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# Solution

1.

m9
(m3n 7 )5

Marks

m9
m15n 35

n 35

15

m m
n 35
6.
m

Remarks

1M

for (a p b q ) r a prb qr

1M

for a p

1
ap

1A
.... (3)

2.

4a 5b 7
8
b
4a 5b 7 8b
3b 7 4a
4a 7
b
.
3

1M
1M

## for putting b on one side

1A

or equivalent

.... (3)
3.

3 2 1
The required probability
45
3
.
10

1M + 1M
1A

## 1M for the nominator

1M for the denominator
or 0.3

.... (3)
4.(a)

x x y 7x
3

x 2 ( x y 7).

4.(b)

1A

x x y 7x x y 7
3

x 2 ( x y 7) x y 7

1M

## for using (a)

( x y 7)[ x ( x y 7)]
( x 1)( x 1)( x y 7).

1M

for factorization

5.(a)

1A
.... (4)

7 3x
2( x 2),
5
7 3x 10x 20
x 1
4 x 13 0
13
x
4
Thus, the required solution is x

or equivalent

1A
1A

13
.
4

5.(b)

6.(a)

x > 3.25

1A
1A
.... (4)

x 250
20%
250

1M

pc

1A

x 300.

6.(b)

7.

## Let \$y be the marked price of the book.

y(1 25%) x
y 400.
Thus, the marked price of the book is \$400 .
Let a be the number of apples owned by Ada and b be that
owned by Billy.
a 4b

a 12 b 12
Solving, a 32 and b 8.
Thus, the total number of apples owned by them is 40 .

8.

BAD 180 2 58 64.
BDC BAC 64 25 39

1M
1A
.... (4)

1M

1A + 1A

1A
.... (4)
1M

1M
1A

BC CE
180 58
61.
2
ABE 61 BAE 22.

BEC

9.(a)

9.(b)

## Let x be the required angle.

x
122
30
360
x 75.
Thus, the angle of the sector is 75 .
75
The required perimeter 2 12
2 12
360
(24 5 ) cm.

1M
1A
.... (5)
1M
1M

for r 2

A
360

1A
1M + 1M
1A

## 1M for the arc length

1M for the radii
in terms of
u1 for missing unit

.... (5)
10.(a)

10.(b)

## Let a and b be two constants.

a 10b 10600

a 6b 9000
Solving, a 6600 and b 400.
Thus, the required income = 6600 20 400 \$14 600.
Let N be the number of handbags sold in a month such that
her income is \$18 000 .
18000 6600
N
28.5
400
Since 28.5 is not an integer, it is not possible that the income
is \$18 000 .

1M
1M
1M + 1A
.... (4)

## u1 for missing unit

1M
1A
.... (2)

11.(a)

f (2) (2 2) 2 (2 h) k 5
f (3) (3 2) (3 h) k 0
2

h 2 and k 5

1M
1A + 1A
.... (3)

## for remainder and factor

theorems

11.(b)

f ( x) 0
( x 2)2 ( x 2) 5 0
( x 2 4 x 4)( x 2) 5 0

x3 2 x 2 4 x 3 0
( x 3)( x 2 x 1) 0

1M

1 5
2
Thus, not all the roots of the equation f(x) = 0 are integers.
The claim is disagreed.
x 3 or

1M
1A
.... (3)

12.(a)

mean = 55 kg
median = 52 kg
range = 79 40 = 39 kg

1A
1A
1A
.... (3)

12.(b)

## Let a and b be the required weights.

a b 55 20
56
22
a b 132
Note that a = 80 kg as the range is increased by 1 kg . Thus,
the weight of each of these students is 80 kg and 52 kg .
13.(a)

AE BF (given)
AB BC (propertie s of square)

ABE BCF (R.H.S.)
13.(b)

AEB 90 FBC

1M
1M
1A + 1A
.... (4)

with reasons

1
.... (2)
1

with reason
using (a)

## BGE 180 GEB GBE

180 (90 GBE ) GBE

90.
BGE is a right-angled triangle.
.... (3)
13.(c)

Note that CF = BE = 15 cm

1M

BG 15 9 12 cm .

1A

14.(a)

(i)

.... (2)
1M

3
4
L : 3x 4 y 59 0.

1M

Slope of L =

1A

3h 59
Centre of C = h,

## The required equation :

(ii)

3h 59

2 3h 59
( x h) 2 y
(1)
(h 4)
4
4

1M

using (a)(i)

1A

(x h)2 + (y k)2 = r2

2 x 2 4 xh 2h 2 2 y 2 3hy 59 y

3481 9h 2 177

h
8
8
4

25h 2 125
4225

h
8
4
8
2 x2 2 y 2 4hx (3h 59) y 13h 93 0.

1M

.... (6)
14.(b)

1M

1M

1A

using (a)(ii)
or otherwise

.... (3)
15.(a)

## Let x marks be the required score.

x 66
0.5
12
x 60
The score of David in the Mathematics examination is 60
marks .

1M

1A
.... (2)

15.(b)

## Note that 60 66 0.5 12 and 49 52 0.3 10 .

As 0.3 > 0.5 , he performs better in the Science
examination than in the Mathematics examination.
The claim is agreed.

1M

for x z

1A
.... (2)

16.(a)

C 5C 9
The required probability = 2 142
C4

1M

360
.
1001

1A

.... (2)
16.(b)

C05C49
C414

C15C39

5
.
11

1M

for p2 = 1 p1

1A
.... (2)

17.(a)

## A(1) A(2) A(3) A(n)

[1 (4n 5)]n
2
2
2n 3n.

17.(b)

logB(1)B(2)B(3)B(n)

log 10A( n)

log(102n

1M
1A
.... (2)

3n

1M

2n2 3n.

2n2 3n 8000
(n 64)( 2n 125) 0

125
n 64
2

1M

using (a)

## The greatest value of n is 64 .

18.(a)

1A
.... (3)

(4k )2 4(2)(3k 2 5)
8(k 2 5)

1M

## Since 8(k 2 5) 0 for all real k , the equation f(x) = 0

does not have real roots and the graph of y = f(x) does not cut
the x-axis.

1A

for discriminant

.... (2)
18.(b)

2 x2 4kx 3k 2 5
2( x 2 2kx k 2 ) k 2 5

1M

2( x k ) k 5.
Thus, the required vertex = (k, k2 + 5) .
2

18.(c)

1M
1A
.... (3)

## Note that when S and T are nearest to each other, they

coincide with the vertices of the graphs of y = f(x) and
y = 2 f(x) respectively.
The coordinates of S is (k, k2 + 5) and that of T is
(k, k2 3) .
Thus, the y-coordinate of the midpoint of ST is 1.
The perpendicular bisector of ST is y 1 = 0 .
The claim is disagreed.

1M
1M

using (b)

1M
1A
.... (4)

19.(a)

(i)

## AC 40 24 2(40)( 24) cos 80

42.92546446
42.9 cm.
2

24 42.92546446 40
2(24)( 42.92546446)
ACB 66.59081487
66.6.
(iii) The area of ABC and that of ABD are fixed.
1
The area of ACD = AC 2 sin CAD
2
When BCD increases from 105 to 145 , CAD
decreases from 103 to 23.2 . The required area increases
when BCD increases from 105 to 111.6 but decreases
when it increases from 111.6 to 145 .
(ii)

cos ACB

1M

cosine law

1A

to 3 sig. fig.

or sine law
1A
1M
1M

2A
.... (7)

19.(b)

## Let M be the projection of B onto CD and N be that of B

onto the plane ACD .

## CM AC cos ACD 17.86278929 cm

1M

BM BC 2 CM 2 16.02874788 cm

1M

cos BMA

to 3 sig. fig.

AM 2 BM 2 AB
2( AM )( BM )

BMA 0.144202402

1M

1A
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