COMMONWEALTH SECONDARY SCHOOL
END OF YEAR EXAMINATION 2022
MATHEMATICS
SECTION A
Name: ____________________________ (
)
Class: ___________________
Tuesday 4 October 2022
SECONDARY 1
EXPRESS
08 00 – 10 30
Total Duration for Sections A & B: 2 hr 30 mins
READ THESE INSTRUCTIONS FIRST
Write in dark blue or black pen.
You may use an HB pencil for any diagrams or graphs.
Do not use staples, paper clips, highlighters, glue or correction fluid.
Answer all questions.
If working is needed for any question it must be shown with the answer.
Omission of essential working will result in loss of marks.
The use of an approved scientific calculator is expected, where appropriate.
If the degree of accuracy is not specified in the question, and if the answer is not exact, give the
answer to three significant figures. Give answers in degrees to one decimal place.
For π , use either your calculator value or 3.142, unless the question requires the answer in terms
of π .
At the end of the examination, hand in Sections A and B separately.
The number of marks is given in brackets [ ] at the end of each question or part question.
The total number of marks for this section is 50.
For Examiner’s Use
Presentation
Accuracy
Total
50
2
This paper consists of 17 printed pages including the cover page.
[Turn over
CWSS / 2022 / EOY / 1E / Math / Section A
3
1
(a)
Round 48.6794 to 2 decimal places.
48.68
Answer ……………..…………………… [1]
(b)
Calculate
(i)
1.12  9.47  3.02 
.
3.1  24.67
Write down the first 5 digits on your calculator display.
3
0.2987
Answer ……………..…………………… [1]
(ii)
Write your answer in part (i), correct to 2 significant figures.
0.30 (2.s.f)
Answer ……………..…………………… [1]
2
Write the following numbers in order of size, starting with the largest.

1
 , 1.4,  2, 1.41,
3
2
1

2
1.4
1.41
3
2
Answer …………, …………, …………, …………, ………… [1]

3
Written as the product of its prime factors, 1400 = 23 × 52 × 7.
(a)
Express 490 as the product of its prime factors.
2 × 5 × 72
Answer 490 = ……………..………… [1]
(b)
Hence, write down the greatest integer that will divide both 1400 and 490 exactly.
1400 = 23 × 52 × 7
490 = 2 × 5 × 72
Greatest integer
=2×5×7
= 70
CWSS / 2022 / EOY / 1E / Math / Section A
70
Answer ……………..…………………… [1]
4
4
A number has exactly 12 factors. Two of the factors are 8 and 12.
List all the factors of the smallest possible number.
LCM of 8 and 12
= 23 × 3
= 24
Since the number needs to have 12 factors, and 24 only has 8 factors, we will
need to multiply 24 by a prime number to find the smallest possible number.
Case 1: 24 × 2 = 48
48 is not the number as it only has 10 factors.
Case 2: 24 × 3 = 72
1 × 72 = 72
2 × 36 = 72
3 × 24 = 72
4 × 18 = 72
6 × 12 = 72
8 × 9 = 72
B
∴72 is the smallest possible number as it has 12 factors.
1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72 [A1]
Answer ………………………..…………………………..……………………………… [2]
CWSS / 2022 / EOY / 1E / Math / Section A
5
5
(a)
Express
8
% as a fraction in its simplest form.
13
8
%
13
8
  100
13
8
1
 
13 100
(b)
8
1300
2

325

2
325
Answer …………………..………… [1]
Express 35 minutes : 2 hours : 20 seconds in its simplest form.
35 mins = 35 × 60
= 2100 secs
2 hours = 2 × 60 × 60
= 7200 secs
35 minutes : 2 hours : 20 seconds
for
= 2100 : 7200 : 20 [M1] Awarded
converting all into
= 105 : 360 : 1
same units.
1
105
360
Answer ……… : ……… : ……… [2]
(c)
A Dokémon Trainer Club has 120 members.
At first, there are 20 female members in the club.
After 25 new members join the club, the ratio of females to males became 2 : 3.
Calculate the percentage increase in the number of female members in the club.
Let 2y and 3y be the new number of
females and males in the Dokémon
Trainer Club respectively.
% increase in female members
58  20
100%
=
20
= 190%
Total number of members = 145
2y + 3y = 145
5y = 145
y = 29
Number of female members
= 2(29)
= 58
190
Answer ……………………..…… % [2]
CWSS / 2022 / EOY / 1E / Math / Section A
6
6
Xin En drew the following graph to illustrate the level of customer satisfaction for her café over
the past 3 months.
State one aspect of the graph that may be misleading and explain how this may lead to a
misinterpretation of the graph.
Answer
The area used to indicate the level of customer satisfaction is misleading. Readers may assume
………….…….………………….…….………………….…….…………………………………
that customer satisfaction from July to August 2022 has doubled. But in reality, customer
………….…….………………….…….………………….…….…………………………………
satisfaction only increase by a factor of 1.4 from July to August 2022.
………….…….………………….…….………………….…….………………………………[1]
Other possible answers:
1. The scale of the graph on the vertical axis is not equally spaced out. This will cause
readers to obtain an inaccurate magnitude of increase in the number of customer
satisfaction over the period from June to August 2022.
2. There is no origin indicated in the diagram. Readers may assume that the number of
customer satisfaction starts from 0, which may be an incorrect assumption to
determine the magnitude of increase of the number of customer satisfaction.
3. As the size of face increases rapidly, the increase in satisfaction is fast. However, they
may not be accurate when compared to the vertical axis.
CWSS / 2022 / EOY / 1E / Math / Section A
7
4. We are unsure whether to compare the height or size of the smiley face to determine
how much the customer satisfaction has increased over the past months.
5. The number of customer satisfaction between months has increased or decreased.
Based on the diagram, it has increased over months. However, it may not necessarily
be true.
6. There is no exact date of comparison. The time period used for collection of data may
not be the same.
CWSS / 2022 / EOY / 1E / Math / Section A
8
7
(a)
In the figure, AOB and COD are straight lines.
DOB = 50° and AOD  5b .
D
A
5b°
O
50°
B
C
Find
(i)
AOC ,
50
Answer AOC = ……………..…  [1]
(ii)
the value of b.
180  50
5
= 26°
b 
CWSS / 2022 / EOY / 1E / Math / Section A
26
Answer b = ……………..………… [1]
9
8
(a)
Expand and simplify 2rq  3r  5  qr  2r  .
2rq  3r  5  qr  2r 
 2rq  3r  5qr  10r
 7qr  13r
7 qr  13r (or 7 rq  13r )
Answer ……………..…………………… [2]
(b)
Factorise completely 45 xy 2  15 x3 y .
45 xy 2  15 x3 y  15 xy  3 y  x 2 


15 xy 3 y  x 2
Answer ……………..…………………… [1]
(c)
3r  s 2
.
5rs
Find the value of p when r  7 and s  2 .
It is given that p 
Given that r  7 and s  2 ,
3  7    2 
p
5  7  2 
2
21  4
70
17
p
70
p
17
70
Answer p = …………..…………………
[2]

CWSS / 2022 / EOY / 1E / Math / Section A
10
9
(a)
Express as a single fraction in its simplest form
2r  p 3 p  r
.

3
6
2r  p 3 p  r

3
6
2  2r  p    3 p  r 

6
4r  2 p  3 p  r

6
5r  p

6
5r  p
6
Answer ……………..……………… [2]
(b)
Solve
1
 5.
2x  3
1
5
2x  3
5  2 x  3  1
10 x  15  1
10 x  14
x  1
2
or −1.4
5
2
1 or −1.4
5
Answer x = ……………..…………
[2]
CWSS / 2022 / EOY / 1E / Math / Section A
11
10
× C  3, 4
×A
× B  3,  2 
(a)
(b)
On the grid above, plot and label the points B  3,  2  and C  3, 4 .
[1]
Write down the equation of the straight line that passes through point A, as shown in the
grid, and is parallel to the y-axis.
x 1
Answer ……………..…………………… [1]
(c)
Calculate the area of the triangle ABC.
Height of triangle ABC = 2 units
Base of triangle ABC = 6 units
Area of triangle ABC
1
=  6 2
2
= 6 units2
CWSS / 2022 / EOY / 1E / Math / Section A
6
Answer ……………..……………… units2 [1]
12
11
Parallelogram ABCD and trapezium EFGH have the same area.
The height of trapezium EFGH is twice the height of parallelogram ABCD.
E
A
2 cm
F
B
5 cm
D
24 cm
C
H
G
Find the length of GH.
Area of parallelogram ABCD
= 5 × 24
= 120 cm2
Height of trapezium EFGH = 10 cm
Area of trapezium EFGH = 120 cm2
1
× (2 + GH) × 10 = 120
2
5 × (2 + GH) = 120
2 + GH = 24
GH = 22 cm
22
Answer GH = ……………..………… cm [3]
CWSS / 2022 / EOY / 1E / Math / Section A
13
12
In the diagram, QP is parallel to GH.
Reflex RGH  320 and PQR  55
Calculate QRG , stating your reasons clearly.
The diagram is not drawn to scale.
P
Q
B
R
A
H
G
320°
By drawing the line AB passing through point R on the diagram, such that it is parallel to
lines GH and PQ.
Method 1
Method 2
RGH = 360° − 320°
RGH = 360° − 320°
= 40° (∡ at a point)
ARG = RGH
= 40° (alt ∡s, AB / /GH )
QRA = PQR
= 55° (alt ∡s, AB / / QP )
QRG = 40° + 55°
= 95°
= 40° (∡ at a point)
QRB = 180° − 55°
= 125° (int ∡, QP / / AB )
BRG = 180° − 40° (int ∡s, AB / /GH )
= 140°
QRG = 360° − 140° − 125°
= 95° (∡s at a point)
95
Answer QRG = ……………..…………  [3]
CWSS / 2022 / EOY / 1E / Math / Section A
14
P
Q
R
H
O
G
320°
Method 3
By drawing the line RO passing through point R on the diagram, such that it forms a
triangle ROG.
RGH = 360° − 320°
= 40° (∡ at a point)
ROG = PQR
= 40° (alt ∡s, PQ / / GH )
ORG = 180° − 44° − 55° (adj ∡s on str. line)
= 85°
QRG = 180° − 85° (adj ∡s on str. line)
= 95°
CWSS / 2022 / EOY / 1E / Math / Section A
15
13
The nth term of a sequence is given by Tn =
(a)
1
1

.
n n 1
Show that sum of the first n terms, S n , is given by 1 
1
.
n 1
Answer
T1  T2  T3  ...  Tn  2  Tn 1  Tn
1   1
1 1
1 
1 1   1 1   1 1 
 1
             ...  

  


1 2   2 3   3 4 
 n  2 n 1   n 1 n   n n  1 
1
 1
n 1
[1]
(b)
Hence, find the sum of the first 200 terms.
1
200  1
200

201
S 200  1 
(c)
200
or 0.995 (3sf)
201
Answer ……………..…………………… [1]
Explain why the sum of the first n terms, S n , is always less than 1.
Answer
1
1
 1 because 0 
 1.
For n > 0, 1 
n 1
n 1
………….…….………………….…….………………….…….…………………………
………….…….………………….…….………………….…….…………………………
………….…….………………….…….………………….…….……………………… [1]
CWSS / 2022 / EOY / 1E / Math / Section A
16
14
The pie chart shows the favourite local food of the students from class 2A.
(a)
If 35 students choose Nasi Lemak and Roti Prata, calculate the total number of students
from Class 2A.
Number of degrees for students who choose Nasi Lemak and Roti Prata
= 200° + 80°
= 280°
Using proportion, the total number of students
35

 360
280
= 45
45
Answer Number of students = ………….……….…… [2]
CWSS / 2022 / EOY / 1E / Math / Section A
17

(b)
(i)
Express 0.08 in the form of
a
, where a and b are integers and b ≠ 0.
b
Let x  0.088888... − (1)
(1) × 100 = (2):
100 x  0.088888... 100
100 x  8.8888...
(1) × 10 = (3):
10 x  0.088888... 10
10 x  0.88888...
(2) – (3):
100 x  10 x  8.8888...  0.88888...
90 x  8
8
90
4

45
x 

(ii)
4
45
Answer ……………..……………… [1]
If 0.08 of the students’ favourite food is Durian Chendol, find the value of x.
4
 360
45
 32
x 
∴Value of x = 32
32
Answer x = ………….……….…… [1]
CWSS / 2022 / EOY / 1E / Math / Section A
18
15
(a)
Construct a quadrilateral WXYZ where WX = 7 cm, WXY = 120°, XY = 8 cm,
WZ = 3 cm and ZWX = 60°.
Answer
[2]
(b)
Measure and write down the length of YZ.
10.5 ± 0.2
Answer YZ = ………….……….…… cm [1]
CWSS / 2022 / EOY / 1E / Math / Section A
19
16
Water flows into an empty cylindrical container at a rate of 6.3 litres per minute.
(a)
Convert 6.3 litres per minute to cm3 per second.
6.3l 6.3 1000cm3

1min
60sec
= 105 cm3/sec
105
Answer ………………….…… cm3/s [1]
(b)
The container has a uniform thickness of 0.4 cm, and an exterior radius of 3 cm.
0.4 cm
0.4 cm
3 cm
The container can be fully filled in 5 minutes.
Calculate the height of water when the container is full.
5 mins = 300 secs
Let x denote the height of water in container.
Amount of water in container
= 105 × 300
= 31 500 cm3
Volume of water in container = 31 500
π × 2.62 × x = 31 500
31500
x 
  2.6 2
= 1480 cm (3s.f.)
Inner radius of container
= 3 cm – 0.4 cm
= 2.6 cm
1480
Answer Height of water = ………….………. cm [2]
CWSS / 2022 / EOY / 1E / Math / Section A
20
17
Mrs Wong is deciding which brand of laundry detergent she should purchase at her neighbourhood
supermarket.
A
B
B B
Option A
Option B
1 bottle of 2.6 litres
$13.30
1 large bottle of 1 litres and
2 small bottles of 125 ml each
$8.80
Use estimation to determine which option is more value for money.
You must show all your workings and explain your answer.
Method 1
Option A:
$13.30 ≈ $13
Cost of 1 litre ≈ $13 ÷ 2.6
= $5
[Method 2] Calculating Option A:
Cost of 1 litre ≈ $13 ÷ 2.6
= $5
Cost of 1.25 litres ≈ $5 × 1.25
= $6.25
Option B:
1 litres + (2 × 125 ml) = 1 litres 250 ml
= 1.25 litres
[Method 3] Calculating Option B:
Cost of 1 litre ≈ $9 ÷ 1.25
= $7.20
$8.80 ≈ $9
Cost of 1 litre ≈ $9 ÷ 1.25
= $7.20
Cost of 2.6 litres ≈ $7.20 × 2.6
= $18.72
Answer
A
it is cheaper as compared to Option B.
Option …………
is more value for money because .…….………………….…….………………
OR Option B is more value for money as Mrs Wong only needs to pay slightly more to receive
………….…….………………….…….………………….…….……………………………… [3]
more bottles of detergent.
CWSS / 2022 / EOY / 1E / Math / Section A