Ganga Academy
[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
CBSE PATTERN SPECIALIST SINCE 2002
SUMMATIVE ASESSMENT- I
CBSE BOARD SAMPLE
20012 RESULTS X & XII CBSE
CLASS X
PAPERS 2012- 13 2012:
CLASS X
1) ASMITA A GAIND
ENGLISH COMM.:
10 ,
SCIENCE: 10
MATHEMATICS :
09
CGPA 9.8
2)RAJ HANMANT KATKAR
ENGLISH COMM.:
10
SCIENCE :10
MATHEMATICS :
09
CGPA 9.8
3)Hemang Pancholi
MATHEMATICS:
10
CLASS XII (Sci)
1) KIRAN ASHOK SALUNKE
PHYSICS : 088 ; CHEMISTRY :086 ; BIOLOGY: 090
2)
Total:85.2%
PRABODHINI .B. BHUMKAR
MATHEMATICS : 077 ; PHYSICS : 094 ;CHEMISTRY : 084
Total:84.8%
3)INDERJEET SINGH
PHYSICS: 84 ;Total:83.2%
! Thax. to all dedicated
OVER ALL 100% Result
teacher
SYLLABUS MATHEMATICS X SAI
I. NUMBER SYSTEMS 11 ;II. ALGEBRA 23 ;
III. G EOMETRY 17 ;IV TRIGONOMETRY 22 ; V STATISTICS 17 TOTAL 90
BY;CBSE EXPERT AJAY KUMAR MISHRA
Note : This booklet contains mathematics Important formulae on next page.
Algebra
1
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
SOLUTIONS MATHEMATICS 2012-13
SAMPLE PAPER –I
Solutions:
1]
2]
3]
One decimal place
An odd integer
Since -3 is the root of quadratic polynomial, we have
(K - 1)(-3)2 + 1 = 0
4]
5]
6]
Median
x = 3 sec2
- 1, y = tan2
x - 3y = 3sec2
= 3(sec2
-2
- 1 - 3tan2
+6
- tan2 ) + 5
=3+5
=8
7]
cos
+ cos2
sin2
+ sin4
= cos
=1
=
1 - cos2
= cos
+ (1 - cos2
)2
+ cos2
=1
8]
2
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9]
Since, 870 = 225
225 = 195
195 = 30
30 = 15
[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
3 + 195
1 + 30
6 + 15
2+0
HCF (870,225) = 15
10]
x + y = 3 -x + y = 1
OR
Subtracting equation (1) from 3x -
= 5, we get
From equation (1), x = 3
11]
are roots of x2 - (k + 6)x + 2(2k - 1)
12]
From
3
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1+
3=
[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
1+
2
3=
2
13]
Cot
=
=
(given)
=
=
14]
C.I
f
c.f.
135 - 140
4
4
Here, n = 40
140 - 145
7
11
Median class is 145 - 150
145 - 150
11
22
150 - 155
6
28
Also, since highest frequency is 11, Modal
class is 145 - 150.
155 - 160
7
35
160 - 165
5
40
15]
To prove 5 +
is irrational, let us assume 5 +
is rational.
We can find integers a and b where a, b are co-prime, b
4
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
Such that,
Now a, b are integers,
is rational.
is rational.
Which is a contradiction. So 5 +
is irrational.
OR
Let us assume to the contrary, that
is a rational number.
is rational.
(n-1)+(n+1)-2
2n+2
is rational
is rational
But we know that
So 2n+2
is an irrational number
is also an irrational number
So our basic assumption that the given number is rational is wrong.
Hence,
16]
is an irrational number.
f(x) = x2 - 2x + 1
Zeroes of f(x) are , 
Sum of zeroes  +  = 2 and . = 1
Required polynomial = k(x2 - 4x + 4), where k is any integer.
17]
Let the length and breadth of the rectangle be x and y respectively.
So the original area of the rectangle=xy
5
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
According to question,
(x+2)(y-2)=xy-28
i.e. xy-2x+2y-4=xy-28
2x-2y=24 ...(i)
Next, (x-1)(y+2)=xy+33
i.e. xy+2x-y-2=xy+33
2x-y=35 ..(ii)
Now we need to solve (i) and (ii)
From (ii) we get,
y=2x-35
substituting this value in (i) we get,
2x-4x+70=24
-2x=-46
x=23
substituting this value in (ii)
we get,
y=11
So the length and breadth of the rectangle are 23 metres and 11
metres respectively.
OR
Let 40 % acids in the solution be x litres
Let 60 % of other solution be y litres
Total Volume in the mixture = x + y
Given volume is 10 litres
x + y = 10 ---(i)
Also,
So, 40 x + 60 y =500 or 2x+3y =25 …(ii)
6
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
Solving (i) and (ii) we get x = y = 5 litres
18]
The system has infinitely many solution
Equating (1) and (2), we get a = 5b
Equating (2) and (3), we get 2a - 4b = 6 On solving, we get b = 1 and a = 5.
19]
By BPT
20]
To prove AB² = AC² = 2AD² +
Draw AE  BC
In ?ABD since D > 90°
7
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
 AB² = AD² + BD² + 2BD x DE …(1) (using Obtuse angle property) ?ACD = since D < 90°
AC² = AD² + DC² - 2DC x DE …(2) (using acute angle property)
Adding (1) and (2)
AB² + AC² = 2(AD² + BD²)
=2(AD² +
)
Or AB² + AC² = 2(AD² + BD²)
Hence proved.
21]
Given:
L.H.S. = (m² + n²) cos²
OR
8
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
22]
, we have
23]
CI
5060
6070
7080
80-90 90100
100110
Total
fi
5
3
4
p
2
13
27+p
xi
55
65
75
85
95
105
fi xi
275
195
300
85p
190
1365
Mean
=
2325+85p
Subs
tituti
ng
the values we get
86 =
86p + 2322 = 2325 + 85p
p = 3
24]
Age in
yrs. (
more than
or equal
to)
010
1020
2030
3040
4050
5060
6070
No. of
persons
(fi)
10
15
25
22
13
10
5
modal class = 20 - 30
Since the
maximum
frequency
is 25 and it
lies in the
class
interval 2030.
Therefore,
??= 20, h = 10, f0 = 15, f1 = 25, f2 = 22
mode = ??+
= 20 +
= 20+7.69 = 27.69 years (approx.)
25]
If the number 15n where n N, were to end with a zero, then its prime
factorisation must have 2 and 5 as its factors.But 15=5  3
9
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
15n = (5  3)n = 5n  3n
So Prime factors of 15n includes only 5 but not 2
Also from the Fundamental theorem of Arithmetic, the prime
factorisation of a number is unique.
Hence a number of the form 15n where n
zero.
26]
N, will never end with a
To solve the equations, make the table corresponding to each
equation.
x
?1
?2
?3
y
4
2
0
x
4
?1
y
0
4
Now plot the points and draw the graph.
Since the lines intersect at the point (?1, 4), so x = ?1 and y = 4 be
the solution.
10
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
Also by observation vertices of triangle formed by lines and x-axis are
A (?1, 4),
B (?3, 0) and C (4, 0).
27]
Let p(x) = x3 - 6x2 - 15x + 80
Let say that we subtracted ax + b so that it is exactly divisible by x2 +
x - 12
s(x) = x3 - 6x2 - 15x + 80 - (ax + b)
= x3 - 6x2 - (15 + a)x + (80 - b)
Dividend = Divisor x Quotient + Remainder
But remainder = 0
 Dividend = Divisor x Quotient
s(x) = (x2 + x -12) x quotient
s(x) = x3 - 6x2 - (15 + a)x + (80 - b)
x (x2 + x - 12) - 7(x2 + x - 12)
= x3 + x2 - 7x2 - 12x - 7x + 84
= x3 - 6x2 - 19x + 84
Hence, x3 - 6x2 - 19x + 84 = x3 - 6x2 - (15 + a)x + (80 - b)
-15 - a = - 19  a = +4
and 80 - b = 84  b = -4
Hence if in p(x) we subtracted 4x - 4 then it is exactly divisible by
x2 + x -12.
28]
AD is the median of ABC since D is mid-point of BC
BD = DC=
11
....(i)
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
In right triangle AEB,
...Pythagoras theorem
=(
)+
Using Pythagoras theorem for right triangle AED and BE=BD-DE
=
+(
AB2=
....from (i)
+
+
-2
-BC x DE +
Hence proved.
OR
Given: A right triangle ABC right angled at B.
To prove: that AC2 = AB2 + BC2
Construction:Let us draw BD
AC (See fig.)
Proof :
Now,
So,
ADB
ABC (Using Theorem:If a perpendicular is drawn
from the vertex of the right angle of a right triangle to the
hypotenuse ,then triangles on both sides of the perpendicular
are similar to the whole triangle and to each other)
(Sides are proportional)
Or, AD.AC = AB2
12
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Also,
BDC
[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
ABC (Theorem)
So,
Or, CD. AC = BC2
Adding (1) and (2),
AD. AC + CD. AC = AB2 + BC2
OR, AC (AD + CD) = AB2 + BC2
OR, AC.AC = AB2 + BC2
OR AC2 = AB2 + BC2
Hence Proved.
29]
ABC
ADE (by AA Similarity)
In right
AB2 = BC2 + AC2 (by PT)
AB2 = 52 + 122 = 25 + 144 = 169
AB = 13
Subsisting AB = 13 cm, BC = 12 cm and AC = 5 cm in (1) and
Getting DE =
and AE =
30]
To prove:
Using the identity cosec2 A = 1 + cot2A
L.H.S =
13
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
= cosecA + cotA
= R.H.S
OR
=RHS
14
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
31]
LHS =
= sin = RHS
32]
Hence, LHS = RHS.
33]
We can find frequency distribution table of less than type as following Daily income
(in Rs)
Cumulative
frequency
(upper class
limits)
15
Less than
120
12
Less than
140
12 + 14 = 26
Less than
160
26 + 8 = 34
Less than
180
34 + 6 = 40
Less than
200
40 + 10 = 50
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
Now taking upper class limits of class intervals on x-axis and their
respective frequencies on y-axis we can draw its ogive as following -
34]
Class Interval
0-6
6-12
12-18 18-24
24-30
Frequency
4
x
5
y
1
Cumulative
frequency
4
4+x
9+x
9+x+y 10+x+y
It is given
that total
frequency
N is 20
So,
10+x+y = 20 i.e. x + y = 10 ….(i)
Given 50% of the observations are greater than 14.4.
So median = 14.4, which lies in the class interval 12-18.
= 12, cf = 4 + x, h = 6, f = 5, N = 20
Median =
14.4 = 12 +
14.4 - 12 =
x6
x6
=6-x
x =4
Now using equation, 10+x+y = 20, we get y = 6.
Hence x =4 and y = 6.
16
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
SAMPLE PAPER –II
Solutions:
1]
tan B
2]
3]
tan c =
Number of zeros is one as the graph touches the x-axis at one point.
3Median = Mode + 2 mean
Median =
4]
=1
=
=
Since
Therefore, lines are parallel.
5]
BC =
17
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
6]
Therefore, PN = 4.8 cm
7]
8]
Let
Then, x=0.7777…… … (1) Here ,the number of digits recurring is only
1,so we multiply both sides of the equation by 10.
… (2)
Subtracting(1) from(2),we get
9x=7
9]
-3 and -4 are zeroes of the polynomial
Sum of zeros = -3 - 4 = -7=
Product of zeros = (-3) (-4) =
10]
=
Let it possible 6n ends with digit 0
6n = 10
(2
3)n = 2
2n
3n = 2
q
5
5
q
q
5 is a prime factor of 2n
3n
Which is not possible 2n
3n can have only 2 and 3 are prime factors.
18
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
Hence, it is not possible the number ends with digit 5.
11]
Marks
f
cf
0 - 10
5
5
10 - 30
15
20
30 - 60
30
50
60 - 80
8
58
80 - 100
2
6
N=
Here, N=60 So, N/2=30
The cumulative frequency is just greater than N/2=30 is 50 and the
corresponding class is 30-60.
Hence,30-60 is the median class.
Therefore, l=30,f=30 ,F=20,h=30
Now, Median=
Median = 40
12]
Condition for infinitely many solution
(½)
similarly,
k + 2 = 4 3k + 2 = 8
k=2k=2
k = 2 is the common solution.
13]
Sin(45 + 30) = sin 45o cos 30o + cos 45o sin 30o
Sin 75o =
19
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
=
=
OR
Cosec =
AB =
Cot
+ tan
=
=
=
14]
In
ABD and
D=
F = 90o
B=
C
By AA similarity,
15]
ECF
ABD
ECF
Let Rekha's Age be 'x'years
And her mother's age be 'y' years
y = 5x as per given data … (1)
20
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
After 5 years
y + 5 = 3(x+5)
y - 3x = 10 … (2)
Solving (1) and (2) equation.
Rekha's age = 5 years
Mother's age = 25 years
OR
Let the two number be 5x, 6x
Two numbers are 40,48.
16]
Sin
17]
=
tan
=
a=12 a+b=25
b=13 c=25+10=35 c+d=43
35=8 43+e=48
e=5
d=43-
f=48+2=50
18]
To calculate the mean, first obtain the column of mid value and then
multiply the corresponding values of frequency and mid value.
C.I.
21
f
Mid
value
fx
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
(x)
0-100
2
50
100
100200
3
150
450
200300
5
250
1250
300400
2
350
700
400500
3
450
1350
15
Here
and
3850
, so the mean is given as
.
19]
=
=
=
Simplifying we get, 1 + sec
. cosec
OR
22
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20]
ADE
[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
ABC by AA similarity
…(1)
21]
Given p(x) = 6x3 + 3x2 - 5x + 1
a = 6, b = 3, c = -5 d = 1
,r are zero.
=
=
22]
Let
then
23
be rational and equal to
where a and b are co primes,
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
here a,b are integers
is rational .Therefore,
contradiction
6+
is rational
is rational which is a
is an irrational number
OR
Let 5 -
be rational equal to
Then 5 -
=
a,b are integers
is rational which is a contradiction
Hence 5 -
is an irrational number
23]
In fig,
PQD,
Using Pythagoras thm.
PD2 = a2 - c2 …(1)
24
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Similarly in
PD2 = b2 - d2 …(2)
From (1) and (2) a2 - c2 = b2 - d2
a2 - b2 = c2 - d2
(a + b) (a - b) = (c + d) (c - d)
24]
Multiply equation (1) by 3 and add in equation (2), we get
Using equation (1),
y-2=3
y=5
Hence x = 4, y = 5.
25]
Let 5q + 2, 5q + 3 be any positive integers
(5q + 2)2 = 25q2 + 20q + 4
= 5q (5q + 4) + 4 is not of the form 5q + 2
Similarly for 2nd
(5q + 3)2 = 25q2 + 30q + 9
=5q(5q+6)+ 9 is not of the form 5q+3
So, the square of any positive integer cannot be of the form5q+2 or 5q+3
For any integer q
26]
Statement If a line is drawn parallel of one side of a triangle to intersect the
other two sides in distinct points, the other two sides are divided in the same
ration.
25
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
Given:A triangle ABC in which a line parallel to side BC intersects other
two sides AB and AC at D and E respectively (see fig.)
To prove that
Construction:Let us join BE and CD and then draw DM
AB.
AC and EN
Proof:Now, area of
Note that BDE
and DEC are on
the same base
DE and between
the same
parallels BC and
DE.
So, ar(BDE) =
ar(DEG)
we have:
Therefore, from
(1), (2) and (3),
OR
Statement: The
ratio of the areas of two similar triangles is equal to the square of the
26
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
ratio of their corresponding sides.
Given: ?ABC ~ ?PQR To Prove:
Construction: Draw
AD?BC and PS?QR Proof:
?ADB
~ ?PSQ (AA) Therefore,
… (iii) But ?ABC
~ ?PQR Therefore,
… (iv) Therefore,
Therefore,
27]
2 sec2
From (iii)
- sec4
- 2sec2
+ cosec4
= 2 (1 + tan2 ) - (1 + tan2 )2 - 2(1 + cot2 ) + (1 + cot2 )2
= (1 + tan2 ) (1 - tan2 ) - (1 + cot2 ) (1 - cot2 )
= 1 - tan4
-1+cot4
= cot4 - tan4
28]
We have,
x-y=1
2x+y=8
Graph of the equation x-y=1:
We have,
x-y=1 =>y=x-1 and x=y+1
Putting x=0,we get y=-1
Putting y=0,we get x=1
Thus, we have the following table for the points on the line x-y=1:
x
0
1
y
-1
0
Plotting points A(0,-1),B(1,0) on the graph paper and drawing a line passing
through them,we obtain the graph of the line represented by the equation xy=1 as shown in fig.
27
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
Graph of eqn 2x+y=8:
We have,
2x+y=8 =>y=8-2x
Putting x=0,we get y=8
Putting y=0,we get x=4
Thus,we have the following table giving two points on the line represented by
the equation 2x+y=8.
x
0
4
y
8
0
Plotting points C(0,8) and D(4,0) on the same graph paper and rawing a line
passing through them, we obtain the graph of the line represented by the
equation 2x+y=8 as shown in fig.
Clearly ,the 2 line intersect at P(3,2).The area bounded by these 2
lines and y-axis is shaded in the given fig.
29]
28
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
=
=
OR
30]
Marks
Frequency
25 - 35
5
35 - 45
10
45 - 55
20
55 - 65
9
65 - 75
6
75 - 85
2
Total
52
Here the maximum frequency is 20 and the corresponding class is 4555.So,45-55 is the modal class.
We have,l=45,h=10,f=20,
Mode =
+
Mode=49.7
31]
Let p(x) = x3 + 2x2 + kx + 3
Then using Remainder theorem
p(3) = 33 + 2
32 + 3k + 3 = 21
k = -9
29
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
Quotient of p(x) is x2 + 5x + 6
Hence, x3 + 2x2 - 9x + 3 = (x2 + 5x + 6) (x - 3) + 21
x3 + 2x2 -9x - 18 = (x - 3) (x + 2) (x + 3)
All the zeros of p(x) are 3,-2,-3.
32]
Given: ABC is a right angled triangle, B = 900
To prove: AB2 + BC2 = AC2
Construction: Drop a perpendicular BD on the side AC.
Proof: From triangle ADB and triangle ABC,
We can re-write as, AC × AD = AB2
Also, triangle BDC is similar to triangle ABC.
Equating the proportional sides of the similar triangles BDC and ABC,
AC × CD = BC2
Now adding this to the equation that we had obtained,
AC × AD + AC × CD = AB2 + BC2
 AC × (AD + CD) = AB2 + BC2
30
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
 AC × AC = AB2 + BC2
 AC2 = AB2 + BC2
33]
LHS=
= Cos A + sin A = RHS
34]
We first prepare the cumulative frequency distribution table by less
than method as given below:
Marks no. of students marks less than cumulative frequency
0-10 7 10 7
10-20 10 20 17
20-30 23 30 40
30-40 51 40 91
40-50 6 50 97
50-60 2 60 100
Other than the given class intervals ,we assume a class-10-0 before
the first class interval 0-10 with zero frequency.
Now, we mark the upper class limits along X-axis on a suitable scale and the
cumulative frequencies along Y-axis on a suitable scale.
Thus, we plot the
points(0,0),(10,7),(20,17),(30,40),(40,91),(50,97)and(60,100).
Now, we join the plotted points by a free hand curve to obtain the required
ogive.
31
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
SAMPLE PAPER –III
Solutions:
1]
2]
3]
4]
5]
6]
Zeroes of a polynomial are the x- coordinates of the points where its
graph crosses or touches the X- axis. Graph of y = p(x) intersects the
X-axis at 4 points. Therefore, the polynomial p(x) has 4 zeroes
Range is not a measure of central tendency.
7×11×13+13 +13 x 2 = 13(7×11 + 1+2) = 13(80)
Smallest composite number= 4 and the smallest prime
number=2. HCF of 2 and 4 = 2.
For the system of linear equations: a1x + b1y + c1 = 0, and
a2x + b2y + c2 = 0 to have no solution
For the system of equations 3x + y = 1,
(2K - 1) x + (K - 1) y = 2K + 1 a1 =3, b1 = 1, c1 = 1, and
32
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
a2 = 2K-1 b2 = K-1 c2 = 2K+1
This gives K=2.
7]
Given that 3cot A=4, we have, cot A=
Then, tan A=
Now, (1-tan2A)= 1-
=
And, ( 1+tan2A) = 1+
=
THerefore,
8]
tan A=cot B
tan A=tan(900-B)
A=900-B or A+B=900
9]
In
ADC, AD2 = AC2 - CD2
In
ABD, AD2 = AB2 - DB2
AB2 - B2 = AC2 - CD2
AB2 + CD2 = AC2 + BD2
10]
11]
120 = 23
3
5
144 = 24
32
LCM = 24
32
HCF = 23
3 = 24
5 = 720
3x + 4y = 10
4x - 4y = 4
x=2
33
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x-y=1
[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
y=x-1=2-1=1
The solution x = 2, y = 1
12]
We know that sin2 + cos2 =1 Taking cube of both sides (sin2 +
cos2 )3 =13
(sin2 )3 + (cos2)3 + 3 sin2 .cos2 (sin2 + cos2)= 1
Therefore, sin6 + cos6 + 3sin2cos2=1
13]
14]
Marks
No. of students
0 - 10
2
10 - 20
7
20 - 30
6
30 - 40
5
40 - 50
10
Total
30
Let p(x) = x3 + ax2 + b x + 16
g(x) = x - 2 is a factor of p(x)
 p(2) = 0
so, p(2) = 8 + 4a + 2b + 16 = 0
= 4a + 2b = -24
= 2a + b = -12 (i)
Also a -b = 16 (ii)
Solving (i) and (ii)
3a = 4  a =
OR
Given polynomial p(x) = ax2 - 5x + c
Sum of zeroes m + n =
Product of zeroes mn =
34
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
Given: m + n = mn = 10
15]
Modal Class = 20-30 ? = 20, f0 = 12, f1 = 32, f2 = 20, h = 10 Mode
=
Mode =
= 26.25
16]
Construction - draw AE
BC
In right triangle AEB and AEC
AB2 + AC2 = BC2 + AE2 + EC2 + AC2 = 2AE2 + (BD - ED)2 + (ED +
DC)2
= 2AE2 + 2ED2 + BD2 + DC2
AB2 + AC2 = 2AE2 + 2ED2 + 2BD2
= 2 [AE2 + ED2] + 2BD2
= 2 (AD2 + BD2)
17]
LHS =
35
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=
[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
= 2 cosec A = RHS
18]
Let the fraction be:
According to the question
x+y=8
On solving we get x = 3, y = 5
The fraction is
OR
Let the tens and units digits of the number be x and y respectively then
7(10x + y) = 4 (10y + x)
70x + 7y = 40y + 4x
66x = 33y
y = 2x
Also, y - x = 3
On solving, x = 3, y = 6
Number = 36.
19]
Let
be a rational number
Let
=
=p
11q2 = p2
11 divides p2 hence 11 divides P.
36
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
Let p = 11c
11q2 = 121 c2
Or q2 = 11c2
11 divides q2 Hence 11 divides q
From (1) and (2) p and q have a common factor 11 which contradicts our
assumption.
is irrational
OR
Let 2
- 7 is rational
=
Since p and q are integer
is rational
is rational
But we have that
is irrational
our assumption is wrong. Hence 2
- 7 is irrational.
20]
Class
Interval
Frequency
0 - 20
7
7
20 - 40
8
15
40 - 60
12
27
60 - 80
10
37
80 - 100
8
45
100 - 120
5
50
Total
50
Median class 40 - 60
37
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
= 40, f = 12 CF = 15 h = 20
Median =
+
= 40 +
= 40 +
= 40 +
= 56.7
OR
Age (in
Years)
No. of
people fi
Value
xi
fixi
0 - 20
15
10
150
20 - 40
f1
30
30f1
40 - 60
21
50
1050
60 - 80
f2
70
70f2
80 - 100
17
90
1530
Total
100
2730 + 30f1 +
70 f2
53 + f1 + f2 = 100
f1 + f2 = 47 …(1)
53 =
5300 - 2730 = 30f1 + 70f2
30f1 + 70f2 = 2570
Or 3f1 + 7f2 = 257 …(2)
f1 + f2 = 47
f2 = 47 - 29 = 18
38
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21]
In
ABC and
BAC =
C=
[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
DAC
ADC
C
22]
LHS =
= 1 + cos A
=
=
23]
OR
p(x) = x3 - 3x2 + x + 2
g(x) = ?
q(x) = x - 2 r(x) = -2x + 4
p(x) = g(x) + r(x)
x3 - 3x2 + x + 2 = g(x) (x - 2) - 2x + 4
g(x) =
=
= x2 - x + 1
24]
The system has infinitely many solutions:
39
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
Equating (1) and (2), we get a = 5b
Equating (2) and (3), we get 2a - 4b = 6
On solving, we get b = 1 and a = 5.
25]
If a and b are one two positive integers. Then a = bq + r, 0
r
b
Let b = 3 Therefore, r = 0, 1, 2 Therefore, a = 3q or a = 3q + 1 or a =
3q + 2 If a = 3q a2 = 9q2 = 3(3q2) = 3m or where m = 3q2 a = 3q + 1
a2 = 9q2 + 6q + 1 = 3(3q2+ 2q) + 1 = 3m + 1 where m = 3q2 + 2q or
a = 3q + 2 a2 = 9q2 + 12q + 4 = 3(3q2 + 4q + 1) + 1
= 3m + 1, where m 3q2 + 4q + 1
Therefore, the squares of any positive integer is either of the form 3m
or 3m + 1.
26]
CI
1-4
4-7
7-10
10-13 13-16 16-19
fi
6
30
40
16
4
4
xi
2.5
5.5
8.5
11.5
14.5
17.5
Average most suitable here is the Mode because we are interested in
knowing the length of surname for maximum no. of people
Since the maximum frequency is 40 and it lies in the class interval 710.
Therefore, modal class = 7-10
?= 7, h=3, f0=30 , f1=40 , f2=16
Mode =
= 7+
= 7 +.88 = 7.88 years(approx.)
40
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
27]
LHS =
= tan? + cot? + 1 = RHS
OR
LHS=(cosec A - sin A) (sec A - cos A) (tan A + cot A)
=
=1
= RHS
28]
Basic proportionality theorem
Statement If a line is drawn parallel of one side of a triangle to intersect the
other two sides in distinct points, the other two sides are divided in the same
ratio.
Given:A triangle ABC in which a line parallel to side BC intersects other
two sides AB and AC at D and E respectively (see fig.)
To prove that
Construction: Let us join BE and CD and then draw DM
AC and EN
AB.
Proof:
41
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
Now, area of
Note that BDE and DEC are on the same base DE and between the
same parallels BC and DE.
So, ar(BDE) = ar(DEG)
Therefore, from (1), (2) and (3), we have :
Hence proved.
OR
Pythagoras Theorem : Statement:In a right angled triangle,the square
of the hypotenuse is equal to the sum of squares of the other two
sides.
Given: A right triangle ABC right angled at B.
To prove: that AC2 = AB2 + BC2
Construction:Let us draw BD
AC (See fig.)
Proof :
Now,
ADB
ABC (Using Theorem: If a perpendicular is drawn
from the vertex of the right angle of a right triangle to the
hypotenuse ,then triangles on both sides of the perpendicular
are similar to the whole triangle and to each other)
So,
(Sides are proportional)
Or, AD.AC = AB2 (1)
Also,
BDC
ABC (Theorem)
So,
Or, CD. AC = BC2
42
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
Adding (1) and (2),
AD. AC + CD. AC = AB2 + BC2
OR, AC (AD + CD) = AB2 + BC2
OR, AC.AC = AB2 + BC2
OR, AC2 = AB2 + BC2
Hence proved.
29]
m2 - n2 = (m + n) (m - n)
=
= 2 tan
. 2 sin
= 4 tan
. sin
=4
=4
=4
=4
=4
30]
Let p(x) = x3 - 6x2 - 15x + 80
Let say that we subtracted ax + b so that it is exactly divisible by
x2 + x - 12
s(x) = x3 - 6x2 - 15x + 80 - (ax + b)
= x3 - 6x2 - (15 + a)x + (80 - b)
Dividend = Divisor x Quotient + Remainder
But remainder = 0
 Dividend = Divisor x Quotient
s(x) = (x2 + x -12) x quotient
s(x) = x3 - 6x2 - (15 + a)x + (80 - b)
43
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
x (x2 + x - 12) - 7(x2 + x - 12)
= x3 + x2 - 7x2 - 12x - 7x + 84
= x3 - 6x2 - 19x + 84
Hence, x3 - 6x2 - 19x + 84 = x3 - 6x2 - (15 + a)x + (80 - b)
-15 - a = - 19  a = +4
and 80 - b = 84  b = -4
Hence if in p(x) we subtracted 4x - 4 then it is exactly divisible by
x2 + x -12.
31]
We have to solve the pair of equations graphically
2x + 3y = 11 … (1)
2x - 4y = -24 … (2)
For (1)
X
1
4
-2
y
3
1
5
For (2)
X
-12
0
-10
y
0
6
1
point of intersection x = -2, = 5
The triangle formed is shaded as
ABC coordinates are
A (-2,5) B (-12,0) C(5.5,0).
44
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
32]
Class Interval
Frequency
0 - 15
6
15 - 30
7
30 - 45
f1
45 - 60
15
60 - 75
10
75 - 90
f2
Total
51
Mode = 55 (Given)
Modal Class 45 - 60
= 45, fo = f1 and f1 = 15
f2 = 10 h = 15
38 + f1 + f2 = 51
f1 + f2 = 51 - 38
f1 + f2 = 13 …(1)
55 = 45 +
10 =
200 - f1 = 225 - 15f1
45
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
5f1 = 25
f1 = 5
f1+f2=13
f2=13-5=8
The missing frequencies are 5 and 8.
33]
Statement: The
ratio of the areas of two similar triangles is equal to the square of the
ratio of their corresponding sides.
Given: ?ABC ~ ?PQR To Prove:
Construction: Draw
AD?BC and PS?QR Proof:
~ ?PSQ (AA) Therefore,
~ ?PQR Therefore,
Therefore,
?ADB
… (iii) But ?ABC
… (iv) Therefore,
From (iii)
34]
LHS =
RHS =
46
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
= tan
Hence, LHS = RHS.
SAMPLE PAPER –IV
olutions:
1]
Mode = 3 median - 2 mean
2]
3]
4]
A rational number can be expressed as a terminating decimal if the denominator has factors 2
or 5.
5]
9 cm 5 cm 7cm cannot form the sides of a right triangle as the Pythagoras theorem is not
satisfied in this case.
6]
For the system of equations: 2x + 3y = 7
4x + 6y = 5we have
Since
is the condition for no solution and hence
inconsistent system of equations.
7]
Since
+ q)2
is a root of the equation (p + q)2 x2 - 2 ( p + q) x + k =0 So, (p
-2(p+q).
+k=0
25 - 10 + k = 0
k = -15
47
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
8]
9]
HCF
9
LCM = Product of the number
90 = 18
x
x=
10]
Monthly income
range
No. of
families
(In Rs.)
10000-13000
5
13000-16000
16
16000-19000
19
19000-22000
17
22000-25000
18
25000-28000
15
No. of families having income range (in Rs.) 16000-19000 is 19. From the graph it is clear that
median is 4.
11]
LHS = 1 +
=1+
= 1 + cosec
- 1 = cosec
=
= RHS
12]
Since
48
A=
B, AC = BC … (1)
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
Also AD = BE … (2)
Subtracting (2) from (1),
AC - AD = BC - BE
DC = EC … (3)
From (2) and (3), we have
Therefore, DE || AB by converse of BPT.
13]
One of the zero =- 15
Sum of the zeroes = 42
Other zero = 42 + 15 = 57 Product of the zeroes = 57
-15 = 855
The quadratic polynomial is x2 - 42x - 855
OR
Let p(x) = 2x2 - 4x + 5
?² + ?²= (? + ?)² - 2??
Substituting the values, we get = ?² + ?²= -1
14]
For no solution:
3p - 3 = 2p - 1
p=2
15]
Modal class - 30 - 40
= 30 fo = 12 fi = 20 f2 = 11 h = 10
49
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Mode =
[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
+
= 30 +
= 30 +
16]
Construction: Draw PA
QR and SB
GR
(As one angle is 90 degrees and one is vertically opposite
angles)
From (1) and (2), we get
17]
Let 5 + 7
5+7
7
is rational number
=
=
=
Since p and q are integers
50
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
a rational number
is rational
But we know that
is rational
Out assumption is wrong
5+
is irrational. OR
Let
be a rational number
Let
= p/q where q
0, p and q are integers and coprime.
7 q2 = p2
7 divides p
Let p = 7m
7q2 = 49 m2
Q2 = 7m2
7 divides q2
7 divides q
7 divides p and q both.
Which is a contradiction for the that p and q are co-prime.
18]
LHS =
=
= sin A cos A
Hence, LHS = RHS.
19]
The polynomial p(x) can be divided by the polynomial g(x) as follows:
51
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
Quotient = x - 3
Remainder = 7x - 9
20]
Class
Interval
Fi frequency
Mid value xi
Fixi
30 - 40
2
35
70
40 - 50
3
45
135
50 - 60
8
55
440
60 - 70
6
65
390
70 - 80
6
75
450
80 - 90
3
85
255
90 - 100
2
95
190
Total
30
1930
Mean =
OR
Daily
Rs)
expenses
(in No,
families
of C.F
20 - 40
6
6
40 - 60
9
15
60 - 80
11
26
80 - 100
14
40
100 - 120
20
60
120 - 140
15
75
140 - 160
10
85
52
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
160 - 180
8
93
180 - 200
7
100
Total
100
Median class - 100 - 120
f = 20 cf = 40 h = 20 l = 100
Median =
= 100 +
= 100 + 10 = 110
21]
Construction: - Draw AE
BC
In right triangle AEB
AB2 = AE2 + BE2
= AE2 + (BD + DE)2
= AC2 - DE2 + BD2 + DE2 + 2BD.DE
= AD2 +
BC2 + 2
= AD2 +
BC2
BC
BC
9AB2 = 9AD2 + 2BC2
53
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9AB2 - 2AB2 = 9AD2
[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
AB = BC
Or 9AD2 = 7AB2
22]
Let the speed of Ritu in still water and the speed of stream be x km/h and y km/h respectively.
Speed of Ritu while rowing upstream =
Speed of Ritu while rowing downstream =
km/h
km/h
According to the question,
Adding equations (1) and (2), we obtain:
Putting the value of x in equation (1), we obtain:
y=4
Thus, Ritu’s speed in still water is 6 km/h and the speed of the current is 4 km/h.
OR
Let the speed of train and bus be u km/h and v km/h respectively.
According to the question,
Let
and
The given equations reduce to:
Multiplying equation (3) by 10, we obtain:
54
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
Subtracting equation (4) from equation (5), we obtain:
Substituting the value of q in equation (3), we obtain:
Thus, the speed of train and the speed of bus are 60 km/h and 80 km/h respectively.
23]
Given: sin  + cos  = m and
sec +cosec  = n
Consider,
L.H.S. = n
=
=
=
24]
Let the cost of one T-shirt be Rs x and that of one jacket be Rs y.
According to given condition
2x+y=625 …(i)
3x+2y=1125 …(ii)
Multiplying (i) by 2 we get
4x+2y=1250 …(iii)
55
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
Subtracting (ii) from (iii) we get,
x=125
Substituting this value of x in (i) we get
250+y=625  y=375
Therefore cost of one T-shirt is Rs125 and the cost of one jacket is Rs 375.
25]
Let a and b be any positive Integers
a = b + r, 0
r <b
Let b = 6 Thes r = 0,1,2,3,4,5
Where r = 0, a = 6m + 0 = 6m. which is even
Where r = 1 a = 6m + 1 odd
Where r = 2 a = 6m + 2 even
Where = 3 a - 6m + 3 odd
Where r = 4 a = 6m + 4 even
Where = 5 a - 6m + 5 odd
All positive even integers are of the from 6m, 6m + 2 or 6m + 4.
26]
We have
5 + f1 + 10 + f2 + 7 + 8 = 50
f1 + f2 = 20
f1 = 20 - f2
C.I
fi
Xi
fixi
0 - 20
5
10
50
20 - 40
fi
30
30fi
40 - 60
10
50
500
60 - 80
20 - fi
70
1400 - 70fi
80 - 100
7
90
630
100 - 120
8
110
882
Mean =
56
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Therefore, f2 = 20 - 8 =12.
27]
In
…(1)
Similarly
…(2)
From (1) and (2)
xy = xz + yz
Dividing by xyz
28]
q=
Consider,
q (p2 - 1)
57
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=
= 2p = RHS
29]
Since =
Zeroes of the given polynomial
(x +
) (x -
) will be a factor
Or x2 = 2 will be a factor
Long division.
2x2 - 3x+ 1 = 2x2 - 2x - 2x + 1
= 2x (x- 1) - 1(x - 1)
= (2x - 1) (x- 1 )
The other zeroes are
and 1.
30]
On dividing the numerator and denominator of lts by cos
58
, we get
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OR
=
31]
Pythagoras Theorem: Statement: In a right angled triangle, the square of the hypotenuse is
equal to the sum of squares of the other two sides.
Given: A right triangle ABC right angled at B.
To prove: that AC2 = AB2 + BC2
Construction: Let us draw BD
AC (See fig.)
Proof :
Now,
ADB
ABC (Using Theorem:If a perpendicular is drawn from the vertex of the
right angle of a right triangle to the hypotenuse ,then triangles on both sides of the
perpendicular are similar to the whole triangle and to each other)
So,
(Sides are proportional)
Or, AD.AC = AB2
Also,
BDC
ABC (Theorem)
So,
Or, CD. AC = BC2
Adding (1) and (2),
59
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AD. AC + CD. AC = AB2 + BC2
OR, AC (AD + CD) = AB2 + BC2
OR, AC.AC = AB2 + BC2
OR AC2 = AB2 + BC2
OR
Statement:Ratio of the areas of two similar triangles is equal to the square of the ratio of
their corresponding sides.
Given: Two triangles ABC and PQR such that
To prove :
Proof For finding the areas of the two triangles, we draw altitudes AM and PN of the triangles.
Now, ar
And ar (PQR) =
So,
60
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Now, in
B=
And
Q (As
m=
n (Each is of 90o)
So,
(AA similarity criterion)
Therefore,
Also,
So,
Therefore,
=
[from (1) and (3)]
[From (2)]
=
Now using (3), we get
32]
The given cumulative frequency distributions of less than type is -
Weight
Number of students
(in kg)
(cumulative
frequency)
upper class
limits
61
Less than 38
0
Less than 40
3
Less than 42
5
Less than 44
9
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Less than 46
14
Less than 48
28
Less than 50
32
Less than 52
35
Now taking upper class limits on x-axis and their respective cumulative frequency on y-axis we
may draw its ogive as following -
Now mark the point A whose ordinate is 17.5 its x-coordinate is 46.5. So median of this data
is 46.5.
33]
x - y + 1 = 0  x = y - 1
Three solutions of this equation can be written in a table as follows:
x
0
1
2
y
1
2
3
3x + 2y - 12 = 0 
Three solutions of this equation can be written in a table as follows:
x
4
2
0
y
0
3
6
Now, these equations can be drawn on a graph. The triangle formed by the two lines and the
x-axis can be shown by the shaded part as:
62
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34]
We have
= Cos (90o - 50o -
) - cos (40o -
)+
SAMPLE PAPER –V
Solutions:
1] Let x1,x2,x3……..,x12 be the 12 values of the given data. Let the 13 th observation
be x13.
x1+x2+x3……..+x12 = 12x19.25 = 231
x1+x2+x3……..,x12+x13= 13x20=260
(x1+x2+x3……..+x12)+x13= 260
63
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x13=260-231 = 29
2] Given, triangle ABC is right angled at C. Therefore,
A+B=90o or A=90o-B
1+cot2A = 1 + cot2(90o-B) = 1+tan2B = sec2B
3]
A real number is an irrational number when it has a non terminating non repeating
decimal representation.
4] x2 +2x+1= (x+1)2
x = -1
? = ?= -1
1/ ? and 1/? are also -1. 1/ ? + 1/? = -2
5]
Since the x-axis y=0 does not intersect y=-7 at any point.
6]
Since
.
7]
Because cosec 90°=1, others are not defined.
8]
9] If a and b are one two positive integers. Then a = bq + r, 0
r
b Let b = 3
Therefore, r = 0, 1, 2 Therefore, a = 3q or a = 3q + 1 or a = 3q + 2 If a = 3q a2 =
9q2 = 3(3q2) = 3m or where m = 3q2 a = 3q + 1 a2 = 9q2 + 6q + 1 = 3(3q2 + 2q) +
1 = 3m + 1 where m = 3q2 + 2q or a = 3q + 2 a2 = 9q2 + 12q + 4 = 3(3q2 + 4q +
1) + 1 = 3m + 1 where m 3q2 + 4q + 1 Therefore, the squares of any positive
integer is either of the form 3m or 3m + 1.
10] Given polynomial P(x) = x4 + 2x3 - 2x2 + x -1
Let g(x) must be added to it.
So, number to be added=-(-x+ 2) = x - 2
11] For infinite number of solution,
Consider
64
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Again,
12]
AC2 = AD2 + CD2 … (2)
[By Pythagoras theorem]
(1) - (2) gives,
Hence proved.
13]
OR
Consider,
7 sin2  + 3 cos2  = 4
7Sin2
+ 3 (1 - sin2 ) = 4
2
7Sin
+ 3 - 3sin2 = 4
4Sin2
=1
Sin
=
Thus, Sec 30o + Cosec30o =
14] Class Interval
More
More
More
More
More
More
15]
Let 3 3-
then
then
then
then
then
then
Cumulative Frequency
108
95
80
63
42
19
50
60
70
80
90
102
be a rational number.
=
[ p,q are integers, 2
0]
Here,
LHS = Rational No.
RHS = irrational No.
But, Irrational no
Rational no
our assumption is wrong
is an irrational.
OR
65
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Let us assume to the contrary, that
is a rational number.
is rational.
(n-1)+(n+1)-2
2n+2
is rational
is rational
But we know that
is an irrational number
So 2n+2
is also an irrational number
So our basic assumption that the given number is rational is wrong.
Hence,
is an irrational number.
16]
… (1)
… (2)
Multiplying (1) with a and (2) with b, we get
From (1), bx + ab = 2ab
bx = ab
x=a
Hence, x = a and y = b.
17] C.I
10
20
30
40
50
60
70
-
Fi
5
3
4
F
2
6
13
33+f
20
30
40
50
60
70
80
Xi
15
25
35
45
55
65
75
Fi. .Xi
75
75
140
45f
110
390
975
1765+45f
Mean =
OR
C.I
100
150
200
250
300
66
-
150
200
250
300
350
fi
4
5
12
2
2
xi
125
175
225
275
325
di
-2
-1
0
1
2
fidi
-8
-5
0
2
4
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-7
Where:
18] Let us assume that Prema invests Rs x @10% and Rs y @8% in the first year.
We know that
Interest =
ATQ,
+
=1640
 10x + 8y = 164000 …(i)
After interchanging,
+
=1600
we get 10y+8x=160000
8x+10y=160000 ...(ii)
Adding (i) and (ii)
18x+18y=324000
 x + y = 18000 ... (iii)
Subtracting (ii) from (i),
2x-2y=4000
 x - y = 2000 ...(iv)
Adding (iii) and (iv)
2x=20000
 x = 10000.
Substituting this value of x in (iii)
y=8000
So the sums invested in the first year at the rate 10% and 8% are Rs 10000 and Rs
8000 respectively.
OR
Solving (1) and (2), we get
x = 30 y = 6.
19]
Let
is one zero.
+7
8
Now,
67
=
=7
is another zero then
=
=
and
=
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20]
[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
The two angles
and  being the acute angles of a right triangle, must be
complementary angles.
So,
Substituting,
in above equation
21]
Now, AP + AQ = 120 + 60 = 180 cm
22]
68
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23]
Let the area of triangle =x sq units
Area of trapezium = 3x sq units
Area triangle ABC = x + 3x = 4x sq units
Now,
Consider triangles AED and ABC,
ED ll BC...given
AED = ABC Corresponding angles
A = A Common
?AED ~ ?ABC [By AA rule]
=
(since Ratio of areas of two similar triangles is equal to
ratio of square of corresponding sides)
So
24] C.I
9.5 - 19.5
19.5 - 29.5
29.5 - 39.5
32.5 - 49.5
49.5 - 59.5
59.5 - 69.5
69.5 - 79.5
Here, l = 39.5
F
Cf
2
2
4
6
8
14
9
23
4
27
2
29
1
30
c.f = 14 f = 9 h = 10
M = 39.5 +
25] Given 2 and 3 are the zeroes of the polynomial.
Thus(x - 2) (x - 3) are factors of this polynomial.
69
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4x4 - 20x3 + 23x2 =5x - 6 = (x2 - 5x + 6) (4x2 - 1)
Thus, 4x4 - 20x3 + 23x2 +5x-6=(x - 2) (x - 3) (2x - 1) (2x + 1)
Therefore, 2,3,
26]
Given: A triangle ABC in which a line parallel to side BC intersects other two sides AB
and AC at D and E respectively
To prove that
Construction: Let us join BE and CD and then draw DM
AC and EN
AB.
Proof: Now, area
of
Note that
BDE and
DEC are on the
same base DE and
between the same
parallels BC and DE.
So, ar(BDE) =
ar(DEG)
Therefore, from (1),
(2) and (3), we have
:
OR
70
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Given: A right triangle ABC right angled at B.
To prove: that AC2 = AB2 + BC2
Construction: Let us draw BD
AC (See fig.)
Proof:
Now,
ADB
ABC (Using Theorem: If a perpendicular is drawn from the vertex
of the right angle of a right triangle to the hypotenuse ,then triangles on both sides
of the perpendicular are similar to the whole triangle and to each other)
So,
(Sides are proportional)
Or, AD.AC = AB2 … (1)
Also,
BDC
ABC (By Theorem)
So,
Or, CD. AC = BC2 … (2)
Adding (1) and (2),
AD. AC + CD. AC = AB2 + BC2
OR, AC (AD + CD) = AB2 + BC2
OR, AC.AC = AB2 + BC2
OR AC2 = AB2 + BC2
Hence proved.
71
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27]
OR
Dividing numerator and denominator of LHS by sin , we get
28]
29] Let the number of girls and boys in the class be x and y respectively.
According to the given conditions, we have:
x + y = 10
x-y=4
x + y = 10  x = 10 - y
Three solutions of this equation can be written in a table as follows:
x
5
4
6
72
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y
5
6
4
x-y=4x=4+y
Three solutions of this equation can be written in a table as follows:
x
5
4
3
y
1
0
-1
The graphical representation is as follows:
From the graph, it can be observed that the two lines intersect each other at the
point (7, 3).
So, x = 7 and y = 3.
30] We can obtain cumulative frequency distribution of more than type as following:
Production yield
Cumulative frequency
(lower class limits)
more than or equal to 50
100
more than or equal to 55
100 - 2 = 98
more than or equal to 60
98 - 8 = 90
more than or equal to 65
90 - 12 = 78
more than or equal to 70
78 - 24 = 54
more than or equal to 75
54 - 38 = 16
Now taking lower class limits on x-axis and their respective cumulative frequencies
on y-axis we can obtain its ogive as following.
73
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31]
Statement: The ratio
of the areas of two similar triangles is equal to the square of the ratio of their
corresponding sides.
Given: ABC ~ PQR To Prove:
Construction: Draw ADBC and
PSQR Proof:
ADB
~ PSQ (AA) Therefore,
… (iii) But ABC
~ PQR Therefore,
… (iv) Therefore,
Therefore,
From (iii)
32]
Hence, L.H.S = R.H.S
33] Let 5q + 2, 5q + 3 be any positive integers
(5q + 2)2 = 25q2 + 20q + 4
= 5q (5q + 4) + 4 is not of the form 5q + 2
Similarly for 2nd
74
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(5q + 3)2 = 25q2 + 30q + 9
=5q(5q+6)+ 9 is not of the form 5q+3
So, the square of any positive integer cannot be of the form5q+2 or 5q+3
For any integer q
34] Marks
Frequency
25 - 35
5
35 - 45
10
45 - 55
20
55 - 65
9
65 - 75
6
75 - 85
2
Total
52
Here the maximum frequency is 20 and the corresponding class is 45-55.So,45-55 is
the modal class.
We have,l=45,h=10,f=20,
Mode =
+
Mode=49.7
SOLUTIONS SCIENCE 2012-13
SAMPLE PAPER –I
Solutions:
1]
Y – Diamond and Z – Graphite
2]
1/25
3]
Sometimes there is a lack of oxygen in our muscle cells hence the pyruvate breaks
down to form lactic acid which builds up in our muscles during sudden activity and
causes cramps.
4]
Plaster of Paris is calcium sulphate hemihydrates. It is obtained by heating gypsum
Uses:- (1) Plastering fractured bones
(2) statues, toys etc.
5]
It is a double displacement and precipitation reaction.
75
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6]
Plant kept in continuous light will live longer, because it will be able to produce
oxygen required for its respiration by the process of photosynthesis.
7]
Maximum current through resistor
A = 3A
Thus the maximum current through resistors B and C each
8]
The following events occur during photosynthesis:
(i) Absorption of light energy by chlorophyll.
(ii) Conversion of light energy to chemical energy and splitting of water molecules
into hydrogen and oxygen.
(iii) Reduction of carbon dioxide to carbohydrates
9]
The steps involved in reflex action are:
(i) Receptors receive the stimulus
(ii) Nerve cells takes the stimulus, converts it into electrical signal and passes it
through sensory nerve.
(iii) The impulse passes through spinal cord where it gets transmitted through relay
neuron and it passes further through motor neuron to the effector organ, i.e.
muscles.
10]
11]
(a) Silver chloride on exposure to sunlight may decompose as per the following
reaction.
Therefore, it is stored in dark coloured bottles.
(b) Calcium oxide reacts vigorously with water to produce slaked lime (calcium
hydroxide) releasing a large amount of heat
12]
(a)
(b)
13]
In absence of litmus paper a reagent phenolphthalein can be used. The solution
which gets pink colour with phenolphthalein is base other is acid.
76
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14]
A stimulus received by a neuron travels through it in the form of an electrochemical
disturbance. When an electrical signal reaches the axonal end of one neuron, it
releases chemical substance called neurotransmitter that cross the synapse and
move towards the dendritic end of next neuron generating another electrical singal.
A narrow fluid filled space, called synaptic cleft occurs between the two neurons. An
impulse travels through the neurons only in one direction because the
neurotransmitter releases only on one side of the synapse.
15]
(a)
(b)
16]
The various plant hormones are auxin, gibberellin, cytokinin and abscisic acid.
Functions of plant hormones:
(a) Auxin - It helps the cells to grow longer at shoot tips.
(b) Gibberellin – Growth of stem.
(c) Cytokinin – Promotes cell division.
(d) Abscisic acid – Inhibits growth in plants
17]
The thumb in right hand rule indicates the direction of current in the straight
conductor held by curled fingers, whereas the Fleming's left-hand rule gives the
direction of force experienced by current carrying conductor placed in an external
magnetic field.
18]
Hints – Biomass: Plant and animal wastes. Give description of biogas plant with the
following diagram.
19]
(a)A current carrying conductor produces a magnetic field around it. When it
77
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interacts with magnetic field it experiences a force. Direction of the force depends
upon direction of current and that of magnetic field.
Fleming's left hand rule. State the rule.
(b)The deflection will increase. As the magnetic field produced around the conductor
is proportional to the current flowing through the wire.
OR
a) The earth wire, which has insulation of green colour, is
usually connected to a metal plate deep in the earth near the house.
This is used as a safety measure, especially for those appliances that
have a metallic body, for example, electric press, toaster, table fan,
refrigerator, etc. The metallic body is connected to the earth wire, which
provides a low-resistance conducting path for the current. Thus, it
ensures that any leakage of current to the metallic body of the appliance
keeps its potential to that of the earth, and the user may not get a severe
electric shock.
b) due to electromagnetic induction, there would be change in the magnetic field
around the coil A and so will be around B . So current will be induced in coil B.
20]
(a) Baking powder (NaHCO3), salt A is commonly used in bakery products. On
heating it forms sodium carbonate (Na2CO3), B and CO2 gas, C is evolved. When
CO2 gas is passed through lime water it forms calcium carbonate (CaCO 3), which is
slightly soluble in water making it milky.
A – NaHCO3
B – Na2CO3
C – CO2 gas
(b) Bleaching powder is used –
(i) for bleaching wood pulp in paper factories.
(ii) for disinfecting drinking water to make it free of germs.
OR
(a) (i) Sulphuric acid + Zinc
Zn(s)
Zinc sulphate + Hydrogen H2SO4 (aq) +
ZnSO4 (aq) + H2 (g) (ii) Hydrochloric acid + Magnesium
Magnesium chloride + Hydrogen 2HCl (aq) + Mg (s)
MgCl2(aq) + H2 (g)
(b) A reaction in which an acid and base react with each other to give a salt and
water is termed as neutralisation reaction. In this reaction, energy is evolved in the
form of heat.
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(c) Uses of washing soda:(i) Sodium carbonate (washing soda) is used in glass, soap and paper industries.
(ii) It is used in the manufacture of sodium compounds such as borax.
21]
(a)
(b)
(c) Na+ cation, Mg+2 cation O-2 anion
(d) Strong inter ionic attraction.
OR
(a) Noble, highly malleable, ductile shinglustre
(b) Forms a protective layer of aluminium oxide
(c) Strong oxidizing agent oxides H2 to H2O
(d) Easy to reduce metallic oxide to metal
(e) Good conductor of heat, Non corrosive.
22]
(a)
(b)
(c) 4V. Hint – V = IR =
= 4V
(d) No difference.
Hint – Same current flows through each element in a series circuit.
OR
Current carrying loops behave like bar magnets and both have their associated lines
of field. This modifies the already existing earth's magnetic field and a
deflection result. Magnetic field has both direction and magnitude. Magnetic
field lines emerge from N-pole and enter Spole. The magnetic field strength is
represented diagrammatically by the degree of closeness of the field lines.
Field lines cannot cross each other as two values of net field at a single point
cannot exist. Only one value a unique net value, can exist. If in a given
region, lines of field are shown to be parallel and equispaced, the field is
79
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understood to be uniform.
23]
(a)
(b) The digestion of carbohydrates (starch) starts in the mouth. The saliva has
salivary amylase or ptyalin enzyme which acts on the starch and converts it into
maltose sugar. The partially digested food reaches the stomach where the food gets
mixed with gastric juice secreted by gastric glands. The gastric juice contains large
quantities of mucus, HCl and pepsin. The enzyme pepsin acts on proteins and
coverts them into peptones and proteoses. The gastric lipase emulsifies fats.
Bile juice secreted from liver lipids into small droplets.
Pancreas secretes pancreatic juice which contains the pancreatic amylase, trypsin
and lipase. Pancreatic amylase digests the leftover starches into maltose. Trypsin
converts peptones and proteoses into small peptides. Lipase acts on small droplets
and converts them into triglycerides.
The food reaches the small intestine where complete digestion of food takes place in
the presence of an enzyme called succus entericus. The maltose converts into
80
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glucose, peptide emulsifies into amino acids and the triglycerides emulsifies into
fatty acids and glycerol. [
OR
(a) In four chambered heart, left half is completely separated from right half by
septa.This prevents oxygenated and deoxygenated blood from mixing. This allows a
highly efficient supply of oxygenated blood to all parts of the body. This is useful in
animals that have energy needs, such as birds and mammals.
(b)
24]
(d) in set up A and (a) in set up B
25]
Sodium bicarbonate > Water > Fruit juice
26]
I
27]
A, C, D, B
28]
A
29]
An antacid
30]
The final solution becomes light green
31]
Metal
Al2(SO4)3
CuSO4
FeSO4
(a) Al
x
?
?
32]
all the three cases
33]
(ii)
34]
(-4mA, +0.2V) and (2mA, 0.1V) respectively
35]
Student A will determine the equivalent resistance of series combination while
student B will determine the equivalent resistance of parallel combination of the two
resistors
81
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36]
epidermal cells, stomata, guard cells each with one nucleus and many chloroplasts
37]
I Cytoplasm II nucleus
III stoma IV chloroplast
38]
39]
Absorbs CO2 released by germinating seeds
40]
Thin layer of fleshy leaf of onion
41]
The glow of the bulbs B2 and B3 will remain the same
SAMPLE PAPER –II
solutions:
1]
Gallium or Cesium (any one).
2]
Cerebellum
3]
Since ventricles have to pump blood into various organs, therefore, they have thicker
muscular walls than the auricles.
4]
Commercial unit of energy is Kilowatt hour or KWh.
5]
Chemical reaction in which one reactant gets oxidized while the other gets reduced is
called a redox reaction.
Mg gets oxidized on burning in air.
6]
(i) Fresh milk is acidic and gets spoiled easily. In presence of baking soda, milk
becomes alkaline and can be stored for a longer time.
(ii) When milk sets to curd, pH decreases. The alkali does not allow it to become more
acidic easily. So it takes more time.
7]
Name of gland Secretion
Salivary gland Saliva
Gastric gland Gastric juice
82
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Liver Bile
Pancreas Pancreatic Juice
(any two)
8]
(a) When Iron nails are kept into a copper sulphate solution, Iron displaces copper
from CuSO4 solution and blue colour of CuSO4 solution becomes light green.
(b)
9]
(a) Ionic compounds are formed by the transfer of electrons from metal to non metal.
(b)
10
] (a) Respiration is an exothermic process since energy is released during the
process.
(b) Energy in the form of heat, light or electricity is required for decomposition
Reactions hence these are endothermic.
(c)
The solution becomes colourless because of copper sulphate formed which is
colourless.
83
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11
]
(i) Test the three solutions with blue litmus paper, the one solution will change blue
litmus into red it is Acidic solution
(ii) Test the remaining two solution with the red litmus (Changed in activity (i)) one
solution will change it again to blue It is basic solution. (iii) Remaining third solutions is
distilled water
12
]
13
] In series (i) R1 = 5+5 =10
(ii) R2 = 6+4 = 10
In pararell
(i)
14
]
(i) Glottis gets covered by a small cartilaginous flap of skin called epiglottis which
prevents the entry of food particles into wind pipe while swallowing.
(ii) The lung alveoli are richly supplied with blood capillaries for the exchange of gases.
From the thin walls of alveoli, oxygen diffuses into blood and is supplied to the tissues
while carbon dioxide is absorbed by blood from the tissues and is carried to the alveoli
of lungs for exhalation.
(iii) The walls of trachea is supported by C – shaped cartilage rings which does allow
the trachea to collapse in the presence of less air in it.
15
] Hind brain has three parts:
(i) Cerebellum: It is the regulating centre for swallowing, coughing, sneezing and
vomiting and helps to maintain the balance of the body and coordinate the muscular
activities.
(ii) Pons: It takes part in the regulation of respiration.
(iii) Medulla oblongata: It is responsible for the coordination and adjustment of
movement and posture.
84
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16
]
(a) Diagram
of Neuron:
Functions of Neuron:
(i) To carry information from receptors to brain and spinal cord.
(ii) To transfer respose from brain and spinal cord to effectors.
17
]
(a) Auxin.
When light is coming from one side of the plant auxin diffuses towards shady sides.
This concentration stimulates the cells to grow longer on the side of the shoot which is
away from the light and shoot bends.
18
] (i) Magnetic field will increase
(ii) Magnetic field will decrease
(iii) Magnetic field will increase
19
] (i) Wave energy be a viable proposition only when waves are very strong. [1x3=3]
(ii) A tidal power station at the mouth of a river blocks the flow of polluted water into
the sea. It creates health and pollution hazards.
(iii) Efficient commercial exploitation is difficult or any other.
20
] (a)
(b) Functions of two chambers of human heart:
85
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
Left Atrium – Receive oxygenated blood from pulmonary vein
Right Atrium – Receive deoxygenated blood from vena cava
Left Ventricle – Pumps oxygenated blood to all parts of body
Right Ventricle – Pumps deoxygenated blood to lungs. (Any two)
OR
(a) Diagram of stomata:
(b) Functions of stomata:
(a) Exchange of gases
(b) Transpiration
Raw materials for photosynthesis CO2, H2O
21
] (a) The chemical name of washing soda is Sodium carbonate Decahydrate.
The formula is Na2CO3.10H2O
It is obtained by heating baking Soda and then recrystallization.
(b) Distilled water is a pure form of water and is devoid of any ionic species. Therefore,
it does not conduct electricity. Rain water, being an impure form of water, contains
many ionic species such as acids and therefore it conducts electricity.
OR
(a)
(i) Neutral- Solution D with pH 7 (ii) Strongly alkaline- Solution C with pH 11 (iii)
86
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Strongly acidic- Solution B with pH 1 (iv) Weakly acidic- Solution A with pH 4 (v)
Weakly alkaline- Solution E with pH 9
The pH can be arranged in the increasing order of the concentration of hydrogen ions
as: 11 < 9 < 7 < 4 < 1
(b) The fizzing will occur strongly in test tube A, in which hydrochloric acid (HCl) is
added.
This is because HCl is a stronger acid than CH3COOH and therefore produces hydrogen
gas at a faster speed due to which fizzing occurs.
22
] (i) Metal – sodium
(ii)
(iii)Electrolysis of molten chloride (NaCl) is the process.
i.e. we can write:
OR
Definition: Homogeneous mixture of two or more metals or a metal and a non metal is
called an alloy.
Prepared: (i) By melting the primary metal.
(ii) By dissolving the other elements in a definite proportion and cooled to room
temperature.
23
] A coil of many circular turns of insulated copper wire wrapped closely in the shape of a
cylinder.
87
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The magnetic field is uniform inside the solenoid
Used to magnetise a piece of magnetic material.
OR
(a)
(i) Right hand Thumb rule
(ii) Fleming's Left hand rule
(iii) Fleming's Right hand rule
(b)
(a) To avoid the risk of electric shocks
(b)To save the electrical appliances from damage.
24
]
(a)
(i) Momentary deflection in the galvanometer to one side.
(ii) Momentary deflection in the galvanometer now in opposite direction.
(iii) No deflection in the galvanometer. Phenomenon involved is electromagnetic
induction.
(b) An energy source that can be replenished in a short period of time and state one
use each.
Eg. Solar Energy, Wind energy, Ocean energy etc. Any use.
OR
(a)Argon or Neon gas is filled in the electric bulbs.
These are used since these are inactive or inert gases to prolong life of the
filament. (b)
(i) Length of the conductor: proportional
88
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(ii) Area of cross section: inversely proportional
(iii) Material of conductor
25
]
26
] II
27
] 1A in Circuit I and 0A in Circuit II
28
] 9
29
] Parallel in circuit I and in series in circuit II
30
] A=Digestor,B=Gas outlet
31
] Safrainin
32
] Alcohol
33
] Alcohol is inflammable
34
] Germinating seeds
35
] To absorb CO2 released by germinating seeds
36
] Covered portion of any of the above
37
] Green
38
] Sodium bicarbonate > Water > Fruit juice
39
] A
40
] Light green
41
] NaOH
42
] (i) and (ii)
SAMPLE PAPER –III
Solutions:
89
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
1]
2]
The receptors present in the specific regions of the tongue are called gustatory
receptors.
3]
A large number solar cell combined in an arrangement.
4]
(i) Human kidney performs the excretion by cleaning the blood of metabolic wastes.
(ii) It performs the function of osmoregulation by maintaining normal levels of water
and mineral ions in the body fluids.
5]
'Zn' being more reactive displaces 'Cu' from CuSO4 solution CuSO4 solution becomes
Colourless.
(blue)
(colourless)
6]
The pH of milk decreases from '6' as it turns into curd. That is curd is more acidic
than milk.
7]
Resistance depends upon:
(i) Length of the conductor
(ii) Area of cross section
(iii) Material of conductor
Resistivity depends on:
(i) Material
(ii) Temperature
SI Unit of resistance is 'Ohm' or
8]
(i) Each appliance have equal potential differenc
(ii) Each appliance have a separate switch to on/off
(iii) Each appliances can be operated on different current
9]
(a) Momentary deflection in the galvanometer to one side.
(b)Momentary deflection in the galvanometer now in opposite direction.
(c)No deflection in the galvanometer.
(d)Phenomenon involved is electromagnetic induction.
(e) Momentary deflection in the galvanometer now in opposite to the direction of the
first case.
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
(f) No deflection in the galvanometer.
10]
(i) Xylem – water and minerals in plants
(ii) Pulmonary Artery – Deoxygenated blood from heart to lungs.
(iii) Pulmonary Vein – Oxygenated blood from lungs to heart.
Phloem – Synthesised food in plants
11]
1)Methane or CH4
2) Hydrogen
3) Hydrogen sulphide
4) CO2
Uses Any two:
. For the production of electricity ii. As a fuel to produce heat and light. iii. Biogas
generation is an effective method of the disposal of bio wastes like, animal wastes
and sewage.
12]
(i) Ionic Compounds which are held together by strong ionic bonds so high amount
of energy is required to break these bonds.
(ii)Ionic compounds are very hard solids due to strong force of attraction between
positive and negative ions.
(iii)Hydrogen gas is evolved when dilute hydrochloric acid is added to a reactive
metal.
13]
(a) Chemical reaction in which one reactant gets oxidized while the other get
reduced is known as a redox reaction.
When Magnesium ribbon burns in air, magnesium is oxidized and the white residue
formed is of magnesium oxide.
(b) Chemical equations must always be balanced to follow the law of conservation of
mass, according to which "Mass can neither be created nor destroyed in a chemical
reaction."
14]
(i) This is an example of decomposition reaction and it is endothermic in nature.
(ii) This is an example of combination reaction and it is also a type of endothermic
reaction.
(iii) This is an example of displacement reaction and it is exothermic reaction.
15]
(i) The chemical name of 'Plaster of Paris' is 'Calcium sulphate Hemi hydrate'
91
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(ii) It is prepared on heating gypsum at 373 K
(iii)
16]
(i)
(ii) In series R2 =7
R3 =6
+4
+3
=10
= 10
In Parallal
OR
17]
The gastric glands in the stomach secrete:
(i) HCl – It kills the bacteria ingested with food and makes the medium of food acidic
92
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
so as to facilitate the action of the enzyme pepsin.
(ii) Pepsin – It helps to digest proteins.
(iii) Mucus – It protects the inner lining of the stomach from the action of HCl
18]
The functions of fore brain are:
(i) It has sensory area where information is received from sense organs.
(ii) It has motor area where impulses are sent to muscles or effector organs.
(iii) It has centres for visual reception, touch, smell, temperature and muscular
activities.
19]
(a) Diagram of Neuron:
Functions:
(i) To carry information from receptors to brain and spinal cord.
(ii) To transfer respose from brain and spinal cord to effectors.
20]
(a) The glands which does not have duct and secretes it's product directly into blood
stream is called an endocrine gland.
Two glands are: Thyroid and pancreas.
Thyroid secretes thyroxin and pancreas secretes insulin.
21]
(a)
(i) Test the three solutions with blue litmus paper, the solution that changes l blue
litmus into red is Acidic solution.
(ii) Test the remaining two solution with the red litmus one solution will change it
again to blue –then that solution is basic solution.
(iii) Remaining third solutions is distilled water which being neutral does not cause
any colour change of either the blue or red litmus paper.
(b) Plaster of Paris should be stored in a moisture-proof container because Plaster
of Paris, a powdery mass absorbs water (moisture) to form a hard solid known as
gypsum.
93
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OR
(a) Two important used of washing soda and baking soda are as follows: Washing
soda:
(i) It is used in glass, soap, and paper industries. (ii) It is used to remove permanent
hardness of water.
Baking soda:- (i) It is used as baking powder. Baking powder is a mixture of baking
soda and a mild acid known as tartaric acid. When it is heated or mixed in water, it
releases CO2 gas that makes bread or cake fluffy. (ii) It is used in soda-acid fire
extinguishers.
(b) Acids dissociate in the presence of water to give free hydrogen ions. It is the
hydrogen ions that are responsible for the acidic behaviour.
22]
(i) Metal – sodium
(ii)
(ii) Electrolysis of molten chloride (NaCl)
OR
Definition: Homogeneous mixture of two or more metals or a metal and a non metal
Prepared:
(i) Melting the primary metal.
(ii) Dissolving the other elements in a definite proportion and then cooling them to
room temperature .
23]
A coil of many circular turns of insulated copper wire wrapped closely in the shape of
94
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
a cylinder.
The magnetic field is uniform inside the solenoid
Used to magnetise a piece of magnetic material.
OR
(a) (i) Right hand Thumb rule
(ii) Fleming's Left hand rule
(iii) Fleming's Right hand rule
(b) (a) To avoid the risk of electric shocks (b) To save the electrical appliances from
damage.
24]
(a)
(b) Functions of chambers of human heart:
95
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
Left Atrium – Receive oxygenated blood from pulmonary vein
Right Atrium – Receive deoxygenated blood from vena cava
Left Ventricle – Pumps oxygenated blood to all parts of body
Right Ventricle – Pumps deoxygenated blood to lungs. (Any two)
OR
(a) Diagram of Stomata:
Functions
(a) Exchange of gases
(b) Transpiration
Raw materials for photosynthesis are CO2, H2O
25]
Green
26]
Sodium bicarbonate > Water > Fruit juice
27]
A
28]
Hydrogen
29]
Red
30]
Calcium
31]
0.25
32]
0A in Circuit I and 1A in Circuit II
96
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
33]
9
34]
3
35]
Parallel in circuit I and in series in circuit II
36]
37]
Glycerine
38]
Alcohol
39]
(I) Guard cell
(II) Nucleus
(III) Stoma
(IV) Chloroplast
40]
Germinating seeds
41]
To absorb CO2 released by germinating seeds
42]
Covered portion of any of the above
SAMPLE PAPER –IV
Solutions:
1]
Metal: Mercury: Non metal: Bromine
2]
The brain box called cranium protects the brain and has shock absorbing fluid in it
which prevents it from shock and injuries.
3]
(i) Volmeter (ii) Ammeter
4]
It is a reaction in which there is an exchange of ions between the reactants
Na2SO4(aq) + BaCl2(aq) ? BaSO4(s) +2NaCl(aq) or any other
97
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
5]
HCl and HNO3 produce H+ ions in aqueous solution. Hence, they show acidic
character.
Alcohol and Glucose do not produce H+ ions in aqueous solution, hence, they do not
show acidic characters.
6]
Glucose that enters the nephron along with the filtrate after passing through the
glomerulus, passes from the tubule of nephron where it is selectively reabsorbed and
sent back to blood.
7]
8]
(a)
Let Resistance of a wire = R
When cut into 3 equal parts
Resistance of each part =
When connected in Parallel, net resistance Rp:
(b) Resistivity will remain same. (c) Current through each part will be triple the
previous and total current in the circuit will be 9 times.
9]
(a) Strength of magnetic field (B) is inversely proportional to radius of the coil (r)
(b) Directly proportional to the Number of turns in the coil (N)
( c) The magnetic field lines will be like:
98
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10]
At the roots, cells in contact with the soil actively take up ions. It creates a difference
in the concentration of these ions between roots of soil. To eliminate this difference,
water moves in to the roots from the soil. Thus water of minerals are absorbed by
the plants.
11]
(i) Buring of coal or Petroleum Products lead to air pollution
(ii) The oxides of C, N and S are released on buring fossil fuels as acidic oxides it
leads to acidic rain and affect water soil.
(iii) CO2, a green house gas causes global warming (or any other relevant point)
i. Source of electricity in artificial satellites ii. Used as source of electricity in radio
and wireless transmissions, at TV relay stations, traffic lights and research centers.
12]
(a) Oxidation: gain of oxygen by a substance
or any other example
Reduction: loss oxygen by a substance
example
or any
(b) Sodium (Na) is oxidised to sodium oxide as it gains oxygen and oxygen gets
reduced.
13]
(a) The compounds formed by transfer of electrons from metal to non metal are
called ionic compounds
In ionic compounds, there are strong electrostatic forces of attraction between the
oppositely charged ions. Due to this, movement of ions is not possible due to their
rigid structure.
In molten state, the electrostatic forces of attraction between the oppositively
charged ions are overcome due to heat.
(b) Ionic compounds have strong electrostatic forces of attraction between the ions.
Therefore, it requires a lot of energy to overcome these forces. That is why ionic
compounds have high melting points.
14]
Exothermic: Reactions in which heat is released.
99
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Endothermic: Reactions in which heat is absorbed.
or any other
15]
When the pH of mouth of a person is lower than 5.5 tooth decay starts
Tooth enamel made up of calcium phosphate is corroded.
Measures:
(i) Clean the mouth after eating food
(ii) Use of toothpaste (generally basic in nature)
16]
The functions of fore brain are:
(i) It has sensory area where information is received from sense organs.
(ii) It has motor area where impulses are sent to muscles or effector organs.
(iii) It has centres for visual reception, touch, smell, temperature and muscular
activities.
17]
The following events occur during photosynthesis:
(i) Absorption of light energy by chlorophyll.
(ii) Conversion of light energy to chemical energy and splitting of water molecules
into hydrogen and oxygen.
(iii) Reduction of carbon dioxide to carbohydrates.
18]
Gland
Harmone
Function
Thyroid
Thyroxin
Regulates fat protein and carbohydrate
metabolisms
Pancrease
Insulin
Regular blood sugar level
Adrenal gland
Adrenaline
Prepage the body to face the emergency
situation and blood pressure
19]
Auxin diffuses to the other part which is away from the support. Part of the tendril in
contact from the support does not grow rapidly
Part of the tendril away from the support grow rapidly and cause the tendril to grow
around a support.
20]
(a) Bleaching Powder
(b) CaOCl2
(c)
100
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(d) Bleaching powder is used:
(i) as an oxidising agent in many chemical industries; and
(ii) for disinfecting drinking water to make it free of germs.
OR
(a) The chemical equation for the reaction of Plaster of Paris and water can be
represented as:
(b) Calcium hydroxide [Ca(OH)2], on treatment with chlorine, yields bleaching
powder.
(c) Washing soda (Na2CO3.10 H2O) is used for softening hard water.
(d) When a solution of sodium hydrogencarbonate is heated, sodium carbonate and
water are formed with the evolution of carbon dioxide gas.
21]
Corrosion: Process of slowly eating up of metals due to action of atmospheric gases
Metal does not corrode ? Gold
Which corrode ? Iron or any other
Prevention:
(i) By painting
(ii) By greasing or oiling
(iii) Galvinization
(iv) Tin plating/ Chromium plating etc. (any three)
OR
(i)
(ii)
[1]
(iii)
101
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(iv)
(v)
22]
(b) Exchange of gases take place in the alveoli
OR
(a)
It receive secretion from liver and pancreas in the form of juice and trypsin
102
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Trysin digest proteins.
Lipase digest fats
Bile juice makes basic medium and emulsifies fats
Intestinal Juice converts proteins to amino acids Carbohydrates to glucose and fats
to fatty acids and glycerol.
23]
(b)
(i) Over loading is caused when too many short circuit appliances are connected to a
singl socket
(ii) When live wire and neutral wire come into direct contact
OR
(i) North Pole to South Pole
(ii) Tesla
(iii) Strong magnetic field
(iv) 50 Hz
(v) Heart and brain
24]
(i) Separately: Current
(ii) Series R=R1+R2=44
(iii) In Parallel
103
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
Advantages of solar cooker:(any two) [1/2x4=2]
(1) Free of cost
(2) Does not cause any pollution
(3) Nutritive value of food is retained
Disadvantages:
(i) Can not used in night
(ii) Can not be used in cloudy day
(iii) Direction of sunlight is need and to be adjusted frequently or any other
OR
(a) Deflection increases
(b) Direction of deflection changes
(c) Deflection decreases. [1]
(d)
Looking at a crossection lying in the plane of the
paper.
At a distance d from the wire, the magnetic field
direction is as shown and the magnitude is constant for all points shown.
25]
sodium bicarbonate > water > lemon juice
26]
HCl
27]
II and IV
28]
Set
104
HCl+Zn
HCl+Na2CO3
NaOH+Zn
NaOH+Na2CO3
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(d)
?
[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
?
?
X
29]
R>P>Q
30]
Displacement reaction
31]
B
32]
A
33]
34]
0.5
35]
Ammeter and rheostat
36]
1
37]
Covered portion of any of the above
38]
Entry of air bubbles
39]
40]
Dark room for 03 days
41]
I cytoplasm II nucleus III stoma IV chloroplast
42]
105
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[SAMPLE QUESTION PAPER FOR CLASS X – MATHEMATICS & SCIENCE ]
SAMPLE PAPER –V
1]
Adrenaline controls the rate of respiration. Pons is the part of the brain which
controls the rate of respiration.
2]
Acid present in pickles or curd will react with metals like copper and poisonous salts
are produced.
3]
60 W bulb because
ie resistance is inversely proportional to the power.
4]
When aluminium powder is heated with manganese dioxide the following reaction
takes place:
When Mn powder is heated with aluminum dioxide no reaction takes place because
Mn is less reactive than Al and hence there will be no displacement reaction.
5]
Baking powder is added to make breads soft and fluffy. Baking soda and an edible
acid like tartaric acid are its main ingredients.
Baking soda (Sodium hydrogen carbonate) is added to release CO 2 gas when heated.
Tartaric acid is added to avoid the bitter taste by reacting with the Na 2CO3 which
is formed by the heating of NaHCO3.
6]
(i) Kidney infection or injury to kidneys.
(ii) Restricted blood flow to kidneys.
7]
The lining of canal has muscles that contract rhythmically in order to push the food
forward.
Peristaltic movement is necessary to move the food in a regulated manner along the
digestive tube so that the food can be processed properly in each part.
8]
Solar cooker with concave mirror reflector is more efficient.
Because concave mirror can focus the heat radiations to the material kept inside to
increase the temperature.
Concave mirror is used in headlight of vehicles or as shaving mirror.
9]
Ohm's law relates current and potential difference.
According to this law:
V/I = constant = R
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10]
Ammeter, A1 will show higher reading.
Because, as wire A is thicker than B, A has lesser resistance. So higher current will
be drawn by A from the battery and hence ammeter A1 will show higher reading.
11]
Observation a is correct.
Because force experienced by a current carrying conductor in a magnetic field is
proportional to the strength of the current. Resistance of a conductor is due to the
obstruction to the flow of electrons due to the collisions with atoms and other
electrons.
12]
(i) The substance 'X' is calcium oxide. Its chemical formula is CaO. (ii) Calcium oxide
reacts vigorously with water to form calcium hydroxide (slaked lime).
(iii)
13]
Main ore is cinnarbar, HgS.
When it is heated in air, it is first converted into mercuric oxide (HgO). Mercuric
oxide is then reduced to mercury on further heating.
14]
(a)
Water of crystallisation is the number of water molecules that combine chemically in
definite molecular proportion, with the concerned salt in the crystalline state.
(b) Two correct examples are:
Copper sulphate ,chemical formula = CuSO4.5H2O
Washing soda, chemical name = Na2CO3.10 H2O.
15]
(a) we can show it by an experiment of displacement reaction in which Mg rod is
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dipped in ZnSO4 solution. Magnesium displaces zinc from zinc sulphate solution.
(b)Cu is less reactive then hydrogen and thus place below hydrogen in the reactive
series of metals, hence cannot displace hydrogen in dilute HCl.
16]
(a)
Functions:
(i) Cerebrum:It controls the conscious sensations.
(ii) Mid brain: It controls the auditory impulses.
17]
Chemotropism – Movement of plant parts due to chemical stimulus is called
chemotropism.
Example - Growth of pollen tube.
Plant Hormones:
(i) Auxin - It helps the cells to grow longer at shoot tips.
(ii) Gibberellin – Growth of stem.
18]
(i) Glycogen and starch – Both are carbohydrates and stored food products.
(ii) Chlorophyll and haemoglobin – Both are pigments.
(iii) Arteries and veins – Both are blood vessels.
19]
Involuntary actions are the actions which cannot be controlled by us if we want to do
so. There is no external stimulus involved. The action takes place on its own like
digestion, heart beat, etc.
Reflex action is a kind of involuntary action controlled by and takes due to external
stimulus. Its respond to the stimulus is quick. Example – sneezing, blinking of eyes,
etc.
20]
Series:
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OR
For maximum resistance, resistors are to be connected in series.
For maximum current, resistance has to be minimum and so resistors are to be
connected in parallel.
I series = V/R series = 6V/9
= 0.67 A
I parallel = V/R parallel = 6V/2
= 3A
Parallel connection is preferred over series for house hold circuits.
21]
(a) Chloralkali process is used to prepare sodium hydroxide using sodium chloride.
When electricity is passed through an aqueous solution of sodium chloride (called
brine), it decomposes to form sodium hydroxide. The process is called the chloralkali process because of the products formed– chlor for chlorine and alkali for
sodium hydroxide.
(b) (i) 7 (ii) less than 7
OR
(a)
(i) The milkman shifts the pH of the fresh milk from 6 to slightly alkaline because in
alkaline condition, milk does not set as curd easily.
(ii) Since this milk is slightly basic than usual milk, acids produced to set the curd are
neutralised by the base. Therefore, it takes a longer time for the curd to set.
(b) A reaction in which an acid and base react with each other to give a salt and
water is
termed as neutralisation reaction. In this reaction, energy is evolved in the form
of heat.
For example:
(i) NaOH + HCl NaCl + H2O
(ii) During indigestion (caused due to the production of excess of hydrochloric acid in
the
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stomach), we administer an antacid (generally milk of magnesia, Mg(OH) 2 which is
basic in nature). The antacid neutralizes the excess of acids and thus gives relief
from
indigestion. Mg(OH)2 + 2 HCl MgCl2 + 2 H2O
22]
(a)Amphoteric oxides are those oxides which show both acids as well as bases to
form salts and water.
Ex: Aluminium metal reacts in this manner with acids and bases.
SnO2 is also an example of amphoteric oxides.
(b) Metals such as Sodium and Potassium are kept immersed is Kerosene
becausethey are very reactive and have high affinity towards oxygen and will
violently react with atmospheric oxygen on contact with air.
(c) Aluminium reacts readily with steam to give aluminium oxide and hydrogen gas,
the reaction does not always occur. This is due to a thin but strong layer of
aluminium oxide being coated onto the metal, thus preventing it from the reaction.
aluminium + steam
Al2O3(s) + 3H2(g) [1]
aluminium oxide + hydrogen 2Al(s) + 3H2O(g)
(d) (i) Bromine (ii) Iodine
OR
(a) Diamond and graphite are the two allotropes of carbon.
Diamond –
 hardest substance
 electrical insulator
Graphite –
 comparatively soft, it is slippery over layers
 good electrical conductor
(b) Aluminium articles have a longer life and attractive finish compared to many
other metals because of the formation of a thin transparent protective film cover of
Aluminium oxide on the surface of Al formed due to its spontaneous reaction with
oxygen.
(c) (i) Ore : An ore is a type of rock that contains minerals with important elements
including metals. The ores are extracted through mining; these are then refined to
extract the valuable element(s).
(ii) Gangue: In mining, gangue is the commercially worthless material that
surrounds, or is closely mixed with, a wanted mineral in an ore deposit.
(d) Electronic configuration of metal atom is significant to know about the kind of
bond that the metal will be forming for example in the formation of calcium chloride,
chlorine only needs one electron to complete its octet so two atoms of chlorine
accept one electron each lost by calcium ion.
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23]
(a)
(i) Part in which starch digestion starts – Mouth
(ii) Part in which bile is stored – Gall Bladder
(iii) Part in which nutrients are absorbed – Small intestine
(iv) Part in which water is absorbed – Large intestine
(b) Kills bacteria in the stomach (1/2) provide acidic medium for the action of
pepsin.
(c) (i) Controls the release of food from the stomach to small intestine.
(ii) Controls the release of undigested waste from the rectum through the anus.
OR
(a)
(b) Within the lungs, the passage divides into smaller and smaller tubes which finally
terminate in balloon – like structures which are called alveoli.
The walls of the alveoli contain an extensive network of blood-vessels which provides
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surface where the exchange of gases can take place.
24]
Brief explanation of activity:
Connect the circuit as shown in the figure below. Switch on the battery so that the
current begins to flow. Sprinkle some fine iron filings around the current carrying
wire. Tap the surface gently. The iron filings get arranged in concentric circles.
When current is decreased, field gets decrease
When the current is reversed, field also gets reversed
OR
A magnet is placed on a sheet of paper. A compass needle is placed near the North
Pole. The position of its two ends is marked with the help of a sharp pencil. Now the
compass is moved in such a way that its south end occupies position occupied by
north end previously. Again the two ends are marked with sharp pencil. In this way,
process goes on step by step till the south pole of the magnet is reached. Now all
points are joined to get a smooth curve which represents a field line. In this way
many field lines can be drawn
Region A has stronger magnetic field
Because the strength of the field is proportional to the relative closeness of field
lines.
25]
Less than 3.5
26]
Blue to red in A and no change in B
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27]
Metal
Al2(SO4)3
CuSO4
FeSO4
ZnSO4
(a) Al
X
?
?
?
28]
Carbon dioxide
29]
sodium bicarbonate > water > lemon juice
30]
Hydrogen gas and iron chloride are produced
31]
Student II only
32]
Both (ii) and (iii)
33]
Correct reading for current I incorrect reading for potential difference V
34]
17.5
35]
A1 = A 3
36]
1.37 ?
37]
To make the plant free of starch, it is:
38]
The part not covered turns blue
39]
(ii), (iii), (iv), (i)
40]
4
41]
(i), (iii), (iv), (ii)
42]
A is KOH solution and B is water
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