1. (a) Express
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Answer all the questions.
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Mathematics
Section 4
Paper 1
as a percentage
(b) Express 12 % as a fraction in its simplest form.
.
√
× .
.
3. Factorise 27
.
− 12
4. (a) Given that 5
(b) Factorise
, correct to 2 significant figures.
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2. Calculate
.
÷ 25 = 5 , find the value of n.
−
−
−
.
5. The petrol consumption of a car is 12 kilometres per litre.
(a) Express 12 kilometres per litre in centimetre per cubic centimetre.
(b) Find how many litres of petrol are required for the car to travel 102 km.
AL
6. Peter, Marc and Tom shared a bottle of marbles so that Peter received of the marbles
and the remainder was shared between Marc and Tom in the ratio2 ∶ 3. Find the ratio
which the marbles was shared between Peter, Marc and Tom.
7. (a) Factorise 6" − 7" − 5.
NY
(b) Hence solve 6" − 7" − 5 = 0.
8. (a) Solve the inequality
$
≥4
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(b) Hence find the largest integer that satisfies the inequality.
(a) angle ,*'
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9. Four points J, K, L and M lie on the circle of centre O. '() is a straight line, angle
*'( = 120° and angle ,() = 98°. Stating your reasons clearly, find
J
(b) angle *,(
240°
K
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O
L
M
N
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10. Given as the product of its prime factors, 132 = 2 × 3 × 11.
(a) Write 756 as the product of its prime factors.
(b) Find the largest integer that is a factor of both 132 and 756.
(c) Find the smallest positive integer value d, such that 756d is a perfect cube.
11. There are 5 green counters and 4 yellow counters in a bag. A counter is drawn at random
replaced before another counter is drawn. Find the probability that
(a) both counters are green
(b) at least one counter is yellow
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(c) both counter are of the same colour.
12. / = 0 : 3456367898:56;1 ≤
≤ 15=
> = 0 : 34;3?343 @8 "3=
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A = 0 : 34;3?343 @8 "4=
Find
(a) > ∪ A
C
(b) 6 >C
DA
(c) > ∩ A
13. It is given that y is inversely proportional to the square root of x. the difference between
the values of y when = 1 and = 16 is 18.
(a) Find as equation connecting x and y.
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(b) Find the value of x when " = 216.
14. The average mass of hemoglobin in a red blood cell is about 30 picograms. An adult
human has about 2.5 × 10 red blood cells at any time.
(a) Express 30 picograms in grams, giving your answers in standard form.
(b) Calculate the average mass, in grams, of hemoglobin in 2.5 × 10
red blood cells.
15. Two coins, X and Y, were each thrown 10 times. The matrices show the numbers of head
and tail obtained and the scores awarded.
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Number of times Number of times
Tail is thrown
X
3
a
Y
6
4
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Head is thrown
Scores
Head
4
Tail
1
(a) Write down the value of a.
3
(b) Using your answer in (a), evaluate E
6
5 4
FE F
4 1
(c) Explain what your result obtained in b) represents.
G
÷
IJ
AL
16. (a) Simply
(c) Simply
H
K
$K
−
GH
K$
NY
17. The speed-time graph shows how a motorist changed speed of his car after he say a
hump. He decelerated his car at a constant rate of 1.4L4 $ for the first 5 seconds.
(a) the value of y
(b) the total distance travelled for the first 12 seconds.
DA
18. A survey was conducted to find the numbers of hours a group of students spent to
complete their homework on a particular Monday afternoon. The results are shown in the
table below.
Number of hours
1 2 3 4
Number of students 6 9 3 X
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(a) Write down the largest possible value of x if the mode is 2.
(b) Write down the largest possible value of x if the median is 2.
(c) Calculate the value of x given the mean is 2.5
19. (a) Sketch the graph of " =
−3
(b) Sketch the graph of " = −
+2
−1
−3
(c) Write down the equation of the line of symmetry of " = −
−1
−3
20. The coordinates of three points are > 3, −1 , A 3, 1 56;O −1,4
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(a) Calculate the length BC
(b) Calculate the gradient of the line AC
(c) Calculate the area of triangle ABC
21.
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(d) Find the value of sin ∠>AO
22. A rectangle has a perimeter 34 cm. The length is 2x cm and the diagonal is
+ 7 L.
(a) Express the breadth in term of x.
(b) Form an equation in x and show that it reduces to 7
(c) Find the value of x.
− 82 + 240 = 0
23. In the diagram, the line CF and DE intersects at point O. DC is parallel to EF, ∠TOU =
22° and VW = 12 L.
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C
22°
O
E
12 cm
D
F
(a) Explain why triangles OCD and OFE are similar.
(b) Write down the value of ∠TWV.
DA
The areas of ΔTOU and ΔTWV are 72 L and 18 L respectively.
Find
(c) The values of OD; OE in its simplest integer form.
(d) The length of CD
24.
(b) Construct the perpendicular bisector of AB.
(c) Construct an angle bisector of ∠>AO.
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(a) Draw the triangle ABC in which >A = 10.9 L, ∠>AO = 55° and AO = 7.4 L.
The line AB is given below.
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NY
AL
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(d) P is equidistant from points A and B and also equidistant from the lines AB and BC.
Draw a circle with centre P and radius AP.
AL
NY
DA
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CA
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AL
NY
DA
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CA
ED
U
AL
NY
DA
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CA
ED
U
AL
NY
DA
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ON
CA
ED
U
AL
NY
DA
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CA
ED
U
AL
NY
DA
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CA
ED
U
AL
NY
DA
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CA
ED
U
× 100% = 1.25%
.
b)
2)
√
.
=
.
× .
.
= 0.090237929 ≈ 0.090
− 12
3) 27
÷5
4a) 5
=3
9
−4
=3
= 5$$
#
∴ 125 − 2& = 99
& = 13
+( (−)
b) ( − )
= (−) (−)+(
= ( − ) 2( − )
5a)
×
×
= 1200
b) 12*+ → 1-
102
× 1- = 8.52
AL
102*+ →
6) /: 1: 2 → 53&456
1 5
→ 3&456
3 2
NY
1 2
:
3 3
/: 1: 2
5
: 2: 3
2
DA
5: 4: 6
3 −2
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U
−2& = −26
3 +2
CA
1a)
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Answer Sheet
7a)
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37 − 5 −107
271
37
67 − 5 −77
∴ 37 − 5 27 + 1
b) 37 − 5 27 + 1 = 0
∴ 37 − 5 = 08927 + 1 = 0
5
3
7 = −
8a) 1 − 2& ≥ 12
1
2
CA
7=
&≤−
b) & = −6
11
2
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−2& ≥ 11
9a) ∠21= = 180° − 98° ∠8&?659. -&.
= 82°
∠2@= = 180° − 82° ∠4&8)). 6AB
= 98°
= 60°
10a) 2 |756
2 |378
NY
3 |189
AL
b) ∠@C1 = 360° − 120° − 98° − 82° ∠63+8DE3?F
3 |63
3 |21
DA
7 |7
1
∴ 756 = 2 × 3 × 7
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b) HIJ = 2 × 3 = 12
c) F = 2 × 7 = 98
11a) × =
$
$
b) K × L + K × L + K × L =
$
$
$
$
$
$
12a) M ∪ O
P
= Q1, 2, 5, 7, 10, 11, 13, 14S
b) & MP = 10
U
13a) 7 =
√V
U
7=
*
√1
√
−
,7 =
*
√16
U
√
= 18
4* − *
= 18
4
3* = 72
* = 24
24
√(
NY
∴7=
AL
* *
− = 18
1 4
b) 7 =
24
√(
DA
216 =
√V
√( =
(=
24
216
1
81
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c) M ∩ O = Q12S
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c) K × L + K × L =
$
$
$
$
= 3 × 10
∴ 3 × 10
b) 2.5 × 10
B
× 3 × 10
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14a) 30 × 10
= 750B
15a) ? = 10 − 3 = 7
b) K
37 4
19
LK L = K L
64 1
28
16a)
b)
W
×
X
[V
V
WX
YZ
V
=
W
YZ X
+
V
V
=
V ]
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19( + 4
3 1+( 1−(
=
^
[V
15( + 4 + 4(
3 1+( 1−(
=
17a)
[
\ [V
= −1.4
10 − _ = 7
−_ = −3
∴_=3
3 + 10 × 5 = 32.5
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b)
1
3 + 10 × 7 = 45.5
2
NY
32.5 + 45.5 = 78
∴ 78+
18a) ( = 8
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b) ( = 11
c)
× [ $× [ × [ V
[$[ [V
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c) The total score obtained by each coin.
= 2.5
6 + 18 + 9 + 4(
= 2.5
18 + (
33 + 4( = 2.5 18 + (
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33 + 4( = 45 + 2.5(
1.5( = 12
(=8
19a) Turning pt: (3, 2)
7 = 0−3
+ 2 = 11
CA
y-coordinate: y = 11
12
10
6
4
2
-2
0
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8
2
4
b) 7 = − ( − 3( − ( + 3 = − ( − 4( + 3 = −( + 4( − 3
1+3
=2
2
7=− 2
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( − 889F4&?5A:( = 1, ( = 3
+4 2 −3= 1
NY
∴ 239&4&B)5: 2,1
7=− 0
+ 4 0 − 3 = −3
DA
7 − 889F4&?5A: 7 = −3
1
-2
2
4
-1
-3
c) ( = −2
+ 4−1
=−
b) a9?F4A&5MI =
c) b
−133 − 1
b
41 − 14
= 53&456
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20a) OI = ` −1 − 3
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-2
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2
1
= |(−1 × 1) + (3 × −1) + (3 × 4) − (4 × 3) − (1 × 3) − (−1 × −1)|
2
1
= |−8|
2
22a)
NY
d) sin ∠MfI =
AL
= 43&456
V
V
=
V
= 17 − 2(
DA
∴ (17 − 2() +
b) (2() + (17 − 2() = (( + 7)
4( + 289 − 68( + 4( = ( + 14( + 49
7( − 82( + 240 = 0
c)
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7( − 40 −40(
( − 6 −42(
7( 240 −82(
7( − 40 ( − 6 = 0
40
,( = 6
7
∴
23a) ∠gIf = ∠gJh ?-5∠
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∠Igf = ∠Jgh _A95. 8)). ∠
∠gfI = ∠ghJ ?-5. ∠
∴ ΔgIf ∼ ΔgJh MMM
c) k
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b) ∠gJh = 22°
=
∴ gf = gh
2:1
d)
lm
no
=
pm
po
If = 24 +
DA
NY
24d)
AL
If 2
=
12 1
C
P
A
B
Springfield 4
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Answer Sheet
1 a i) 1ℎ → 12
3ℎ → 12 × 3
∴ = 36
3ℎ → 18 × 3
∴ = 54
iii) ∠ = 135° − 50° = 85°
= 54 + 36 − 25436cos 85°
∴ = 62.23454 … ≈ 62.2
!"# ∠$%&
'(
=
!"# )*°
(.'
ED
iv)
UC
A
ii) 1ℎ → 18
∴ ∠ = 35.1905 … ≈ 35.2°
180° − 50° = 130°,-.∠
∴ 360° − 130° − 35.2° = 194.8°∠/./0.
∴ -1: 194.8°
3
NY
A
cos 14.8° =
L
b) 194.8° − 180° = 14.8°
cos 14.8° =
3
62.23
∴ 3 = 60.165419 … ≈ 60.2
*4
'
2 a) 5) = 4
DA
56 = 54
6 = 10.8
6 = ±3.28633 … ≈ 3.2989 − 3.29
b) /: − / = ;: − ; /: − ;: = / − ; / + ;/ − ;
/ − ;
: =/+;
c i) = 30 +
∴ $48.75
<**
5==
= 48.75
UC
A
:=
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:/ − ; = / + ;/ − ;
<*@
ii) 63.75 = 30 + 5==
75= 33.75
100
- = 45
(*@
iii) = 28 + 5==
iv) = 28 +
(*A*
5==
= 57.25
∴ 86.50
=
L
63.75 − 57.25 = 6.5
ED
75- = 3375
NY
A
3 a i) 5== × 22500 = 4500
∴ $4500
ii) 4500 + 24 × 810 = 23940
∴ $23940
iii) 23940 − 22500 = 1440
1440
2
× 100% = 6
%
23940
133
DA
∴
F
b) C8./D$ = E1 + 5==G
@
5 '
= 22500 H1 +
I
100
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= 26046.5625
≈ 26046,56
26046.56 − 22500 = 3546.56
5==
∴ $3546.56
c) 55= × 800 = 727.2727 … ≈ 727.27
UC
A
∴ $727.27
KKKKKL − M
KKKKKL
KKKKKL = M
4 a i) = 0 + 3N − 30 − N
= 0 + 3N − 30 + N
= 2−0 + 2N
KKKKKL − M
KKKKKL
KKKKKL = MO
ii) O
ED
= −20 + 4N
= −20 + 9N − 0 + 3N
= −30 + 6N
L
= −20 + 9N − 0 − 3N
= 3−0 + 2N
NY
A
KKKKKL = KKKKKL + O
KKKKKL
iii) O
= −20 + 4N − 30 + 6N
= −50 + 10N
= 5−0 + 2N
DA
KKKKKKL = M
KKKKKL + P
KKKKKL
b) MP
1
KKKKKL
KKKKKL + = M
2
1
= 30 − N + −20 + 4N
2
= 30 − N − 0 + 2N
= 20 + N
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KKKKKL + KQ
KKKKL = M
KKKKL
c i) KMQ
= 30 − N + 0 + 3N
= 40 + 2N
= 220 + N
KKKKKL = 2MP
KKKKKKL
ii) MQ
UC
A
M, P, Q/9RS8D,-R/9
5
d) i) ii)
5 a i)
&%
&T
=*
5*U×5)=°
5*
= 156°
ii)
5)=°U5*(°
ED
∴ ∠OP = 156°
= 12°,181Δ
∴ ∠OP = 12°
iii) ∠PQ = 156° − 12° = 144°,-.∠
L
∠QP = 180° − 144° = 36°
NY
A
b) ∠WO = 180° = 156° = 24°∠8-1.9. D-. ∠WPO = 180° − 156° = 24°∠8-1.9. D-. ∠OP9RXDR6 = 360° − 156° = 204°
∠WP = 360° − 204° − 24° − 24° = 108°∠1Y8XNY/Z. 6 a) [$6 = \[$1
DA
3730
I
[$3730 = \[$ H
6
b) [$6 + 0.50 = \[$1
1500
I
[$1500 = \[$ H
6 + 0.50
= 3200
37306 + 0.5 + 15006 = 320066 + 0.5
37306 + 1865 + 15006 = 32006 + 16006
32006 − 36306 − 1865 = 0
6406 − 7266 − 373 = 01ℎ8^-
=
−−726 ± d−726 − 4640−373
2640
UC
A
=
U_±√_a UAbc
b
726 ± √1481956
1280
∴6=
726 + √1481956
726 − √1481956
896 =
1280
1280
= 1.51824 … ≈ 1.52
e) \[$1 = [$6 + 0.50
\[$1 = [$1.52 + 0.50
\[$1 = [$2.02
= −0.38387 …
ED
d) 6 =
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'<'=
5*==
+ 4]=.*
4
c)
9Re
!"# ∠%&T
)
=
!"# '°
5=
NY
A
b)
L
7 a) P = √120 + 57 = 132.8495 … ≈ 133
∴ ∠O = 21.22992 … ≈ 21.2°
5
5
c) 9R/OP = E × 120 × 57G + E × 120 × 82 × sin180° − 21.2° − 32°G
= 7359.598 … ≈ 7360
DA
= 120 + 82 − 212082cos180° − 21.2° − 32°
∴ = 181.4188 … ≈ 181
,-,YDR-h.ℎ = 132.8 + 57 + 82 + 181.4 = 453.2
d) ∠O = 180° − 21.2° − 32° = 126.8°∠1Y8XΔ
∴ = 181.4188 … ≈ 181
iR.6;R.ℎR0R90R-Z,SYD/9Z,1./-SRX98.8O
1
1
× 181.4 × 6 = × 120 × 82 × sin 126.8°
2
2
∴ 6 = 43.43548 … ≈ 43.4
iR.j;R.ℎR/-hDR8XRDRk/.,8-.
50
43.44
UC
A
tan j =
∴ j = 49.01588 … ≈ 49.0
8 a i) 1NY/9R: 89
S,9SDR: 2n9
=
=
4+n
8 + 2n
4+n
24 + n
1
2
NY
A
= 0.5S
L
9=
ED
∴ 89 + 2n9 = 4 + n
98 + 2n = 4 + n
ii) 9R/ = 1NY/9R − S,9SDR
= o20.5 × 20.5p − n × 0.5 = 1 − 0.25n
DA
n
∴ E1 − G S
4
b) i) n9 ℎ = n4 25 = 1256.637 … ≈ 1260S'
ii) 2n9ℎ = 2 × n × 4 × 25 = 628.3185 … ≈ 628S
iii) {a} 4 × 2 = 0
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= 120 + 82 − 212082cos 126.8°
80 ÷ 8 = 10
40 ÷ 8 = 5
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10 × 5 = 50
{b} k8DYR8X;86 = 80 × 40 × 25 = 80000
k8DYR8XS/-1 = 50 × 1256.6 = 62830
k8DYR8XR0.:10/SR = 80000 − 62830 = 17170 ≈ 17200S'
b)
200
100
- 100
c) i) 4170
10
20
30
40
50
ED
0
UC
A
a) / = 225
1
: = 6 − 1070 − 6
4
L
ii) 40 articles
iii) -85. They did not gain any profit, but lost some money instead.
NY
A
d) : = 36
DA
-x 0 20 40
y 0 60 120
60
70
: = 36
10 a)
=
∑u4
∑u
3 × 2.5 + 4 × 7.5 + 8 × 12.5 + 5 × 17.5
20
= 11.25
ii) 1./-Z/9ZZRk,/.,8- = v
∑u4 a
∑u
∑u4 − E ∑u G
UC
A
b) i) R/- =
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2.5
7.5
12.5
17.5
Mark (m) 0 < ≤ 5 5 < ≤ 10 10 < ≤ 15 15 < ≤ 20
Frequency
3
4
8
5
3 × 2.5 + 4 × 7.5 + 8 × 12.5 + 5 × 17.5 =w
− 11.25
20
= 4.9686 …
)
<
× 5x
=
5A
= x*
ED
iii)
≈ 4.97
c) School B did better as their mean mark is higher.
DA
NY
A
L
School B’s students are more consistent as their standard deviation is lower.