Certificate Mathematics in Action Full Solutions 4B
10 Variations
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Activity
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4.
Activity 10.1 (p. 263)
1.
(a)
$25x
• • • • • • • • • •
(a) ∵ x : 4  5 : 2
x 5

∴
4 2
2 x  20
x  10
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• • • •
•
(b) $30y
2.
(b)
(a) T  25 x  30 y
(b) (i) No, T does not vary directly as x. It varies
partly as x only.
(ii) No, T does not vary directly as y. It varies
partly as y only.
∵
∴
8 : 6  16 : x
8 16

6
x
8 x  96
x  12
(c)
Follow-up Exercise
∵
( x  1) : ( x  1)  3 : 2
x 1 3

x 1 2
2 x  2  3x  3
x5
∴
p. 238
1.
(a)
21
/ cup
6
 $3.5 / cup
Cost rate  $
p. 242
1.
(a)
20 km
20
h
60
 60 km/h
(b) Travel rate 
(c)
2.
(a)
$24
30
dozen
12
 $9.6 /dozen
30 cm
50 cm
3

5
 3:5
(b)
2.
1
The required ratio
 35 min :1 hr :1 hr 15 min
 35 min : (1  60) min : (1  60  15) min
 35 min : 60 min : 75 min
 35 : 60 : 75
 7 : 12 : 15
∵
∴
zx
z  kx, k  0
By substituting x  4 and z  12 into the
equation, we have
12  k  4
k 3
∴
z  3x
When x  32,
z  3  32
 96

3.
When x  6,
y  3 6
 18
(b) 1.25 L : 500 mL  1250 mL : 500 mL
1250 mL
500 mL
5

2
 5:2
yx
y  kx, k  0
By substituting x  8 and y  24 into the
equation, we have
24  k  8
k 3
∴
y  3x
Cost rate 
30 cm : 50 cm 
∵
∴
3.
∵
∴
( y  4)  x
y  4  kx, k  0
By substituting x  8 and y  8 into the
equation, we have
10 Variations
8  4  k 8
4
k
8
1

2
1
∴
y4  x
2
1
y  x4
2
When y  10,
1
10  x  4
2
1
6 x
2
x  12
∴

∴
3.
(a)
∵
y  x2
y  kx2 , k  0
∴
Let x0 and y0 be the original values of x and y
respectively.
∴ New value of x  (1  10%) x0
 1.1x0
(b)
2
New value of y  k (1.1x 0 )
 1.21kx0
2
 1.21 y 0
∴
Percentage change in y 
1.21 y0  y0
 100%
y0

(1.21  1) y0
 100%
y0
 21%
2.
∵
∴
p. 243
1.
Percentage change in y
1.3 y0  y0

 100%
y0
∴
y is increased by 21%.
(a)
∵
y
∴
y  k x, k  0
1.
(a)
By substituting x  16 and y  8 into the
equation, we have
8  k ( 16 )
8  4k
k2
y2 x
∴
(b) Let x0 and y0 be the original values of x and y
respectively.
∴ New value of x  (1  69%) x0
∵
y varies inversely as x.
k
∴ y ,k 0
x
By substituting x  1 and y  3 into the
equation, we have
k
3
1
k 3
3
x
When x  3,
(b)
y
3
3
1
y
 1.69 x0
 1.3 y 0
C  r2
C  kr 2 , k  0
By substituting r  6 and C  20 into the
equation, we have
20  k (6) 2
20
k
36
5

9
5
∴
C  r2
9
(i) When C  45,
5
45  r 2
9
r 2  81
∴
r 9
∴ The radius of the hemisphere is 9 cm.
(ii) When r  0.18  100  18,
5
C  (18) 2
9
 180
∴ The cost of painting a hemisphere is
$180.
∴
 1.3( 2 x 0 )
 30%
y is increased by 30%.
p. 251
x
New value of y  2 1.69 x 0
(1.3  1) y0
 100%
y0
2.
∵
∴
z varies inversely as x.
k
z ,k 0
x
By substituting x  1 and z 
1
into the equation,
3
we have
2
Certificate Mathematics in Action Full Solutions 4B
∴
1 k

3 1
1
k
3
1
∴
z
3x
When z  2,
Percentage change in y
0.64 y0  y0

 100%
y0

 36%
y is decreased by 36%.
∴
1
3x
6x  1
2
2.
1
x
6
3.
y2 varies inversely as x.
k
y2  , k  0
∴
x
By substituting x  4 and y  3 into the equation,
we have
k
32 
4
k  36
∴
36
x
25 x  36
52 
36
25
∴
3.
(a)
1
x2
k
y 2,k 0
∴
x
By substituting x  2 and y  4 into the
equation, we have
k
4 2
2
k  16
∵
y
∴
y
16
x2
New value of y 
16
(1.25 x0 ) 2
 16 
 0.64 2 
 x0 
 0.64 y0
3
0.64 x0
(1.25  1) y0
 100%
y0
 25%
y is increased by 25%.
Let T be the number of days that the food can last and
n be the number of people trapped on an isolated
island.
1
∵ T
n
k
∴ T  ,k 0
n
By substituting n  10 and T  30 into the
equation, we have
k
30 
10
k  300
300
n
When n  12,
300
T 
12
 25
∴ The food can last for 25 days if 2 more people
go ashore on the island.
∴
(b) Let x0 and y0 be the original values of x and y
respectively.
∴ New value of x  (1  25%) x0
 1.25 x0
k
 k 

 1.25
 x 
 0 
 1.25 y0
Percentage change in y
1.25 y0  y0

 100%
y0

p. 253
x
k
New value of y 
36
x
When y  5,
1.
y
1
 0.64 x0
y2 
x
y
,k 0
x
Let x0 and y0 be the original values of x and y
respectively.
∴ New value of x  (1  36%) x0
∴
∵
∴
∵
(0.64  1) y0
 100%
y0
T 
10 Variations
∴
p. 261
1.
(a)
(b)
(c)
2.
(a)
z  kx2 y 3
kd
s
t
kb3
a2 
c
∴
Percentage change in z
0.5z 0  z 0

 100%
z0
 50%
z is decreased by 50%.
p. 269
∵
z varies jointly as x2 and
∴
z  kx2 y , k  0
y.
1.
By substituting x  2 , y  9 and z  24
into the equation, we have
24  k ( 2) 2 9
k2
2.
(a)
y  k1 x  k2 x 2
(b)
a  k1 
(c)
A  k1B 2 
(a)
z  2x y
∴
(b) When x  3 and y  16 ,
∵
2
∴
4.
∵
∵
z
z
 0.75 x0
New value of y  (1  50%) y0
 1 .5 y 0
New value of z  k (0.75 x 0 )
1.5 y 0
kx
 0.5  0 
 y0 
 0.5z 0
z partly varies directly as x and partly
varies directly as y.
z  k1 x  k 2 y, where k1 , k 2  0
(2)  (1)  2, k2  3
By substituting k2  3 into (1), we have
2k1  3  4
1
2
1
z  x  3y
2
k1 
∴
(b)
When x  3 and y  2 ,
1
z  (3)  3( 2)
2
15

2
(a)
∵
x
y
kx
,k 0
y
Let x0, y0 and z0 be the original values of x, y and z
respectively.
∴ New value of x  (1  25%) x0
∴
B
By substituting x  4 , y  3 and z  11
into the equation, we have
11  k1 (4)  k2 (3)
…… (2)
4k1  3k2  11
x varies directly as y and inversely as z2.
ky
∴ x 2,k 0
z
By substituting x  4 , y  2 and z  1 into
the equation, we have
k ( 2)
4 2
1
k2
2y
∴ x 2
z
(b) When y  3 and z  2 ,
2(3)
x 2
2
3

2
(a)
k2
By substituting x  2 , y  1 and z  4 into
the equation, we have
4  k1 (2)  k2 (1)
…… (1)
2k1  k2  4
z  2(3)2 16
 72
3.
k2
c2
3.
P partly varies directly as x and partly
varies inversely as x.
k
∴ P  k 1 x  2 , where k 1 , k 2  0
x
By substituting x  1 and P  2 into the
equation, we have
k
2  k1 (1)  2
1
…… (1)
k1  k2  2
By substituting x  2 and P  7 into the
equation, we have
k
7  k 1 ( 2)  2
2
…… (2)
4k 1  k 2  14
( 2)  (1), 3k1  12
k1  4
4
Certificate Mathematics in Action Full Solutions 4B
By substituting k1  4 into (1), we have
When y  6,
4  k2  2
1
x
4
x  24
6
k 2  2
P  4x 
∴
2
x
When x  40,
1
(40)
4
 10
y
(b) When x  4 ,
2
P  4( 4) 
4
31

2
4.
(a)
∵
∴
When x  60,
1
(60)
4
 15
y
C is partly constant and partly varies
directly as n.
C  k1  k 2 n, where k1 , k 2  0
(b)
By substituting n  10 and C  1750 into
the equation, we have
1750  k1  k 2 (10)
k1  10k 2  1750
…… (1)
When y  2.2,
2.2  2.2 x
( 2)  (1), 15k 2  1875
x 1
k 2  125
When x  6,
y  2.2(6)
By substituting k2  125 into (1), we have
k1  10(125)  1750
 13.2
k1  500
When y  17.6,
17.6  2.2 x
C  500  125n
(b) When n  20 ,
C  500  125(20)
 3000
3000
The cost per head  $
20
 $150
y varies directly as x.
y  kx, k  0
By substituting x  3 and y  6.6 into the
equation, we have
6 .6  k  3
k  2 .2
y  2 .2 x
∴
By substituting n  25 and C  3625 into
the equation, we have
3625  k1  k2 (25)
k1  25k2  3625
…… (2)
∴
∵
∴
x  8
When y  19.8,
19.8  2.2 x
x9
2.
(a)
(i)
∵
y varies directly as x.
∴
Variation constant 
Exercise
Exercise 10A (p. 245)
Level 1
1.
(a)
∵
∴
1
4
1
∴ y x
4
When y  3,
k
1
x
4
x  12
3
5
(ii) The equation connecting x and y is
2
y x.
3
y varies directly as x.
y  kx, k  0
By substituting x  32 and y  8 into the
equation, we have
8  k  32
20
30
2

3
(b)
(i)
∵
∴
y varies directly as x.
Variation constant
20

10
 2
(ii) The equation connecting x and y is
y  2 x .
10 Variations
3.
(a)
2
∴
(b)
When x  9 ,
2
y
9
3
2
 (3)
3
2
∵
( y  1)  3 x
(b) From the graph, when x  8 , y  12
4.
(a)
∵
A varies directly as l 2.
∴
A  kl2 , k  0
7.
y 1  k3 x , k  0
y 1
k
3
x
For any two pairs of x and y, say (x1, y1) and
(x2, y2), we have
y1  1
y 1
 2
3
3
x1
x2
∴
By substituting l  4 and A  48 into the
equation, we have
48  k ( 4) 2
48  16k
k 3
∴
A  3l 2
By substituting x1  27 , y1  7 and x2  64 into
the equation, we have
(b) When l  3 ,
A  3( 3 ) 2
 3(3)
9
5.
(a)
∵
V varies directly as d 3 .
∴
V  kd 3 , k  0
y 1
7 1
 32
3
27
64
y2  1
6

3
4
8  y2  1
By substituting d  5 and V  62
y2  9
1
into the
2
equation, we have
1
62  k (5)3
2
125
 k (125)
2
1
k
2
1 3
∴
V  d
2
i.e. When x  64 , y  9
8.
∵
y  x2
y  kx 2 , k  0
∴
Let x0 and y0 be the original values of x and y
respectively.
∴ New value of x  (1  75%) x 0
 0.25 x 0
2
New value of y  k (0.25 x 0 )
 0.0625kx0
(b) When V  32 ,
1
32  d 3
2
d 3  64
∴
d 4
(a)
∵
y
∴
y  k x, k  0
∴
x
By substituting x  16 and y  2
2
 0.0625 y 0
Percentage change in y
0.0625 y 0  y 0

 100%
y0

6.
2
 k 16
3
8
 k (4)
3
2
k
3
2
y
x
3
∴
(0.0625  1) y 0
 100%
y0
 93.75%
y is decreased by 93.75%.
2
into the
3
equation, we have
6
Certificate Mathematics in Action Full Solutions 4B
9.
∵
∴
S  t
S  kt, k  0
3
 60  45 and S  45
4
into the equation, we have
Level 2
13. (a)
By substituting t 
45  k (45)
k 1
∴
S  t
When t  25 ,
S  25
∴ The train travels 25 km in 25 minutes.
10. ∵
∴
(b)
(c)
By substituting x  4 and C  96 into the
equation, we have
96  k (4) 2
k  6
∴
C  6x 2
When x  3 ,
∴
11. ∵
∴
14. (a)
y  1  kx 3 , k  0
∵
y 2  ( x  b)
∴
y 2  k ( x  b), k  0
(2) ÷ (1),
k ( 9  b)
16

4
k ( 3  b)
9 b
4 
3b
12  4b  9  b
3b  3
b 1
(b)
y varies directly as x.
y  kx, k  0
2
kx  2

2
x  2
k ( x  2)  2k  2

x  2
2  2k
 k 
x  2
2  2k
Obviously, k 
is not a constant.
x  2
∴ ( y  2) does not vary directly as ( x  2) .
When y  3 x ,
3x  5 x  2
By substituting x  9 and y  4 into the
equation, we have
4 2  k ( 9  b)
16  k (9  b) ...... (2)
( y  1)  x 3
y  2
By substituting k  1 and x  2 into the
equation, we have
y  1  (1)( 2) 3
12. ∵
∴
y 
x 
y  5x  2
i.e.
When x  2 ,
y  5( 2)  2
 8
By substituting x  3 and y  2 into the
equation, we have
2 2  k ( 3  b)
4  k (3  b) ...... (1)
C  6(3) 2
 54
The cost of painting a cube of side 3 cm is $54.
y  9
The two possible pairs of x and y are x  1 ,
y  2 or x  2 , y  9 . (or any other
reasonable answers)
y  2  kx, k  0
2x  2
x 1
By substituting k  1 and x  1 into the
equation, we have
y  1  (1)(1) 3
∴
( y  2)  x
By substituting x  3 and y  13 into the
equation, we have
13  2  k (3)
k  5
y  2  5x
∴
C  x2
C  kx 2 , k  0
∵
∴
From (1), 4  k (3  1)
2k  4
∴
(c)
y
2
k  2
 2( x  1)
When y  6 ,
6 2  2( x  1)
18  x  1
x  19
15. (a)
∵
y  x2
∴
y  kx 2 , k  0
By substituting x  t and y  2 into the
equation, we have
...... (1)
2  kt 2
7
10 Variations
By substituting x  t  1 and y  8 into
the equation, we have
8  k (t  1) 2 ...... (2)
k (t  1) 2
8

2
kt 2
(t  1) 2
4 
t2
4t 2  t 2  2t  1
(2) ÷ (1),
∴
18. (a)
k  800
∴
V  800W 3
When W  3.5 ,
(b) When t  1 ,
from (1), 2  k (1) 2
k  2
y  2x 2
∴
v t
v  kt, k  0
By substituting t  2.5 and v  25 into the
equation, we have
25  k ( 2.5)
k  10
∴
v  10t
When t  5.5 ,
v  10(5.5)
 55
∴ The speed of the body is 55 m/s after it has
fallen for 5.5 s.
17. (a)
∵
∴
V  800(3.5) 3
∴
(b)
19. (a)
(b)
When d  3 ,
t  3( 3)
 9
∴ It takes 9 seconds for you to hear the
thunder from lightning that is 3 km
away.
(ii) When d  8 ,
t  3(8)
 24
∴ It takes 24 seconds for you to hear
the thunder from lightning that is
8 km away.
When t  1 ,
1  3d
1
d 
3
The number of kilocalories and the weight of
the potato chips are in direct variation.
∵
∴
∴
(c)
(b) (i)
(c)
Original value  $6400
New value  $[2  800(1)]
E W
E  kW, k  0
By substituting W  200 and E  1000
into the equation, we have
1000  k ( 200)
By substituting d  5 and t  15 into the
equation, we have
15  k (5)
∴
 34 300
The value of a diamond weighing
3.5 carats is $34 300.
 $1600
$(6400  1600)
Percentage loss 
 100%
$6400
 75%
t  d
t  kd, k  0
k  3
t  3d
V  kW 3 , k  0
By substituting W  2 and V  6400 into
the equation, we have
6400  k (2) 3
1
or t  1
3
16. ∵
∴
Let $V be the value of the diamond and W carats
be the weight of the diamond.
∵ V  W3
∴
3t 2  2t  1  0
(3t  1)(t  1)  0
t  
The distance apart from the lightning is
1
km when the time between the thunder
3
and lightning is only 1 second.
20. (a)
k  5
E  5W
When W  70 ,
E  5(70)
 350
∴ If Sally has eaten 70 g of potato chips, she
has taken 350 kcal.
∵
∴
∵
∴
V m
V  k1 m, k1  0
mt
m  k 2 t, k 2  0
V  k1 m
 k1 (k 2 t )
 k1 k 2 t
 k 3 t (where k 3  k 1 k 2  0)
∴
V varies directly as t.
8
Certificate Mathematics in Action Full Solutions 4B
(b)
When x  28,
V  m  k1 k 2 t  k 2 t
 ( k 1 k 2  k 2 )t
56
28
 2
y 
 k 4 t (where k 4  k 1 k 2  k 2  0)
∴
(V  m ) varies directly as t.
∵
∴
V  (m  t )
When x  56,
21. (a)
∵
∴
56
56
1
y 
V  k1 (m  t ), where k1  0
m  (t  V )
m  k 2 (t  V ), where k 2  0
V  k 1 (m  t )
(b)
 k 1 {[(k 2 (t  V )]  t}
 k 1 k 2 t  k 1 k 2V  k 1 t
(1  k 1 k 2 )V  ( k 1 k 2  k 1 )t
k k  k1
V   1 2
 1  k1 k 2
t


V  k 3 t (where k 3 
∴
k1 k 2  k1
 0)
1  k1 k 2
∵
y varies inversely as x.
k
∴
y  , k  0
x
By substituting x  12 and y  5 into the
equation, we have
k
5 
12
k  60
60
x
When x  2,
∴
V varies directly as t.
y 
m  k 2 (t  V )
(b)
60
2
 30
y 
m  k 2 [t  k 1 ( m  t )]
m  k 2 t  k1 k 2 m  k1 k 2 t
When y  15,
(1  k 1 k 2 )m  (k 2  k 1 k 2 ) t
k  k1 k 2
m   2
 1  k1 k 2
m  k 4 t (where k 4 
∴
60
x
x  4
t


15 
k 2  k1 k 2
 0)
1  k1 k 2
When x  6,
60
6
 10
m varies directly as t.
y 
Exercise 10B (p. 254)
Level 1
1.
(a)
When y  6,
∵
y varies inversely as x.
k
y  , k  0
∴
x
By substituting x  4 and y  14 into the
equation, we have
k
14 
4
k  56
56
x
When x  2,
∴
60
x
x  10
6 
When y  4,
60
x
x  15
4 
y 
2.
(a)
56
2
 28
(i)
∵
∴
 360
y 
(ii) The equation connecting x and y is
360
y 
.
x
When x  7,
56
7
 8
y 
When x  8,
56
8
 7
y 
9
y varies inversely as x.
Variation constant  30  12
(b)
(i)
∵
∴
y varies inversely as x.
Variation constant  5  5
 25
(ii) The equation connecting x and y is
25
y 
.
x
10 Variations
3.
(a)
6.
(a)
( y  2) varies inversely as x2.
∵
k
, k  0
x2
1
By substituting x 
and y  25 into the
3
equation, we have
k
25  2 
2
1
 
 3
27  9k
y  2 
∴
k  3
3
y  2  2
x
3
y  2  2
x
∴
(b) From the graph, when x  24 , y  7.5
i.e.
4.
(a)
∵
∴
F varies inversely as d
k
F  2 , k  0
d
2.
(b)
When y  x 2 ,
x4
x 4  2x 2  3  0
(x
2
By substituting x  1 into y  x 2 , we have
y  12  1
By substituting x  1 into y  x 2 , we
have
y  ( 1) 2  1
The value of y is 1 when y  x 2 .
∴
(a)
∵
∴
y varies inversely as 3 x .
k
y  3 , k  0
x
By substituting x  27 and y 
equation, we have
1
k
 3
6
27
1
k 
2
1
y  3
∴
2 x
(b) When x 
y 
1
,
8
1
1
2
8
1

1
2  
2
 1
 1)( x 2  3)  0
x 2  1 or x 2  3 (rejected)
x  1
(b) When d  2 ,
4
F  2
2
1
5.
3
 2
x2
 3  2x 2
x2 
1
By substituting d  4 and F 
into the
4
equation, we have
1
k
 2
4
4
k  4
∴ F  42
d
7.
1
into the
6
1
x2
k
y  2 , k  0
∴
x
Let x0 and y0 be the original values of x and y
respectively.
∴ New value of x  (1  20%) x 0
∵
y 
 0 .8 x 0
New value of y 
∴
3
 k
 1.5625 2
 x0
 1.5625 y 0
Percentage change in y
1.5625 y 0  y 0

 100%
y0

∴
k
(0.8 x 0 ) 2



(1.5625  1) y 0
 100%
y0
 56.25%
y is increased by 56.25%.
10
Certificate Mathematics in Action Full Solutions 4B
8.
Let h cm be the height of the cylinder and A cm2 be
the base area of the cylinder.
1
∵ h 
A
k
∴ h 
, k  0
A
(c)
When y  4 ,
25
x  9
2 x  18  25
4  
2 x  7
x  
By substituting A  2 and h  30 into the
equation, we have
k
30 
2
k  60
∴
h 
10. (a)
60
A
When h  15 ,
60
15 
A
A  4
∴
The base area of the cylinder is 4 cm2 when the
∴
Level 2
(a)
1
x  a
k
∴
y 
, k  0
x  a
By substituting x  4 and y  25 into the
equation, we have
k
25 
4  a
k
5 
...... (1)
4  a
By substituting x  8 and y  625 into the
equation, we have
k
625 
8 a
k
..... (2)
25 
8 a
∵
k
25
8

a

5
k
4  a
4  a
5 
8 a
40  5a  4  a
4a  36
a  9
V 
∴
(b)
The value of the flat in 2004 is $750 000.
When V  2 000 000 ,
6 000 000
A
A  3
2 000 000 
∴
The flat was worth $2 000 000 in 1999.
1
P
k
, k  0
∴ V 
P
k
, k  0
i.e. P 
V
Let V0 and P0 be the original volume and pressure of
the gas respectively.
∴ New volume  (1  10%)V 0
11. ∵
V 
 0.9V 0
New pressure 
(b)
11
k
0.9V 0
10  k 
 
9  V0 
10

P
9 0

k
From (1), 5 
4  ( 9 )
k  25
25
y  
∴
x 9
6 000 000
A
The age of the flat in 2004 is ( 4  4) years old,
i.e. 8 years old.
When A  8 ,
6 000 000
V 
8
 750 000
y 
(2) ÷ (1),
Let $V be the value of the flat and A years old
be the age of the flat.
1
∵ V 
A
k
∴ V 
, k  0
A
By substituting A  4 and V  1 500 000
into the equation, we have
k
1 500 000 
4
k  6 000 000
height is 15 cm.
9.
7
2
10 Variations
∴
Percentage change in pressure
10
P0  P0
 9
 100%
P0
(b)
 10  1 P
 9
 0

 100%
P0
 11
∴
12. (a)
1
%
9
The pressure is increased by 11
1
%.
9
1
n
k
p 
, k  0
∴
n
By substituting n  1600 and p  2.5 into
the equation, we have
k
2.5 
1600
k  100
∵
∴
p 
(c)
13. (a)
14. Let n be the number of workers and T be the number
of months needed to complete the job.
1
∵ n 
T
k
∴ n 
, k  0
T
By substituting T  21 and n  56 into the
equation, we have
k
56 
21
k  1176
1176
T
When T  7 ,
∴
∴
15. (a)
When n  2500 ,
100
p 
2500
 2
∴ The retail price is $2 when the number of
bottles available on the island is 2500.
1
T
k
, k  0
∴ V 
T
By substituting T  1.8 and V  265 into
the equation, we have
k
265 
1.8
k  477
477
∴
V 
T
When T  2.5 ,
477
V 
2.5
 190.8
∴ The average speed of the helicopter is
190.8 km/h when its travelling time is
2.5 hours.
∵
V 
n 
1176
7
 168
168  56  112
An additional 112 workers has to be employed
to complete the same job in 7 months.
n 
100
p 
n
(b) When p  4 ,
100
4 
n
n  25
n  625
∴ 625 bottles are available on the island
when the retail price is $4.
When V  210 ,
477
210 
T
T  2.27 (cor. to 3 sig. fig.)
∴ The travelling time of the helicopter is
2.27 h when its average speed is 210 km/h.
1
d
k
∴ n 
, k  0
d
By substituting d  40 and n  800 into
the equation, we have
k
800 
40
k  32 000
32 000
∴
n 
d
∵
n 
(b)
When d  80 ,
32 000
n 
80
 400
∴ The front wheel of a bicycle with a
diameter of 80 cm would make 400
revolutions in travelling 1 km.
(c)
When n  200 ,
32 000
200 
d
d  160
∴ The diameter of the front wheel of a
bicycle is 160 cm if it makes only 200
revolutions in travelling 1 km.
12
Certificate Mathematics in Action Full Solutions 4B
16. (a)
The length of the rectangle and the width of the
rectangle are in inverse variation.
1
(b) ∵ l 
w
k
∴ l 
, k  0
w
By substituting w  20 and l  45 into the
equation, we have
k
45 
20
k  900
∴
(c)
l 
Exercise 10C (p. 261)
Level 1
1.
∴
(b)
When m = 4 and n = 6,
P  8(4)(6)
 192
When w  18 ,
900
l 
18
 50
∴ The length of the rectangle is 50 cm when
its width is 18 cm.
∵
∵ P varies jointly as m and n.
∴ P = kmn, k  0
By substituting m = 3, n = 5 and P = 120 into
the equation, we have
120  k (3)(5)
k 8
∴ P = 8mn
900
w
2.
(d) When the rectangle becomes a square, l  w .
When l  w ,
900
w 
w
w 2  900
w  30 or w  30 (rejected)
∴ The length of the rectangle is 30 cm when
it becomes a square.
17. (a)
(a)
1
1
( x  y )    
y
x
k
x  y 
,k  0
1
1

x
y
kxy
x  y 
x  y
3.
(a)
∵ z varies jointly as x and y2.
∴ z = kxy2, k  0
By substituting x = 3, y = 2 and z = 3 into the
equation, we have
3  k (3)(2) 2
1
k
4
1 2
∴ z  xy
4
(b)
When x = 5 and y = 4,
1
z  (5)(4) 2
4
 20
(a)
∵
( x  y ) 2  kxy
∴
( x  y ) 2  xy
(b)
(b)
x
2
( x  y)
2
 kxy, k  0
 2 xy  y
2
 kxy
x 2  y 2  (k  2) xy
1
xy 
(x 2  y 2 )
k 2
∵ k  0
1
 0
∴
k 2
1
By letting k  
, we have
k  2
xy  k ( x 2  y 2 ), k   0
∴
13
xy  ( x 2  y 2 )
4.
(a)
a varies directly as b and inversely as c2.
kb
∴ a  2 ,k  0
c
By substituting a = 1, b = 3 and c = 3 into the
equation, we have
k (3)
1 2
3
k 3
3b
∴ a 2
c
When b = 2 and c = 2,
3( 2)
a 2
2
3

2
∵
y varies directly as u2 and inversely as
t .
y
ku2
,k  0
t
By substituting u = 2, t = 9 and y = 4 into the
equation, we have
k ( 2) 2
4
9
k 3
3u 2
∴ y
t
∴
10 Variations
(b) When u = 4 and y = 12,
3( 4) 2
12 
t
t 4
∴
t  16
(b)
5.
(a)
3
u
v2
ku3
∴ w 2 ,k  0
v
By substituting u = 6, v = 3 and w = 2 into the
equation, we have
k (6)3
2 2
3
1
k
12
u3
∴ w
12v 2
∵
w
8.
(b) When w  3u3 ,
u3
3u 3 
12v 2
2
36v  1
1
v2 
36
1
v
6
6.
∵
k (4)(3)
22
k 5
5 pq
W  2
r
15 
When W = 6, p = 3 and r = 5,
5(3)q
6 2
5
q  10
T  wn
T = kwn, k  0
T
i.e. w  , k  0
kn
Let T0, n0 and w0 be the original values of T, n and w
respectively.
∴ New value of T = 2T0
New value of n = 3n0
2T0
New value of w 
k (3n0 )
∵
∴
∴

2  T0

3  kn0

2
w0
3




2
w0  w0
 100%
Percentage change in w  3
w0
1
 33 %
3
Cd t
2
∴ C  kd 2t , k  0
Let d0, t0 and C0 be the original values of d, t and C
respectively.
∴ New value of d  (1  40%)d 0
9.
∴
1
w is decreased by 33 % .
3
(a)
∵
V  r 2h
∴ V  kr2 h, k  0
By substituting r = 3, h = 4 and V = 108 into the
equation, we have
108  k (3) 2 (4)
k 3
∴
V  3r 2h
 1.4d 0
New value of t  (1  20%)t 0
 0.8t 0
New value of C  k (1.4d 0 ) 2 (0.8t 0 )
 1.568kd0 2 t 0
 1.568C0
∴
Percentage change in C 
∴
C is increased by 56.8%.
1.568C 0  C0
C0
 100%
 56.8%
(b)
Let r0, h0 and V0 be the original values of r, h
and V respectively.
∴ New value of r = 0.5r0
New value of h = 2h0
2
New value of V  3(0.5r0 ) (2h0 )
 0.5(3r0 h0 )
2
Level 2
7.
(a)
pq
r2
kpq
∴ W  2 ,k  0
r
By substituting p = 4, q = 3, r = 2 and W = 15
into the equation, we have
∵
 0.5V0
W 
∴
new value of V
original value of V
0.5V0

V0
The required ratio 
1
2
 1: 2

14
Certificate Mathematics in Action Full Solutions 4B
M 1M 2
d2
kM1M 2
,k  0
∴ F
d2
Let (M1)0, (M2)0, d0 and F0 be the original values of
M1, M2, d and F respectively.
∴ New value of M 1  0.5( M 1 )0
10. ∵
F
New value of M 2  0.5( M 2 )0
New value of d  2d 0
New value of F 
∴
k[0.5( M 1 ) 0 ][0.5( M 2 ) 0 ]
( 2d 0 ) 2
 k (M 1 ) 0 (M 2 ) 0 
 0.0625

2
d0


 0.0625F0
new value of F
The required ratio 
original value of F
0.0625F0

F0
1
16
 1 : 16

11. (a)
A
d
kA
,k  0
∴ n
d
By substituting A = 1500, d = 50 and n = 12 000
into the equation, we have
k (1500)
12 000 
50
k  400
400 A
n
∴
d
∵
n
(b) When A = 1200 and d = 80,
400(1200)
n
80
 6000
V
P
kV
,k  0
∴ x
P
Let x1 min, V1 cm3 and P1 W be the time required to
boil the first kettle of water, the volume of water in
the first kettle and the power of the stove to boil the
first kettle of water, and x2 min, V2 cm3 and P2 W be
the time required to boil the second kettle of water,
the volume of water in the second kettle and the
power of the stove to boil the second kettle of water
respectively.
kV
∴ x1  1  (1)
P1
12. ∵
x
x2 
15
kV2
P2
 (2)
x1 kV1 P2


x2
P1 kV2
(1) ÷ (2),
x1 V1 P2


x2 V2 P1
9
9 13
 
x2 13 12
x2  12
∴
The time required to boil the second kettle of
water is 12 min.
y
x
k1 y
∴ z
, where k1 is a non-zero constant  (1)
x
1
y
∵
x
k2
y
∴
, where k2 is a non-zero constant.
x
k
i.e. x  2 , where k2 is a non-zero constant  (2)
y
By substituting (2) into (1), we have
ky
z 1
k2
y
13. ∵
z
k1 2
(y )
k2
k1 and k2 are non-zero constants.
k1
is also a non-zero constant.
k2

∵
∴
By letting k3 
k1
, we have
k2
z  k3 y 2 , where k3 is a non-zero constant.
i.e.
z  y2
14. ∵ z 
3
x
y2
∴ z
k1 3 x
, where k1 is a non-zero constant  (1)
y2
∵ x
1
y
∴ x
k2
, where k2 is a non-zero constant.
y
k2
, where k2 is a non-zero constant
x
By substituting (2) into (1), we have
i.e. y 
z
k1 3 x
 k2 
 
 x 
2
1
 k1 x 3 

k1
k2
2
x2
2
k2
7
(x 3 )
 (2)
10 Variations
∵
∴
(2) – (1), 324k2  81
k1 and k2 are non-zero constants.
k1
is also a non-zero constant.
2
k2
By letting k 3 
k1
k2
2
k2 
By substituting k2 
, we have
z  k3 x , where k3 is a non-zero constant.
z 3  k3 x 7 , where k3 is a non-zero constant.
i.e. z3  x7
Exercise 10D (p. 270)
Level 1
(a)
∵
z is partly constant and partly varies
directly as x2.
∴ z  k1  k2 x 2 , where k1, k2  0
By substituting x = 2 and z = 4 into the equation,
we have
4  k1  k2 (2)2
k1  4k2  4
 (1)
13
By substituting x = 3 and z 
into the
2
equation, we have
13
 k1  k 2 (3) 2
2
13
k1  9k 2 
 (2)
2
5
(2) – (1), 5k 2 
2
1
k2 
2
1
By substituting k2 
into (1), we have
2
1
k1  4   4
2
k1  2
1
∴ z  2  x2
2
1
into (1), we have
4
1
3k1  108   28
4
1
k1 
3
1
1
∴ w  s  s3
3
4
7
3
1.
1
4
3.
(b)
When s = 2,
1
1
w  (2)  (2)3
3
4
8

3
(a)
∵
∴
z is partly constant and partly varies
inversely as the square root of y.
k
z  k1  2 , where k1, k2  0
y
By substituting y = 1 and z = 8 into the equation,
we have
k
8  k1  2
1
k1  k 2  8
 (1)
By substituting y = 4 and z = 7 into the equation,
we have
k
7  k1  2
4
2k1  k 2  14
 (2)
(2) – (1), k1 = 6
By substituting k1 = 6 into (1), we have
6  k2  8
k2  2
∴
z  6
2
y
(b) When x = 4,
1
z  2  (4) 2
2
 10
2.
(a)
(b)
When y = 25,
2
z  6
25
32

5
(a)
∵
∵
w partly varies directly as s and partly
varies directly as s3.
∴ w = k1s + k2s3, where k1, k2  0
By substituting s = 6 and w = 56 into the
equation, we have
56  k1 (6)  k 2 (6) 3
6k1  216k 2  56
3k1  108k 2  28
 (1)
By substituting s = 12 and w = 436 into the
equation, we have
436  k1 (12)  k2 (12)3
12k1  1728k2  436
3k1  432k2  109
4.
a partly varies directly as the square root
of b and partly varies inversely as the
square of c.
k
∴ a  k1 b  22 , where k1, k2  0
c
By substituting b = 1, c = 1 and a = 3 into the
equation, we have
k
3  k1 1  22
1
k1  k 2  3
 (1)
 (2)
16
Certificate Mathematics in Action Full Solutions 4B
By substituting b = 4, c = 2 and a 
equation, we have
5
k
 k1 4  22
2
2
8k1  k2  10
(1) – (2), 2 k 2  60
k 2  30
5
into the
2
By substituting k2 = 30 into (2), we have
k1  38(30)  1940
 (2)
k1  800
(2) – (1), 7 k1  7
∴ C = 800 + 30n
When n = 43,
C  800  30(43)
 2090
∴ The cost for a class of 43 students is $2090.
k1  1
By substituting k1 = 1 into (1), we have
1  k2  3
k2  2
∴
a b
2
c2
7.
(a)
(b) When b = 16 and c = 3,
2
a  16  2
3
38

9
5.
(a)
∵
E is partly constant and partly varies
directly as x.
∴ E = k1 + k2x, where k1, k2  0
By substituting x = 300 and E = 680 into the
equation, we have
680  k1  k2 (300)
k1  300k 2  680
 (1)
By substituting x = 450 and E = 845 into the
equation, we have
845  k1  k2 (450)
k1  450k2  845
 (2)
(2) – (1), 150 k 2  165
k 2  1 .1
By substituting k2 = 1.1 into (1), we have
k1  300(1.1)  680
∴
(b)
E = 350 + 1.1x
(b) When x = 380,
E  350  1.1(380)
 768
∴ The daily expenditure of hiring a taxi is
$768 for travelling a distance of 380 km.
6.
Let $C be the cost of holding a class picnic and n be
the number of students in the class.
∵ C is partly constant and partly varies directly as
n.
∴ C = k1 + k2n, where k1, k2  0
By substituting n = 40 and C = 2000 into the equation,
we have
2000  k1  k 2 (40)
k1  40k2  2000
 (1)
By substituting n = 38 and C = 1940 into the equation,
we have
1940  k1  k2 (38)
k1  38k2  1940
 (2)
17
C is partly constant and partly varies
inversely as n.
k
∴ C  k1  2 , where k1, k2  0
n
By substituting n = 2000 and C = 225 into the
equation, we have
k
225  k1  2
2000
 (1)
2000k1  k 2  450 000
By substituting n = 5000 and C = 210 into the
equation, we have
k
210  k1  2
5000
5000k1  k2  1 050 000
 (2)
(2) – (1), 3000 k1  600 000
k1  200
By substituting k1 = 200 into (1), we have
2000(200)  k2  450 000
k2  50 000
k1  350
∴
∵
Level 2
8. (a)
C  200 
50 000
n
When n = 8000,
50 000
C  200 
8000
 206.25
∴ The cost of producing a copy of the video
game is $206.25.
∵
y is partly constant, partly varies directly
as x and partly varies directly as x2.
∴ y = k1 + k2x + k3x2, where k1, k2 , k3  0
By substituting x = 0 and y = 5 into the equation,
we have
5  k1  k2 (0)  k3 (0)2
k1  5
By substituting x = –1 and y = 2 into the
equation, we have
2  5  k2 (1)  k3 (1)2
k2  k3  3
 (1)
By substituting x = 2 and y = 5 into the equation,
we have
10 Variations
5  5  k2 (2)  k3 (2) 2
10. (a)
2 k 2  4 k3  0
k 2  2 k3  0
 (2)
(2) – (1), 3k3  3
k 3  1
By substituting k3 = –1 into (1), we have
k 2  ( 1)  3
k2  2
y  5  2x  x2
∴
(b)
S is partly constant and partly varies
directly as A.
∴ S = k1 + k2A, where k1, k2  0
By substituting A = 300 000 and S = 7000 into
the equation, we have
7000  k1  k2 (300 000)
k1  300 000k2  7000
 (1)
By substituting A = 400 000 and S = 9000 into
the equation, we have
9000  k1  k2 (400 000)
k1  400 000k2  9000
 (2)
(2) – (1), 100 000k2  2000
When y = –3,
 3  5  2x  x2
1
50
1
By substituting k 2 
into (1), we have
50
k2 
x  2x  8  0
( x  2)( x  4)  0
x   2 or x  4
2
(c)
∵
 1 
k1  300 000   7000
 50 
k1  1000
1
A
∴ S  1000 
50
y  5  2x  x2
 ( x 2  2 x  5)
 ( x 2  2 x  12  12  5)
 ( x 2  2 x  1)  6
 6  ( x  1) 2
∴
9. (a)
(b)
When A = 350 000,
1
S  1000  (350 000)
50
 8000
∴ Her income for that month is $8000.
(c)
When S = 20 000,
The maximum value of y is 6.
∵
C partly varies directly as l and partly
varies directly as w2.
∴ C = k1l + k2w2, where k1, k2  0
By substituting l = 1, w = 20 and C = 200 000
into the equation, we have
200 000  k1 (1)  k 2 (20) 2
 (1)
k1  400k 2  200 000
20 000  1000 
∴
150 000  1.5k1  100k 2
200
 (2)
k 2  100 000
3
(1) – (2), 1000 k2  100 000
3
k2  300
By substituting k2 = 300 into (2), we have
200
k1 
(300)  100 000
3
k1  80 000
∴ C = 80 000l + 300w2
(b) When l = 2 and w = 15,
C  80 000(2)  300(15)2
∴
 227 500
The cost of building a road is $227 500
when the length is 2 km and the width is
15 m.
1
A
50
A  950 000
The amount of her sales should be
$950 000.
19 000 
By substituting l = 1.5, w = 10 and C = 150 000
into the equation, we have
150 000  k1 (1.5)  k 2 (10) 2
k1 
1
A
50
11. (a)
∵
C is partly constant and partly varies
inversely as n.
k
∴ C  k1  2 , where k1, k2  0
n
By substituting n = 1000 and C = 53 000 into
the equation, we have
k
53 000  k1  2
1000
 (1)
1000k1  k 2  53 000 000
By substituting n = 2000 and C = 58 000 into
the equation, we have
k
58 000  k1  2
2000
 (2)
2000k1  k 2  116 000 000
(2) – (1), 1000k1  63 000 000
k1  63 000
By substituting k1 = 63 000 into (1), we have
1000(63 000)  k2  53 000 000
k2  10 000 000
18
Certificate Mathematics in Action Full Solutions 4B
∴
C  63 000 
(b)
10 000 000
n
When x = 20,
P  800(20)  5(20) 2
 14 000
(b) When n = 4000,
C  63 000 
∴
12. (a)
10 000 000
4000
(c)
 5( x 2  160 x)
 60 500
The cost of printing 4000 copies of the
school magazine is $60 500.
Let L0 cm and Le cm be the original length and
the extended length of the spring respectively.
∵ Le  W
∴ Le = kW, k  0
10.1  L0  k ( 2)
 (1)
11.3  L0  k (3.5)
 (2)
11.3  L0
(2)  (1),
 1.75
10.1  L0
 5( x 2  160 x  802  802 )
 32 000  5( x  80) 2
When x = 80, P attains its maximum value.
∴ The price of the magazine is $80 in order
to obtain the maximum profit.
Revision Exercise 10 (p. 274)
Level 1
1.
L0  8.5
The original length of the spring is 8.5 cm.
∴
(b) By substituting L0 = 8.5 into (1), we have
10.1  8.5  2k
k  0.8
∴ Le = 0.8W
∵
L = L0 + Le
∴
L = 8.5 + 0.8W
13. (a)
When L = 2(8.5) = 17,
17  8.5  0.8W
W  10.625
∴ The weight of the load is 10.625 kg when
the length of the spring is double that of
the original length.
2.
19
(a) ∵
∴
( y  2) 
x
y  2  k x, k  0
By substituting x  9 and y  7 into the
equation, we have
72  k 9
9  3k
k 3
∴
i.e.
k1  160k2  0
 (1)
By substituting x = 80 and P = 32 000 into the
equation, we have
32 000  k1 (80)  k2 (80)2
P = 800x – 5x2
4
,
25
4
 4x2
25
1
x2 
25
1
x
5
y varies directly as x and z varies directly
as x2.
∴ y = k1x and z = k2x2, where k1, k2  0
∵ P=y+z
∴ P = k1x + k2x2, where k1, k2  0
By substituting x = 160 and P = 0 into the
equation, we have
0  k1 (160)  k2 (160)2
∴
y  4x2
(b) When y 
∵
k1  80k2  400
 (2)
(1) – (2), 80k2  400
k2  5
By substituting k2 = –5 into (1), we have
k1  160(5)  0
k1  800
y  kx2 , k  0
By substituting x  3 and y  36 into the
equation, we have
36  k (3) 2
k4
0.75L0  6.375
(c)
(a) ∵ y varies directly as x2.
∴
11.3  L0  17.675  1.75L0
∴
P  800 x  5 x 2
y23 x
y 3 x 2
(b) When y  10 ,
10  3 x  2
3 x  12
x 4
x  16
3.
(a) ∵ y varies inversely as x.
k
∴ y ,k 0
x
By substituting x  4 and y  3 into the
equation, we have
10 Variations
(b) When a  3 and c  4 ,
3b
3
2 ( 4) 2
3b  96
k
4
k  12
12
y
x
3
∴
(b) When y  5 ,
12
5
x
12
x
5
4.
b  32
7.
(a)
∵
varies directly as
∴
18  k1 9  k2 16
3k1  4k2  18
∴
z  2x2 y
(b) When x  2 and y  3 ,
z  2( 2) 2 (3)
 24
6.
(a) ∵ a varies directly as b and inversely as c2.
kb
∴ a 2,k 0
c
By substituting a  1 , b  6 and c  3 into
the equation, we have
k ( 6)
1 2
3
3
k
2
3b
∴ a 2
2c
...... (2)
(1)  3  (2)  2, k2  3
By substituting k2  3 into (1), we have
2k1  3(3)  13
k1  2
∴ w  2 x 3 y
(b)
When x  16 and y  25 ,
w  2 16  3 25
 23
8.
By substituting x  5 , y  2 and z  100
into the equation, we have
100  k (5) 2 ( 2)
k2
...... (1)
By substituting x  9 , y  16 and w  18
into the equation, we have
57
y
8
z  kx2 y , k  0
w  k1 x  k 2 y , where k1 , k 2  0
2k1  3k2  13
16 y  114
∴
y.
13  k1 4  k2 9
(b) When x  8 ,
138
8
2y  3
16 y  24  138
(a) ∵ z varies jointly as x2 and y.
x and partly
By substituting x  4 , y  9 and w  13
into the equation, we have
(a) ∵ x varies inversely as (2 y  3) .
k
∴ x
,k0
2y  3
By substituting x  6 and y  10 into the
equation, we have
k
6
2(10)  3
k  138
138
∴ x
2y  3
5.
w partly varies directly as
(a)
∵ y is partly constant and partly varies
inversely as x.
k
∴ y  k1  2 , where k1 , k2  0
x
By substituting x  10 and y  251 into the
equation, we have
k
251  k1  2
10
10k1  k2  2510 ...... (1)
By substituting x  1 and y  260 into the
equation, we have
k
260  k1  2
1
...... (2)
k1  k2  260
(1)  (2), 9k1  2250
k1  250
By substituting k1  250 into (2), we have
250  k 2  260
k 2  10
∴
y  250 
10
x
20
Certificate Mathematics in Action Full Solutions 4B
(b) When x  5 ,
y  250 
10
5
 kx 2 
 1 .6  0 
 y 
 0 
 1 .6 z 0
 252
9.
∵
y
x
∴ y  k x, k  0
Let x0 and y0 be the original values of x and y
respectively.
∴ New value of x  (1  69%) x0
 1.69 x0
New value of y  k 1.69 x0
 1.3k x0
 1 .3 y 0
∴ Percentage change in z 
1.3 y0  y0
 100%
y0
 30%
∴ y is increased by 30%.
 60%
12. ∵
z
 1.5 x0
New value of y 
k
(1.5 x0 ) 2

4  k 
9  x0 2 

4
y0
9
kx2
,k  0
y
Let x0, y0 and z0 be the original values of x, y and z
respectively.
∴ New value of x  2x0
z
5
 55 %
9
5
∴ y is decreased by 55 % .
9
 kx
 8 0
 y0
 8z 0
kx2
,k  0
y
Let x0, y0 and z0 be the original values of x, y and z
respectively.
∴ New value of x  (1  20%) x0
z
 1.2 x0
New value of y  (1  10%) y0
 0 .9 y 0
21
2
∴ Percentage change in z 



8 z0  z0
 100%
z0
 700%
∴ z is increased by 700%.
13. Let $C be the value of the circular gold plate and d be
the diameter of the circular gold plate.
∵ C  d2
C  kd 2 , k  0
Let $C1 and d1 be the value and the diameter of the
smaller circular gold plate, and $C2 and d2 be the
value and the diameter of the larger circular gold
plate.
2
∴ C1  kd1
...... (1)
C2  kd2
2
...... (2)
2
(1) ÷(2),
x2
y
k (2 x 0 ) 2
0 .5 y 0
New value of z 
∴
4
y0  y0
9
 100%
∴ Percentage change in y 
y0
∴
x2
y
New value of y  0.5 y0
1
10. ∵ y  2
x
k
∴ y  2 ,k  0
x
Let x0 and y0 be the original values of x and y
respectively.
∴ New value of x  (1  50%) x0
z
1.6 z0  z0
 100%
z0
∴ z is increased by 60%.
∴
∴ Percentage change in y 
11. ∵
k (1.2 x0 ) 2
( 0 .9 y 0 )
New value of z 
C1
kd
 1
C2 kd2 2
C1  d1 
 
C2  d 2 
2000  2 
 
C2
3
2000 4

C2
9
2
2
C2  4500
∴ The value of the larger plate is $4500.
10 Variations
7590  k1  k 2 (750)
14. Let T be the time taken to drink a bottle of cola and d
be the diameter of the straw.
1
∵ T  2
d
k
∴ T  2 ,k  0
d
k1  750k 2  7590
(2)  (1), 600k2  3840
32
5
32
By substituting k 2 
into (1), we have
5
k2 
Let T0 and d0 be the original values of T and d
respectively.
∴ New value of d  2d0
New value of T 
k
( 2d 0 ) 2
 k
 0.25 2
 d0
 0.25T0
∴ Percentage change in T 
∴



C  2790 
C  kTd 2 , k  0
Let $C1, T1 mm and d1 cm be the cost, the thickness
and the diameter of the first gold coin, and $C2,
T2 mm and d2 cm be the cost, the thickness and the
diameter of the second gold coin respectively.
2
∴ C1  kT1d1 ...... (1)
C2  kT2 d 2
2
17. ∵ y varies inversely as x.
k
∴ y ,k 0
x
By substituting x  15 and y  20 into the
equation, we have
k
20 
15
k  300
300
x
Take x  5 ,
300
y
5
 60
Take x  10 ,
300
y
10
 30
∴ The ordered pairs (5, 60) and (10, 30) lie on the
graph. (or any other reasonable answers)
∴
...... (2)
2
(1) ÷(2),
C1
kT d
 1 12
C2 kT2 d 2
C1 T1  d1 
  
C2 T2  d 2 
16 3  4 
  
27 4  d 2 
16
64

2
81
d2
2
2
81  16
64
 20.25
d 2  4.5 or  4.5 (rejected)
d2 
18.
y
(a) ∵
2
∴ The diameter of the second coin is 4.5 cm.
16. (a)
∵
∴
C is partly constant and partly varies
directly as n.
C  k1  k 2 n, where k1 , k 2  0
By substituting n  150 and C  3750 into
the equation, we have
3750  k1  k 2 (150)
...... (1)
k1  150k 2  3750
32
(200)
5
 4070
∴ The cost of making 200 coats is $4070.
 75%
∴ The time taken is decreased by 75%.
∴
 32 
k1  150   3750
 5 
k1  2790
32
C  2790 
n
5
(b) When n  200 ,
0.25T0  T0
 100%
T0
15. Let $C be the cost of the gold coin, T mm be the
thickness of the gold coin and d cm be the diameter
of the gold coin.
∵ C  Td 2
...... (2)
∴
(b) ∵
∴
(c) ∵
∴
The expenditure ($E) of the party includes
buying gifts and food. $400 is spent for the
gifts and the expenditure of food is $F.
E  400  F and it is in partial variation.
12 L of drink is provided for N classmates
and each classmate is supposed to
consume x mL of the drink.
12 000
x 
and it is in inverse variation.
x
Each classmate has to pay $40 for the food
and the expenditure of food is $F.
F  40N and it is in direct variation.
By substituting n  750 and C  7590 into
the equation, we have
22
Certificate Mathematics in Action Full Solutions 4B
Level 2
19. (a)
∵
∴
y  (ax  3)
(b)
By substituting a  2 into (1), we have
4
k

3 2(2)  1
k4
4
∴ y
2x  1
(c)
When y  4 x ,
y  k (ax  3), k  0
By substituting x  4 and
equation, we have
3  k (4a  3) ...... (1)
By substituting x  5 and
equation, we have
6  k (5a  3) ...... (2)
k (5a 
6
(2) ÷(1),

3
k ( 4a 
5a  3
2 
4a  3
8a  6  5a  3
y  3 into the
y  6 into the
4
2x  1
2x2  x  1  0
4x 
3)
3)
(2 x  1)( x  1)  0
x
3a  3
a 1
21. (a)
(b)
(c)
By substituting a  1 into (1), we have
3  k[4(1)  3]
k 3
y  3( x  3)
∴
y  3( x  3)
 3x  9
When 2 y  3 x ,
y  2y  9
∵
∵
∴
w
∴
w
1
ax  1
k
y
,k0
ax  1
(b)
2a  4
a2
23
z3
,k 0
w
2x2 y
z3
When y  2w  2 x  2 z , i.e. y  2 x ,
w  x and z  x ,
w
4
into the
3
x
equation, we have
4
k

3 a ( 2)  1
4
k
...... (1)

3 2a  1
equation, we have
4
k

7 a ( 4)  1
4
k
...... (2)

7 4a  1
4
k
3  2a  1
(1) ÷(2),
4
k
7
4a  1
7 4a  1

3 2a  1
14a  7  12a  3
kx2 y
k (1) 2 1
13
k  2
y
By substituting x  4 and y 
z3
2 
∴
By substituting x  2 and y 
x2 y
By substituting x  1 , y  1 , z  1 and
w  2 into the equation, we have
y9
20. (a)
∵
1
or x  1
2
2x2 y
z3
2x2 2x
x3
x4  2x2 2x
x2
 2x
2
x4
 2x
4
x4  8x
4
into the
7
x4  8x  0
x( x 3  8)  0
x  0 (rejected) or x 3  8
x2
22. (a)
∵
∴
y is partly constant, partly varies directly
as x and partly varies directly as x2.
y  k 1  k 2 x  k 3 x 2 , where
k1 , k 2 , k 3  0
By substituting x  0 and y  5 into the
equation, we have
5  k1  k2 (0)  k3 (0) 2
k1  5
10 Variations
By substituting x  2 and y  15 into the
equation, we have
15  5  k2 (2)  k3 (2)2
...... (1)
k2  2k3  5
By substituting x  2 and y  11 into the
equation, we have
11  5  k2 (2)  k3 (2)2
...... (2)
k2  2k3  3
(1) (2), 4k3  8
0.81h0
∴
k3  2
By substituting k3  2 into (1), we have
∴
k 2  2( 2)  5
k2  1
y  5  x  2x2
∴
(b)
25. ∵
When y  6 ,
∴
6  5  x  2x2
2x2  x  1  0
( x  1)(2 x  1)  0
x   1 or x 
1
2
k
New value of r 

10  k  
 
9  h0 

10
r0
9
10
r0  r0
Percentage change in r  9
 100%
r0
1
 11 %
9
The base radius of the cone is increased by
1
11 %.
9
E  mv2
E  kmv2 , k  0
Let m0, v0 and E0 be the original values of m, v and E
respectively.
∴
New value of m  2m0
New value of v  (1  20%)v0
 0.8v0
23. Let V cm3 be the volume of the metal cube and x cm
be the length of the side of the cube.
∵
V  x3
∴
2
New value of E  k (2m0 )(0.8v0 )
 1.28 km0v0
 1.28 E0
V  kx3 , k  0
∴
Total volume of the three metal cubes
 [k (3)3  k (4)3  k (5)3 ] cm 3
 k (27  64  125) cm
3
∴
 216k cm 3
Let VL cm3 and xL cm be the volume and the length of
the side of the larger cube.
3
VL  kxL
∵
∴
216k  kxL
3
xL  216
3
xL  6
∴
24. ∵
∴
The length of the side of the new cube is 6 cm.
1
h
k
r2  , k  0
h
r2 
k
,k 0
h
k
r
, k  0
h
r
2
26. (a)
Percentage change in E  1.28E0  E0  100%
E0
 28%
The kinetic energy of the moving body is
increased by 28%.
∵
V  r 2h
∴
V  kr 2 h, k  0
By substituting r  4 , h  30 and
V  160 into the equation, we have
160  k (4) 2 (30)
1
k
3
1 2
V  r h
∴
3
When V  540 and h  27 ,
1
540  r 2 (27)
3
r 2  60
r  60
 2 15
Let r0 and h0 be the original values of r and h
respectively.
∴
New value of h  (1  19%) h0
 0.81h0
24
Certificate Mathematics in Action Full Solutions 4B
(b) Let r0, h0 and V0 be the original values of r, h
and V respectively.
∴ New value of r  2r0
New value of h  0.5h0
∴
(b)
1
(2r0 ) 2 (0.5h0 )
3
1 2 
 2 r0 h0 
3

 2V0
New value of V 
∴
Percentage change in V  2V0  V0  100%
V0
∴
V is increased by 100%.
∵
V 
(c)
 100%
(c)
1 2
r h
3
New value of r 




 1.36 r0
Percentage change in r 
∴
28. (a)
1.36 r0  r0
 100%
r0
 16.6% (cor. to 1 d.p.)
∴
r is increased by 16.6%.
∵
T is partly constant and partly varies
directly as n.
T  k1  k 2 n, where k1 , k 2  0
The airtime used in this month is
(150  2n0 ) minutes.
∵
C is partly constant and partly varies
inversely as n.
k
∴ C  k1  2 , where k1 , k2  0
n
By substituting n  200 and C  250 into
the equation, we have
k
250  k1  2
200
...... (1)
200k1  k 2  50 000
By substituting n  400 and T  110 into
the equation, we have
110  k1  k 2 (400)
k1  400k 2  110
...... (1)
By substituting n  400 and C  200 into
the equation, we have
k
200  k1  2
400
400k1  k 2  80 000 ...... (2)
By substituting n  600 and T  150 into
the equation, we have
150  k1  k 2 (600)
(2)  (1), 200k1  30 000
k1  150
∴
k1  600k 2  150
By substituting k1  150 into (1), we have
...... (2)
200(150)  k 2  50 000
(2)  (1), 200k 2  40
k2 
k 2  20 000
1
5
By substituting k 2 
1
into (1), we have
5
1
k1  400   110
5
k1  30
25
Let $T0 be the mobile phone charge in last
month.
1
∵ T  30  n
5
1
∴ T0  30  n0
5
1
n
5
1 
1

2 30  n0   30  n
5 
5

2
1
60  n0  30  n
5
5
2
1
30  n0  n
5
5
n  150  2n0
3(1.36V0 )
h0
 3V0
 1.36 
 h0

27. (a)
When n  300 ,
1
T  30  (300)
5
 90
∴ The monthly charge is $90 if the airtime
used is 300 minutes.
2T0  30 
r
New value of h  h0
∴
1
n
5
When T  2T0 ,
3V
h
Let V0, h0 and r0 be the original values of V, h
and r respectively.
∴ New value of V  (1  36%)V0
 1.36V0
∴
T  30 
(b)
20 000
∴ C  150 
n
When n  500 ,
20 000
C  150 
500
 190
10 Variations
25 000
500
 $50
The selling price of each rice cooker
 $(190  50)
(2)  (1), k2  1
 $240
∴
Profit of a rice cooker  $
∴
29. (a)
∵
∴
By substituting k2  1 into (1), we have
k1  4( 1)  2
k1  6
(b)
When r  2 ,
C  6(2)  (2) 2
8
(c)
C  6r  r 2
P is partly varies directly as x and partly
varies directly as x2.
P  k1 x  k2 x 2 , where k1 , k2  0
By substituting x  20 and P  60 000 into
the equation, we have
60 000  k1 (20)  k2 (20) 2
 ( r 2  6r )
 ( r 2  6r  3 2  3 2 )
k1  20k2  3000
...... (1)
By substituting x  30 and P  75 000 into
the equation, we have
75 000  k1 (30)  k2 (30)2
k1  30k2  2500
(2)  (1),
 ( r 2  6r  3 2 )  9
 ( r  3) 2  9
When r  3 , C attains its maximum value.
∴ The radius of the model is 3 cm such that
the cost is a maximum.
...... (2)
10k 2  500
31. (a)
k 2  50
By substituting k2  50 into (1), we have
k1  20(50)  3000
k1  4000
∴
(b)
C  6r  r 2
P  4000 x  50 x2
∵
y is partly constant and partly varies
directly as x.
∴ y  k  k x, where k , k   0
From the graph, when x  0 , y  3000 .
By substituting x  0 and y  3000 into the
equation, we have
3000  k  k (0)
k  3000
When x  35 ,
P  4000(35)  50(35)
 78 750
The total profit is $78 750 if the selling
price of each VCD is $35.
2
∴
(c)
(b)
From the graph, when x  120 , y  6000 .
By substituting x  120 and y  6000 into
the equation, we have
6000  3000  k (120)
k   25
y  3000  25 x
∴
(c)
When x  700 ,
y  3000  25(700)
 20 500
∴ The monthly income of the salesperson is
$20 500 if the number of items sold is 700.
P  4000 x  50 x 2
 50( x 2  80 x)
 50( x 2  80 x  402  402 )
 50( x 2  80 x  402 )  80 000
 50( x  40) 2  80 000
When x  40 , P attains its maximum value.
∴ The maximum profit from selling the
VCDs is $80 000 and the corresponding
selling price of each VCD is $40.
30. (a)
∵
∴
C partly varies directly as r and partly
varies directly as the square of r.
C  k1r  k 2 r 2 , where k1 , k 2  0
32. (a)
∵
( x2  y 2 )  ( x2  y 2 )
∴
x 2  y 2  k ( x 2  y 2 ), k  0
x 2  y 2  kx2  ky2
(k  1) y 2  ( k  1) x 2
By substituting r  4 and C  8 into the
equation, we have
8  k1 (4)  k2 (4)2
k1  4k2  2
...... (1)
By substituting r  5 and C  5 into the
equation, we have
5  k1 (5)  k 2 (5) 2
k1  5k 2  1
...... (2)
∵
∴
y2 
k 1 2
x
k 1
y
k 1
x
k 1
k0
k 1
0
k 1
By letting k  
k 1
, we have
k 1
26
Certificate Mathematics in Action Full Solutions 4B
i.e.
(b)
∵
∴
y  k x, k   0
yx
(b)
y  k x, where k   0
x  y  x  k x
 (1  k ) x
x  k x
(1  k ) x
(1  k ) x
(1  k ) x
1  k 

1  k 
1  k  
x  y  
 ( x  y)
 1  k  
1  k 
 0 , provided that k   1 and
∵
1  k 
k   1 .
1  k 
By letting k  
, we have
1  k 
x  y  k ( x  y), k   0
x y 

x y

x y
33. (a)
( x  y)  ( x  y)
∵
 3 1   3 1 
 x  3    x  3 
y  
y 

∴

1
1 
 k  x 3  3 , k  0
y3
y


k 1
3
(1  k ) x  3
y
k
1 1
y3 

1  k x3
 2  k  2
 k  
 x    x   1 
x
x 
 k   2




 x 2  k   x  k   1 
  x 
 x  

1
 x  
y

2
 k   k   1 
1
 x  
 k  2



k

k

1
y 


 k 2  k   1 
 , we have
By letting k   k  2

 k  k  1 
x
x3 
y3
By letting k   3
∴

1
1
 k  x  , k   0
y
y


1 
1
 x     x  
y 
y

Multiple Choice Questions (p. 278)
1.
k 1 1

1 k x
Answer: C
∵ x varies directly as y2.
x  ky2 , k  0
∴
By substituting y  2 and x  20 into the
equation, we have
20  k ( 2) 2
k 5
k 1
, we have
1 k
k
, k  0
x
1
y
x
y
∴

1
1 
 k  x 3  3 , k  0
3
y
y 



1 
x 1 
1 
x 1 
 x   x 2   2   k  x   x 2   2 
y
y
y
y
y y 





 2 x 1 
x   2 
1
1
y y 
 x  
x   k
 2 x 1 
y
y
 x   2 
y
y


 x 2 y 2  xy  1 
1
 x  
 k  2 2


y 
 x y  xy  1 
yx
i.e.
x3 
x  5y2
∴
When x  80 ,
80  5 y 2
y 2  16
y 4
2.
27
Answer: A
∵ (3x  5 y )  (4 x  6 y )
3x  5 y  k (4 x  6 y ), k  0
∴
(5  6k ) y  ( 4k  3) x
4k  3 
y  
x
 5  6k 
4k  3
By letting k  
, we have
5  6k
y  k x, k   0
i.e.
yx
10 Variations
3.
equation, we have
 115  k1 (5)  k2 (5)3
Answer: B
∵ y varies inversely as ( x  1) 2 .
k
∴
y
, k0
( x  1) 2
By substituting x  2 and y 
1
into the equation,
3
we have
1
k

3 ( 2  1) 2
1 k

3 9
k 3
3
y
∴
( x  1) 2
When y  12 ,
12 
By substituting k2  1 into (1), we have
k1  9(1)  7
k1  2
∴
8.
3
( x  1) 2
y  2 x  x3
Answer: A
∵ z varies directly as x2 and inversely as y.
kx2
,k 0
∴ z
y
Let x0, y0 and z0 be the original values of x, y and z
respectively.
∴ New value of x  (1  30%) x0
 1.3 x0
4( x  1) 2  1
New value of y  (1  30%) y0
 1.3 y0
4x  8x  3  0
( 2 x  1)(2 x  3)  0
2
x
4.
∴
5.
1
3
or x  
2
2
New value of z 
∴
9.
Answer: B
∵ x varies inversely as y and directly as z2.
kz2
x
∴
, where k is a non-zero constant
y
Answer: D
∵ z  ( y  2) and y  ( x  1)
∴ z  k1 ( y  2) and y  k2 ( x  1) , where
k1, k2  0
z  k1 (k2 x  k2  2)
 k1k2 x  k1k2  2k1
7.
Answer: A
∵ y partly varies directly as x and partly varies
directly as the cube of x.
y  k1 x  k2 x 3 , where k1 , k2  0
∴
By substituting x  3 and y  21 into the
equation, we have
 21  k1 (3)  k2 (3)3
…… (1)
k1  9k2  7
2



1.3z0  z0
 100%
z0
 30%
xy
 k , where k is a non-zero constant
z2
6.
Percentage increase in z 
k
, k0
x y
( x  y ) varies inversely as ( x  y ) .
k (1.3x 0 ) 2
1.3 y 0
 kx
 1.3 0
 y0
 1.3 z 0
Answer: D
∵ ( x  y ) varies inversely as ( x  y ) .
k
x y 
, k0
∴
x y
x y 
…… (2)
k1  25k2  23
(2) – (1), 16 k 2  16
k 2  1
Answer: B
∵ xy  1  315  3  105  5  63  7  45
 9  35  315
315
y
∴
x
1
y 
∴
x
10. Answer: D
∵ y is partly constant and partly varies directly as
x.
∴
y  k1  k2 x, where k1, k2  0
Let’s Discuss
p. 244
1.
2.
Only graph number  shows the relation y varies
directly as x. It is because the graph of direct variation
is a straight line passing through the origin.
Yes
By substituting x  5 and y  115 into the
28