If a and b are any two real numbers with opposite
signs, which of the following is the greatest ?
[KVPY 2008]
(A) (a–b)2
(B) (|a| – |b|)2
(C) |a 2 – b 2|
(D) a 2 + b 2
79. The number (1024)1024 is obtained by raising (16)16
to the power n. What is the value of n ?
[KVPY 2008]
(A) 64
(B) 64 2
(C) 64 64
(D) 160
80. The number of integers a such that 1  a  100 and
a a is a perfect square is :
[KVPY 2008]
(A) 50
(B) 53
(C) 55
(D) 56
81. The sum of 7 consecutive positive integers is equal
to the sum of the next five consecutive integers. What
is the largest among the 12 numbers ? [KVPY 2008]
(A) 24
(B) 23
(C) 22
(D 21
PAGE # 5656
82. Let 0 < a < b < c be 3 distinct digits. The sum of all
3-digit numbers formed by using all the 3 digits
once each is 1554. The value of c is : [KVPY 2008]
(A) 3
(B) 4
(C) 5
(D) 6
83. If the decimal 0.d25d25d25 —— is expressible in the
form n/27, then d + n must be :
[KVPY 2008]
(A) 9
(B) 28
(C) 30
(D) 34
84. Let S n be the sum of all integers k such that
2n < k < 2n+1, for n  1. Then 9 divides Sn if and only if :
[KVPY 2009]
(A) n is odd
(B) n is of the form 3k+1
(C) n is even
(D) n is of the form 3k + 2
85. The number of ways in which 1440 can be expressed
as a product of two positive integers is : [IAO 2007]
(A) 17
(B) 18
(C) 35
(D) 36
86. Let N = 28 , the sum of all distinct factors of N is :
[IAO 2008]
(A) 27
(B) 28
(C) 55
(D) 56
2
3
2009
87. The units digit of (1 + 9 + 9 + 9 + --------- + 9 ) is :
[IAO 2009]
(A) 0
(B) 1
(C) 9
(D) 3
88. The biggest among the following is :
[IAO 2009]
(A) 21/2
(B) 31/3
(C) 61/6
(D) 81/8
89. W hen a positive integer x is divided by 47, the
remainder is 11. Therefore, when x2 is divided by 47,
the remainder will be :
[IAO 2009]
(A) 7
(B) 17
(C) 27
(D) 37
90. When the number 72010 is divided by 25, the remainder
will be :
[IAO 2009]
(A) 1
(B) 7
(C) 18
(D) 24
91. P, Q and R are three natural numbers such that P and
Q are primes and Q divides PR. Then out of the
following the correct statement is :
[IJSO 2008]
(A) Q divides R
(B) P divides R
(C) P divides QR
(D) P divides PQ
92. The expression (5a – 3b)3 + (3b – 7c)3 – (5a – 7c)3 is
divisible by :
[IJSO 2008]
(A) (5a + 3b + 7c)
(B) (5a – 3b – 7c)
(C) (3b – 7c)
(D) (7c – 5a)
93. If a2 + 2b = 7, b2 + 4c = – 7 and c2 + 6a = – 14, then the
value of (a2 + b2 + c2) is :
[IJSO 2009]
(A) 14
(B) 25
(C) 36
(D) 47
94. In the familiar decimal number system the base is 10.
In another number system using base 4, the counting
proceeds as 1, 2, 3, 10, 11, 12, 13, 20, 21 .... The
twentieth number in this system will be:
[IJSO 2010]
(A) 40
(B) 320
(C) 210
(D) 110
95. If the eight digit number 2575d568 is divisible by 54
and 87, the value of the digit ‘d’ is:
[IJSO 2011]
(A) 4.
(B) 7.
(C) 0.
(D) 8.
96. The number of distinct prime divisors of the number
5123 – 2533 – 2593 is :
[KVPY 2011]
(A) 4
(B) 5
(C) 6
(D) 7

PAGE # 5757
LO G A R I T H M
(ii) If 0 < a < 1, then
The logarithm of the number N to the base ' a ' is the
exponent indicating the power to which the base
' a ' must be raised to obtain the number N. This
number is designated as log a N. Hence:
logaN = x  a x = N , a > 0, a  1 & N > 0
(a) loga x < 0 for all x > 1
Systems of Logarithm :
There are two systems of logarithm which are generally
used.
function.
(b) loga x = 0 for x = 1
(c) logax > 0 for all x satisfying 0 < x < 1
(d) x > y

logax < loga y i.e. loga x is a decreasing
 graph of y = loga x, 0 < a < 1.
y
(i) Common logarithm : In this system base is always
taken as 10.
If a = 10, then we write log b rather than log 10 b.
y = logax, 0 < a < 1.
(1,0)
y'
a  1, b > 0, b  1 and  is any real number then ;
log 102 = 0.3010 ; log 103 = 0.4771 ;
n 2 = 0.693 ;
n 10 = 2.303
REMARK :
The existence and uniqueness of the number
log a N can be determined with the help of set of
conditions, a > 0 and a  1 and N > 0.
(i)
loga (mn) = loga m + logan
[Where m and n are +ve numbers]
in general log a(x1 x2 ......xn) = log ax1 + log a x2 + ........+
log a xn
m
(ii) loga   = logam – logan
n
(iii) loga(m)n = n logam
Some Useful Results :
(i) If a > 1 then
(a) loga x < 0 for all x satisfying 0 < x < 1
(b) loga x = 0 for x = 1
(iv) logam 
log m
b
log a
b
(v) logam . logma = 1
(vi) If ‘a’ is a positive real number and ‘n’ is a positive
rational number, then
(c) loga x > 0 for x > 1
 loga x > loga y i.e. logax is an increasing
a loga n  n
function.
(vii) If ‘a’ is a positive real number and ‘n’ is a positive
 graph of y = loga x, a > 1
rational number, then
y
logaq np  p loga n
q
y = logax, a > 1
x'
x
Let m & n are arbitrary positive numbers, a > 0,
REMEMBER
(d) x > y
0
x'
(ii) Natural logarithm : In this system the base of the
logarithm is taken as ‘e’. Where ‘e’ is an irrational
number lying between 2 and 3.
If a = e, we write nb rather than log e b. Here ' e ' is
called as Napiers base & has numerical value equal
to 2.7182.
0
(1,0)
x
(viii) p loga q  qloga p
Ex.1 If log3a = 4, find value of a.
Sol. 

y'

log3a = 4
a = 34
a = 81.
PAGE # 5858
Ex.2 Find the value of log
Sol. Given log
Ex.7 Find the value of the following :
9
27
3
– log
 log
8
32
4
3
 9 27 
9
27
3
  log
– log
 log  log 
4
8
32
4
 8 32 



(x – 2)2 = 0 
log4
Ex.4 Evaluate : 3
2 – log 3 5
x=2
= 9. 3 log3 5
[am + n = am.an]
(2  3 ) x =

x = –1.
(2  3 ) (2  3 )
1
= (2  3 )
2 3

10 –3x = 10 –4  x = 4/3.
(sin 2 60   cos 2 60  )
tan 45  cot 45 
sec 30 
(iii) log 20 1
alogaN = N, a > 0, a  1 & N > 0
REMARKS :
[ loga a  1 ]
(i)
log a1 = 0
[As a1 = a]
(ii) loga a = 1
log11 13
and B = log9125 + 13
log11 13
4
= log 3 3 5 + 7
4
log35 + 7 log11 13
3
log11 13
...(i)
3
log35 + 7 log11 13
2
By (i) and (ii) we have,
3
4
log35 = B – log35
2
3
4
3
log35 < log35
3
2
A < B.
,
(iii) loga 0 = not defined
[As an = 0 is not possible, where n is any number]
(iv) loga (–ve no.) = not defined.
[As in loga N, N will always be (+ ve)]
(v) log 1/a a =  1
(vi)
1
loga b
log ba =
ex na
Ex.9 Find the value of the following
B = log 3 2 5 3 + 7 log11 13
B=
log11 7
(vii) a x =
and, B = log9125 + 13 log11 7


Fun dame ntal Log arit hmic Ide ntit y :
then find the relation between A and B.

(2  3 )
Sol. (i) For (i) and (ii) log is not defined because base is
equal to 1.
(iii) log 201 = 0.
3 2
–
2 3
A–
3)
(2  3 ) x = 2  3
(ii) log
5
= .
6
or,
 x = 20/3.

(i) log
3
2
= log55 – log2 2
2
3
or,
1024
(2 2 ) = 1024
2 3x/2 = 2 10
9
= .
5
= log 5 2 5 3 – log 23 2 2
A=
2
x
Ex.8 Find the value of the following :
–1
Ex.5 Find the value of log25125 – log84.
Sol. Given log25125 – log84
or,
(ii) Let x = log2
= 9 × 5–1
Sol. A = log27625 + 7
(2  3 )
3)
(v) Let x = log 0.001 0.0001  (0.001)x = 0.0001
.
Ex.6 If A = log27625 + 7
1024
 4x = 8
 x = 3/2.
Sol. (i) Let x = log 48
 2 2x = 2 3
(iv) Let x = log( 2
2 – log 3 5
Sol. Given, 3 2 – log3 5 = 3 .3
=
(iv) log( 2
2
(iii) Let x = log 1/49 343
 7 –2x = 7 3
 x = – 3/2.
2

(iii) log 1/49343


Ex.3 If 2log4x = 1 + log4(x – 1), find the value of x.
Sol. Given 2log4x = 1 + log4(x – 1)
 log4x2 – log4(x – 1) = 1
x
=1
x –1
x2
41 =
x –1
x2 = 4x – 4
x2 – 4x + 4 = 0
(ii) log
2
(v) log 0.001 0.0001
 9 32 3 
 log 
 
 8 27 4 
= log1 = 0. [ loga1 = 0]

(i) log 48
(i) 81
...(ii)
1
log 53
Sol. (i) Let x = 81
= 3
log3 5 
(ii) Let x = 8
(ii) 8
1
log 53
1

 log 2 3 121  
3

= (3 4 )log3 5
= 5 4 = 625.
1

 log 2 3 121  
3

= 2 3 log 2
3

121  1
= 2.2log2 121 = 2(121) = 242.
PAGE # 5959
Ex.10 Find the value of
(i) log 3 4. log 4 5. log 5 6. log 67 . log 78. log 89.
The equality log a x = log a y is possible if and only if
x = y i.e. loga x = loga y x = y
(ii) log tan1° + log tan2° + log tan3° + .... + log tan89°.
Sol. (i) log 3 4. log 4 5. log 5 6. log 67 . log 78. log 89
log 4 log 5 log 6 log 7 log 8 log 9
log 3 . log 4 . log 5 . log 6 . log 7 . log 8
log 9
= log 3 = 2.
(ii) log tan1° + log tan2° +.......+ log tan89°
= log(tan1°. tan2°. tan3° .... tan45° ..... tan89°)
= log (tan1° tan2° tan3° ... tan45° ... cot3° cot2° cot1°)
= log (1.1.1 ....... 1) = log 1 = 0.
Ex.11 Find the value of :
(i) 2log3 5 – 5 log3 2
(ii) log 2
75
32
5
– 2 log 2
+ log 2
16
243
9
Always check that the solutions should satisfy x > 0,
y > 0, a > 0, a  1.
Ex.13 Solve the following equations :
(i) log 3(x + 1) + log 3 (x + 3) = 1
(ii) log 2 log 4 log 5 x = 0
(iii) log 3 [5 + 4 log 3 (x – 1)] = 2
Sol. (i) log 3(x + 1) + log 3 (x + 3) = log 3 (x + 1) (x + 3) = 1
 (x + 1) (x + 3) = 3
 x2 + 4x = 0
 x = 0, – 4.
Also x + 1 > 0
or
x>–1
And x + 3 > 0
or
x>–3
Hence solution is x = 0.
(ii)



(iii) log3 0. 1
Sol. (i) 2log3 5 – 5 log 3 2
log 2 log 4 log 5x = 0
log 4 log 5x = 1
log 5x = 4
x = 5 4 = 625.
log 25
(iii)



Here 2log3 5 = 2 log 2 3
= 2log2 5

log3 2

= 5log3
2
Hence 2log3 5 – 5 log3 2 = 0.
32
75
5
(ii) log 2
– 2 log 2
+ log 2
243
16
9
 75 32 81 
 = log 22 = 1
= log 2  .
 16 243 25 
(iii) log3 0. 1 = log 31/9 = log 33 –2 = – 2
Ex.12 If log 615 = a, log 1218 = b, then find log 2524 in terms
of a and b.
log3 15
1  log3 5
Sol. log 615 = log 6 = 1  log 2 = a
...(i)
3
3
log3 18
2  log3 2
log 1218 = log 12 = 1  2 log 2 = b
...(ii)
3
3
From (i) and (ii)
2b
log 32 =
2b  1
ab  a  2b  1
And log 35 =
2b  1
log3 24
1  3 log3 2
log 2524 = log 25 = 2 log 5
3
3
 2b 
1 3 

 2b  1 
=
 ab  a  2b  1 
2

2b  1


5b
= 2 (a  2b  ab  1) .
log 3 [5 + 4log 3 (x – 1)] = 2
5 + 4 log 3 (x – 1) = 9
log 3 (x – 1) = 1
x – 1 = 3  x = 4.
Ex.14 Find the total number of digits in the number
6 100 (Given log 102 = 0.3010 ; log 103 = 0.4771).
Sol. Let x = 6 100

log 10x = 100 log106

log 10x = 100 [log 102 + log 103]
= 100 [0.3010 + 0.4771]
= 100 (0.7781)
= 77.81
 The number of digits = 78.
Ex.15 Prove that log 35 is irrational.
Sol. Let log 35 is rational.

log35 =
p
[where p and q are co-prime numbers]
q
 3 p/q = 5  3 p = 5 q.
which is not possible, hence our assumption is
wrong and log 35 is irrational.
1.
Given log2 = 0.3010, then log 16 is :
(A) 2.4080
(C) 0.2408
2.
(B) 1.2040
(D) 1.9030
If logxy = 100 and log2x = 10, then the value of y is :
(A) 21000
(B) 2100
(C) 22000
(D) 210000
PAGE # 6060
3.
4.
(A) 0
(B) 1
(C) 2
(D) 10
7.
9
N
9
(B) N 
9
M
3
(C) M 
3
N
9
(D) N 
3
M
(A) 243
(B) 27
(C) 343
(D) 64
Find all the real values of x, which satisfy the equation
1
(B) 2,
2
1
(C)
 2.
1
(A) 2,
,
1
2 2 2
,1
1
and
2
2 2
(D) 2
(C)
1

s 
of
(C)
3
(1  x  y )
summation
1
1
1

 ... 
log2 x log3 x
log43 x i s e q u a l t o
1 x  y
3
1
5
(C) 6
(D)
1
6
14. If log 2 = a and log3 = b then [log(1) + log(1 + 3) + log
(1 + 3 + 5) + .......+ .....+ log (1 + 3 + 5 + 7 + ..... + 19)]
– 2[log 1 + log2 + log3 + ....... log7] = p + qa + rb where
p,q,r are constants. What is the value of p + 2q + 3r if all
logs are in base 10 ?
(A) 12
(B) 26
(C) 18
(D) Cannot be determined
2n  1
n  1
(B)
(C) 1 – n
(D) 0
following
(D)
(B)
(A)
the
(B) 3 (1 - x - y)
(A) 5
(B) 1
1 1 1
 
q r s
Va l u e
1
3(1  x  y )
15. The sum to 2n terms of the
log3 1 – log33 + log3 9 – log3 27 ...... is :
If ap = bq = cr = ds, then loga (bcd) is equal to :
1 1
(A) p   
q r
(A)
13. If logyx = 10, then logx 3 (y6) is equal to :
The value of x, when log3(log2 x) + 2 log9(log7 8) = 2, is :
2
5
2
 log2 x  log2 x  4 

9.
12. If log30 3 = x and log30 5 = y, then log8 30 is equal to :
1
If log 3 M  3 log 3 N = 1+ log0.0085, then :
3
9
(A) M 

(A) 1
(B) 2
(C) 3
(D) No real value of x exists
q
1 q
q
(D) p =
1 q
x3
8.

(B) p =
(C) p = q = 1
6.
log log x
equation: 2 27 3 4 = (log4x)2 + 5(log4 x) + 2.
log10 p + log10 q = log10 (p–q), then :
(A) p = q = 0
5.
11. Find the number of different values of x that satisfy the
The value of [log10 (5 log10 100)]2 is :
:
series,
n  1
2n  1
(D) – n
16. When the curves y = log10x and y = x – 1 are drawn in the
x-y plane. How many times do they intersect for values
x  1 ?
(A) Never
(B) Once
(C) Twice
(D) More than twice
17. Solve : log (x + 4) +log (x – 4) = log 3.
1
(A) log x
e
1
(B)
e
(A) x = – 19
(C) x = 
1
(C) log
( 43!) x
43!
(D) log e
x
10. Solve for ‘x’ if 2logx 7 + log7x7+3log49x7 = 0.
(A) x 
4
3
(C) x  7 4 / 3
(B) x  7 1/ 2
(D) (B) or (C) only
(B) x = + 19
19
(D) Cannot be determined
18. Find x from the equation ax  c– 2x = b3x + 1.
2 log b
(A) log a  2 log c  3 log b
log b
(B) log a  2 log c  3 log b
log b
(C) log a  2 log c  3 log b
(D) Cannot be determined
PAGE # 6161
19. If log 2 = 0.3010 and log 3 = 0.4771, then find the
number of digits in 617 .
(A) 12
(B) 14
(C) 18
(D) 17
20. Express 3 log 105 – 4 log10 x + log10y2 as the log of a
single number.
(A) log10
 25 y 2 


 x4 


(B) log10
 125 y 2 


(C) log10 
3

 x

 125 y 3

 x4





 125 y 2 


(D) log10 
4

 x

 a3 
 in terms of log a, log b and
 c 5b 2 


21. Express the log 
log c.
31. [(log x – log y)(logx2 + logy2)]  [(logx2 – logy2)(logx + log y)]
is equal to :
(A) 0
(B) 1
(C) log x/y
(D) log xy
32. If log10 [log10(log10x)] = 0.
(A) x = 103
(C) x = 155
(B) x = 1010
(D) None
(B) 2
(D) none of these
34. Find the number of integral pairs for (x, y) from the
following equations :
22. Find x if log10 1250 + log1080 = x.
(A) 5
(B) 4
(C) 8
(D) 7
23. If logx2 = a, logx3 = b, logx5 = c, then find the value of the
following in terms of a, b and c.
(i) logx16
(ii) logx75
(iii) logx60
(A) 4a, 2a + b, 2a + b + c
(B) 4a, 2c + b, 2a + b + c
(C) 4a, 2c + b, 2c + b + a
(D) 4a, 2c + b, 2a + 3b + c
24. Find the value of log (1 + 2 + 3) =
(A) log1 + log2 – log3
(B) log1 + log2 + log3
(C) log1 – log2 + log3
(D) – log1 + log2 + log3
(B) a rational number
(D) whole number
xy
1
= (log x + log y), then :
2
2
(B) x = – y
(D) 2x = yZ
27. Find the value of (yz)logy – logz × (zx)logz – logx × (xy)logx – logy = ?
(A) 3
(B) 2
(C) 0
(D) 1
28. The number of solutions for the equation in
logex + x – 1 = 0 is :
(A) 1
(B) 2
(C) 4
(D) None of the above
29. If log 10N  2.5 then, find out total number of digits in
N.
(A) 3
(C) 5
(D) Cannot be determined
(B) 3
(D) 100
(A) 1
(C) 4
3
(D) log a – 5 log c – log b
2
(A) x = y
(C) x = 2y
(A) 1
(C) 10
equation 2log 2 log 2 x + log 1/2 log 2 (2 2 x ) = 1 :
2
3
(C) log a – 5 log c –
log b
3
2
26. If log
1 

 . When simplified has the value equal to
log10 1 
 1999 
33. The number of solutions for real x, which satisfy the
3
log a – 5 log c – 2 log b
2
(B) 3 log a – 5 log c – 2 log b
(A)
25. log418 is :
(A) an irrational number
(C) natural number
1
1
1



30. log 10  1   + log 10  1   + log 10  1   + ... +
2
3
4



log 100|x + y| =
(A) 0
(C) 2
1
& log 10 y – log 10|x| = log 100 4
2
(B) 1
(D) none of these
35. How many positive real numbers x are there such that
xx
x
(A) 1
(C) 4
x
 ?
 x x
[KVPY 2007]
(B) 2
(D) infinite
36. Let logab = 4, logcd = 2 where a, b, c, d are natural
numbers. Given that b – d = 7, the value of c – a is :
[KVPY 2009]
(A) 1
(B) – 1
(C) 2
(D) – 2
37. The value of log10(3/2) + log10 (4/3) + ------ up to 99
terms.
[IAO 2008]
(A) 0
(B) 2
(C) 2.5
(D) none of the above
38. If a, b  1, ab > 0, a  b and logba = logab, then ab = ?
[IAO 2008]
(A) 1/2
(B) 1
(C) 2
(D) 10
39. Find x, if 100.3010 = 2, 100.4771 = 3 and 10x = 45.
[IAO 2009]
(A) 0.6532
(B) 1.6532
(C) 1.6570
(D) 1.7781
40. If x < 0 and log7 (x2 – 5x – 65) = 0, then x is :
[IJSO-2011]
(A) –13
(B) –11
(C) – 6
(D) – 5
(B) 4
PAGE # 6262
NUTRITION
It helps in mechanical churning and chemical
digestion of food.
The stomach of ruminant (cud chewing) animals have
compound stomach and consists of 4 chambers
rumen, reticulum; omasum and abomasum.
Rumen and reticulum harbour numerous bacteria
and Protozoa which help in digestion of cellulose
called symbiotic digestion.
Abomasum is true stomach as it secretes gastric juice
containing HCl and pepsin.
DIGESTIVE SYSTEM
It includes alimentary canal and digestive glands.
Alimentary canal starts from mouth and ends into
anus.
•
•
Mouth or Buccal Cavity : An adult has 16 teeth on
each jaw.
In each half of jaws starting from middle to backward
these are, incisors - 2, canine - 1, premolar - 2, molars
- 3 (2 + 1 + 2 + 3).
Dental formula is 2123/2123.
Last molars are called wisdom teeth.
Milk teeth start erupting after 6 months of birth and
appear between 6 to 24 months.
Dental formula of milk teeth is 2102/2102.
Stomach : Inner mucosa of stomach is raised into
larger number of longitudinal folds called gastric
rugae.
•
Small Intestine : Villi and microvilli increase the
surface area of digestion and absorption of food.
Pancreatic duct release few enzymes which act on
carbohydrates, fats and proteins.
•
Large Intestine : In some herbivores (like horse and
ass), caecum is large and is a site of microbial
digestion of cellulose.
In man caecum very small vestigial organ and is called
appendix.
Table : Vitamins Necessary for Normal Cell Functioning
Deficiency Disease
Deficient Nutrient
Deficient Nutrient
1. Xerophthalmia
Vitam in A / Retinol
10. Megaloblastic
anaemia
Folic acid and Vitam in B 12
2. Night-blindness
Vitam in A
11. Pernicious
anaemia
Vitam in B 12 (Cyanocobalam ine)
3. Rickets (in children)
Vitam in D/ Sun-Shine Vitam in 12. Scurvy
4. Osteomalacia (adults) Vitam in D
•
Deficiency Disease
Vitam in C/ As corbic acid
13. Tetany
Calcium
5. Sterility
Vitam in E (Tocopherol)
14. Anaemia
Iron
6. Bleeding disease
Vitam in K (Phylloquinone)
15. Goitre
Iodine
7. Beri beri
Vitam in B 1 (Thiam ine)
16. Fluorosis
Exces s of fluorine
8. Cheilosis
Vitam in B 2 (Riboflavin)
17. Kw ashiorkor
Proteins
9. Pellagra
Vitam in B 3 (Niacin)
18. Marasmus
Proteins and food calories
Knowledge Boosters :
(a) Cardiac glands : secrete an alkaline mucus.
(1) Salivary Glands : It produces saliva. In man only
three pairs of salivary glands are present.
(b) Pyloric glands : secrete an alkaline mucus.
(a) Parotid glands : largest glands present just below
the external ear. In this glands, virus causes mumps
disease.
1. Peptic / Zymogen cells - secrete pepsinogen,
prorennin
(b) Submaxillary glands : these lie beneath the jawangles.
(c) Sublingual glands : smallest glands which lie
beneath the tongue and open at the floor of buccal
cavity.
(2) Gastric Glands: Present in the mucosa of the
stomach. These are of 3 types :
(c) Fundic glands : each gland has 4 types of cells.
2. Oxyntic cells - secrete HCI
3. Goblet cells - secrete mucus
4. Argentaffin cells - produces serotonin, somatostatin
and histamine
(3) Liver : It is the largest gland and consists of a
large right lobe, a small left lobe and two small lobes.
On the right lobe lies gall bladder, which, stores bile
juice secreted by the liver.
PAGE # 63
•
•
Bile juice contains no enzyme but possesses bile
salts and bile pigments (bilirubin-yellow and
biliverdin-green).
Formation of glucose from excess organic acids.
Storage of vitamins : A, D, E, K
Synthesis of vitamin A from carotene.
•
•
•
•
Secretions of blood anticoagulant named heparin.
Synthesis of blood or plasma proteins, (fibrinogen
and prothrombin)
Secretion of bile, detoxification of harmful chemicals.
Elimination of pathogens and foreign particles through
phagocytic cells called Kupffer’s cells.
Table : Various Enzymes Involved in Digestion
5.
Stomach is protected from HCl and gastric juice
because :
(A) The two are dilute
(B) Epithelial lining is resistant to them
(C) Wall has neutralizing agents
(D) Stomach lining is covered by mucus
6.
In a villus, some of the glycerol and fatty acids are
combined to form fats, coated with proteins, and then
transported to the :
(A) Lacteal
(B) Capillaries
(C) Lumen of the small intestine
(D) Lumen of the large intestine
7.
Among mammals, an herbivore has :
(A) More teeth than a carnivore
(B) Fewer teeth than a carnivore
(C) Flatter teeth than a carnivore
(D) Teeth that are more pointed than a carnivore
8.
Gastric juice has a pH of about :
(A) 1
(B) 2
(C) 6
(D) 10
9.
Softness and deformities of bones like bowlegs and
pigeon-chest are symptoms of the disease :
(A) Rickets
(B) Osteomalacia
(C) Goitre
(D) Beri-Beri
EXERCISE
1.
In human digestion is :
(A) Intercellular
(C) Extracellular
(B) Intracellular
(D) Both A & B
2.
The enzyme rennin converts
(A) Proteins to proteoses (B) Fats to fatty acids
(C) Casein to paracasein (D) Proteins to peptones
3.
Digestion in Amoeba is :
(A) Intercellular
(B) Intracellular
(C) Both of these
(D) None of these
4.
Statements true or false is:
(i) No absorption of food takes place in mouth and
oesophagus
(ii) Absorption of H2O, alcohol, simple salts, glucose
and chloride takes place in the stomach to slight
extent.
(iii) Whole protein particles can be absorbed by
pinocytosis.
(A) (i) and (ii) true (iii) rarely true
(B) All true
(C) (i) and (iii) true B false
(D) (i) and (ii) true (iii) false
PAGE # 64
10. In rabbit when a dilute solution of glucose reaches
small intestine is :
(A) Absorbed rapidly by the active transport with
sodium ions
(B) Absorbed by facilitated diffusion
(C) Lost to outside with undigested food
(D) Absorbed rapidly by active transport independent
of sodium ions
18. B.M.R. for a normal human adult is :
(A) 1600 Kcal/day
(B) 2000 Kcal/day
(C) 2500 Kcal/day
(D) 3000 Kcal/day
11. Chymotrypsin acts on :
(A) Carbohydrates
(C) Fats
20. If all bacteria of intestine die, then what will be the
effect on body:
(A) Man will feel tired all the body
(B) It will cause blindness
(C) There will be no synthesis of vitamin -B Complex
(D) Excretion will be effected
(B) Proteins
(D) Starch
12. The vermiform appendix is made up of :
(A) Striated muscles
(B) Excretory tissue
(C) Lymphatic tissue
(D) Stem cells
21. The gall bladder is involved in
13. Digestion is completed in :
(A) Ileum
(B) Duodenum
(C) Stomach
(D) Rectum
[KVPY 2011]
14. One of the following is not an enzyme of digestive
system :
(A) Trypsin
(B) Amylase
(C) Enterogastrone
(D) Enterokinase
15. Fat soluble vitamins are :
(A) A, D and E
(B) B, C and D
(C) B,D and E
(D) A, B and C
16. Sunshine vitamin is :
(A) Vitamin A
(C) Vitamin K
19. Cholesterol rich diet is undesirable because it :
(A) Makes person obese
(B) Causes heart ailments
(C) Is difficult to remove it from body
(D) All of the above
(B) Vitamin D
(D) Vitamin E
17. Which of the vitamins is essential for normal vision :
(A) Folic acid
(B) Biotin
(C) Riboflavin
(D) Retinol
(A)
(B)
(C)
(D)
synthesizing bile
storing and secreting bile
degrading bile
producing insulin
22. In the 16th century, sailors who travelled long
distances had diseases related to malnutrition,
because they were not able to eat fresh vegetables
and fruits for months at a time scurvy is a result of
deficiency of
[KVPY 2011]
(A) carbohydrates
(B) proteins
(C) vitamin C
(D) vitamin D
23. Several mineral such as iron, iodine, calcium and
phosphorous are important nutrients. Iodine is
found in
[KVPY 2011]
(A) thyroxine
(B) adrenaline
(C) insulin
(D) testosterone

PAGE # 65
SERIES COMPLETION
Series completion problems deals with numbers, Ex 5.
alphabets and both together. While attempting to
solve the question, you have to check the pattern of
the series. Series moves with certain mathematical Sol.
operations. You have to check the pattern.
Type of questions asked in the examination :
Ex 6.
(i) Find the missing term(s).
(ii) Find the wrong term(s).
Sol.
NUMBER SERIES
8, 12, 21, 46, 95, ?
(A) 188
(B) 214
(C) 148
(D) 216
(D) The pattern is + 22, + 32, + 52, + 72, .......
missing number = 95 + 112 = 216
3, 9, 36, 180, ?
(A) 1080
(B) 900
(C) 720
(D) None of these
(A) Each term is multiplied by 3, 4, 5 and so on
respectively. Therefore, the next term would be
180 × 6 = 1080.
(a) Some Important Patterns :
(b) Multiple Series :
(i) a, a ± d, a ± 2d, a ± 3d.......(Arithmetic Progression)
A multiple series is a mixture of more than one
series :
(ii) a, ak, ak2, ak3, ................(Geometric Progression)
a a a
(iii) a, , 2 , 3 , .............(Geometric Progression)
k k k
Ex 7.
Sol.
(iv) Series of prime numbers – i.e. 2, 3, 5, 7, 11, ......
(v) Series of composite numbers –
i.e. 4, 6, 8, 9, 10, 12, .................
Directions : (1 to 10) Find the missing numbers :
Ex 1.
Sol.
16, 19, 22, 25, ?
(A) 27
(B) 28
(C) 29
(D) 25
(B) As per series a, a + d, a + 2d,.........
a = 16
d=3
a + 4d = 16 + 4 × 3
Ex 8.
Sol.
Ex 9.
Ex 2.
Sol.
Ex 3.
Sol.
Ex 4.
Sol.
9, 18, 36, ?, 144
(A) 70
(B) 56
(C) 54
(D) 72
(D) As per series, a, ak, ak2, ak3, ........
a = 9, k = 2
ak3 = 9 × 23 = 72
2, 6, 14, 26, ?
(A) 92
(B) 54
(C) 44
(D) 42
(D) The pattern is +4, +8, +12, +16, .......
240, ? , 120, 40, 10, 2
(A) 120
(C) 40
(B) 240
(D) 10
1
1
1
1
(B) The pattern is ×1, × , × , × , ×
3
5
2
4
 missing term = 240 × 1 = 240
Sol.
Ex 10.
Sol.
4, 7, 3, 6, 2, 5, ?
(A) 0
(B) 1
(C) 2
(D) 3
(B) The sequence is a combination of two series
I
4, 3, 2, ?
II 7, 6, 5
The pattern followed in I is – 1, – 1, – 1
 missing number = 2 – 1 = 1
14, 15, 12, 16, 9, 18, 4, 21, ?
(A) 2
(B) 3
(C) – 3
(D) – 5
(C) The sequence is a combination of two series.
 14, 12, 9, 4, (....) and
 15. 16, 18, 21
The pattern followed in  is – 2, – 3, – 5, .......
 missing number = 4 – 7 = – 3
1, 1, 4, 8, 9, ? ,16, 64
(A) 21
(B) 27
(C) 25
(D) 28
(B) (i) 1, 4, 9, 16 [12, 13, 22, 23, 32, 33.............]
(ii) 1, 8, __, 64
mixed combination
3, 6, 24, 30, 63, 72, ?, 132
(A) 58
(B) 42
(C) 90
(D) 120
(D) The difference between the terms is given
below as :
3
6
24 30
Difference 3
18
6
Difference
3
15
63
33
3
72
9
15
?
48
132
?
?
Therefore, alternate difference between the
difference is 3 and 15 respectively.
Hence, the next term would be 72 + 48 = 120.
PAGE # 66
Directions : (11 to 12) Find the wrong term(s) —
Ex 11.
Sol.
Ex 13.
5, 8, 10, 12, 15, 18, 20, 23
(A) 8
(B) 12
(C) 15
(D) 18
(B)
Sol.
12 28 64
140
37 (P) (Q)
(R)
Which number will come
(A) 1412
(C) 696
(A)
Therefore, number 12 is wrong and should be
Similarly
replaced by 13.
Ex 12.
Sol.
37
1, 3, 8, 19, 42, 88, 184
(A) 3
(B) 8
(C) 19
(D) 88
(D)
3
1
2
19
8
11
5
3
6
12
(Q)
(R)
(S)
(T)
78
164
340
696
1412
×2+8
×2+12
×2+16
×2+20
Therefore, the number 1412 will come in place of (T).
Ex 14.
184
89
47
24
(P)
×2+4
42
23
(S)
(T)
in place of (T) ?
(B) 164
(D) 78
95
48
Hence, number 88 is wrong and should be
replaced by 89.
or 1 × 2 + 1, 3 × 2 + 2, 8 × 2 +3, 19 × 2 + 4, 42 × 2 + 5, Sol.
89 × 2 + 6
2
9
57
3
(P)
(Q)
Which number will come
(A) 113
(C) 3912
(T)
(A)
Similarly,
(P)
Directions : (13 to 14) In each of the following questions, a
number series is given. After the series, below it in
the next line, a number is given followed by (P), (Q),
(R), (S) and (T). You have to complete the series
starting with the number given following the
sequence of the given series. Then answer the
question given below it.
337
(R)
(S)
in place of (Q) ?
(B) 17
(D) 8065
(Q)
×8–7
×7–6
(S)
673
113
17
3
(R)
3361
×5–4
×6–5
Therefore, the number 113 will come in place of (Q).
ALPHABET SERIES
(a) Pattern of Alphabets Show Variation Based on :
(i) Position of the letters
(ii) Difference between the alphabets
(i) Position of alphabets :
Alphabets in order :
Alphabets in reverse order :
Directions : (15 to 24) Find the missing term(s) :
Ex 15.
Sol.
Ex 16.
Sol.
Ex 17.
B, E, H, ?
(A) K
(B) L
Sol.
(C) J
(D) M
(A) In the given series, every letter is moved three
steps forward to obtain the corresponding letters
of the next term. So, the missing term is K.
Q, N, K, ?, E
(A) H
(B) I
(C) J
(D) G
(A) In the given series, every letter is moved three Ex 18.
steps backward to obtain the corresponding letters
of the next term. So, the missing term is H.
A, Y, D, W, G, U, J, ?
(A) R
(B) T
(C) S
(D) P
(C) The given sequence consists of two series :
. A, D, G, J in which each letter is moved three
steps forward to obtain the next term
. Y, W, U, ? in which each letter is moved two steps
backward to obtain the next term.
So, the missing term would be S.
AG, LR, WC, HN, ?
(A) SX
(C) SY
(B) RY
(D) TX
PAGE # 67
Sol.
(C) The first letter of each group and the second Ex 24.
letter of each group differs by 11 letters between
them.
L
A
W
H
Sol.
12
1
11
23
11
11
R
18
G
7
Similarly,
Alphabetical
positions
8
11
Difference in
Alphabetical
positions
C
N
3
14
11
11
 A  B  C   6  D  E  F   15  G  H  I   24



, 4 5 6
,


1 2 3
 7 8 9
Alphabetical
positions
Difference in
Alphabetical
positions
Therefore, the next group of letter would be SY.
H
N
S


 10 11 12 
So, the missing term would be  J  K  L   33
Directions : (25 to 27) Find the wrong term (s) :
Ex 25.
Y
And
11
Ex 19.
Sol.
Ex 20.
Sol.
Ex 21.
Sol.
Ex 22.
Sol.
Ex 23.
Sol.
(ABC) – 6, (DEF) – 15, (GHI) – 24, ?
(A) (IJK) – 33
(B) (JKM) – 33
(C) (IJK) – 32
(D) (JKL) – 33
(D) In a given series
Let A = 1, B = 2, C = 3, D = 4, E = 5, F = 6, and so on
11
Sol.
AD, EI, JO, PV, ?
(A) VD
(B) WC
(C) WD
(D) VE
(C) The first letter of subsequent groups have a
difference of 4, 5 and 6 places respectively, whereas
the second letter of the subsequent groups has a
difference of 5, 6, and 7 places respectively,
Therefore, on following the same pattern, we get Ex 26.
‘WD’ as the next term which would replace the
question mark.
Sol.
AB, BA, ABD, DBA, PQRS, ?
(A) SRQP
(B) SRPQ
(C) SQRP
(D) RSQP
(A) The first term is reversed to get second term.
The third term is reversed to get the fourth term.
Similarly, to get the sixth term, we reverse the fifth
Ex 27.
term. So, the missing term would be SRQP.
HEJ, JGL, LIN, NKP, ?
(A) MOR
(B) PNS
Sol.
(C) PMR
(D) NPT
(C) First letter of each group differs by 2 letters.
Second letter of each group differs by 2 letters.
Third letter of each group differs by 2 letters. All the
letters differ in the forward direction. Hence, the
next choice would be PMR.
XYQ, ZAR, BCS, DET, ?
(A) GFU
(B) FUG
(C) FZU
(D) FGU
(D) Here, first two terms of every group of letters
are in continuation, like XY, ZA, BC, DE, and the
third letter of each group is again in forward
continuation,
i.e. Q, R, S, T. Hence, the term replacing the
question mark would be FGU.
ABD, DGK, HMS, NTB, SBL, ZKW
(A) NTB
(B) DGK
(C) SBL
(D) ZKW
(A) First letter of first, second, third,.........terms is
moved three, four, five, ........steps forward
respectively. Similarly, second letter is moved five,
six, seven,......steps forward respectively and third
letter is moved seven, eight, nine,........steps
forward respectively. Hence, NTB is the wrong term
and should be replaced by MTB.
EPV, FQW, GRX, HTY, ITZ
(A) FQW
(B) GRX
(C) HTY
(D) ITZ
(C) In every term, first second and third letter is in
alphabetical order to its next term respectively.
Fourth term is not following the same rule. Hence,
HTY is the wrong term and should be replaced by
HSY.
D4V, G10T, J20R, M43P, P90N
(A) P90N
(B) G10T
(C) J20R
(D) D4V
(B) First letter of every term is moved three steps
forward in each next term. Second number of every
term of the pattern  × 2 + 1, × 2 + 2, × 2 + 3,............and
third letter of every term is moved two steps
backward. Hence, G10T is the wrong term and
should be replaced by G9T.
LETTER REPEATING SERIES
Pattern of such questions is that some letters in
sequence are missing.
(i) The letters may be in cyclic order (clockwise or
anti-clockwise).
(ii) To solve a problem, we have to select one of the
alternative from the given alternatives. The
alternative which gives a sequence form of letters
is the choice.
17Z5, 15X4, 13V3, ?, 9R1
(A) 11S2
(B) 11T2
(C) 11U2
(D) 11T3
Directions : (28 to 32) Find the missing term(s) :
(B) The first number & second letter of every term
is moved two steps backward & the third number Ex 28. a a _ b a a _ b b b _ a
of every term is moved one step backward. So, the
(A) baa
(B) abb
missing term would be 11T2.
(C) bab
(D) aab
PAGE # 68
Sol.
(A) we proceed step by step to solve the above Ex 35.
series:
Steps :
1.
The first blank space should be filled in by 'b' so
that we have two a's followed by two b's.
2.
Second blank space should be filled in by 'a' so
that we have three a's followed by three b' s.
3.
Ex 29.
The last blank space must be filled in by 'a' to keep
the series in sequence.
_ bca _ ca _ c _ b _
Ex 36.
(A) aabbc
(B) abbbc
(C) aabcc
(D) abbac
Sol.
(D)
Sol.
Sol.
Series is abc/ abc/ abc/ abc. So, pattern abc is
repeated.
Ex. 30
a_h_ _c_ ne_ h_ eac_ _ _ _ _
21_4 3_5 __2 54 ____ _ _ _ _
The last five terms in the series are
(A) 32524
(B) 43215
(C) 25314
(D) 32541
(B) By taking a = 2, c = 1, n = 4, h = 5 and e = 3, the
numbers series runs as 21543 15432 54321
43215. If first digit of a group of five digits is placed
as the last digit, we obtain the second group of five
digits and so on.
_ m y e _ _ y l x _ y l m _ _ l _ _ _ _
4 6 _ 5 8 6 _ __ 5 7_ 6 5 8 __ _ _ _
The last five terms of the number series are
(A) 46758
(B) 74658
(C) 76485
(D) 46785
(D) By taking e = 5, l = 4, m = 6, y = 7 and x = 8 the
number series runs as 46758 67485 74658 46785.
By taking the digits in the groups of five, we find
that first digit of the first group (i.e. 4) is the third
digit of the second group and the last two digits
have interchanged their positions. The same rule
applies in others groups also.
a _ abb _ aa _ ba _ a _ b
(A) ababa
(B) aabba
(C) aabab
(D) aaabb
Sol(C) Series is aaabb/ aaabb/ aaabb. So, pattern Direction : (37) In the following question, three sequences
aaabb is repeated.
of letter/numbers are given which correspond to
each other in some way. In the given question, you
Ex 31. a _ c _ abb _ ca _ a
have to find out the letter/numerals that come in
(A) baca
(B) bbca
the vacant places marked by (?). These are given
(C) bacc
(D) bacb
as one of the four alternatives under the question.
Sol(A) Series is abc/ aabbcc/ aaa
Mark your answer as instructed.
Ex 32. bc _ b _ c _ b _ ccb
Ex 37. C B _ _ D _ B A B C C B
(A) cbcb
(B) bbcb
(C) cbbc
(D) bcbc
_ _ 2 3 5 4 _ _ ? ? ? ?
Sol(A) Series is bccb / bccb / bccb. So, pattern bccb is
p _ p q _ r _ q _ _ _ _
repeated
(A) 4 5 5 4
(B) 4 3 3 4
(C)
4
2
2
4
(D) 2 5 5 2
Directions : (33 to 34) The question given below is based
(C) Comparing the positions of the capital letters,
on the letter series, In series, some letters are Sol.
numbers and small letters, we find p corresponds
missing. Select the correct alternative. If more than
to C and 2 corresponds to p. So, p and 2 correspond
five letters are missing, select the last five letters
of the series.
to C. q corresponds to A and 3 corresponds to q.
So, q and 3 corresponds to A. Also, 5 corresponds
Ex 33. xyzu _ yz _ v _ _ uv _ _ _ _ _ _ _
to D. So, the remaining number i.e., 4 corresponds
(A) uvxyz
(B) vuzyx
to B. So, BCCB corresponds to 4, 2, 2, 4.
(C) uvzyx
(D) vuxyz
Sol.
(A) The series is x y z u v / y z u v x/ z u v x y/u v x y z
MISSING TERMS IN FIGURES
Thus the letters are written in a cyclic order.
EX 34.
Sol.
abcd _ bc _ e _ _ de _ _ _ _ _ _ _
(A) deabc
(B) edcba
(C) decba
(D) edabc
(A) The series is a b c d e / b c d e a / c d e a b / b e a b c
Thus the letters are written in a cyclic order.
Directions : (38 to 47) Find the missing number(s) :
Ex 38.
Direction : (35 to 36) There is a letter series in the first row
and a number series in the second row. Each
number in the number series stands for a letter in
the letter series. Since in each of that series some
term are missing you have to find out as to what Sol.
those terms are, and answer the questions based
on these as given below in the series.
6
9
15
8
12
20
4
6
?
(A) 5
(B) 10
(C) 15
(D) 21
(B) In the first row, 6 + 9 = 15
In the second row, 8 + 12 = 20
 In the third row, missing number = 4 + 6 = 10.
PAGE # 69
3
Ex 39.
10 2
6
Ex 43.
9
4
(A) 11
(C) 3
Sol.
(B) 6
(D) 2
(C) Clearly, in the  column,
6 4
8
3
18  3
 27 We take x in place of ?
2
95
15  x
Similarly in the  column,
3
 9 ,x 
15
5
Sol.
In the  column,
Ex 40.
Sol.
3C
27D
9E
7I
21K
3M
4D
?
7J
Ex 41.
Sol.
Ex 42.
2
Sol.
(A) 32
(B) 22
(C) 18
(D) 27
(B) In first figure] 5 × 4 + 6 = 26
In second figure] 8 × 3 + 5 = 29
missing number in third figure] 6 × 3 + 4 = 22
81
18
Sol.
88
? 11
9
(B) 21
(D) 81
In second figure, 9 
6 ? 3
4
3
8 29 3
5
7
6
174
5
3
2 5
?
336
3 2 9
2 7 9
6 4 5
(A) 140
(B) 150
(C) 200
(D) 180
(B) In first figure]8 × 5 × 3 + 3 × 2 × 9 = 120 + 54 =
174
In second figure]6 × 7 × 5 + 2 × 7 × 9= 210 + 126
= 336
 missing number in third figure]
3 × 2 × 5 + 6 × 4 × 5 = 30 + 120 = 150
9
15
7
21
25
14
2
18
2
Ex 47.
= 84.
= 81.
= 88 or x =
88  2
11
= 16.
Sol.
176
?
(A) 184
(B) 210
(C) 241
(D) 425
(A) The number at the bottom is the difference of
squares of two numbers given at top
In first figure] 112 – 92 = 121 – 81 = 40
In second figure] 152 – 72 = 225 – 49 = 176
 In third figure] 252 – 212 = 625 – 441 = 184
3
Let the missing number In third figure be x.
x
2
5
8
40
(A) In first figure, 12
Then, 11
2
5
Ex 46.
(A) 16
(C) 61
Sol.
5
(A) 15
(B) 20
(C) 25
(D) 40
(B) Clearly
In first figure] 6 × 3 – 4 × 2 = 18 – 8 = 10
In second figure] 9 × 5 – 5 × 3 = 45 – 15 = 30
 In third figure] 6 × 5 – 2 × 5 = 30 – 10 = 20
11
(A) 16
(B) 9
(C) 85
(D) 112
(C) Hint ; 42 + 52 = 16 + 25 = 41
12 + 22 = 1 + 4 = 5
62 + 72 = 36 + 49 = 85
84
14 12
?
5 26 4
6
1
7
6
5
41
? 5
6
30 3
Ex 44.
(A) 11E
(B) 28G

(C) 35I
(D) 48F
(B) The letters in the first row form a series C, D, E
(a series consecutive letters). The letters in the
second row form a series I, K, M (a series of
Ex 45.
alternate letters). Similarly, the letters in the third
row will form the series D, G, J (a series in which
each letter is three steps ahead of the previous
one). So, the missing letter is G. Also, the
number in the second column is equal to the
product of the numbers in the first and third
Sol.
columns. So, missing number is (4 × 7) i.e. 28.
Thus, the answer is 28G.
4
5
5
5
33
6 3
4
7
48
5
5
3
?
4
5
(A) 47
(B) 45
(C) 37
(D) 35
(D) In first figure, 6 × 3 + 3 × 5 = 33
In second figure, 5 × 4 + 4 × 7 = 48
 In third figure, 5 × 4 + 3 × 5 = 35
PAGE # 70
4
17.
EXERCISE-1
Directions : (1 to 25) Find the missing numbers :
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
2, 8, 18, 32, ?
(A) 62
(C) 50
16, 54, 195, ?
(A) 780
(C) 816
14, 316, 536, 764, ?
(A) 981
(C) 8110
(B) 60
(D) 46
3, 15, 35, ?, 99, 143
(A) 63
(C) 69
(B) 77
(D) 81
9, 16, 30, 58, ?
(A) 104
(C) 116
(B) 114
(D) 118
3, 12, 27, 48, 75, 108, ?
(A) 192
(C) 162
(B) 183
(D) 147
1, 4, 12, 30, ?
(A) 60
(C) 64
(B) 62
(D) 68
94, 166, 258, ?, 4912
(A) 3610
(C) 1026
(B) 1644
(D) 516
22.
(B) 180
(D) 452
7,19, 55, 163, ?
(A) 387
(C) 527
(B) 329
(D) 487
1, 2, 9, 4, 25, 6, ?
(A) 51
(C) 50
(B) 49
(D) 47
24.
16, 33, 67, 135, ?
(A) 371
(C) 271
(B) 175
(D) 287
25.
101, 100, ?, 87, 71, 46
(A) 92
(C) 89
(B) 88
(D) 96
13.
6, 24, 60, 120, 210, 336, ?, 720
(A) 496
(B) 502
(C) 504
(D) 498
16.
(B) 581
(D) 481
2, 3, 6, 18, ?, 1944
(A) 154
(C) 108
100, 50, 52, 26, 28, ? 16, 8
(A) 30
(B) 36
(C) 14
(D) 32
15.
137, 248, 359, 470, ?
(A) 582
(C) 571
21.
(B) 210
(D) 258
3, 1, 4, 5, 9, 14, 23, ?
(A) 32
(C) 41
(B) 37
(D) 28
3, 6, 18, 72, 360, ?
(A) 720
(C) 1600
(B) 1080
(D) 2160
78, 79, 81, ?, 92, 103, 119
(A) 88
(B) 85
(C) 84
(D) 83
(B) 195
(D) 103
20.
8, 11, 15, 22, 33, 51, ?, 127, 203
(A) 80
(B) 53
(C) 58
(D) 69
2, 12, 36, 80, 150, ?
(A) 194
(C) 252
2, 9, 28, 65, ?
(A) 121
(C) 126
1, 11, ?, 11, 11, 11, 16, 11
(A) 1
(B) 11
(C) 6
(D) 192
(B) 1048
(D) 9100
8, 24, 16, ?, 7, 14, 6, 18, 12, 5, 5, 10
(A) 14
(B) 10
(C) 7
(D) 5
(B) 112
(D) 108
19.
(B) 802
(D) 824
12.
14.
18.
0, 6, 20, 42, 72, ?
(A) 106
(C) 110
23.
Directions : (26 to 28) In each of the following questions, a
number series is given. After the series, below it in
the next line, a number is given followed by (P), (Q),
(R), (S) and (T). You have to complete the series
starting with the number given following the
sequence of the given series. Then answer the
question given below it.
26.
2
3
8
27
5
(P)
(Q)
(R)
(S)
(T)
Which of the following numbers will come in place
of (T) ?
(A) 184
(B) 6
(C) 925
(D) 45
27.
5
18
48
7
(P)
(Q)
Which number will come
(A) 172
(C) 270
28.
112
(R)
(S)
in place of (S) ?
(B) 276
(D) 376
(T)
15
159
259
323
7
(P)
(Q)
(R)
(S)
(T)
Which of the following numbers will come in place
of (R) ?
(A) 251
(B) 315
(C) 176
(D) 151
PAGE # 71
Directions : (29 to 35) Find the wrong term(s) —
29.
30.
31.
32.
33.
9, 11, 15, 23, 39, 70, 135
(A) 23
(B) 39
(C) 70
(D) 135
3, 9, 36, 72, 216, 864, 1728, 3468
(A) 3468
(B) 1728
(C) 864
(D) 216
2, 5, 11, 20, 30, 47, 65
(A) 5
(C) 30
121, 143, 165, 186, 209
(A) 143
(C) 186
(B) 20
(D) 47
(B) 165
(D) 209
9, 15, 24, 34, 51, 69, 90
(A) 15
(B) 24
(C*) 34
(A) 15
(C) 34
8.
9.
10.
11.
35.
(C) 69
(D) 132
105, 85, 60, 30, 0, – 45, – 90
(A) 85
(B) – 45
(C) 105
(D) 0
4.
5.
6.
7.
ZGL, XHN, VIQ, TJU, ?
(A) RKX
(C) RLZ
(B) RKY
(D) RKZ
RML, VIJ, ZFH, DDF, ?
(A) HDC
(C) HCD
(B) CHI
(D) DIC
LRX, DJP, VBH, NTZ, ?
(A) ELS
(C) GKS
(B) FMR
(D) FLR
MAD, OBE, SCH, YDM, ?
(A) HET
(C) GET
(B) HES
(D) UAE
2B, 4C, 8E, 14H, ?
(A) 22L
(C) 22K
(B) 24L
(D) 2M
15.
16.
Directions : (1 to 24) Find the missing term(s) :
3.
(B) OET
(D) PEV
(B) ZVX
(D) VZX
17.
2.
CYD, FTH, IOL, LJP, ?
(A) PET
(C) OEY
ZSD, YTC, XUB, WVA, ?
(A) VZZ
(C) VWZ
EXERCISE-2
1.
(B) ELS
(D) DLS
13.
9, 13, 21, 37, 69, 132, 261
(B) 37
JXG, HTJ, FPN, ?, BHY
(A) EKS
(C) DLR
MTH, QRK, UPN, YNQ, ?
(A) CKT
(B) ELT
(C) CLT
(D) EKT
(D) 51
(B) 24
(D) 51
(A) 21
(B) KBO
(D) LBN
12.
14.
34.
DFK, FEL, HDM, JCN, ?
(A) KBN
(C) LBO
X, U, S, P, N, K, I, ?
(A) J
(C) M
(B) K
(D) F
18.
1 BR, 2 EO, 6 HL, 15 KI, ?
(A) 22 NF
(B) 31 NF
(C) 31 NE
(D) 28 NF
Z, X, U, Q, L, ?
(A) F
(C) G
(B) K
(D) E
19.
P3C, R5F, T8I, V12L, ?
(A) Y17O
(C) X17O
A, H, N, S, W, ?
(A) A
(C) B
(B) Y
(D) Z
20.
Z 15 A, W 13 C, ?, Q 9 G, N 7 I
(A) T 12 E
(B) R 11F
(C) T 11E
(D) R 13 D
Q, T, V, Y, A, ?
(A) B
(C) D
(B) C
(D) F
21.
B3M, E7J, H15G, K31D, ?
(A) N65A
(B) O63A
(C) N63A
(D) N63Z
X, A, D, G, J, ?
(A) N
(C) M
(B) O
(D) P
22.
5X9, 8U12, 11R15, 14O18, ?
(A) 17L21
(B) 17K21
(C) 17M21
(D) 17L23
Z, L, X, J, V, H, T, F, ?, ?
(A) R, D
(C) S, E
(B) R, E
(D) Q, D
23.
6C7, 8F10, 11J14, 15O19, ?
(A) 19U24
(B) 20U25
(C) 19U25
(D) 20U24
AZ, YB, CX, WD, ?
(A) VE
(C) EU
(B) UE
(D) EV
24.
B2E, D5H, F12K, H27N, ?
(A) J58Q
(B) J56Q
(C) J57Q
(D) J56P
(B) X17M
(D) X16O
PAGE # 72
Directions : (25 to 30) Find the wrong term(s) :
25.
26.
27.
28.
29.
30.
ECA, JHF, OMK, TQP, YWU
(A) ECA
(B) JHF
(C) TQP
(D) YWU
9.
10.
a _ cab _ a _ c _ b c
(A) bbac
(C) abba
(B) abab
(D) bcba
ba _ cb _ b _ bab _
(A) acbb
(C) cabb
(B) bcaa
(D) bacc
DKY, FJW, HIT, JHS, LGQ
(A) FJW
(B) LGQ
(C) JHJ
(D) HIT
11.
DVG, FSI, HPK, JNM, LJO
(A) DVG
(B) JNM
(C) HPK
(D) LJO
a _ bc _ a _ bcda _ ccd _ bcd _
(A) abddbd
(B) acbdbb
(C) adbbad
(D) bbbddd
12.
CDF, DEG, EFH, FHI
(A) CDF
(C) FHI
cc _ ccdd _ d _ cc _ ccdd _ dd
(A) dcdcc
(B) dcddc
(C) dccdd
(D) None of these
13.
a_baa_baa _ba
(A) a a b
(C) b b a
(B) b a b
(D) b b b
babbb_b_b_bb
(A) b b a
(C) a b a
(B) b a a
(D) a a a
m _ l _ ml _ m _ llm
(A) lmmm
(C) lmml
(B) lmlm
(D) mllm
(B) DEG
(D) EFH
ZLA, BMY, CNW, FOU, HPS
(A) ZLA
(B) BMY
(C) FOU
(D) CNW
14.
G4T, J10R, M20P, P43N, S90L
(A) G4T
(B) J10R
(C) M20P
(D) P43N
15.
EXERCISE-3
Directions : (16 to 19) The questions given below are based
on the letter series, In each of these series, some
Directions : (1 to 15) Which sequence of letters when placed
letters are missing. Select the correct alternative. If
at the blanks one after the other will complete the
more than five letters are missing, select the last
given letter series ?
five letters of the series.
1.
2.
3.
4.
a_baa_aa__ab
(A) a a a a
(C) b b a a
(B) b a a a
(D) a b b a
_aabb_a_ab_b
(A) b b a a
(C) b a a b
(B) b a b a
(D) a b a b
aab_aaa_bba_
(A) b a a
(C) b a b
(B) a b b
(D) a a b
a__b_a_ab_aa
(A) a b a a b
(C) b b a b b
(B) b b a b a
(D) b a a b a
5.
abc _ d _ bc _ d _ b _ cda
(A) bacdc
(B) cdabc
(C) dacab
(D) dccbd
6.
a _ bbc _ aab _ cca _ bbcc
(A) bacb
(B) acba
(C) abba
(D) caba
7.
_bc__bb_aabc
(A) acac
(B) babc
(C) abab
(D) aacc
_ b c c _ ac _ a a b b _ a b _ c c
(A) aabca
(B) abaca
(C) bacab
(D) bcaca
8.
16.
_ _ r _ ttp _ _ s _ tp _ _ _ s _ _ _
(A) rstqp
(B) tsrqp
(C) rstpq
(D) None
17.
_ _ x _ zbxazyxabyz _ _ _ _ _
(A) abxzy
(B) abzxy
(C) abxyz
(D) bxayz
18.
x _ xxy _ x _ xy _ yxx _ _ yy _ y
(A) xyyyy
(B) xxyyx
(C) yxxyx
(D) xyxyx
19.
_ _ r _ tqrptsrpqst _ _ _ _ _
(A) pqrts
(B) pqtrs
(C) pqrst
(D) qrpst
Directions : (20 to 23) There is a letter series in the first row
and a number series in the second row. Each
number in the number series stands for a letter in
the letter series. Since in each of that series some
term are missing you have to find out as to what
those terms are, and answer the questions based
on these as given below in the series.
20.
ab_cd_a_abd_dba_
1_3_32_1___4____
The last four terms in the series are
(A) 1234
(B) 3112
(C) 3211
(D) 4312
PAGE # 73
21.
22.
23.
_bnt _ _nam _nab_ _a__ _ _
13_25 3__ 5 24_32 5__ _ __
The last five terms in the series are
(A) 13425
(B) 41325
(C) 34125
(D) 13452
n _ g f _ t _ f h t n _ _ t _ b _ f
1 3_ 2 4 5 0 _ 4 _ _ 3 _ _ _ _ _ _
The last five terms of the number series are
(A) 50123
(B) 40321
(C) 40231
(D) 51302
3.
_ m i a x _ i r x a _ _ma _ _ _ _ _ _
4 _ 5 _7 3 _ _ _6 _ _ _ _ _ _ _ _ _ _
The last five term of the letter series are
(A) r m x i a
(B) x m r a i
(C) x r m a i
(D) r m i x a
25.
_ A C _ B D _ CD C D
2 _4 1 _ 1 4 _ _ _ _
r s _ q r _ p ? ? ? ?
(A) p q p q
(B) p r p r
(C) r q r q
(D) r s r s
6.
(A) 235
(C) 144
(B) 141
(D) 188
12
18
30
16
6
(A) 18
(C) 9
5
12
_ A D A C B _ _ B D C C
2 4 _ _2 3 5 3 _ _ _ _
p _ _ q _ _ r s ? ? ? ?
(A) p r s s
(B) p s r r
(C) r p s s
(D) s r p p
36
18 27
?
(B) 12
(D) 6
6
6
21
7
4
5
(A) 14
(C) 32
A _ B A C _ D _ B C D C
_ 4 _ 3 _ 2 _ 5 ? ? ? ?
d c _ _ b a c b _ _ _ _
(A) 2 4 5 4
(B) 2 5 4 5
(C) 3 4 5 4
(D) 4 5 2 5
32 40
8
4
7.
26.
(B) 92
(D) 102
4.
Directions : (24 to 26) In each of the following questions,
three sequences of letter/numbers are given which
5.
correspond to each other in some way. In each
question, you have to find out the letter/numerals
that come in the vacant places marked by (?). These
are given as one of the four alternatives under the
question. Mark your answer as instructed.
24.
(A) 112
(C) 82
?
8
10
(B) 22
(D) 320
5
9
8
5
15
?
3
5
6
(A) 12
(C) 16
(B) 11
(D) 26
(A) 72
(C) 9
(B) 18
(D) 19
(A) 1
(C) 90
(B) 18
(D) 225
(A) 20
(C) 24
(B) 22
(D) 12
EXERCISE-4
8.
Directions : (1 to 39) Find the missing term in the given
figures
1.
9.
(A) 36
(C) 25
(B) 9
(D) 64
2.
10.
(A) 14
(C) 11
(B) 18
(D) 13
PAGE # 74
11.
7
12
15
11 49
8 54
4
?
(A) 36
(C) 25
12.
18 24 32
12 14 16
3
?
4
72 112 128
(A) 28
(C) 81
(B) 3
(D) 5
2
4
10
3
5
27
80
39
45
29
33
42
20.
43
43
(A) 69
(C) 50
30
70
59
31
44
35
43
48
15
56
(A) 1
(C) 3
(B) 142
(D) 198
7
3
4
74
12
3
6
4
1
140
4 10 7
3
(B) 14
(D) 22
4
9
3
24
9
?
(A) 117
(C) 32
(B) 36
(D) 26
(A) 26
(C) 27
(B) 25
(D) 30
4
13
21.
34
184
12 18 30
16 32 40
36 18 34
30
44
?
22.
(A) 48
(C) 44
23.
17.
(B) 145
(D) 18
8
11
20
6
?
8
104
(B) 2
(D) 4
5
20
(B) 9
(D) 64
5
21
16 109 2
22 53 19
6
15
(A) 25
(C) 7
(A) 33
(C) 135
8
?
2
6
10
?
(A) 127
(C) 158
1
3
5
35
80
?
39
38
15.
40
(B) 49
(D) 60
101
8
6
(B) 3
(D) 5
29
72
5
2
(A) 16
(C) 20
?
J90
4
(A) 1
(C) 4
16.
6
6
4
8
H70
5
14.
19.
4
C26
(B) 36
(D) 49
6
(A) 2
(C) 4
13.
18.
(B) 7
(D) 0
24.
17
?
13
(B) 129
(D) 49
2
3
3 33 2
4 54 2
4
5
(A) 78
(C) 94
51
6
3
?
5
4
(B) 82
(D) 86
PAGE # 75
48
2
25.
4
5
28 5
1
7 38 4
3
2
3
?
34.
3
7
(A) 14
(C) 11
3
8
10
2
?
1
6
56
90
2
20
0
(A) 0
(C) 5
(B) 18
(D) 26
(B) 3
(D) 7
2
15
35.
(A) 9
(C) 10
27.
6
HCA - 138 FED - 456 E?H - 87?
BIG - 792
36
49 26 64
25
9
81 21
16
2
25
64
25
?
144
8
11
(A) 48
(C) 35
(B) 72
(D) 120
(A) 38
(C) 4
(B) 64
(D) 16
3
8
4
6
101
36
6
12 ?
8
15
56
(A) 127
(C) 158
184
(B) 142
(D) 198
8
4
(A) 3
(C) 5
34
?
38
35
6
1
48
43
37.
(B) 23
(D) 31
2
3
6
4
36.
(B) I, 9
(D) I, 5
(A) 19
(C) 25
29.
?
(B) 11
(D) 12
(A) G, 6
(C) G, 5
28.
16
13
65
5
26.
7
9
80
(B) 4
(D) 6
12 18 30
16 32 40
6
8
36 18 27
38.
29
30.
39
27
80
45
(A) 69
(C) 50
33
43
29
42
43
30
70
31
44
59
40
80
?
39
(B) 49
(D) 60
?
(A) 18
(C) 9
10
20
39.
(B) 12
(D) 6
4
9
9
6
16
16
12
?
20
31.
(A) 0
(C) 3
(B) 2
(D) 1
(A) 60
(C) 21
40.
(B) 50
(D) 25
Find the value of X in the following figure :
15
32.
4
33
(A) 12
(C) 14
33
27
(B) 9
(D) 10
(B) JNS
(D) KRS
2
36
8
32
Find the missing letters from left to right.
(A) JSN
(C) JRS
2
(A) 3
(C) 8
X
18
9
22
11
12
3
(B) 4
(D) 12
PAGE # 76
PUZZLE TEST
Directions : (1 to 5) Read the following information carefully
and answer the questions given below it.
(i). Five professors (Dr. Joshi, Dr. Davar, Dr.
Natrajan, Dr. Choudhary and Dr. Zia) teach five
different subjects (zoology, physics, botany, geology
and history) in four universities ( Delhi, Gujarat,
Mumbai, and Osmania). Do not assume any
specific order.
(ii). Dr. Choudhary teaches zoology in Mumbai
University .
(iii). Dr. Natrajan is neither in Osmania University
nor in Delhi University and he teaches neither
geology nor history.
(iv). Dr. Zia teaches physics but neither in Mumbai
University nor in Osmania University.
(v). Dr. Joshi teaches history in Delhi University.
(vi). Two professors are from Gujarat University.
(vii). One professor teaches only one subject and
in one University only.
Ex 1.
Ex 2.
Ex 3.
Ex 4.
Ex 5.
Sol. :
Who teaches geology ?
(A) Dr Natrajan
(C) Dr. Davar
(B) Dr. Zia
(D) Dr. Joshi
Which university is Dr. Zia from ?
(A) Gujarat
(B) Mumbai
(C) Delhi
(D) Osmania
Who teaches botany ?
(A) Dr. Zia
(C) Dr. Joshi
(B) Dr. Davar
(D) Dr. Natrajan
Names
Dr. Joshi
Dr. Davar
Dr. Natrajan
Dr. Choudhary
Dr. Zia
University
Delhi
Osmania
Gujarat
Mumbai
Subject
History
Geology
Botany
Zoology
Gujarat
Physics
On the basis of the above table, rest of the questions
can be solved very easily.
1.
(C) Dr. Davar teaches geology.
2.
(A) Dr. Zia is from Gujarat university.
3.
(D) Dr. Natrajan teaches botany.
4.
(B) Dr. Davar is from Osmania University.
5.
(D) Dr. Natranjan - Gujarat University is the correct
combination.
Ex 6.
Ramesh is taller than Vinay who is not as tall as
Karan. Sanjay is taller than Anupam but shorter
than Vinay. Who among them is the tallest ?
(A) Ramesh
(B) Karan
(C) Vinay
(D) Cannot be determined
(D) In this question ranking of Karan is not defined.
Consequently, either Ram or Karan occupies the
top position
with regard to height. Hence,
option (d) is the correct choice.
Sol.
Directions : (7 to 11) Read the following information carefully
and answer the questions given below it :
There are five men A, B, C, D and E and six women
P, Q, R, S, T and U. A, B and R are advocates; C, D,
Which of the following combinations is correct ?
P, Q and S are doctors and the rest are teachers.
(A) Delhi University - Dr. Zia
Some teams are to be selected from amongst
(B) Dr. Choudhary - geology
these eleven persons subject to the following
(C) Dr. Davar - Mumbai University
conditions :
(D) Dr. Natranjan - Gujarat University
A, P and U have to be together.
(1 to 5)
B cannot go with D or R.
From the given information in the question :
E and Q have to be together.
From II, we get Dr. Choudhary teaches zoology in
C and T have to be together.
Mumbai University.
D and P cannot go together.
From III, We get Dr. Natrajan is neither in Osmania
C cannot go with Q.
nor in Delhi University. Therefore, he will be either
If the team is to consist of two male advocates, two
at Mumbai or Gujarat University. Similarly, as he Ex 7.
lady doctors and one teacher, the members of the
teaches neither geology nor history, therefore, he
team are
must be teaching physics or botany. ..........(1)
(A) A B P Q U
(B) A B P U S
From IV, Dr. Zia  Physics but as he is not teaching
(C) A P R S U
(D) B E Q R S
in either Mumbai or Osmania University, he must
(B) The male advocates are A and B, lady doctors
be teaching either in Delhi or Gujarat University...(2) Sol.
are P, Q and S ; teachers are E, T and U.
Form V, we get Dr Joshi teaches history in Delhi
Now, A and B will be selected.
University Form (1) and (2), we conclude that Dr
Natarajan teaches botany. And from (1), (2) and VI,
A, P and U have to be together. Now, we have to
we get both Natarajan and Zia teach in Gujarat
select one lady doctor more. It can be Q or S. But Q
University. Finally, On summarisation we can
and E have to be together. Since E is not selected,
prepare the following table.
so S will be selected. Thus, the team is A B P U S.
Who is from Osmania University ?
(A) Dr. Natrajan
(B) Dr. Davar
(C) Dr. Joshi
(D) Dr. Zia
PAGE # 77
Ex 8.
Sol.
Ex 9.
If the team is to consist of one advocate, two Directions : (12 to 15) Read the following paragraph
carefully :
doctors, three teachers and C may not go with T,
Four women A, B, C and D and three men E, F and
the members of the team are :
G play bridge, a game for four players.
(A) A E P Q S U
(B) A E P Q T U
(i) The group consists of three married couples
and a widow.
(C) B E Q S T U
(D) E Q R S T U
(ii) Spouses are never partners in a game.
(B) The advocates are A, B and R ; doctors are
(iii) No more than one married couple ever plays in
C, D, P, Q, S ; teachers are E, T and U. The team
the same game.
(iv) One day they played four games as follows.
consists of 3 teachers i.e. E, T, U. Now, A, P and U
A and E versus B and F.
have to be together. E and Q have to be together.
A and G versus D and F.
Thus, the team is A E P Q T U.
B and C versus F and G.
C and E versus D and G.
If the team is to consist of one male advocate, one
male doctor, one lady doctor and two teachers, the Ex 12.
members of the team are :
Sol.
(A) A C P T U
(B) A D E P T
(C) A D E P U
(D) B C E Q U
Ex 13.
(A) The male advocates are A and B ; male doctors
are C and D ; lady doctors are P, Q and S ; teachers
are E, T and U. If A is selected, P and U will be Ex 14.
selected. D and P cannot go together. So, a male
doctor C will be selected. C and T have to be
together. Thus, the team is A C P T U. If B is Ex 15.
selected, D will not be selected. So, male doctor C
Whom is E married to ?
(A) A
(C) C
(B) B
(D) D
Whom is F married to ?
(A) A
(C) C
(B) B
(D) D
Whom is G married to ?
(A) A
(C) C
(B) B
(D) D
Which of the following is a widow ?
(A) A
(B) B
(C) C
(D) D
will be chosen. C and T have to be together. Now,
the second teacher to be selected is E or U. But, U
cannot go without A. So, E will be selected. E and Q
have to be together. Thus, the team can also be
B C E Q T.
Ex 10.
If the team is to consist of one advocate, three
doctors and one male teacher, the members of
Sol. : (12 to 15)
From (iv), is married either to A or to C. If F is married
to A, then G is married to B or to C. If G is married to
B, then E is married to D ; if G is married to C, then
E is married to B or to D. If F is married to C, then G
is married to B ; then E is married to D. Hence, the
married couples are : FA, GB, ED or FA, GC, EB or
FA, GC, ED or FC, GB, ED. Of these, only FA, GB,
ED does not contradict any of the statements.
the team are:
Sol.
(A) A D P S U
(B) C D R S T
(C) D E Q R S
(D) D E Q R T
12.
(D) E is married to D.
13.
(A) F is married to A.
(C) The advocates are A, B and R ; the doctors are 14.
C, D, P, Q and S ; male teacher is E. Clearly, E will
15.
be selected. E and Q have to be together. C and Q
cannot be together. So, C will not be selected. P Ex 16.
also cannot be selected because U is not selected.
So, two other doctors D and S will be selected. P is
not selected, so A will not be selected. D is
selected, so B cannot be selected. Thus, the team
is D E Q R S.
Ex 11.
If the team is to consist of two advocates, two Sol.
doctors, two teachers and not more than three
ladies, the members of the team are :
Sol.
(A) A B C P T U
(B) A C P R T U
(C) A E P Q R T
(D) B C E Q R T
(B) G is married to B.
(C) C is a widow.
A vagabond runs out of cigarettes. He searches for
the stubs, having learnt that 7 stubs can make a
new cigarette, good enough to be smoked, he
gathers 49 stubs, If he smokes 1 cigarette every
three - quarters of an hour, how long will his supply
last ?
(A) 5.25 hr
(B) 6 hr
(C) 4.5 hr
(D) 3 hr
(B) He has got =
49
 7 cigarettes.
7
 The duration of time he will take to smoke these
3
7 cigarettes = 7  hr = 5.25 hr (i.e. 5 hr and 15
4
each of these combinations consists of four ladies.
min). Now note that after he has smoked these 7
cigarettes, he will collect 7 more stubs (one form
each), form which he will be able to make another
B C E Q R T is incorrect because B and R cannot
cigarette. This will take him another
(A) A C P R T U and A E P Q R T are wrong because
go together.
3
hr (45 min)
4
to smoke. Therefore, total time taken = 6hr.
PAGE # 78
Directions : (17 to 18) Read the following information and 1.
answer the questions that follow.
There are 70 clerks working with M/s. Jha Lal
Khanna & Co. chartered accountants, of which 30
are female.
2.
(i) 30 clerks are married.
(ii) 24 clerks are above 25 years of age
(iii) 19 Married clerks are above 25 years of age;
among them 7 are males.
3.
(iv) 12 males are above 25 years of age
(v) 15 males are married.
Ex 17.
Ex 18.
Sol.
17.
18.
How many unmarried girls are there ?
(A) 12
(B) 15
(C) 18
(D) 10
4.
Who stays in locality Q ?
(A) A
(C) C
(B) B
(D) E
What is E’s occupation ?
(A) Business
(B) Engineer
(C) Lawyer
(D) Doctor
Agewise who among the following lies between A
and C ?
(A) Lawyer
(B) Doctor
(C) Cloth merchant
(D) Engineer
What is B’s occupation ?
(A) Business
(B) Engineer
(C) Lawyer
(D) Doctor
How many of these unmarried girls are above 25 ?
(A) 12
(B) 15
(C) 4
(D) 0
5.
What is C’s occupation ?
(17 to 18) : From the given data, we can make the
(A) Doctor
(B) Lawyer
following table with the help of which rest of the
(C) Engineer
(D) Business
questions can be solved very easily.
Directions : (6 to 10) Study the given information carefully
Male (40) Female (30)
and answer the questions that follow.
Above 25
There are four people sitting in a row : one each
from India, Japan, USA and Germany, but not in
Married
7
12
that order,
Unmarried
5
0
. They are wearing caps of different colours - green,
Below 25
yellow, red and white, not necessarily in that order.
married
8
3
II. One is wearing a kurta and one a T-shirt.
III. The Indian is wearing a green cap and a jacket.
unmarried
20
15
IV. The American is not seated at either end.
40
30
Total
V. The persons with kurta and T-shirt are sitting
next to each other.
There are 15 unmarried girls.
VI. The persons with kurta wears a red cap and
sits next to the Japanese.
In these 15 unmarried girls no one is above 25.
VII. The Japanese wears a shirt and is not seated
at either end.
VIII. The man with white cap wears T-shirt and is
EXERCISE
seated at one end.
Directions : (1 to 5) Study the following information carefully
and answer the questions given below it :
There are five friends A, B, C, D and E. Two of them
are businessmen while the other three belong to
different occupations viz. medical, engineer and
legal. One businessman and the lawyer stay in
the same locality S, while the other three stay in
three different localities P, Q and R. Two of these
five persons are Hindus while the remaining three
come from three different communities viz. Muslim,
Christian and Shikh. The lawyer is the oldest in
age while one of the businessmen who runs a
factory is the youngest. The other businessman is
a cloth merchant and agewise lies between the
doctor and the lawyer. D is a cloth merchant and
stays in locality S while E is a Muslim and stays in
locality R. The doctor is a Christian and stays in
locality P, B is a Shikh while A is a Hindu and runs
a factory.
6.
Who wears the T-shirt ?
(A) Indian
(C) American
(B) Japanese
(D) German
7.
Who is wearing a kurta ?
(A) Indian
(B) Japanese
(C) American
(D) German
8.
What is the colour of the cap worn by the Japanese?
(A) Red
(B) Green
(C) Yellow
(D) White
9.
Who precedes the man wearing T-shirt ?
(A) Indian
(B) Japanese
(C) American
(D) German
10.
Who precedes the man wearing jacket ?
(A) Indian
(B) German
(C) Japanese
(D) Cannot say
PAGE # 79
Directions : (11 to 15) Read the following information Directions : (19 to 23) Read the information given below
carefully and answer the questions that follow.
and answer the questions.
I. There are six students ( A, B, C, D, E and F) in a
The age and height of six children in a class are as
group. Each student can opt for only three choices
follows :
out of the six which are music, reading, painting,
(i) A is taller and older than B but shorter and
badminton, cricket and tennis.
younger than C.
II. A, C and F like reading.
(ii) D is taller than E who is not as tall as B.
III. D does not like badminton, but likes music.
(iii) The oldest is the shortest.
IV. Both B and E like painting and music.
(iv) The youngest would be fourth if the children
V. A and D do not like painting, but they like cricket.
stood in a line according to their height and one
VI. All student except one like badminton.
started counting from the tallest.
VII. Two students like tennis.
(v) D is younger than F but older than E who is
VIII. F does not like cricket, music and tennis.
older than C.
11.
12.
Which pair of students has the same combination 19.
of choices ?
(A) A and C
(B) C and D
(C) B and E
(D) D and F
20.
Who among the following students likes both
tennis and cricket ?
(A) A and B
(B) C
21.
(C) B and D
(D) D
13.
How many students like painting and badminton ?
(A) 1
(B) 2
(C) 3
(D) 4
14.
Who among the following do not like music ?
(A) A , C and D
(B) A, B and C
(C) A, C and F
(D) B, D and F
15.
Which of the following is the most popular choice?
(A) Tennis
(B) Badminton
23.
(C) Reading
(D) Painting
22.
16.
Who among them is the tallest ?
(A) B
(B) E
(C) C
(D) Data inadequate
Who is older than B but younger than C ?
(A) F
(B) D
(C) A
(D) Data inadequate
Which of the following statements is definitely true?
(A) D is the most old person
(B) B has the max. height
(C) A is older than D
(D) F is the shortest
Which of the following is the correct order of height
in descending order?
(A) A, C, D, B, E, F
(B) F, D, E, C, A, B
(C) D, C, A, B, E, F
(D) C, D, A, B, E, F
W hose Rank in height cannot be positioned
definitely ?
(A) B
(B) D
(C) C
(D) E
R earns more than H but not as much as T, M
earns more than R. Who earns least among
them?
Directions : (24 to 28) Study the information given below
(A) R
(B) T
and answer the questions that follow.
(C) H
(D) M
(i) Six Plays P, Q, R, S, T and U are to be organised
from Monday to Saturday i.e. 10 to 15 one play each
17.
Harish is taller than Manish but shorter than
day.
Suresh. Manish is shorter than Anil but taller than
(ii) There are two plays between R and S and one
Raghu. Who among them is the shortest having
play between P and R.
regard to height ?
(iii) There is one play between U and T and T is to
(A) Anil
(B) Manish
be organised before U.
(C) Raghu
(D) Cannot be determined
(iv) Q is to be organised before P, not necessarily
Direction : (18) Examine the following statements :
immediately.
I. Either A and B are of the same age or A is older
(v) The organisation does not start with Q.
than B.
The organisation would start from which play ?
II. Either C and D are of the same age or D is older 24.
(A) P
(B) S
than C.
(C)
T
(D) None
III. B is older than C.
18.
Which one of the following conclusions can be 25.
drawn from the above statements ?
(A) A is older than B
(B) B and D are of the same age
26.
(C) D is older than C
(D) A is older than C
On which date is play T to be organised ?
(A) 10th
(B) 11th
th
(C) 12
(D) None
The organisation would end with which play ?
(A) P
(B) Q
(C) S
(D) None
PAGE # 80
27.
Which day is play Q organised ?
(A) Tuesday
(B) Wednesday
(C) Thursday
(D) None
28.
Which of the following is the correct sequence of
organising plays ?
32.
(A) PTRUQS
(B) QSTURP
(C) SUTRQP
(D) None
31.
If it is sure that Henna will go to the fair, then who
among the following will definitely go ?
(A) Rama
(B) Shamma
(C) Reena
(D) Rama and Reena
If Tina does not go to the fair, which of the following
statements must be true ?
(i) Henna cannot go
(ii) Shamma cannot go
(iii) Reena cannot go
(iv) Rama cannot go
(A) (i) and (ii)
(B) (iii) and (iv)
(C) (i), (iii) and (iv)
(D) (i) and (iv)
Directions : (29 to 30) Read the following information
carefully and answer the questions given below it.
I. Seven books are placed one above the other in a
particular way .
II. The history book is placed directly above the
civics book.
Directions : (33 to 37) Read the following paragraph
III. The geography book is fourth from the bottom
carefully and choose the correct alternative.
and the English book is fifth from the top.
The office staff of XYZ corporation presently
IV. There are two books in between the civics and
consists of three females A, B, C and five males D,
economics books.
E, F, G and H. The management is planning to
open a new office in another city using three males
29.
To find the number of books between the civics
and two females of the present staff. To do so they
and the science books, which other extra piece of
plan to separate certain individuals who do not
information is required, from the following ?
function well together. The following guidelines
(A) There are two books between the geography
were established
and the science books.
I. Females A and C are not to be together
(B) There are two books between the mathematics
II. C and E should be separated
and the geography books .
III. D and G should be separated
(C) There is one book between the English and
IV. D and F should not be part of a team.
the science books.
(D) The civics book is placed before two books
33.
above the economics book.
30.
To know which three books are kept above the
English book, which of the following additional
pieces of information, if any, is required?
(A) The economics book is between the English 34.
and the science books.
(B) There are two books between the English and
the history books.
(C) The geography book is above the English book.
35.
(D) No other information is required.
Directions : (31 to 32) A five-member team that includes
Rama, Shamma, Henna, Reena, and Tina, is
planning to go to a science fair but each of them
36.
put up certain conditions for going .They are as
follows.
I. If Rama goes, then at least one amongst
Shamma and Henna must go.
II. If Shamma goes, then Reena will not go.
37.
III. If Henna will go, then Tina must go.
IV. If Reena goes, then - Henna must go.
V. If Tina goes, then Rama must go but Shamma
cannot go.
VI. If Reena plans not to go the fair, then Rama will
also not go.
If A is chosen to be moved, which of the following
cannot be a team ?
(A) ABDEH
(B) ABDGH
(C) ABEFH
(D) ABEGH
If C and F are to be moved to the new office, how
many combinations are possible ?
(A) 1
(B) 2
(C) 3
(D) 4
If C is chosen to the new office, which number of
the staff cannot be chosen to go with C ?
(A) B
(B) D
(C) F
(D) G
Under the guidelines, which of the following must
be chosen to go to the new office ?
(A) B
(B) D
(C) E
(D) G
If D goes to the new office, which of the following
is/are true ?
I. C cannot be chosen
II. A cannot be chosen
III. H must be chosen.
(B) II only
(A) I only
(C) I and II only
(D) I and III only
PAGE # 81
Directions : (38 to 42) Study the following information Directions : (45 to 49) Read the following information
carefully and answer the questions given below.
carefully and answer the questions that follow :
(i) There is a family of six persons- L, M, N, O, P
A team of five is to be selected from amongst five
and Q. They are professor, businessman,
boys A, B, C, D and E and four girls P, Q, R and S.
chartered account, bank manager, engineer and
Some criteria for selection are :
medical representative, not necessarily in that
A and S have to be together
order.
(ii) There are two married couples in the family.
P cannot be put with R.
(iii) O, the bank manager is married to the lady
D and Q cannot go together.
professor.
C and E have to be together.
(iv) Q, the medical representative, is the son of M
R cannot be put with B.
and brother of P.
Unless otherwise stated, these criteria are
(v) N, the chartered accountant, is the daughter - in
applicable to all the questions below :
law of L.
(vi) The businessman is married to the chartered
38.
If two of the members have to be boys, the team
acconuntant.
will consist of :
(vii) P is an unmarried engineer.
(viii) L is the grandmother of Q
(A) A B S P Q
(B) A D S Q R
(C) B D S R Q
39.
40.
(D) C E S P Q
45.
How is P related to Q.
(A) Brother
(B) Sister
(C) Cousin
(D) Either brother or sister
46.
Which of the following is the profession of M ?
(A) Professor
(B) Chartered accountant
(C) Businessman
(D) Medical representative
If R be one of the members, the other members of
the team are :
(A) P S A D
(B) Q S A D
(C) Q S C E
(D) S A C E
If two of the members are girls and D is one of the
members, the members of the team other than D
are :
47.
(A) P Q B C
(B) P Q C E
(C) P S A B
(D) P S C E
41.
If A and C are members, the other members of the 48.
team cannot be :
(A) B E S
(C) E S P
42.
(B) D E S
(D) P Q E
If including P at least three members are girls, the
members of the team other than P are :
(A) Q S A B
(C) Q S C E
(B) Q S B D
(D) R S A D
Directions : (43 to 44) Read the given information carefully
and answer the questions that follow :
Ratan, Anil, Pinku and Gaurav are brothers of Rakhi,
Sangeeta, Pooja and Saroj, not necessarily in that
order. Each boy has one sister and the names of
bothers and sisters do not begin with the same
49.
44.
Pooja’s brother is
(A) Ratan
(B) Anil
(C) Pinku
(D) Gaurav
Which of the following are brother and sister ?
(A) Ratan and Pooja
(C) Pinku and Sangeeta
(B) Anil and Saroj
(D) Gaurav and Rakhi
Which of the following is one of the couples ?
(A) QO
(B) OM
(C) PL
(D) None of these
How is O related to Q?
(A) Father
(C) Uncle
(B) Grandfather
(D) Brother
Directions : (50 to 54)
I. There is a group of six persons P,Q, R, S, T and U
from a family. They are Psychologist, Manager,
Lawyer, Jeweller, Doctor and Engineer.
II. The Doctor is grandfather of U, who is a
Psychologist.
III. The Manager S is married to P.
IV. R, the Jeweller is married to the Lawyer.
V. Q is the mother of U and T.
VI. There are two married couples in the family.
letter. Pinku and Gaurav are not Saroj’s or 50.
Sangeeta’s brothers. Saroj is not Ratan’s sister.
43.
Which of the following is the profession of L ?
(A) Professor
(B) Charted accountant
(C) Businessman
(D) Engineer
51.
52.
What is the profession of T ?
(A) Doctor
(B) Jeweller
(C) Manager
(D) None of these
How is P related to T ?
(A) Brother
(C) Father
(B) Uncle
(D) Grandfather
How many male members are their in the family ?
(A) One
(B) Three
(C) Four
(D) Data inadequate
PAGE # 82
53.
What is the profession of P ?
(A) Doctor
(B) Lawyer
(C) Jeweller
(D) Manager
54.
Which of the following is one of the pairs of couples
in the family ?
(A) PQ
(B) PR
(C) PS
(D) Cannot be determined
Direction : (55) The ages of Mandar, Shivku, Pawan and
56.
Chandra are 32, 21, 35 and 29 years, not in order,
Whenever asked they lie of their own age but tell
the truth abut others.
(i) Pawan says, “My age is 32 and Mandar’s age is
not 35”
(ii) Shivku says, “My age is not 2 9 and Pawan’s 57.
age in not 21”
(iii) Mandar says, “My age is 32.”
55.
What is Chandra’s age ?
(A) 32 years
(B) 35 years
(C) 29 years
(D) 21 years
Directions : (56 to 57) Answer the questions on the basis
of the information given below. 5 friends Nitin,
Reema, Jai, Deepti and Ashutosh are playing a
58.
game of crossing the roads. In the beginning, Nitin,
Reema and Ashutosh are on the one side of the
road and Deepti and Jai are on the other side. At
the end of the game, it was found that Reema and
Deepti are on the one side and Nitin, Jai and
Ashutosh are on the other side of the road. Rules
of the game are as follows :
I. One “movement” means only one person crosses
the road from any side to the other side.
II. No two persons can cross the road
simultaneously from any side to the other side.
III. Two persons from the same side of the roads
cannot move in consecutive “movements”.
IV. If one person crosses the road in a particular
movement, he or she cannot immediately move
back to the other side.
V. Jai and Reema did not take part in first 3
movements.
W hat is the minimum possible number of
movements that took place in the entire game ?
(A) 3
(B) 4
(C) 5
(D) 6
If number of movements are minimised in the
game, then which of the following combination of
friends can never be together on one particular
side of the road during the course of the game ?
(A) Nitin, Reema amd Deepti
(B) Nitin, Jai and Deepti
(C) Deepti, Jai and Ashutosh
(D) Ashutosh, Nitin and Deepti
You have 12 similar looking coins. 11 of them weigh
the same. One of them has a different weight, but
you don’t know whether it is heavier or lighter. You
also have a scale. You can put coins on both sides
of the scale and it’ll tell you which side is heavier or
will stay in the middle if both sides weigh the same.
What is the minimum number of weighing required
to find out the odd coin.
(A) 3
(B) 4
(C) 5
(D) 6

PAGE # 83
CALENDAR AND CLOCK TEST
Similarly, 200 years = 10 odd days = 03 odd days

300 years =
= 1 odd day..
7
20  1
400 years =
= 0 odd day (1 is added as 400
7
is a leap year)
Similarly, 800, 1200, 1600, 2000, 2400 years
contain 0 odd days.
We are to find the day of the week on a mentioned
date. Certain concepts are defined as under.

An ordinary year has 365 days.

In an ordinary year, first and last day of the year are
same.

15
After counting the odd days, we find the day
according to the number of odd days.

A leap year has 366 days. Every year which is
divisible by 4 is called a leap year. For example

1200, 1600, 1992, 2004, etc. are all leap years.
Sunday for 0 odd day, Monday for 1 odd day and so
on as shown in the following table.

For a leap year, if first day is Monday than last day
will be Tuesday for the same year.

In a leap year, February is of 29 days but in an
ordinary year, it has only 28 days.

Year ending in 00's but not divisiable by 400 is not
considered a leap year. e.g., 900, 1000, 1100, 1300,
1400, 1500, 1700, 1800, 1900, 2100 are not leap
years.
Ordinary
Year
Days
The day on which calendar started (or the very first
day ) i.e., 1 Jan, 0001 was Monday.
January
31
3
February
28
0
March
31
3
April
30
2
May
31
3
Table : 1 (Odd days for week days)
Days
Sunday Monday Tuesday Wednesday Thursday Friday Saturday
Odd Days


Calendar year is from 1 Jan to 31 Dec. Financial
year is from 1 April to 31 March.
ODD DAYS
The no. of days exceeding the complete no. of
weeks in a duration is the no. of odd days during
that duration.
COUNTING OF ODD DAYS

Every ordinary year has 365 days = 52 weeks +1 day.
 Ordinary year has 1 odd day.

Every leap year 366 days = 52 weeks + 2 days.
 Leap year has 2 odd days.
0
1
2
3
4
5
6
Table : 2 (Odd days for months in a year)
Odd
Leap year
Days
Days
Odd
Days
January
31
3
February
29
1
March
31
3
April
30
2
May
31
3
June
30
2
June
30
2
Total
181 days
6
Total
182 days
0
July
31
3
July
31
3
August
31
3
August
31
3
September
30
2
September
30
2
October
31
3
October
31
3
November
30
2
November
30
2
December
31
3
December
31
3
Total
184 days
1
Total
184 days
2
Table : 3 (Odd days for every quarter)


Odd days of 100 years = 5,
Odd days of 200 years = 3,
Odd days of 300 years = 1,
Odd days of 400 years = 0.
M onths
of
years
I st three
m onths
1 Jan to
31 M arch
Total days
90 / 91
Ord. / Leap
Odd days
6 /0
0
Ord. / Leap O dd day
Explanation :
100 years = 76 ordinary years + 24 leap years Ex 1.
( The year 100 is not a leap year)
= 76 odd days + 2 × 24 odd days = 124 odd days. Sol.
Odd days =
124
7
= 5 odd days.
IInd three IIIrd three
m onths m onths
1 Apr to 1 July to
30 June 30 Sep.
91
92
1
Odd day
Iv th three
Total year
m onths
1 Jan to
1 Oct. to
31 Dec.
31 Dec.
92
365 / 366
Ord. / Leap
1
1 /2
Odd day Ord. / Leap
If it was Saturday on 17th December 1982 what
will be the day on 22nd December 1984 ?
Total number of odd days between 17 Dec.1982
to 17 Dec.1984 the number of odd days = 1+2 = 3.
From 17 to 22 Dec. number of odd days = 5
 3 + 5 = 8 odd days = 1 odd day.
 Saturday + 1 odd day = Sunday.
PAGE # 84
Ex 2.
Sol.
Ans.
Ex 3.
Sol.
Ans.
Ex 4.
Sol.
Ans.
Ex 5.
Sol.
Find the day of the week on 16 January, 1969.
Ex 6.
1600 years have ‘0’ odd day. .....................(A)
300 years have ‘1’ odd day. ......................(B)
Sol.
68 years have 17 leap years and 51 ordinary years.
Thus = (17 × 2 + 51 × 1 ) = 85 odd days
 ' 01' odd day ...(C)
16 January has = ' 02' odd days..(D)
Adding (A) + (B) +(C) +(D),
We get, 0 + 01 +01 +02 = 04 odd days
Thursday
Ex 7.
Find the day of the week on 18 July, 1776 (leap Sol.
year).
Here 1600 years have ‘0’ odd day.....................(A)
100 years have ‘5’ odd days..............................(B)
75 years = (18 leap years + 57 ordinary years)
= (18 × 2 + 57 × 1)
= 93 odd days
= (7 × 13 + 2) = ‘2’ odd days.............................(C)
Now, the no. of days from 1st January to 18 July,
1776
= 182 + 18 = 200 days
= (28 × 7 + 4) days = ‘4’ odd days.....................(D)
Adding (A) + (B) +(C) +(D),
We get, 0 + 5 + 2 + 4 = 04 odd days
Thursday
On what dates of October, 1975 did Tuesday fall ?
For determining the dates, we find the day on 1st
Oct, 1975.
1600 years have ‘0’ odd days.....................(A).
300 years have ‘01’ odd days.....................(B).
74 years have (18 leap years + 56 ordinary years)
2 × 18 + 1 × 56 = 92 odd days
= ‘01’ odd days.............(C)
Days from 1st January to 1st Oct., 1975
1st Jan – 30 June + 1st July to 1st Oct.
181 + 31 + 31 + 30 + 1 = 274 days
= ‘01’ odd days......(D) (274/7= 01 days)
Adding (A) + (B) +(C) +(D) = 0 + 01 +01 +01
= '03' odd days
Wednesday( 1st Oct), hence 7,14,21,28 Oct. will
Tuesday fall.
The year next to 1996 having the same Calendar
will be 1996 1997 1998 1999 2000 2001 2002 2003
2
1
1
1
2
Total = 2 + 1 + 1 + 1 + 2 = 7= 0 odd days
Hence, year 2001 will have the same calendar as
year 1996.
Prove that last day of a century cannot be Tuesday,
Thursday or Saturday.
100 years have = 5 odd days
 Last day of st century is Friday
200 years have = 10 odd days
 Last day of IInd century is Wednesday
= 3 odd days
300 years have = 15 odd days
 Last day of rd century is Monday
= 01 odd day
400 years have = (5 × 4 + 1)
Last day of 4th century is Sunday
= 21 odd days
= 0 odd days
Since the order keeps on cycling, we see that the
last day of the century cannot be Tuesday, Thursday
or Saturday.
Important Notes :

Minute hand and hour hand coincides once in every
hour. They coincide 11 times in 12 hours and 22
times in 24 hours.

They coincide only one time between 11 to 1 O’
clock. at 12 O’ clock.

Minute hand and hour hand are opposite once in

Calendar for 1995 will serve for 2006, prove ?
The Calendar for 1995 and 2006 will be the same
,if day on 1st January of both the years is the same. 
This is possible only if the total odd days between
31st Dec. 1994 and 31st Dec.2005 is 0. [one day
before both the years as we want to know the day
on 1st January of both the years i.e. same]

During this period, we have
3 leap years
(1996, 2000, 2004)
and
08 ordinary years

(1995,1997,1998,1999, 2001, 2002, 2003,2005)
Total odd days = (2 × 3 + 1 × 8) = 14 = 0 odd
days (Thus Proved)
every hour. They do it 11 times in 12 hours and 22
times in 24 hours.
They opposite only one time between 5 to 7 O’
clock. at 6 O’ clock.
Both hands (minute and hour) are perpendicular
twice in every hour. 22 times in 12 hours and 44
times in 24 hours.
In one minute, hour hand moves 1/2º and minute
hand moves 6º. In one hour, hour hand moves 30º
and minute hand moves 360º.
In an hour, minute hand moves 55 minutes ahead
of hour hand.
PAGE # 85
Case-II When the time taken (20 + 15) = 35 min.
HANDS COINCIDE
Ex.8
Sol.
 Minute hand is 55 min. ahead of hour hand in
At what time between 3 O’Clock and 4 O’Clock will
the two hands coincide ?
At 3 O’clock the distance between the two hands is
15 minutes when they coincide with each other the
distance between the two hands will be 0 min.
So, the time taken (15 + 0 ) = 15 minutes.
 Minute hand is 55 min. ahead of hour hand in
60 min.
 Minute hand is 1 min. ahead of hour hand in
60 min.
55
60 min.
Minute hand is 35 min. ahead of hour hand in
60  35
420
2
=
= 38
min.
55
11
11
2
Hence, the right time is 38 min. past 4.
11
MIRROR IMAGE OF CLOCK

to find the mirror image, time is subtracted from
11 : 60.
 Minute hand is 15 min. ahead of hour hand in
60  15
180
4
=
= 16 min.
55
11
11

4
Hence the right time is 16 minute past 3.
11
HANDS ARE OPPOSITE
Ex.9
Sol.
60  40
480
7
=
= 43
min.
55
11
11
Hence, the right time is 43
23 : 60.
Ex.11
The time in the clock is 4 : 46, what is the mirror image ?
Sol.
(11 : 60) – (4 : 46) = 7 : 14.
7
min. past 2.
11

Case-I
At what time between 4 O’clock and 5 O’clock will
the hands are perpendicular ?
At 4 O’clock the distance between the two hands is
20 min. When they are at 15 minutes distance,
they are perpendicular to each other.
When the time taken (20 – 15) = 5 min.
 Minute hand is 55 min. ahead of hour hand in
60 min.
Minute hand is 5 min. ahead of hour hand in
Both types of angles are 360º in total. If one angle
is known, other can be obtained by subtracting from
360º.
Ex.13
At 4 : 30, what is the angle formed between hour
Sol.
hand and minute hand ?
At 4 O’ clock angle between hour and min. hand is
HANDS ARE PERPENDICULAR
Sol.
If the time is between 11 O’clock to 1 O’clock, then
to find the mirror image, time is subtracted from
At what time between 2 O’clock and 3 O’clock will
Ex.12 The time in the clock is 12 : 35, then find its mirror
the two hands be opposite ?
image.
At 2 O’clock the distance between the two hands is
Sol.
(23 : 60) – (12 : 35) = 11 : 25.
10 minutes. When they are at 30 minutes distance,
they are opposite to each other. The time taken
TO FIND THE ANGLE BETWEEN TWO HANDS
(30 + 10 ) = 40 min.
 Minute hand is 55 min. ahead of hour hand in

Angle are of two types :
60 min.
Positive angle : It is obtained by moving from hour
 Minute hand is 1 min. ahead of hour hand in
hand to minute hand moving in clockwise direction.
60
min.
Negative angle : It is obtained by moving from
55
minute hand to hour hand.
 Minute hand is 40 minutes ahead of hour hand
in
Ex.10
If the time is between 1 O’clock to 11 O’clock, then
of 120º.
In 30 min. minute hand make an angle of 180º.
So, the resultant angle is 180º – 120º = 60º.
But in 30 min. hour hand will also cover an angle of 15º.
Hence, the final angle between both hands is
60º – 15º = 45º.
Short trick
60  5
60
5
=
= 5
min.
55
11
11
5
Hence, the right time is 5 min. past 4.
11
PAGE # 86
9.
Ex.14
Sol.
A bus for Delhi leaves every thirty minutes from a bus
stand. An enquiry clerk told a passenger that the bus
had already left ten minutes ago and the next bus will
leave at 9.35 A.M. At what time did the enquiry clerk 10.
give this information to the passenger ?
Bus leaves after every 30 minutes.
The next bus will leave at 9 : 35 A.M.
The last bus left at 9 : 35 – 0 : 30 = 9 : 05 A.M.
but clerk said that bus had left 10 minutes earlier. 11.
9 : 05 + 0 : 10 = 9 : 15 A.M.
EXERCISE
1.
Find the day of the week on 26 January, 1950.
(A) Tuesday
(B) Friday
(C) Wednesday
(D) Thursday
2.
W hich two months in a year have the same
calendar ?
(A) June, October
(B) April, November
13.
(C) April, July
(D) October, December
3.
4.
5.
12.
Are the years 900 and 1000 leap years ?
(A) Yes
(B) No
(C) Can't say
(D) None of these
If it was Saturday on 17th November, 1962 what
will be the day on 22nd November, 1964 ?
(A) Monday
(B) Tuesday
14.
(C) Wednesday
(D) Sunday
Sangeeta remembers that her father's birthday
was certainly after eighth but before thirteenth of
December. Her sister Natasha remembers that
their father's birthday was definitely after ninth but
before fourteenth of December. On which date of
December was their father's birthday ?
(A) 10th
(B) 11th
(C) 12th
(D) Data inadequate
15.
6.
Find the day of the week on 15 August, 1947.
(A) Tuesday
(B) Friday
(C) Wednesday
(D) Thursday
7.
Karan was born on Saturday 22nd March 1982. On
what day of the week was he 14 years 7 months
and 8 days of age ?
(A) Sunday
(B) Tuesday
(C) Wednesday
(D) Monday
8.
If on 14th day after 5th March be Wednesday, what
day of the week will fall on 10th Dec. of the same
year ?
(A) Friday
(B) Wednesday
(C) Thursday
(D) Tuesday
16.
If the day before yesterday was Saturday, what day
will fall on the day after tomorrow ?
(A) Friday
(B) Thursday
(C) Wednesday
(D) Tuesday
If February 1, 1996 is Wednesday, what day is March
10, 1996 ?
(A) Monday
(B) Sunday
(C) Saturday
(D) Friday
If the seventh day of a month is three days earlier
than Friday, what day will it be on the nineteenth
day of the month ?
(A) Sunday
(B) Monday
(C) Wednesday
(D) Friday
Mohini went to the movies nine days ago. She goes
to the movies only on Thursday. What day of the
week is today ?
(A) Thursday
(B) Saturday
(C) Sunday
(D) Tuesday
At what time are the hands of a clock together
between 5 and 6 ?
(A) 33
3
min. past 5
11
(B) 28
3
min. past 5
11
(C) 27
3
min. past 5
11
(D) 26
3
min. past 5
11
At what time between 9 and 10 will the hands of a
clock be in the straight line, but not together ?
(A) 16 minutes past 9
4
(B) 16 minutes past 9
11
6
(C) 16 minutes past 9
11
9
(D) 16
minutes past 9
11
At what time between 5 & 5 : 30 will the hands of a
clock be at right angle ?
10
(A) 10
minutes past 5
11
5
(B) 11
minutes past 5
11
10
(C) 9
minutes past 5
11
9
(D) 10
minutes past 5
11
Ajay left home for the bus stop 15 minutes earlier
than usual. It takes 10 minutes to reach the stop.
He reached the stop at 8.40 a.m. What time does
he usually leave home for the bus stop ?
(A) 8.30 a.m.
(B) 8.45 a.m.
(C) 8.55 a.m.
(D) Data inadequate
PAGE # 87
17.
The priest told the devotee, "The temple bell is 26.
rung at regular intervals of 45 minutes. The last
bell was rung five minutes ago. The next bell is
due to be rung at 7.45 a.m." At what time did the
priest give this information to the devotee ?
(A) 7.40 a.m.
(B) 7.05 a.m.
(C) 6.55 a.m.
(D) None of these
18.
There are twenty people working in an office. The
first group of five works between 8.00 A.M. and 2.00
P.M. The second group of ten works between 10.00 27.
A.M. and 4.00 P.M. And the third group of five works
between 12 noon and 6.00 P.M. There are three
computers in the office which all the employees
frequently use. During which of the following hours
the computers are likely to be used most ?
(A) 10.00 A.M. –– 12 noon
(B) 12 noon –– 2.00 P.M.
(C) 1.00 P.M. –– 3.00 P.M.
(D) 2.00 P.M. –– 4.00 P.M.
28.
19.
A tired worker slept at 7.45 p.m.. If he rose at 12
noon, for how many hours did he sleep ?
(A) 5 hours 15 min.
(B) 16 hours 15 min.
(C) 12 hours
(D) 6 hours 45 min.
20.
How many times are the hands of a clocks
perpendicular in a day ?
(A) 42
(B) 48
(C) 44
(D) 46
21.
If a clock shows 04: 28 then its mirror image will
be ?
(A) 07: 42
(B) 07: 32
(C) 08: 32
(D) 08: 42
22.
A watch, which gains uniformly, is 3 minutes slow
at noon on Monday and is 3 minutes 48 seconds
fast at 2 p.m. on the following Monday. What time it
was correct ?
(A) 2 p.m. On Tuesday
30.
(B) 2 p.m. On Wednesday
(C) 3 p.m. On Thursday
(D) 1 p.m. On Friday.
29.
23.
How many times are the hands of a clocks coincide
in a day ?
(A) 10
(B) 11
(C) 12
(D) 22
24.
At what time between 2 and 3 O’ clock the hands of
a clock will make an angle of 160º ?
(A) 20 minutes past 2
(B) 30 minutes past 2
(C) 40 minutes past 2
(D) 50 minutes past 2
25.
Ashish leaves his house at 20 minutes to seven in
the morning, reaches Kunal’s house in 25 minutes,
they finish their breakfast in another 15 minutes
and leave for their office which takes another 35 32.
minutes. At what time do they leave Kunal’s house
to reach their office ?
(A) 7.40 am
(B) 7.20 am
(C) 7.45 am
(D) 8.15 am
31.
The train for Lucknow leaves every two and a half
hours from New Delhi Railway Station. An
announcement was made at the station that the
train for Lucknow had left 40 minutes ago and the
next train will leave at 18. 00 hrs. At what time was
the announcement made ?
(A) 15.30 hrs
(B) 17.10 hrs
(C) 16.00 hrs
(D) None of these
A monkey climbs 30 feet at the beginning of each
hour and rests for a while when he slips back 20
feet before he again starts climbing in the
beginning of the next hour. If he begins his ascent
at 8.00 a.m., at what time will he first touch a flag at
120 feet from the ground ?
(A) 4 p.m.
(B) 5 p.m.
(C) 6 p.m.
(D) None of these
If the two incorrect watches are set at 12 : 00 noon
at correct time, when will both the watches show
the correct time for the first time given that the first
watch gains 1 min in 1 hour and second watch
loses 4 min in 2 hours :
(A) 6 pm, 25 days later
(B) 12 : 00 noon, 30 days later
(C) 12 noon, 15 days later
(D) 6 am 45 days later
Rajeev and Sanjeev are too close friends Rajeev's
watch gains 1 minute in an hour and Sanjeev's
watch loses 2 minutes in an hour. Once they set
both the watches at 12 : 00 noon, with my correct
watch. When will the two incorrect watches of
Rajeev and Sanjeev show the same time together?
(A) 8 days later (B) 10 days later
(C) 6 days later (D) can't be determined
At a railway station a 24 hour watch loses 3 minutes
in 4 hours. If it is set correctly on Sunday noon
when will the watch show the correct time ?
(A) 6 pm after 40 days
(B) 12 noon after 75 days
(C) 12 pm after 100 days
(D) 12 noon after 80 days
A swiss watch is being shown in a museum which
has a very peculiar property. It gains as much in
the day as it loses during night between 8 pm to 8
am. In a week how many times will the clock show
the correct time ?
(A) 6 times
(B) 14 times
(C) 7 times
(D) 8 times
A wrist watch which is running 12 minutes late on
a Sunday noon is 16 minutes ahead of the correct
time at 12 noon on the next Sunday. When is the
clock 8 minutes ahead of time ?
(A) Thursday 10 am
(B) Friday noon
(C) Friday 8 pm
(D) Tuesday noon
PAGE # 88
33.
34.
35.
A clock loses 2 minutes in a hour and another clock 38.
gains 2 minutes in every 2 hours. Both these clocks
are set correctly at a certain time on Sunday and
both the clocks stop simultaneously on the next
day with the time shown being 9 am and 10 : 06
AM. What is the correct time at which they stopped?
(A) 9 : 54 am
(B) 9 : 44 pm
(C) 9 : 46 am
(D) 9 : 44 am
David sets his watch at 6 : 10 am on Sunday, which
gains 12 minutes in a day. On Wednesday if this
watch is showing 2 : 50 pm. What is the correct
time ?
39.
(A) 1 : 50 pm
(B) 2 : 10 pm
(C) 2 : 30 pm
(D) 3 : 30 pm
Ramu purchased a second hand Swiss watch
which is very costly. In this watch the minute-hand
and hour hand coincide after every 65
36
37.
3
minutes.
11
Kumbhakarna starts sleeping between 1 pm and
2 pm and he wakes up when his watch shows
such a time that the two hands (i.e., hour-hand
and minute-hand) interchange the respective
places. He wakes up between 2 pm and 3 PM on
the same night. How long does he sleep ?
(A) 55
5
min
13
(B) 110
10
min
13
(C) 54
6
min
13
(D) None of these
A clock loses 3% time during the first week and
then gains 2% time during the next one week. If the
clock was set right at 12 noon on a Sunday, what
will be the time that the clock will show exactly 14
days from the time it was set right ?
(A) 1 : 36 : 48
(B) 1 : 40 : 48
(C) 1 : 41 : 24
(D) 10 : 19 : 12
How much time does the watch lose or gain per
day ?
Direction : (40 to 41) A 12 dial clock has its minute hand
(A) 4 min
(B) 5 min
defective. W henever it touches dial 12, it
(C) 4 min, 20 sec
(D) none of these
immediately falls down to 6 instead of running
My watch was 8 minutes behind at 8 pm on Sunday
smoothly (the hour hand remains unaffected during
but within a week at 8 pm on Wednesday it was 7
that fall). It was set right at 12 ‘O’ clock in the noon.
minutes ahead of time. During this period at which
time this watch has shown the correct time :
40.
What was the actual time when the minute hand of
(A) Tuesday 10 : 24 am
the clock touched dial 9 for the 5th time?
(B) Wednesday 9 : 16 pm
(A) 2 : 15
(B) 3 : 00
(C) It cannot show the correct time during this period
(C) 5 : 15
(D) 6 : 45
(D) None of the above
Out of the following four choices which does not 41.
show the coinciding of the hour hand and minutehand :
(A) 3 : 16 : 2
(B) 6 : 32 : 43
(C) 9 : 59 : 05
(D) 5 : 27 : 16
If the actual time is 10 : 10, what is the position of
the hour hand in that defective clock ?
(A) Between 2 and 3
(B) Between 4 and 5
(C) Between 10 and 11
(D) Between 3 and 4

PAGE # 89
CUBE AND DICE-TEST
CUBES
A cube is three dimensional figure, having 8
corners, 6 surfaces and 12 edges. If a cube is
painted on all of its surfaces with any colour and
further divided into various smaller cubes, we get
following results. Smaller cubes with three
surfaces painted will be present on the corners of
the big cube.
Here
3
2
2
2
1
1
3
2
3
2
2
3
2
1
1
2
2
1
1
2
3
2
2
3
2 3
1 2
1 2
2 3
3
2
2
3
n=
2
1
1
side of big cube
4
 4
side of small cube
1
3
2
1
1
2
2
Ex 1.
How many smaller cubes have three surfaces
2
painted ?
3
(A) 4
(B) 8
(C) 16
(D) 24
2
Sol.
(B) Number of smaller cubes with three surfaces
Smaller cubes with two surface painted will be
present on the edges of the big cube. Smaller
cubes with one surface painted will be present on Ex 2.
the surfaces of the big cube. Smaller cubes with
no surface painted will be present inside the big
cube.
painted = 8
If a cube is painted on all of its surfaces with a Sol.
colour and then divided into smaller cubes of equal
size then after separation, number of smaller cubes
so obtained will be calculated as under :
Ex 3.
Number of smaller cubes with three surfaces
painted = 8
Number of smaller cubes with two surfaces
painted = (n – 2) × 12
Number of smaller cubes with one surfaces
Sol.
painted = (n – 2)2 × 6
Number of smaller cubes with no surfaces painted
= (n – 2)3
Ex 4.
Where n = No of divisions on the surfaces of the
bigger cube
(D) Number of smaller cubes with two surfaces
length of edge of big cube
=
length of edge of one smaller cube
(D) Number of smaller cubes with no surface
TYPE I
Sol.
How many smaller cubes have two surfaces
painted ?
(A) 4
(B) 8
(C) 16
(D) 24
painted = (n – 2) × 12 = (4 – 2) × 12 = 24
How many smaller cubes have only one surface
painted ?
(A) 8
(B) 16
(C) 24
(D) 32
(C) Number of smaller cubes with one surface
painted = (n – 2)2 × 6 = (4 – 2)2 × 6 = 4 × 6 = 24
How many smaller cubes will have no side painted ?
(A) 18
(B) 16
(C) 22
(D) 8
painted = (n – 2)3 = (4 – 2)3 = (2)3 = 8
TYPE II
If a cube is painted on all of its surfaces with single
colour and then divided into various smaller cubes
of equal size.
If a cube is painted on all of its surfaces with
different colours and then divided into various
smaller cubes of equal size.
Directions : ( 1 to 4) A cube of side 4 cm is painted black on
all of its surfaces and then divided into various
smaller cubes of side 1 cm each. The smaller Directions : ( 5 to 7 ) A cube of side 4 cm is painted black on
the pair of one opposite surfaces, blue on the pair
cubes so obtained are separated.
444
 64
Total cubes of obtained =
1 1 1
of another opposite surfaces and red on remaining
pair of opposite surfaces. The cube is now divided
into smaller cubes of equal side of 1 cm each.
PAGE # 90
Ex 9.
Sol.
Ex 5.
Sol.
Ex 6.
Sol.
How many smaller cubes have two surfaces
painted ?
(A) 4
(B) 8
(C) 16
(D) 24
(C) Number of smaller cubes with two surfaces
painted = Number of cubes present at the corners
+ Numbers of cubes present at 4 edges
= 8 + (n – 2) × 4 = 8 + 8 = 16
How many smaller cubes have three surfaces
painted ?
(A) 4
(B) 8
(C) 16
(D) 24
(B) Number of smaller cubes with three surfaces
painted = 8
(These smaller cubes will have all three surfaces
painted with different colour blue, black and red.)
How many smaller cubes have two surfaces
painted ?
Ex 10.
(A) 4
(B) 8
(C) 16
(D) 24
(D) Number of smaller cubes with two surfaces
painted = 24. And out of this (a) Number of cubes with two surfaces painted Sol.
with black and blue colour = 8.
(b) Number of cubes with two surfaces painted
with blue and red colour = 8.
(c) Number of cubes with two surfaces painted
with black and red color = 8.
Ex 11.
Ex 7.
How many smaller cubes have only one surface
painted ?
(A) 8
(B) 16
(C) 24
(D) 32
(D) Number of smaller cubes with one surface
painted = Number of cubes present at the 8 edges
+ number of cubes present at the four surfaces=
(n – 2) × 8 + (n – 2)2 × 4
= 2 × 8 + 4 × 4 = 16 + 16 = 32
How many smaller cubes will have no side painted
?
(A) 18
(B) 16
(C) 22
(D) 8
(B) Number of smaller cubes with no side painted
= Number of cubes on the two unpainted surfaces +
number of cubes present inside the cube.
= (n – 2)2 × 2 + (n – 2)3 = 4 × 2 + (2)3 = 8 + 8 = 16.
How many smaller cubes have only one surface
painted ?
(A) 8
(B) 16
Sol.
(C) 24
(D) 32
Sol.
(C) Number of smaller cubes with one surface
painted = 24. And out of this (a) Number of cubes with one surface painted
with black colour =8.
TYPE IV
(b) Number of cubes with one surface painted
with blue colour = 8.
If a cube is painted on its surfaces in such a way
(c) Number of cubes with one surface painted
that one pair of adjacent surfaces is left unpainted.
with red colour = 8.
Directions : (12 to 15 )A cube of side 4 cm is painted red on
TYPE III
the pair of one adjacent surfaces, green on the
pair of other adjacent surfaces and two adjacent
If a cube is painted on its surfaces in such a way
surfaces are left unpainted. Now the cube is divided
that one pair of opposite surfaces is left unpainted.
into 64 smaller cubes of side 1 cm each.
Directions : ( 8 to 11 ) A cube of side 4 cm is painted red on
the pair of one opposite surfaces, green on the
pair of another opposite surfaces and one pair of
opposite surfaces is left unpainted. Now the cube
is divided into 64 smaller cubes of side 1 cm each.
Ex 8.
Sol.
How many smaller cubes have three surfaces
painted ?
(A) 0
(B) 8
(C) 16
(D) 20
(A) Number of smaller cubes with three surfaces Ex 12.
painted = 0 (Because each smaller cube at the
corner is attached to a surface which is unpainted.)
How many smaller cubes have three surfaces
painted ?
(A) 2
(B) 4
(C) 8
(D) 6
PAGE # 91
Sol.
Ex 13.
Sol.
Ex 14.
Sol.
Ex 15.
(A) Number of smaller cubes with three surfaces
Type-II
painted = Number of smaller cubes at two corners
=2
Ex 17. The figures given below show the two different
positions of a dice. Which number will appear
How many smaller cubes have two surfaces
opposite to number 2 ?.
painted ?
(A) 4
(B) 8
(C) 16
(D) 14
(D) Number of smaller cubes with two surfaces
painted = Number of smaller cubes at four corners
(A) 3
(B) 4
+ Number of smaller cubes at 5 edges.
(C) 5
(D) 6
= 4 + (n – 2) × 5 = 4 + 2 × 5
Sol.
(C) The above question,
= 4 + 10 = 14
where only two positions of
a dice are given, can easily
How many smaller cubes have only one surface
be solved with the
painted ?
following method.
(A) 8
(B) 16
Step I. The dice, when unfolded, will appear as shown in
(C) 24
(D) 30
the figure given on the right side.
(D) Number of smaller cubes with one surface
painted = Number of smaller cubes at four Step II. Write the common number to both the dice in the
middle block. Since common number is 4, hence
surfaces + Number of smaller cubes at 6 edges +
number 4 will appear in the central block.
Number of smaller cubes at two corners.
= (n – 2)2 × 4 + (n – 2) × 6 + 2
Step III. Consider the figure (i) and write the first number in
= 4 × 4 + 2 × 6 + 2 = 16 + 12 = 28 + 2 = 30
the anti-clockwise direction of number 4,
(common number) in block I and second number
How many smaller cubes will have no side painted
in block II. Therefore, numbers 3 and 2 being the
?
first and second number to 4 in anticlockwise
directions respectively, will appear in block I & II
(A) 18
(B) 16
respectively.
(C) 22
(D) 8
Sol.
(A) Number of smaller cubes with no surfaces Step IV. Consider figure (ii) and wire first and second
number in the anticlock-wise direction to number
painted = Number of smaller cubes from inside
4, (common number) in block (III) & (IV). Hence
the big cube + Number of cubes at two surfaces +
numbers 6 and 5 will appear in the blocks III and IV
Number of cubes at one edge.
respectively.
3
2
= (n – 2) + (n – 2) × 2 + (n – 2)
3
2
Step V. Write remaining number in the remaining block.
= (2) + (2) × + 2
Therefore, number 1 will come in the remaining
= 8 + 8 + 2 = 18
block. Now, from the unfolded figures we find that
DICES
number opposite to 6 is 3, number opposite to 2 is
5 and number opposite to 4 is 1. Therefore, option
(C) is our answer.
Type-I
( Short Trick : From the given dice, we will take the
General Dice : In a general dice the sum of numbers
common number as the base and then in its
on the any two adjacent faces is ‘7’.
respect move clockwise direction and write as
follows : 4 – 2 – 3
Standard Dice : In a standard dice the sum of
4 – 5 – 6.
numbers on the opposite faces is '7'.
Here,we find that number opposite to 6 is 3, number
Ex 16. Which number is opposite 4 in a standard dice
opposite to 2 is 5 and number opposite to 4 is
given below ?
remaining number 1.
Therefore, option (C) is our answer. )
1
5
4
(A) 1
(C) 5
Sol.
Ex 18.
(B) 3
(D) Can’t be determined
Clearly , from the standard dice the sum of
numbers on the opposite faces is '7', so number
opposite to 4 is 3.
On the basis of two figures of dice, you have to tell
what number will be on the opposite face of number
5?
(A) 1
(C) 4
(B) 2
(D) 6
PAGE # 92
Sol.
(D) The above question where only two positions
Type-V
of a dice are given, can easily be solved with the
following method :
If in the given dice, there are two numbers common, Ex 21. Which of the following dices is identical to the
unfolded figure as shown here ?
then uncommon numbers will always be opposite
of each other.
Therefore, option (D) is our answer.
Type-III
Ex 19.
From the following figures of dice, find which
number will come in place of ‘?’
(X)
Sol.
(A) 4
(B) 5
(C) 2
(D) 3
(D) If the above dice is unfolded, it will look like as
the figure (i) given below.
Sol.
Figure (i)
Now the number in place of ‘?’ can be obtained by
making a slight change in the figure as given here.
Now comparing figure (ii) with third dice as above,
we get that number in place of ? is 3.
Figure (ii)
Sol.
(B)
(C)
(D)
(A) From the unfolded figure of dice, we find that
number opposite to 2 is 4, for 5 it is 3 and for 1 it is
6. From this result we can definitely say that figure
(B), (C) and (D) can not be the answer figure as
numbers lying on the opposite pair of surfaces are
present on the adjacent surfaces.
EXERCISE
Directions : (1 to 5) A cube is coloured orange on one face,
pink on the opposite face, brown on one face and
silver on a face adjacent to the brown face. The
other two faces are left uncoloured. It is then cut
into 125 smaller cubes of equal size. Now answer
the following questions based on the above
statements.
1.
Type-IV
Ex 20.
(A)
A dice has been thrown four times and produces
following results.
2.
How many cubes have at least one face coloured
pink ?
(A) 1
(B) 9
(C) 16
(D) 25
How many cubes have all the faces uncoloured ?
(A) 24
(B) 36
(C) 48
(D) 64
3.
How many cubes have at least two faces coloured ?
(A) 19
(B) 20
(C) 21
(D) 23
4.
How many cubes are coloured orange on one face
and have the remaining faces uncoloured ?
(A) 8
(B) 12
(C) 14
(D) 16
Which number will appear opposite to the number
3?
5.
(A) 4
(B) 5
(C) 6
(D) 1
(A) From the figures (i), (ii) and (iv) we find that
numbers 6, 1, 5 and 2 appear on the adjacent
surfaces to the number 3. Therefore, number 4
will be opposite to number 3.
How many cubes one coloured silver on one face,
orange or pink on another face and have four
uncoloured faces ?
(A) 8
(B) 10
(C) 12
(D) 16
PAGE # 93
Directions : (6 to 11) A cube is painted red on two adjacent Directions : (17 to 21) The outer border of width 1 cm of a
surfaces and black on the surfaces opposite to
cube with side 5 cm is painted yellow on each side
red surfaces and green on the remaining faces.
and the remaining space enclosed by this 1 cm
Now the cube is cut into sixty four smaller cubes of
path is painted pink. This cube is now cut into 125
equal size.
smaller cubes of each side 1 cm. The smaller
6.
How many smaller cubes have only one surface
cubes so obtained are now seperated.
painted ?
(A) 8
(B) 16
17.
How many smaller cubes have all the surfaces
(C) 24
(D) 32
uncoloured ?
7.
How many smaller cubes will have no surface
(A) 0
(B) 9
painted ?
(C) 18
(D) 27
(A) 0
(B) 4
(C) 8
(D) 16
18.
How many smaller cubes have three surfaces
8.
How many smaller cubes have less than three
coloured ?
surfaces painted ?
(A) 2
(B) 4
(A) 8
(B) 24
(C) 28
(D) 48
(C) 8
(D) 10
9.
10.
11.
How many smaller cubes have three surfaces
19.
painted ?
(A) 4
(B) 8
(C) 16
(D) 24
How many cubes have at least two surfaces
coloured yellow ?
(A) 24
(B) 44
How many smaller cubes with two surfaces
painted have one face green and one of the
20.
adjacent faces black or red ?
(A) 8
(B) 16
(C) 24
(D) 28
(C) 48
(D) 96
(A) 0
(B) 1
How many smaller cubes have at least one surface
painted with green colour ?
21.
(A) 8
(B) 24
(C) 32
(D) 56
(C) 2
(D) 4
How many cubes have one face coloured pink and
an adjacent face yellow ?
How many cubes have at least one face coloured ?
(A) 27
(B) 98
(C) 48
(D) 121
Directions : (12 to 16) A cube of 4 cm has been painted on
its surfaces in such a way that two opposite Directions : (22 to 31) A solid cube has been painted yellow,
surfaces have been painted blue and two adjacent
blue and black on pairs of opposite faces. The
surfaces have been painted red. Two remaining
cube is then cut into 36 smaller cubes such that
surfaces have been left unpainted. Now the cube
32 cubes are of the same size while 4 others are
is cut into smaller cubes of side 1 cm each.
of bigger sizes. Also no faces of any of the bigger
cubes is painted blue.
12.
How many cubes will have no side painted ?
(A) 18
(B) 16
22.
How many cubes have at least one face painted
(C) 22
(D) 8
blue ?
13.
How many cubes will have at least red colour on
(A) 0
(B) 8
its surfaces ?
(C) 16
(D) 32
(A) 20
(B) 22
(C) 28
(D) 32
23.
How many cubes have only one faces painted ?
(A) 24
(B) 20
14.
How many cubes will have at least blue colour on
(C)
8
(D) 12
its surfaces ?
(A) 20
(C) 24
15.
16.
(B) 8
(D) 32
24.
How many cubes will have only two surfaces
painted with red and blue colour respectively ?
25.
(A) 8
(B) 12
(C) 24
(D) 30
How many cubes will have three surfaces coloured ?
(A) 3
(B) 4
26.
(C) 2
(D) 16
How many cubes have only two faces painted ?
(A) 24
(B) 20
(C) 16
(D) 8
How many cubes have atleast two faces painted ?
(A) 36
(B) 34
(C) 28
(D) 24
How many cubes have only three faces painted ?
(A) 8
(B) 4
(C) 2
(D) 0
PAGE # 94
27.
28.
29.
How many cubes do not have any of their faces 37.
painted yellow ?
(A) 0
(B) 4
(C) 8
(D) 16
38.
How many cubes have at least one of their faces
painted black ?
(A) 0
(B) 8
(C) 16
(D) 20
39.
How many cubes have at least one of their faces
painted yellow or blue ?
(A) 36
(B) 32
(C) 16
(D) 0
40.
30.
How many cubes have no face painted ?
(A) 8
(B) 4
(C) 1
(D) 0
31.
How many cubes have two faces painted yellow
and black respectively ?
(A) 0
(B) 8
41.
(C) 12
(D) 16
Directions : (32 to 35) Some equal
cubes are arranged in the
form of a solid block as
shown in the adjacent
figure. All the visible
sufaces of the block (except
the bottom) are then
painted.
32.
How many cubes have one face painted ?
(A) 9
(B) 24
(C) 22
(D) 20
34.
How many cubes have only two faces painted ?
(A) 0
(B) 16
(C) 20
(D) 24
35.
How many cubes have only three faces painted ?
(A) 4
(B) 12
(C) 6
(D) 20
Directions : (36 to 40) A cuboid of dimensions
(6 cm  4 cm  1 cm) is painted black on both the
surfaces of dimensions (4 cm 1 cm), green on the
surfaces of dimensions (6 cm 4 cm). and red on
the surfaces of dimensions (6 cm 1 cm). Now the
block is divided into various smaller cubes of side
1 cm. each. The smaller cubes so obtained are
separated.
36.
How many cubes will have all three colours black,
green and red each at least on one side?
(A) 16
(B) 12
(C) 10
(D) 8
If cubes having only black as well as green colour
are removed then how many cubes will be left?
(A) 4
(B) 8
(C) 16
(D) 30
How many cubes will have 4 coloured sides and
2 sides without colour?
(A) 8
(B) 4
(C) 16
(D) 10
How many cubes will have two sides with green
colour and remaining sides without any colour?
(A) 12
(B) 10
(C) 8
(D) 4
Which alphabet is opposite D ?
(A) E
(C) F
42.
How many cubes do not have any of the faces
painted ?
(A) 27
(B) 8
(C) 10
(D) 12
33.
How many cubes will be formed?
(A) 6
(B) 12
(C) 16
(D) 24
(B) C
(D) A
What should be the number opposite 4 ?
(i)
(ii)
(iii)
(A) 5
(C) 3
(B) 1
(D) 2
(i)
(ii)
43.
(iii)
(iv)
Which letter will be opposite to letter D ?
(A) A
(B) B
(C) E
(D) F
Directions : (44 to 45) The figure (X) given below is the
unfolded position of a cubical dice. In each of the
following questions this unfolded figure is followed
by four different figures of dice. You have to select
the figure which is identical to the figure (X).
PAGE # 95
50.
44.
(X)
(A)
(B)
51.
(B)
45.
Which symbol will appear on the opposite surface
to the symbol x?
(D)
(A) 
(B) =
(C) 
(D) O
Three positions of the same dice are given below.
Observe the figures carefully and tell which number
will come in place of ‘?’
1
6 3
3
5 4
4
2 ?
(i)
(ii)
(iii)
(A) 1
(C) 3
(X)
52.
(A)
(C)
(B)
(B) 6
(D) 5
On the basis of the following figures you have to
tell which number will come in place of ‘?’
3
6 1
4
2 6
?
1 5
(i)
(ii)
(iii)
(A) 2
(C) 6
(D)
(B) 3
(D) 4
Directions : (53 to 55) Choose from the alternatives, the
boxes that will be formed when figure (X) is folded:
Directions : (46 to 48) In each of the following questions,
select the correct option for the question asked.
53.
(i)
46.
47.
48.
(X)
(ii)
Which number will come opposite to number 2?
(A) 5
(B) 1
(C) 6
(D) 3
Which number will come opposite to number 6?
(A) 1
(B) 5
(C) 4
(D) 3
(A)
(B)
(C)
(D)
Which number will come opposite to number 4?
(A) 3
(B) 5
(C) 1
(D) 2
+
49.
On the basis of two figures of dice, you have to tell what 54.
number will be on the opposite face of number 5?
(i)
(ii)
(A) 1
(C) 4
(B) 2
(D) 6
(X)
(A)
(C)
(B)
+
+
(D)
PAGE # 96
55.
59.
(X)
(i)
(A)
(C)
(ii)
(B)
(iii)
(iv)
Which number is opposite to number 5?
(A) 6
(B) 5
(C) 1
(D) 3
(D)
Directions : (60 to 64) Choose the cube from the options
that will unfold to give the figure on the left
Direction : (56) The six faces of a cube have been marked
with numbers 1, 2, 3, 4, 5 and 6 respectively. This
cube is rolled down three times. The three
60.
positions are given. Choose the figure that will be
formed when the cube is unfolded.
X
M
56.
M
X
(A)
(A)
M
X
M
(B)
(C)
(D)
(E)
(B)
1
4
8
3
61.
7
9
(C)
(D)
9
7
8
1
57.
Which number is opposite 3 in a standard dice
given below ?
(A) 1
(C) 5
58.
(B) 4
(D) Can’t be determined
(A)
4
1
7
(B)
(C)
7
(D)
8
8
7 4
(E)
62.
8
D
Which number is opposite 4 ?
(A) 5
(C) 2
(B) 3
(D) 1
Directions : (59) In the following question four positions of
the same dice have been shown. You have to see
these figures and select the number opposite to
the number as asked in each question.
8
8
(A)
(B)
D
(C)
(D)
(E)
(D)
(E)
63.
B
B
(A)
(B)
(C)
PAGE # 97
66.
Which number/letter is opposite 2 ?
J
3
64.
J
(A)
(B)
(A) A
(C) 1
J
(C)
(D)
Which letter is opposite Q ?
Q
O P L
N
M
(A) L
(C) N
(B) C
(D) 3
(E)
Directions : (65 to 68) In each of the following questions, a 67.
diagram has been given which can be folded into
a cube. The entries given in the squares indicate
the entries on the face of the cube. In each question
a number or a letter has been given . Of the four
alternatives given below it, you have to find the one
that would appear on the face opposite to it in the
cube.
65.
I C
A
B
2
68.
Which number/letter is opposite O?
L
N M 2
I O
(A) L
(C) N
(B) M
(D) 2
Which letter is opposite R?
Q R
S P
U T
(B) M
(D) P
(A) P
(C) T
(B) S
(D) U

PAGE # 98
ANSWER KEY
ELECTRICITY(PHYSICS)
Que.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Ans.
A
A
C
C
C
B
A
D
A
B
D
C
A
B
B
Que.
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Ans.
B
A
A
A
D
D
B
C
B
C
B
C
B
B
A
Que.
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
Ans.
A
D
B
C
D
B
C
D
A
B
D
C
ABCD CD
BD
Que.
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
C
D
C
B
B
C
B
C
C
B
B
D
C
B
62
63
64
65
66
67
68
69
70
71
72
73
C
B
B
D
A
A
B
A
A
C
D
D
14
C
29
A
44
D
59
A
74
B
89
D
15
B
30
D
45
A
60
D
75
D
90
D
Ans. ABD
Que. 61
Ans.
B
MOLE CONCEPT(CHEMISTRY)
Ques.
Ans.
Ques.
Ans.
Ques.
Ans.
Ques.
Ans.
Ques.
Ans.
Ques.
Ans.
1
B
16
A
31
D
46
B
61
C
76
A
2
C
17
A
32
C
47
B
62
D
77
A
3
D
18
A
33
C
48
C
63
A
78
C
4
B
19
A
34
B
49
C
64
B
79
A
5
C
20
A
35
C
50
C
65
A
80
A
6
B
21
B
36
B
51
C
66
B
81
D
7
A
22
A
37
C
52
B
67
C
82
B
8
D
23
A
38
D
53
C
68
A
83
D
9
A
24
D
39
A
54
B
69
D
84
B
10
B
25
A
40
B
55
A
70
B
85
C
11
B
26
A
41
C
56
A
71
C
86
B
12
D
27
A
42
A
57
D
72
B
87
C
13
B
28
D
43
C
58
C
73
C
88
C
NUMBER SYSTEM(MATHEMATICS)
Q.
Ans.
Q.
Ans.
Q.
Ans.
Q.
Ans.
Q.
Ans.
Q.
Ans.
Q.
Ans.
Q.
Ans.
Q.
Ans.
Q.
Ans.
1
B
11
A
21
C
31
B
41
C
51
C
61
B
71
C
81
B
91
A&D
2
B
12
B
22
B
32
A
42
B
52
C
62
B
72
A
82
B
92
C&D
3
A
13
A
23
B
33
A
43
A
53
B
63
D
73
C
83
D
93
A
4
D
14
A
24
A
34
B
44
A
54
D
64
D
74
C
84
C
94
D
5
C
15
B
25
D
35
D
45
D
55
C
65
D
75
B
85
D
95
B
6
B
16
C
26
B
36
D
46
A
56
C
66
A
76
B
86
D
96
C
7
A
17
C
27
D
37
C
47
D
57
C
67
A
77
B
87
A
8
A
18
B
28
B
38
C
48
D
58
C
68
B
78
A
88
B
9
A
19
C
29
C
39
D
49
A
59
C
69
A
79
D
89
C
10
D
20
A
30
C
40
C
50
A
60
A
70
D
80
C
90
D
PAGE # 9999
LOGARITHM (MATHEMATICS)
1
Q.
2
3
4
5
6
7
8
9
10
Ans.
B
A
B
D
B
C
B
A
C
D
Q.
11
12
13
14
15
16
17
18
19
20
Ans.
A
A
B
B
D
D
B
C
B
D
Q.
21
22
23
24
25
26
27
28
29
30
Ans.
A
A
B
B
A
A
D
A
A
B
Q.
31
32
33
34
35
36
37
38
39
40
Ans.
B
B
A
B
B
A
D
B
B
C
NUTRITION(BIOLOGY)
Que s.
1
2
3
4
5
6
7
8
9
10
Ans.
C
C
B
D
D
A
C
B
A
A
Que s.
11
12
13
14
15
16
17
18
19
20
Ans.
Que s.
Ans.
B
21
B
C
22
C
A
23
A
C
A
B
D
A
D
C
SERIES COMPLETION(MENTAL ABILITY)
EXERCISE-1 (Number Series)
Que.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Ans.
C
D
D
A
C
D
B
C
C
C
D
C
C
B
D
Que.
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Ans.
B
C
C
C
B
A
B
D
D
A
C
B
B
C
A
Que.
31
32
33
34
35
Ans.
C
C
C
D
D
EXERCISE- 2 (Alphabet Series)
Que.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Ans.
D
A
D
C
C
A
D
C
D
B
D
C
C
C
D
Que.
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Ans.
C
A
B
C
C
C
A
B
A
C
D
B
C
D
B
EXERCISE- 3 (Letter Repeating Series)
Que.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
D
C
B
D
C
B
25
26
Ans.
D
D
A
A
C
B
A
C
D
Que.
16
17
18
19
20
21
22
23
24
Ans.
C
A
A
A
C
D
D
D
A
B
D
PAGE # 100
100
EXERCISE- 4 (Missing Term In Figure)
Que .
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Ans .
B
D
B
D
C
C
C
D
A
D
D
B
C
A
B
Que .
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
A
C
D
B
A
Ans .
B
C
A
B
B
A
C
A
D
D
Que .
31
32
33
34
35
36
37
38
39
40
Ans .
B
B
A
C
A
C
B
C
D
B
PUZZLE-TEST(MENTAL ABILITY)
Que.
Ans.
Que.
Ans.
Que.
Ans.
Que.
Ans.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
A
B
D
C
A
D
C
C
C
C
C
D
C
C
B
16
17
18
19
20
21
22
23
24
25
26
C
C
D
D
C
D
D
B
B
C
A
31
32
33
34
35
36
37
38
39
40
41
27
A
42
28
D
43
29
C
44
30
D
45
D
C
B
A
B
A
D
A
D
C
D
A
C
B
D
46
47
48
49
50
51
52
53
54
55
56
C
A
D
B
D
D
D
A
C
A
A
57
D
58
B
CALENDAR AND CLOCK-TEST(MENTAL ABILITY)
Que.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Ans.
D
16
B
31
D
C
17
B
32
B
B
18
B
33
D
D
19
B
34
B
D
20
C
35
A
B
21
B
36
A
C
22
C
37
C
B
23
D
38
A
C
24
C
39
D
C
25
B
40
A
A
26
D
41
C
B
27
C
C
28
B
B
29
B
A
30
D
Que.
Ans.
Que.
Ans.
CUBE AND DICE TEST(MENTAL ABILITY)
Que.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Ans.
D
C
C
D
A
C
C
D
B
B
C
A
C
D
B
Que.
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Ans.
C
D
C
B
A
B
D
D
A
D
C
A
D
C
B
Que.
Ans.
Que.
Ans.
Que.
Ans.
31
32
33
34
35
36
37
38
39
40
41
B
56
C
42
B
57
B
43
A
58
A
44
D
59
C
45
B
60
C
C
D
C
D
C
A
D
C
B
C
46
D
61
A
47
A
62
D
48
B
63
E
49
C
64
D
50
D
65
C
51
A
66
A
52
B
67
B
53
D
68
B
54
B
55
D
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