S .
ARITHMETIC
Revision Notes
The three basic
Number systems
The basic unit of
mass
is kiogramrne
(kg)
which is divided
into 1000 gramnes (g)
N ti, 2,3. 4
Length
Set of whole numbers,
0,
quantities in S.I. units.
Mass
Sots of rumbers
Set of nainursl numbrs
(
physical
is
The asic udit of length
1.2 3.4,
Sei of integers,
mece
Time
2={..3,-2, -1,0, 1, , 3,.
ts
Set of raiional numbers,
base unjt is sccond.
Density
The
is mass per unit volume.
The density of a substance
or kg/m3
units of density are gm/cm'
Q=(*: x=c,
d, e,Z, d 0}
d
Set of real numbers,
R = set of all rational and iTetional numbets
Density
Volume
Vhole Numbers
Se of even numbers=
Set of odd numbeers
=
2 , 3, 5. 7. 1t, ... )
Prime rnumbers are numbers that
1. ànd
are
divisible
i 15 n&i considered
by
them
1 km =
1000 m
=
then 2
gives an ever nLnber.
The factrs
of 72
are
:000 mg
Mass,
l tonne
Rational end irratienal nu.abers
The se: of ratfona' nuabers consists of terminat1ng
decimsls (e g. 5625 = 5
Irrationl numbers are nurnhers waich cannot be cx
pressed as a fractiou (e g. rV2 )
=
1000 kg
Average speed
Average Speed
=
distance travelledL
0tal
total time taken
Ratio
frac tion showing
ratio of two similar quantities is a
the first quantity as part of the second quantity.
The
and recurring dezinmals
(e.g.02727
(1
The ratio
where &*0,
of a to b written a: 5 is a+ b or
a, be Z
(ii) Ratio has no units
()Vulgar fractúon and decimal
The vulgar frtctiorn of 0.25 '
1000 g
kg
Integers
The set of integers consists of positive integers,
is nejther positive nor
regative integers and zero. Zero
1000 ml
1000 cm
1, 2, 3,4, 6, 8, 9, 12, 18, 24,
36 and 2
10* m
100 hectares
1 litre =
Volame
10 mm2
10* cm
hectare (ha) =
negative.
d
100 cim
m
The pnme facturs of 72 are 2 and3.
E)
10 m n
=
1 km*
and (2x-1) gives an odd nunbe:
g
=
l m
1 em
Aren,
prime
number
If rE W,
I cm
Length.
= , 3.5, 7,
Set of prine numbers
eives and bj
Conversion of units
2. 4,6, 8,
of multiples of 2
. .
set
= .
cf, is 0-25
The decimal
Percentages
A traction
Square, square root and reciprocal
with
1%
a
denominator 100 is called
100
a
percentage.
66% = /3
The squere sf x is x2
Lhe square rtot uf x is Vx is.r
The
reuiprocal
Anthmetic
of
cis
25%=
33
=
75% =
100%
=
Y4
1
'O' Maths D Paper
ARITHHETT
Decimal places
Proportion
5314
Inverse proportion
If a unit of flat can be completed by 10 workers in 20
days, the number of days to complete a similar unit by
20 workers is
i)
=
531
(correct to 1 dec. place)
07364= 074
(correct to 2 dec. places)
7-1302
7-130
000743= 0-007
(correctto 3 dec. places)
(correct to 3 dec. places)
10 days.
20
Standard form
Direct proportion
Very large or snall nurnbers can be erpressed in standard
A toy costs $20, the cost of 5 similar toys is
form ax 10", where 1 sa< 10, and n é Z.
5 x($20) =$100.
1,350,000 = 1-35x 106
e.g
0-00135 = 1-35x 103
Discount
A discount is a reduction in price, usuelly given as a
percentage of the list price.
Variations
()
Percentage profit or loss
cost price
100%
(i)
S.P.-CPx
100%
C.P.
loss
Percentage loss
=
kr where k is a
constant.
profit
Percentage pront
Direct variation
If y varies directly as z, then y
Inverse variation
If y varies inversely as z, thea y = where k is a
constant.
100%
Cost price
CPSPx 100%
In general,
percentage increase
LRCrease
=
x
100%
X
200%
original amuunt
percentage decrease = - -
ese
onginal amount
Simple interest
Interest, I= FR, where P = principal
100
R
=
Tate per year in percentage
T
=
time
Money
Singapore
$1
I00 certs
British
£l
100 penice
Significant figures
Tbe first non-zero digit ef any nunber is called the firs
significant figure.
236
200
00236
0-024
356
360
03002 0:300
6-049
6-05
6-049 6 - 0
Arithmetic
(corect to 1 sig. fig.)
(corect to 2 s1g. fig
(corect to 2 sig. fig.)
(correct to 3 sig. fig.)
(correct to 3 sig. fig.)
(correct
to 2
sig. Fig.
O
Maths D Paper
erimetertdis
distance al
the
r o und
MENSURATION
Revision Notes
Fqure
2-dimenson obj«cks
Figures
Diagra1s
Area
Perimeter
Square
4
Rectangle
lxb
b
L
P: 2(l+ b)
/ 2 x base x height
Triangle
= Vabh
or
Paralelogram
base x height
= bxh
h
A
:
a+b+c
72bc sin LA
2(a+b
B
Trapezium
V21a+bh
Rhombus
Where p & q
are lengths of diagonals
Circle
Annulus
Mensuratlon
Lr
t
Circumference 27t
T(R> - r)
O' Maths D Paper
MENS RATIoV
Figures
Diagrams
Area
Sector
0XTTr
360
Solid
Caboid
Perimeter
Diagramns
x
360°
Volume
-12
}tr +2r
Total Surface Area
Ixbxh
2CL h+hL)
Cylinder
Curved surface area = 27trh
Trh
Total surface area = 27trh+2t
close
T SA
pea o
Area of cross-section
base area x height
Prisn
x
=
1si/a
length
OR
22Acea
Pyramid
Cone
Sphere
A
base area x h
3 trh
3 Ttr
las iLukoa.
T:st= Sn
a
Curved surface area = trl
Total surface area
Cri + tr
Surface area = 4TTr
A3Ar (ciosed)
Mensuration
O' Maths D
ALGEBRA,
Revision Notes
Approach for its solution
Change of subject of a formula
By factorisatiou
The subject of e formula is the variable written explicitly in
terms of other given variables.
Solve the equation
2
+ 5x-3 = 0
(2x-1)x+3) =
We can have,
Givea that A = 2 (c + d) h,
express h in terms of A, c andd.
A = h (c +d) h
(i)
hor-3.
Using the quadratic formula
2A (c+d)h
=
Yb-4ac
2a
h =2
C
0
=
+d
Solve the equation x - 6r + 4 = 0, giving your ars
corect to 2 decimal place_.
Some useful identities
a= 1, b=-6, c = 4
(a +b=a?+ 2ab + b2
(a-b=a2-2ab +b2
2-
)£Y2-4(1)
(4)
2 (1)
(a + b) (a - b)
*V 20
2
Factorisation
(
20
or
x
or
==-
Common Factors
Consider the expression 4ab +2ab - 6ab?.
=
5.24
(ii) By completing the square
Each tern in the expression contains 2ab.
4ab +2ab - 6ab2= 2ab (2a - 3b+ 1)
Steps: (a) Change the coefficient of the
termtodne
(b) Bring the constant term (c) to the other s/«
() Grouping
Consider:
= 0.76
of the equation.
3x +10xy -5y 6a2
(c) The coefficient of the x term is divide d
2, then introduce he square of it tw
(3x- 6)+ (10y-5)
sides of the equation.
(d) Take the square roots on both sides f éhe
3x(1 2x) + 5y(2x- 1)
equation and simplify the terns.
= (3x-5y) (1 -2x)
Solve the equation 3x2 - Sr - 9 = 0, giving your ans
(iii) Difference of two squares
correct to
25x-4 = (5x2)2 -22
=
2 decimal places.
3x2-5x 9
(5-2) (5x2 +2)
= 0
- a x -3 = 0
1v) Cross Multiplying (Trial & Error)
2- hx = 3
To find the factors of 2x + 5x-3.
a-(
- 3+ ()*
+6x
+3
-3-
-X
+5x
Hence, 2x+ 5x - 3 =(2x- 1) (+ 3)
-
Quadratic equation
An
equation
of the form of ar
bx + c
=
0 is called
quadratic equation; where a,b and c are constants.
Algebra
r x=
+
bs
th
a
X
-1.09
or
z = 2.76
0ORDINATE GEOMETRY
Revision Notes
niices
yT
B(2. Y2'
71-
(")-
.
nidpoinat
A
,
The gradient ot the line joining any two givzn polnls
A
and
)
a#
B2 y) is
The mid-point
of
.0
-m
the line
joining
points AX, "
two
and
)
Inequality
The distance between
two
points A(Ti. y1)
B(X2. Y:)
and
S
Froperties of tnequalities:
given by
Lf
u
AB = Na - x + 0 - y
a
~
b, then
a +k2 b +k
a-k2b-k
(iThe gradient of a horizontal line is
gradient of
vertica! line
1s
7ero.
Ci)
The
i)
The equation of
a
horizontal line is y
i)
Tne
equarion of
a
verti-al line is
a
(ii)
undefined.
x
=
=
Ifa 2b and c20, then
ca
cOnstant.
cb
Constant.
ii)
ifa 2 h and d <0. thea
Paraliel lines have tne same gradient.
dad
ine
o a straight
gradient'intercept forr of the equaiion
intercept on y-ax.s
and c
m r * C , Waer: m
gradien:
ISy
The
=
=
Numter iine
2
Gradient =
-
PorPendicu lar
frod
ha
gradyet of
x M
Coordinate Geonetry
rad en
-I/
rpnJult le
-
GRAPHS,
KINEMATICS
Revision Notes
function:
(vi) Graphs ofexponential
a
is
positive integer and
y a", where a
Graphs of functions
Graphs
(1)
ofy=ar" where
Graphs
n
=
£1, t2, +3.
oflinear functon:
y
=
ar, when n =+i.
a<0
a>0
(iu)
R.
xe
Graphs of quadratic function; y
=
axs,
Distance- time graph (s4 graph)
when
n
=
+2.
at a
The gradient of s t graph
instant (refer to figs. 1 &z 2).
P gives the speec
point
A
ig.1
Spred i
a>0
(iii)
Graphs of y= a r , when n = +3
rest
uniform
speed
A
Fig. 2
ac0
a>0
iv) Graphs
ofreciprocal function:y ,
=
when
n
=
-1
Tu
Speed
time
graph (p.4 graph)
v - i graph
gradient of the
instant (refer
acceleradoi at that
gives the distance
under
The
point P gi
figs. 3 & 4). T
at a
to
travelled.
graph
Acc
VA
a<)
a>0
(v)
Graphs of reciprocal function: y =
chan
chan
Fig. 3
when
n=
-2.
uniform speed
acceleration
Fig 4
a>0
Graphs, Kinematics
Acceenb
Graden
t
cc
Kebord ho n - A a
qnde
Vistana travel/l AAa
vt
GEOMETRY
Revision Notes
(alt. s,)
a=b
1.
Basic Angle Properties
(corr. 2s.)
c=b
b+d= 180" (nt. s between il lines)
a+b 180° (adj. <s on straight line)
a+e= 180°
2
aa
A
b
R
a+b+ C +d+e = 360° (Zs at a point)
angles if a
u
and b
are
complermentary
a
and b
are
supplementary angles
if a
+
+
b
=
90*
b= 180°
a b (vert. opp. <s)
c d (vert opp. Zs)
a
2.
d
Triangles
Acute-angled
Right-angled
Obtuse-angled
Tiiangle
Equileteral
Triangle
Triangle
Triangle
Triangle
All sides are of
Two sides are of
All sides are of
All angles are
One angle is
unequal lengths
equal length
equal length
acute angles
à nght angle
Scalane
Isosceles
Triengle
Geometry
One angle s
an obtuse angte
GEOMETR Y
3.
Quadrilaterals
Both pairs of opposite
sides are equal and parallel.
Both pairS of opposite
Diagonals bisect each other.
angles are equal
parallelogram
All tour angles are rnght
Both pairsS of opposite
SIdes are equal and
Diagonals
Angles
Sides
Diagonals bisect each other
and are equal.
angles.
parallel.
rectangle
All four sides are equal.
Opposite sides are parallel.
All four angles are right
Diagonals bisect
angles
Square
All four sides are equal.
Opposite sides are parallei.
each other
at right angles and are equal.
Both pairs of opposite
angles are equal.
Diagonals
bisect each other
at right angle,
Thombus
Two pairs of adjacent sides
are equal but not al four
sides are equal.
Only one pair of opposite
angles are equal.
Only
one
diagonal bisects
the other diagonal at right
angles.
kite
Only one pair of opposite
sides ae parallel.
trapezium
D Paper
O ' Maths
Gemetry
GEoHETRY
Tangents
length.
of Polygons
Angle Properties
4.
to
a circle froi
an
external
point
are
equal
s/
Angle
surn
of triangle: p
+q
+ r=180°
Exterior angle of triangle: s = p + r
tp+4
6.
of circle
Angle properties
(b)
a)
a
Equilateral triangle:
Isoceles triangle:
a = b=c = 60°
a = b
Sum
n-sided polygon
of interior angles ofan
Each interior
angle of a regular
n-sided
=
(n -2)
Angles in the
Angle in a semicircle = 9
same
segment are equal
180
(d)
(c
polygon
7-2)x 180
Sum of exterior angles of a polygon = 360
Each exteror
n-sided polygon
angle of a regular
of a quadrilateral
Sum of interior angles
=
=
360
360°
|Angle at the
centre
is twice
the angle at the
Angle between tangent and
chord equals the angle
subtended in the alternate
circurnference
5.
Symmetry Properties
ine from
centre
to
LA = 2 x LB
of a circle
mid-point of
segment
chord is perpendicular
(e)
to the chord.
Cyclic quadrilateral
Equal
chords
Geometry
are
equidistant from the
centre.
Angle beween tangent
LA + L C 180°
and radius = 90°
180°
ZB +2D =
GEOMETAY
7.
Congruent triangles
A set of triangles are congruent
(i)
Sane
S
5 at
8. Similar tríangles
Sumt hase
shapt
difkm t
if two triangles have equal corresponding angles, they
are sinilar.
If all three sides of one triangle are cqual to the
corresponding sides of another triangle, the triangles
2.
If two tiangles have their corresponding sides in the
same ratio, they are similar.
are conguent.
Abbreviation: SSS
.
If two pairs of corresponding sides of two triangles
are in the same ratio, and the included angies are
equal, these triangles are sirnilar
Exanple:
Similarity
i)
F two sides and the included angle of one triangle are
equal to the corresponding sides and angle of a
second triangle, then the triangles are congruent.
Abbreviation: SAS
D
Corresponding sides form equal ratüos
DE
B
(ii)
If two angles and a side of one triangle are equal to
When two figures are similar.
the ratuo of the areas = (the ratio of the corresponding
the corresponding angles and side of a second
triangle, then the triangles are congruent.
dimnsions)
When two solids are similar,
Abbreviation: ASA, AAS or SAA
the ratio of the volumes = (the ratio of the comesponding
dimensions)
Example:
lf X and Y are two similar solids, then
B
C
R
i ) f two triangles are right-angled, and the hypotenuse
and a second side of one are equal to the hypotenuse
and a second side of the other, the triangles are
congruent.
Abbreviation. RHSs
Gil)
T
B
Geometry
O' Maths D Paper
TRIGONOMETRY
Revision Notes
3nyth s gm
eabhs
and
ncudd
Cosine Rule:
Ang
Right-angled Triangles
side AAC
right-angled at B, the longest
is the hypoteruse.
(opposite the righi angle)
In
a
triangle
a=b+c - 2bc cosA
+ C - 2ac cosB
ABC
c=a+ b -2ab cosC
When i s obruse (90° < 8< 180°).
sin (180-6)
cos
(180°-6)
sin
=
6
-cos
Area of a triangle
A b a s e x height
Choosing one of the angles:
2
BC
side oppositeor
or =2
sin 6
AC
hypotenuse
=bc sin A
side adjacent or = AB
cos
hypotenuse
sideopOSt
tan
or
2
BC
a c sin B
AB
side adjacent
BC2 (Pythagoras'
AC2 AR2
ab sin C
AC
2
Theorem)
Bearings
CHypAngles of Elevation and Depression
Bearings
wise
are
always
measured
direction and stated as
a
from the North and in
a
cl*
three-digit nurmber
B2-
North
a
A
North
level OA,
from a horizontal reference
of point P from 0; the angle
is called the angle of elevation
from P.
O
of
of
depression
the angle
The
angle
a, measured
240
B, is
Non-right-angled Triangles
A
The bearing of B from A is 060
The
bearing of A
from B is 240°.
Ba
B
e
Add 160
Sine Rule:
b
sinA
sincB
Trigonometry
les
sin4
and
eth
Po ssi hle
6+13)
(
3o)
Mins
LOCI & SIMPLE CONSTRUCTIONS
Revision Notes
(d)
Loci in Two Dimensions
equdistant from two intersecting lines is a pair of lines
and
The locus of
(a)
a
bisecting the angles between 1 and iz
moving point P
at a fixed distance r from a given point A, is a circle,
centre A.
-
(b)
at a fixed distance r from a straight line 1, is a pair of
lirnes parallel to h one on each side of .
Locus
io3-o )
Poin ts
(a)
whrch a r e
A
equidistant from two given points A and B is the
Poinks
perpendicular bisector of the line AB.
6cm
Fromm
sphre wn
A S-
(c)
l
radins 5 em
5m
wich
fam
AB
nder
its
s,
radiS
A
*B
Points
trom
which
A
are
edia tas
and
B
A
Loci &Simple ConstructionsS
Scm
PoSSin
B
t AB
prptndicular
oF
i d point
throuh
the
O' Maths D Paper
STATISTICS
Revision Notes
consist of a
or bar
series of bars all of the same width. The bars may be drawn
diagrams
1Simple bar charts
vertically
or
horizontally. Fig.
1 shows
simple
bar charts for
the informmation of Table l.
3
The histogram
This is a diagram which is used to represent a frequency
distribution. It consists of a
Type of personnel
Number employed
Unskilled workers
90
Craftsmen
50
10
20
Designers
Cierical staff
of rectangles whose areas are
frequencies of the various classes. If all
the rectangles will
the classes have the same width then all
be the same width and the frequencies are then represented
3 shows the histogram
by the heights of the rectangles. Fig.
for Table 2.
proportional
Table 1
set
to thc
20
15
170
Total
Designers
6
Clerical staff
100
M
Fig
8
A histogram for a grouped frequency distribotãon
drawn by using the mid-points of the class intervals as
the
the centres of the rectangles. The histogram for
distribution of Table 3 is showa in Fig. 4. Note that the
Craftsnen
80
60
Unskilled workers
40
extrernes of each rectangle rrpresent the lower md
upper ciass boundaries.
20
80
0
Fig. 1
20 40 60 80 1000
Number employed
40
2
Pie charts dispiay
Designers
the proportions of the
whole by means of
sector areas. Fig. 2
Clerical
angles
2-5 6.5 |10.5 14.5 18.5
Mass (kg)
420
Class width|
=4 kg
4.5 8.5
the information of
Table 1 and the
calculations
that
follow show how the
sector
21
stali
shows a pie chart for
Fig.4
191°
Unskulled
106
are
Mark
Frequcncy
workers
Craftsmen
obtained.
18
5
8
Fig.2
Unskilled workers
= 90 360 = 191°
170
Craftsmen 6 =
50
x 360° =106
Mass (kg
Frequency
x 360° =42
14
5-8
50
54
9-12
13-16
17-20
26
170
Clencal staff 6=
20
170
Designers 6 =
170
Table 2
Total
x 360°
21
Table 3
' ' M a t h s D Papar
Statistics
SIATIs1T¢s
(ii)
For Unequsl Class Intervals
Table 4
Rectangles are drawn so that the area of each rectangle
is
t.
proportional to the frequency
1.e. area
frequency.
Height (cm)
in the range covered by
Exanple:
The frequency distribution gi ves the marks of ó60
0-9 10-14 15-19 20-24
Frequency
13
19
12
6
+
165-169
29
4
Total
325-34
7
I00
9
50
Solution:
Step 1.
155-159
160-164
170-174
pupiis in s test
Mark
Frequency
150--154
«
+0
The class boundaries are
0. 9.5, 14.5, 19.5, 24.5, 34.5
30
Step 2.
The corresponding class widths are
Step 3.
We choose a class widh of 5 as the standard
20
9.5, 5, 5, 5, 10.
10
width.
152 157 162 167 172
Step 4.
f class width
=
n
x
ofrectangle = x
standard width, then height
coresponding frequency.
Height (cm)
Fig.
5
Step 5.
5
Marks
Ciass width
(nx Standard width)
0-9
-14
15-19
x5
x5
There are threc main statistical measures which atternpt to
locate a 'typical' value. These are (a) The median, (6) the
(stundard frequency)
3 1 3 =6.8
20-24
Measures of central tendency
Height of rectangle
1x
19
= 19
1 x12
12
1 x7
25-34
x9
mode, and (c) the arithmetic mean.
(a)
The median
If
a set of values is
arranged in ascending (or
descending) order of size, the median is the value
= 7
which lies half-way along the series.
4.5
The median of the numbers 3, 4, 6, 7, 8, 8, 9 is 7
because there are three numbers below this value and
three numbers above it.
When there are an even number of values, tde mediann
is found by obtaining the mean of the two
middle
values
E.g. The median of 5,7,7,8,'0,11,I1,13 is V2(8+ 10) = 9
0
9.5 14.519.5 24.5
(b)
The mode of a set of values is the value which occurs
Marks
most frequently.
Histogram to show the marks of 60 pupils.
The mode of 3, 4, 4, 5, 5, 5, 6, 7, 7, 8 is 5 because ihis
Frequency polygon
This
dis
is
a
ribution.
second way of
It is
drawn by
The nmode
34.5
representing
value occurs three times (a fiequency of 3) which is
more times than any othe value.
a
frequency
connecting the mid-points
tops of thee rectangles by
straight lines as shown in Fig.
at the
5.
The modal class
Example:
When data has been grouped into classes, the class
DTaw frequency
for the informalion given in
Table 4 which relatespolygon
to the
heights of men.
which has the largest standard frequency is called the
a
modal class. An estimate of the mode can be obtaine
from the modal class.
Statistics
O' Maths D Paper
STeL
The
(C)
mean
arithmetic
arihmetic
The
vaues
in
IICs
a set
found
is
mean
(raw
all
by adding up
sum
divinding this
datal and
the
values mak1ng up
the number of
the number
805
1620
2608
640
161
by
162
set
163
64
Sum of all the values
ArithmeTic mean=
160
60
the
165
of values
332
166
The
8155
50
Example:
heights
of
are
people
some
173
170, 181, 175, 179 and
mean height of these people.
as
172.
foll ows.
Calculate
cm.
the
L
We have
=
8155 and
mean
=
Therefore the
1050
The
(iv)
The arithmetic
an
assurmeed mean
(Fa).
i.e. T
a
=
Let
an
assumed
mean
Where d=and d =z-fa
example in (i)
(ha)
( ) of an ungrouped
the Assune
Distribution, using
be 173
The calculations
Example
: Refer to
Solution a
=
ha -
can
be made easier by using
an
assumed mean.
above.
cm
Deviation d
Height, h
mean
za.
n:number of values in the set given.
Refer to the
cm.
Note: =a +
+
reading from
where d: deviation of each
Example:
heightis 163.I
Mean (a)
of the set of values. using
mean
mean
arithmetic
Frequency
= 175 cm
(ü)
50. Hence
=
163.1
+I/0+181+175+179+173
Arithmetic
n
example above
= 163
173
d = - 163
170
172
173
160
175
179
6
8
18
E d = 4 +16
162
163
10
164
10
6
2
165
12
166
The
mean
height
h
=
173
175 (cm).
+
= 1630
= 163+01
= 163.1
where x are tle measured observations and j
Therefore the
the corresponding frequencies.
of the products
corresponding values off and x and
total
the
making
tlie
is the
of
su n
number o f items
2fa-23+28-5
= 163+d
Efx
is
6
50
frequency
(i) The arithmetic m e a n of an ungrouped
distribution by using the formula
2fx
-10
-10
161
n
uP
cm
Frequency
Mean (T) of a grouped
Distribution, using the Assumed Mean (Fa)
Note
Whep data has
been
grouped iato
of
Example:
a t d, where =
distribution:
Bxample:
Heighe th cm)
Nuinber of students
heights of 20
shown.
The
60<h
7
s
70
70
students
are
h $ 80 80 <h S
8
O'
Statistics
t
nid-point, x,
intervals,
the interva
nterval is taken to represent
the
Find the mean value of the following frequency
Height (cm) 160 161 | 162 163 164 165 166
6
5
1016 10
Frequency
l63.1
height is
(V)The
the
frequency distribution.
mean
Maths
D Pape
T
i-
Let Ag 75, then d =x- 75,
where
t
is the
mid-point
Heght (cm) Mid-point, x
60« h 7 0
80 <h90
an
Examination Questions for Unit 9
interval
I
The pie chart represents the distribution of population in e
village.
d -75
6
75
S5
70<h s880
of
-10
i
-
Calcuiate the percentage
of the population
0
50
10
20
Z.fd
(ii)
70+50
The
Measures of
6
75
+
mean
20
Given that there are
is 74
a
meas ure
cm.
of the
following figure
shows
a
cumulative
had read in the last month.
The results of the survey are
shown in the bar hart.
frequency
Find
curve.
(1)
the modal number of
Cumuiarive Frequency Curve
books,
004
Tpper
75
2uarile
25
3
3
4
(iiu
the proportion of children who read more than } books.
Answer the whole of this
question
or
June 90//19
graph paper.
Mathematics and English
Mark
2, 2
60
2
80
OSrS20 20<s40 40ks60 60s80 80«s100
1
MathematicS
Marks
values
(i)
is calied the lower quartile or 25th percentile. ( 4 n a "
25
12
2
English
Tbe range is the aifference between the highest and west
,
2
The following table gives the frequency distriburion o*
marks obtained by 80 candidates in exarninatuons in
-
40
1
Nunber of bocks
the total nuraber of books.
giving your answer as a fraction.
ure
20
O
(ii)
-
MMedian_
50
June 90/U11
2 A group of 25 children were
asked how many books they
Cumulative frequency curve
The
Women
tii) Given also that the
number of boys is 432.
calculate the total
population of the viliage.
dispersion
There are several ways of
obtaining
spread' of a set of observations.
Boys
as men calculatex.
=74
height of the students
Crirls
twIce as many WoTTien
-20
I=
Men
who are giris.
10
33
3
Copy and complete the fllowing table showing the
curnulative frequency distribution in each subject.
22 is called the middle quartile or median r 50th percentile.
Nunber of candidares with
this mark or less
2 is called the upper quartile or 75th percentile.
2-2, is called the interquartile range.
Mark
English
Mathemancs
20
40
20
60
80
100
1
Using
a
scale of 2
iorizontal axis and 2
e
80
cm to
represent 20 marks
On uie
c n to represent 20 candidates on
axis. draw separare cumulative
requ
tor each of the subjects Mathematics
vertical
Clagrams
ana
English
9 Statistics
O
Maths D Paper
SIMPLE PROBABILITY
Revision Notes
Definition
Red
2
ff a event A can happen in a ways out of n equally likely
ways,
then the probability or chance of its happening
2
2
denoted by piA) is defined asp(4) =
2
number of outcomes favourable
i.e. Probability = -
Blue
to the event of 4
equally possible
33
34
6
total number of
outcomes
3
4
5
4
7
8
67 8 8 9
Properties of probsbility
()
If A is an impossible event, then plA) =0
(i)
fA i5 a sure event, then p(A) = 1|
Tree diagram
Tne outcomes are written at the ends of the branches, and the
coresponding probabilities by the side of each branch.
i ) IE A
is any event, and 0
p(A) S
1, then
PA) = l -p(A), where A' is the event does not occur.
(iv)
Example
fA and B are two muially exclusive events, then
An Australian has three 50 cent co:ns and wo 10 cent coins
pA or B) =p{A) + p(B).
n his pocket. He takes coins out nf his pocket, at random.
one after the other. The coins re not replaced.
i.e bothA and B events cannot occur at the same time.
The tree diagram below shows the possible outecmes and
( ) I f A and B are independent events, then
PA and B) =plA) xp(B).
their probabilities
2d coin
1 coin
i.e. both events bear no relation to each other. One
50c
event has no effect whatsoever on the other.
Possibility diagramn
50c
t involves the representation of outccmes by points on grid.
10c
Example
The four faces of a red tetrahedral die are marked 1, 2, 2 and
50c
B.
The four faces of a blue tetrahedral die are marked 1, 2. 5
andd6.
10c
When such a die is thrown, the 8core i5 the number on the
face on which it lands.
10c
h e two dice are thrown together and their scores added.
The possibility diagram shows alil of the totals.
Simple Probability
°
Maths D Paper
TRANSFORMATIONS
Revision Notes
Geometrical transformations (using coordinates)
angie of rotation is . (clociwise 4). LAOA
BOB= LCOC = 9 0is the only invariant point. The
It the
Reflection in the x-axis. p(a b)p\a, -b)
order of the letters is preserved,
Reflection in the y-axis. p(a. b) a
Congruent
p'(-a b)
the triangies are directly
Reflection n the liney =z pla, b) +pib, a)
Relection in the line y = -a, p(a. b) ++ p'(-b, -2)
Reflection in the line x =C, pla, b) ++ p (2c-a b)
Reflection in the line y= c, pla, b) +pla 20b)
The rotati on of 90
anti-clockwise about the origin is grven
by p(a b)p'(-b. a)
The rotation of 90
clockwise about the origin is given by
pla b)F'(b, -a)
The
rotation
of 180° about the
origin
is
given by p(a.
b
Relectionn
AA 'B'C' is the image of AABC under a reflection in the line
OP. OP is called the axis of reflection which is the perpen-
Enlargement
dicular bisector of AA', BB' or cC
Cnaer a reflection, the figure and its image are congruent.
mediator ('nrTor)
-H---dT-+IE - -
-
(a) k= -1/a
4
(b)
k= 2
(c)k =2
Under an enlargement with cente O and scale factor &. e
Transiation
area of the imaze is K2 times that ofthe ubject. IfK> 0, he
given figure and its image are on the sarne side of tho cenae
ot enlargement O. f
B
R < 0, the given figure and its im4
are n opposite sides of the cente of enlargemeat O.
Under an enlargement, the odject and its inmage are SLIa
and their corresponding sides are parallel The scale factor
OB
C
ere are no invar1ant points. The order of the letteis is
preserved, the triangles are directiy
congrue nt.
Rotation about a point
RA)=A', R(B) =B, RIC)= C.
The centre of rotation O is a the point of intersectjon of the
perpendicular bisetors of AA, BB', CC
Transtormations
Shearing
Scale
AOPQ' is the image of AOPQ uader a shear witn
factor & along the X-axis. The x-axis is the invariant u
Scale factor
K=
r 2
O P or
QR
and
TRANS foR MA 1Tw
MaLaa
a
shear,
arta ol any
tiRure
remains
the
ame
C
clocKwr
st
abt
abc kuis
./ 40
()
-
RePieeFo
Stretching (*)
a
stretch with scale
is the image of OPOR unde a
QR
is the invariant line and
actor X rom the -axis. The y-axis
OPPROThe area of the imags
ne scale factor is K
Under
stretch is K times the
a
area
of
RePle..
original figure.
in
Refler tian
Elarstmel, cnk
S
Ombination of Trunsformations
then XY denotes
and Y are two. transfermations,
X.
transfomation
Tansformation Yfollowed by
T
denc
es transformation X
followed
by
transformation Y.
s hear
.
shear
RANSi AT]
dett
wtu
-atis
invais
Pacr K.
A***
shew
inv
shar
paralei
ais
+
man's
I)
-
S trtkh
(.
(,
Tw
t
invarian, Scale fach
PArallel t
nrarm', S P
sbe rh
w
sereteh
13/.
-1
strh panllel
Rleebioa
in
.
-ava, sf
iAvarih
anthe
ias
th
g
VECTORS
IN T v O
DIMENSIONS
Revision Notes
(f
(a)
A
vector
may be
(b)
rerresenied
A
vector
whose initial
or
4
or in
point (h. k)
If P is the
tion
A
a
and
then the
origin,
o the
posi-
iector
OP
:qual
Two vecbrs e
direction and megnitude
when
they
have the
sarne
of
(g) Magnitude
CD means
AB
called
to O is
of P with respect
Equal Vectors
1)
O is
point is the origin,
position v e c t o r
in the forn AB
)
components
Position vectors
both
quantity that has
is
A
magnitude and direction.
a
vector
vector
a
the magniude of ABAB= .
cD nd
(h) Mid-point
IiMis the rud-pointof AB.
V ect»rs
(c) Negative
OM=
then
04 0 B
AB and Cb have bhe
same magnitude but in
oPposite
directions
(i)
we
Multiplication
by
a number
wite
(1)
d)
k =
Addition
Itk
Vectors are
ndded by
the
ringle
law
or
the
parallelo-
is
positive, then
If k is
grain l a
negative,
u and
then u
same sene
have the
ku
and
ku
have
opposte
senses.
the
sa
is positive, ka is a vector háving
equal to
direction as that of g and "magnitude
times and magnitude of a.
:! i t k
It kis negative, ka is a vector hav1ng the opfos
direction to that of a and
umes the magninude of 4.
Fig
Fy
In fi2.
AB tBC AC
In fig. Gi
AB+AD AB BC = AC
(ii) a=mblal =m ibl and a' b.
iV
If
V
tvector
(
(e) Subtraction
magn1tude equal
vector a
then h
ra
parallel
is
a
is not
to
vector b then a k
parallel
to
vector ò
and ha
0 and k = 0
+
mb
=
ha + kb
and
a X b, then
n
-*|
h and
"*
k.
ABB=AZ
ko
B
- AB
D
Maths
O
Pap
Section
Form
Index No.
Date
Paper.
Do not
write in
ject.
S l l b IS
Additicn
OLaodinate
Arg the
this margin
geometny
th2
quodient o f
2)
t h e p o i n t _CL
line_passinO
Hhrougb
anc C 6 ) .
qralien ta
6-2 E
m
Hence,
OL }ne
hndtho cr CAdionE
line
_perperdicuO
5_OA line 2m.
qrcdien
recentY-2iperdicular lioe
hindthe
equction
2e2CD tc
2
2t
C2
of
petpordiculQ
m
Do not
write in
this margin
statistks
Range
a ni n t e r v a l
A anges the
anme z
i
differen.ce
s0-10: Yo
2elatve feCIENCY elotive fequery
Plative
isro
toHhetota
ekative
etween
yequenCy Deing
cequency
equenc
Recik
Toaie P
A
COlour
frequeray
BIue
reed
C
NLedvellou
Back
-
1ot
3
y
2elcrave_frequemy_oc
blue
relative_pequeaCuo biackE
dicqrQm and
Corelabion.
Scatter
_A_SCatkr diaQYAm
vsuai labIShip
On thg
ocSatter
plot Shouos a
measue¢_
pehuoEDN_
SaNeSetof
Civaunis
Camp2
mavk in
ShoLos Hh2 lest
table
Tho
and p n y s i c s r loSdents.
MctthemaiC
Section
Form
T
Index No.
Date
Dject_
Stter
rehlieashep
UiSual-
Shows
Masutd
the
on
this margin
S_a ller p l t
iaam
SCa bte
VNinses
Do not
write in
Sarre laian
hehwes
S a p 1 es e t
L a d iv i i u a l s .
Tht
table
Shas
test_macks La Ma hemabes
the
studeats
hyscs
Helieaahe3o|5o3s|25 S |34|H8 HoaS
26 l39 3s|28 1 l31|us |33|Ll12
Phgsics
StterolieqLam
to
epessat
Machs _for
mathemuti es a A s
Che
e
lostudent6
30
2
5
s
3oSS
n.tlols
Ho
HS
-
Marh_
CirAatisn
T
he
thert
Uariables
Pants
Lhesatter dinqamappeozta t
A
straight Ane
A
Called_a
pcel bo a. between_the
two
osve
NegabeCutlab4
Corhben
Do not
write in
this margin
decege
No
Corlabiia
abare he
auung Us many pamt
line a s below
I t shoudd also pass hrzugh
ardCnd * *
mean
and
2
ne
Drav
he
22
in
malhan
.a r k FC
bes
.
heace
".. slet
estiaale
JCOeA