Book 4B Chapter 9 Variations
Exam Focus Plus (Full Solutions)
9 Variations
Exam-type Questions
Suggested Solutions
Paper I
k
1. (a) Let y  3 , where k is a non-zero constant.
x
When x  6 and y  90 ,
k
90  3
6
k  19 440
19 440
∴ y
x3
Marks
1A
1M
1A
19 440
(by (a))
33
 720
The increase in the value of y  720  90
(b) The final value of y 
1M
 630
2.
Remarks
1A
(a) Let f ( x)  k1 x 2  k2 x , where k1 and k2 are non-zero constants.
f (4)  80
1M
k1 (4)2  k2 (4)  80
16k1  4k2  80
f (3)  3
......(1)
1M for either one*
3k1  k2  1
......(2)
(1) + (2)  4: 28k1  84
for either one*
k1 (3) 2  k2 (3)  3
9k1  3k2  3
k1  3
By substituting k1  3 into (2), we have
3(3)  k2  1
k2  8
∴
(b)
f ( x)  3 x 2  8 x
1A
f ( x)  11
3x 2  8 x  11  0
(3x  11)( x  1)  0
11
x
or x  1
3
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Book 4B Chapter 9 Variations
3.
Exam Focus Plus (Full Solutions)
Suggested Solutions
Marks
(a) Let g ( x)  k1 x  k2 x 2 , where k1 and k2 are non-zero constants.
g (1)  7
1M
Remarks
k1 (1)  k2 (1) 2  7
k1  k2  7
g (1)  11
......(1)
1M for either one*
......(2)
for either one*
k1 (1)  k2 (1) 2  11
k1  k2  11
(1) + (2): 2k2  18
k2  9
By substituting k2  9 into (2), we have
k1  9  11
k1  2
∴
g ( x)  9 x 2  2 x
1A
g ( x)  5  2 x
(b)
9x  2x  5  2x
2
1M
9 x2  4 x  5  0
( x  1)(9 x  5)  0
x  1 or x 
4.
5
9
1A
(a) Let f ( x)  k1  k2 x 2 , where k1 and k2 are non-zero constants.
f (4)  69
1M
k1  k2 (4)2  69
k1  16k2  69
f (2)  21
......(1)
1M for either one*
k1  4k2  21 ......(2)
(1)  (2): 12k2  48
for either one*
k1  k2 (2) 2  21
k2  4
By substituting k2  4 into (2), we have
k1  4(4)  21
k1  5
∴
f ( x)  5  4 x 2
1A
f (3)  5  4(3)2  41
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Book 4B Chapter 9 Variations
Exam Focus Plus (Full Solutions)
Suggested Solutions
Marks
(b) By (a), we have a  41.
∵
1M for either one*
f ( x)  5  4 x 2
∴ b  f (3)  f (3)  41
AB  3  (3)  6
for either one*
1M
1
Area of ABC  (6)(41)
2
 123
5.
Remarks
1M
1A
(a) Let g ( x)  k1 x  k2 x3 , where k1 and k2 are non-zero constants.
g (3)  72
1M
k1 (3)  k2 (3)3  72
3k1  27k2  72
1M for either one*
k1  9k2  24 ......(1)
g (2)  8
k1 (2)  k2 (2)3  8
2k1  8k2  8
k1  4k2  4
(1)  (2): 5k2  20
for either one*
......(2)
k2  4
By substituting k2  4 into (2), we have
k1  4(4)  4
k1  12
∴
g ( x)  12 x  4 x3
1A
g (2)  12(2)  4(2)  8
3
1A
(b) By (a), we have p  8 .
1M for either one*
∵
g ( x)  12 x  4 x3
∴
q  g (1)  12(1)  4(1)3  8
PQ  2  (1)  3
1M
1
Area of PQR  (3)(72  8)
2
 96
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for either one*
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Book 4B Chapter 9 Variations
Exam Focus Plus (Full Solutions)
Suggested Solutions
6.
Marks
(a) Let C  k1 x  k2 x 2 , where k1 and k2 are non-zero constants.
2.28  k1 (15)  k2 (15)
Remarks
1M
2
15k1  225k2  2.28 ......(1)
9.06  k1 (30)  k2 (30)
1M for either one*
2
30k1  900k2  9.06 ......(2)
for either one*
(2)  (1)  2: 450k2  4.5
k2  0.01
By substituting k2  0.01 into (1), we have
15k1  225(0.01)  2.28
k1  0.002
C  0.002 x  0.01x 2
1A
(b) When C  25.1,
25.1  0.002 x  0.01x 2
1M
∴
5 x 2  x  12 550  0
( x  50)(5 x  251)  0
251
(rejected)
5
∴ The length of the ruler is 50 cm.
x  50 or x  
7.
1A
(a) Let C  k1  k2 w , where k1 and k2 are non-zero constants.
1M
1480  k1  k2 16
k1  4k2  1480 ......(1)
1M for either one*
1600  k1  k2 25
k1  5k2  1600 ......(2)
(2)  (1): k2  120
By substituting k2  120 into (1), we have
k1  4(120)  1480
for either one*
k1  1000
C  1000  120 w
The required cost  $(1000  120 36)  $1720
∴
1A
1A
(b) The total cost of three wooden wardrobes of weight 10.24 kg
 3  $(1000  120 10.24)
 $4152
 2  $1720
∴ The claim is incorrect.
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Book 4B Chapter 9 Variations
Exam Focus Plus (Full Solutions)
Suggested Solutions
8.
Marks
(a) Let E  k1n  k2 , where k1 and k2 are non-zero constants.
Remarks
1M
52 500  k1 (5)  k2
5k1  k2  52 500 ......(1)
78 500  k1 (9)  k2
1M for either one*
9k1  k2  78 500 ......(2)
(2)  (1): 4k1  26 000
for either one*
k1  6500
By substituting k1  6500 into (1), we have
5(6500)  k2  52 500
k2  20 000
∴ E  6500n  20 000
The required expenditure  $[6500(16)  20 000]  $124 000
1A
1A
(b) Suppose the expenditure of the company in that month is $200 000.
200 000  6500n  20 000
1M
180 000  6500n
360
n
, which is not an integer.
13
∴ It is not possible that when the company employs a certain
number of workers in a month, its expenditure in that month
is $200 000.
1A
9.
(a)
f ( x)  4 x 3  79 x  105
 (2 x  3)(2 x 2  3 x  35)
 (2 x  3)(2 x  7)( x  5)
1M
1A
(b) (i) Let P  k1a 3  k2 a , where k1 and k2 are non-zero constants. 1M
78  k1 (0.5)3  k2 (0.5)
0.125k1  0.5k2  78 ......(1)
1M for either one*
150  k1 (1)  k2 (1)
3
k1  k2  150
......(2)
for either one*
(2)  (1)  2: 0.75k1  6
k1  8
By substituting k1  8 into (2), we have
8  k2  150
k2  158
∴
P  8a 3  158a
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Book 4B Chapter 9 Variations
Exam Focus Plus (Full Solutions)
Suggested Solutions
Marks
Remarks
(ii) When P  210 ,
210  8a3  158a
4a  79a  105  0
(2a  3)(2a  7)(a  5)  0
a  1.5 or a  3.5 (rejected) or a  5 (rejected)
1M
3
∴
The area of the carpet is 1.5 m2.
1M
1A
Paper II
10. C
Let t  kr 2 s , where k is a non-zero constant.
108  k (3) 2 9
k 4
∴ t  4r 2 s
When r  6 and s  25 ,
t  4(6) 2 25
 720
11.
C
ky 2
Let x 
, where k is a non-zero constant.
z
1 k (2) 2

2
6
3
k
4
3y2
∴ x
4z
When x  6 and y  12 ,
3(12) 2
6
4z
z  18
12.
C
Let z  kx2 3 y , where k is a non-zero constant.
72  k (4) 2 3 27
k  1.5
∴ z  1.5x2 3 y
When x  9 and y  64 ,
z  1.5(9) 2 3 64
 486
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Book 4B Chapter 9 Variations
Exam Focus Plus (Full Solutions)
Suggested Solutions
13.
C
Let d  k1  k2c 3 , where k1 and k2 are non-zero constants.
42  k1  k2 (2)3
k1  8k2  42
......(1)
357  k1  k2 (5)3
k1  125k2  357 ......(2)
(2)  (1): 133k2  399
k2  3
By substituting k2  3 into (1), we have
k1  8(3)  42
k1  18
∴ d  18  3c3
When c  4 ,
d  18  3(4)3
 210
14.
A
k2
, where k1 and k2 are non-zero constants.
q
k
12  k1  2
2
2k1  k2  24
......(1)
k
4  k1  2
10
10k1  k2  40 ......(2)
(2)  (1): 8k1  64
Let p  k1 
k1  8
By substituting k1  8 into (1), we have
2(8)  k2  24
k2  40
40
∴ p  8 
q
When p  10 ,
40
10  8 
q
q  20
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Book 4B Chapter 9 Variations
Exam Focus Plus (Full Solutions)
Suggested Solutions
15.
C
k
, where k is a non-zero constant.
v2
Let v0 be the original value of v.
Let u 
New value of v  (1  40%)v0  0.6v0
k
k
New value of u 

2
(0.6v0 )
0.36v0 2
k
k
 2
2
0.36v0 v0
100%
Increase % of u 
k
v0 2
7
 177 %
9
16.
C
k a
, where k is a non-zero constant.
b2
Let a0 and b0 be the original values of a and b respectively.
Let c 
New value of a  (1  84%)a0  0.16a0
New value of b  (1  50%)b0  0.5b0
New value of c 
k 0.16a0 0.4k a0 1.6k a0


(0.5b0 )2
0.25b0 2
b0 2
1.6k a0 k a0

b0 2
b0 2
% change of u 
 100%
k a0
b0 2
∴
 60%
c is increased by 60%.
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Book 4B Chapter 9 Variations
Exam Focus Plus (Full Solutions)
Suggested Solutions
17.
D
z
.
kx 2
be the original values of x and z respectively.
Let z  kx 2 y , where k is a non-zero constant. Then y 
Let x0 and z0
New value of x  (1  20%) x0  0.8x0
New value of z  (1  20%) z0  1.2 z0
1.2 z0
1.875 z0
New value of y 

2
k (0.8 x0 )
kx0 2
1.875 z0
z
 02
2
kx0
kx0
% change of y 
 100%
z0
kx0 2
∴
18.
 87.5%
y is increased by 87.5%.
A
kq3
, where k is a non-zero constant. Then
r
p r
k 3
q
Let p 
k2 
p2r
q6
1
q6

k 2 p2r
∴
19.
q6
must be constant.
p2r
D
Let s 
k
t 2 u3
, where k is a non-zero constant. Then
k  st 2 u 3
k 2  s 2t 4u 3
∴ s 2t 4u 3 must be constant.
20.
C
∵ The equation of the straight line in the figure is
y  mx  c , where m and c are non-zero constants.
∴ y is partly constant and partly varies directly as x.
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