Certificate Mathematics in Action Full Solutions 4B
10 Variations
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • =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
Activity Activity 10.1 (p. 263) 1.
(a)
$25x
(b) $30y 2.
(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.
4.
(a)
∴
Cost rate
21 / cup 6 =$3.5 / cup =$
20 km 20 (b) Travel rate h 60 = 60 km/h
(c)
$24 30 Cost rate dozen 12 = $9.6 /dozen =
1.
30 cm 50 cm (a) 30 cm : 50 cm 3 = 5 =3 : 5
=1250 mL : 500 mL 1250 mL = 500 mL (b) 1.25 L : 500 mL 5 = 2 =5 : 2 3.
(a)
8 16 = 6 x 8 x = 96 x = 12
( x +1) : ( x −1) = 3 : 2 x +1 3 = x −1 2 2 x + 2 = 3 x −3 x =5
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
(b) When x = 6, y =3 ×6 =18 2.
z∝x ∵ z = kx , k ≠ 0 ∴ By substituting x = 4 and equation, we have
z = 12 into the
12 = k ×4
∴
k =3 z = 3x
x = 32 , When z = 3 ×32 = 96
The required ratio
3.
1
8 : 6 = 16 : x
p. 242
=
2.
∵ ∴
=
(c)
∴
x 5 = 4 2 2 x = 20
(b) ∵
p. 238 (a)
x:4 =5:2
x =10
Follow-up Exercise
1.
∵
( y − 4) ∝ x ∵ y −4 = kx , k ≠ 0 ∴ By substituting x = 8 and y = 8 into the
10 Variations equation, we have
8− 4 = k ×8
New value of y
4 8 1 = 2 1 y−4= x 2 1 y = x+4 2 k=
∴
= 2 1.69 x 0 = 1.3( 2 x 0 ) = 1.3 y 0
When y =10 ,
1 x +4 2 1 6= x 2 x = 12
10 =
p. 243 1.
y ∝ x2
∵
y = kx 2 , k ≠ 0 ∴ Let x0 and y0 be the original values of x and y respectively.
∴
New value of
New value of
x = (1 + 10 %) x0 = 1.1x0
y = k (1.1x 0 ) 2 = 1.21kx 0
2
= 1.21 y 0 ∴
y=
1.21 y0 − y0 × 100 % y0
=
(1.21 − 1) y0 × 100 % y0
∴ 2.
Percentage change in
(a)
= 21%
y is increased by 21%. ∵
y∝
∴
y =k
x x , k ≠0
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 = 1.69 x0
2
Certificate Mathematics in Action Full Solutions 4B ∴
Percentage change in y
∴
1.3 y0 − y0 = ×100 % y0 =
3.
(a)
∵
C ∝ r2
∴
C = kr 2 , k ≠ 0
∴
By substituting r = 6 and C = 20 into the equation, we have
∴
(b) (i)
20 = k (6) 2 20 k = 36 5 = 9 5 C = r2 9 When C = 45 ,
1 3x 6x = 1 2=
3.
∵
(a)
∵
∴
y2 =
∴
k 3= 1 k =3 y=
3 x
(b) When x = 3, 3 y = 3 =1 2.
∵
z varies inversely as x.
36 x
When y = 5,
36 x 25 x = 36 36 x= 25 52 =
y varies inversely as x.
equation, we have
k 4 k = 36
32 =
The cost of painting a hemisphere is $180.
y=
y2 =
equation, we have
5 C = (18 ) 2 9 = 180
k , k ≠0 x By substituting x =1 and y = 3 into the ∴
y2 varies inversely as x.
k , k ≠0 x By substituting x = 4 and y = 3 into the ∴
p. 251 1.
1 6
x=
∴ The radius of the hemisphere is 9 cm. (ii) When r = 0.18 ×100 =18 ,
∴
1 k = 3 1 1 k= 3 1 z= 3x
When z = 2,
5 45 = r 2 9 2 r = 81 r =9
∴
1 into the 3
equation, we have
= 30 %
y is increased by 30%.
k , k ≠0 x
By substituting x =1 and z =
(1.3 − 1) y0 ×100 % y0
∴
z=
p. 253 1.
(a)
∵ ∴
1 x2 k y= 2,k ≠0 x
y∝
By substituting x = 2 and y = 4 into the equation, we have
k 22 k = 16 16 y= 2 x 4=
∴
(b) Let x0 and y0 be the original values of x and y
3
10 Variations ∴
respectively. ∴
New value of x
= (1 + 25 %) x0 = 1.25 x0
= New value of y
Percentage change in y
16 (1.25 x0 ) 2
16 = 0.64 x 2 0 = 0.64 y0
0.64 y0 − y0 ×100 % y0
=
(0.64 − 1) y0 ×100 % y0
= −36 %
∴ 2.
=
y is decreased by 36%.
∵
y∝
∴
y=
1 x k x
, k ≠0
Let x0 and y0 be the original values of x and y respectively. ∴
New value of
x = (1 − 36 %) x0 = 0.64 x0 y=
New value of
∴
∴ 3.
k 0.64 x0
k = 1.25 x 0 = 1.25 y0
Percentage change in y
=
1.25 y0 − y0 ×100 % y0
=
(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 n k T = , k ≠0 ∴ n By substituting n = 10 and T = 30 into the ∵
T ∝
equation, we have
k 10 k = 300
30 = ∴
T =
300 n
When n =12 ,
300 12 = 25
T = ∴
The food can last for 25 days if 2 more people go ashore on the island.
4
Certificate Mathematics in Action Full Solutions 4B p. 261 1.
(a) (b) (c)
2.
(a)
z = kx y 2
z =
3
New value of
kd t kb 3 a2 = c s=
∵ ∴
z varies jointly as x2 and z = kx
y
.
y , k ≠0
2
By substituting x = 2 , y = 9 and
z = 24 into the equation, we have
∴
24 = k ( 2) 2 k =2 z = 2 x2
9 y
(b) When x = 3 and y =16 , z =2(3) 2 =72
3.
(a)
∵ ∴
16
x varies directly as y and inversely as z2.
x=
ky ,k ≠0 z2
By substituting x = 4 , y = 2 and into the equation, we have
z =1
k ( 2) 12 k =2 4=
∴
x=
2y z2
(b) When y = 3 and
z =2,
2(3) x= 2 2 3 = 2
4.
∵
z∝
x y
kx , k ≠0 y Let x0, y0 and z0 be the original values of x, y and z respectively. ∴
∴
z=
New value of x
New value of
5
= (1 − 25 %) x0 = 0.75 x0
y = (1 + 50 %) y0 = 1.5 y0
k ( 0.75 x 0 ) 1.5 y 0
kx = 0.5 0 y0 = 0.5z 0
10 Variations ∴
equation, we have
Percentage change in z
0.5z 0 − z 0 = × 100% z0 ∴
7 = k 1 (2) + 4k 1 + k 2 = 1 4
= − 50%
(a)
y = k1 x + k 2 x 2
(b)
k a = k1 + 22 c
(c) 2.
(a)
A = k1B + 2
∵
k1 = 4 By substituting k1 = 4 into (1), we have
4 + k2 = 2 k2 = − 2 ∴
B
(b) When x = 4 ,
P = 4( 4) −
∴
z = k 1 x + k 2 y , where k 1 , k 2 ≠ 0 By substituting x = 2 , y =1 and z = 4 into the equation, we have
2k1 + k2 = 4
= 4.
4 = k1 (2) + k2 (1)
(a)
…… (1)
By substituting x = 4 , y = 3 and z = 11
∵
C is partly constant and partly varies directly as n.
By substituting n = 10 and C = 1750 into the equation, we have
1750 = k1 + k 2 (10)
…… (2)
k1 + 10k 2 = 1750
…… (1) By substituting n = 25 and C = 3625 into the equation, we have
( 2) − (1) ×2, k 2 = 3 By substituting k 2 = 3 into (1), we have
2k1 + 3 = 4
3625 = k1 + k2 ( 25)
1 k1 = 2 1 z = x + 3y 2
k1 + 25k 2 = 3625
k 2 = 125 k By substituting 2 = 125 into (1), we have
k1 + 10(125) = 1750
1 (3) +3(2) 2 15 = 2
z =
∵
k1 = 500
∴
C = 500 + 125n
(b) When n = 20 ,
C = 500 +125 ( 20 )
P partly varies directly as x and partly varies inversely as x.
= 3000
∴
The cost per head
k2 , where k 1 , k 2 ≠ 0 x By substituting x =1 and P = 2 into the
P = k1 x +
equation, we have
2 = k1 (1) + k1 + k2 = 2
k2 1
…… (2)
( 2) − (1), 15k 2 = 1875
(b) When x = 3 and y = 2 ,
(a)
31 2
∴
11 = k1 (4) + k2 (3)
3.
2 4
C = k 1 + k 2 n, where k 1 , k 2 ≠ 0
into the equation, we have
4k1 + 3k2 = 11
2 x
P = 4x −
k2
z partly varies directly as x and partly varies directly as y.
∴
…… (2)
( 2) −(1), 3k1 =12
z is decreased by 50%.
p. 269 1.
k2 2
3000 20 = $150 =$
Exercise …… (1)
By substituting x = 2 and P = 7 into the
Exercise 10A (p. 245) Level 1 1.
(a)
∵
y varies directly as x.
6
Certificate Mathematics in Action Full Solutions 4B y = kx , k ≠ 0 ∴ By substituting x = 32 and y = 8 into the equation, we have
∴
8 = k × 32 1 k = 4 1 y= x 4
(b) (i)
1 x 4 x =12 3=
When y = 6,
1 x 4 x = 24
6=
When x = 40 ,
1 ( 40 ) 4 =10
y =
When x = 60 ,
1 (60 ) 4 =15
y =
(b) ∵ 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 When y = 2.2, ∴
2.2 =2.2 x x =1
When x = 6, y =2.2(6) =13 .2 When y = 17 .6, 17 .6 = 2.2 x x =8
When y =19 .8, 19 .8 =2.2 x x =9 (a)
(i)
∵
y varies directly as x.
∴
20 30 Variation constant 2 = 3 =
(ii) The equation connecting x and y is
7
2 x. 3
∵
y varies directly as x.
∴
−20 Variation constant = 10 =−2
(ii) The equation connecting x and y is y = −2 x .
When y = 3,
2.
y=
10 Variations 3.
(b) When V = 32 ,
(a)
1 3 d 2 3 ∴ d = 64 d =4 32 =
6.
(a)
∵
y∝
∴
y =k
x x , k ≠0
By substituting x = 16 and y = 2
2 into 3
the equation, we have
2
∴
2 = k 16 3 8 = k ( 4) 3 2 k = 3 2 y= x 3
(b) When x = 9 ,
2 9 3 2 = (3) 3 =2
y=
(b) From the graph, when x = 8 , y =12 4.
(a)
∵ ∴
A varies directly as l 2. A = kl 2 , k ≠ 0
By substituting l = 4 and A = 48 into the equation, we have
48 = k ( 4)
2
7.
48 =16 k ∴
∵ ∴
k =3 A = 3l
y1 − 1 3
1 into 2
the equation, we have
62
∴
x1
=
y2 − 1 3
x2
y −1 7−1 = 32 3 27 64 y −1 6 = 2 3 4 8 = y2 − 1
V = kd 3 , k ≠ 0
1 = k (5)3 2 125 = k (125 ) 2 1 k = 2 1 3 V = d 2
=k
By substituting x1 = 27 , y1 = 7 and x2 = 64 into the equation, we have
V varies directly as d 3 .
By substituting d = 5 and V = 62
x
For any two pairs of x and y, say (x1, y1) and (x2, y2), we have
=9
∵ ∴
y −1 3
A =3( 3 ) 2 =3(3)
(a)
y −1 = k 3 x , k ≠ 0
2
(b) When l = 3 ,
5.
( y −1) ∝ 3 x
y2 = 9 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.
8
Certificate Mathematics in Action Full Solutions 4B
∴
New value of
New value of
x = (1 − 75%) x 0 = 0.25x 0
y = k ( 0.25 x 0 ) 2 = 0.0625 kx 0 = 0.0625 y 0
∴
Percentage change in y
=
0.0625y 0 − y 0 × 100% y0
=
(0.0625 − 1) y 0 × 100% y0
= −93.75% ∴
9
y is decreased by 93.75%.
2
10 Variations 9.
∵ ∴
S ∝ t
Level 2
S = kt , k ≠ 0
By substituting t =
3 × 60 = 45 and S = 45 4
13. (a)
into the equation, we have
45 = k ( 45 ) k =1 S = t When t = 25 , S = 25 10. ∵ ∴
k = 5 y + 2 = 5x y = 5x − 2 i.e. (b) When x = 2 , ∴
∴
∴
( y + 2) ∝ x ∵ y + 2 = kx , k ≠ 0 ∴ By substituting x = 3 and y = 13 into the equation, we have 13 + 2 = k ( 3)
y = 5( 2) − 2 =8
The train travels 25 km in 25 minutes.
C ∝ x2
(c)
When y = 3x , 3x = 5 x − 2 2x = 2 x =1
C = kx , k ≠ 0 2
By substituting x = 4 and C = 96 into the equation, we have
∴
96 = k ( 4) 2 k = 6 C = 6x
14. (a)
2
11.
C = 6( 3) 2 = 54 The cost of painting a cube of side 3 cm is $54.
∵
( y − 1) ∝ x 3
∴
y − 1 = kx 3 , k ≠ 0
y = 2
y = 9
12. ∵ ∴
b = 1
(b) From (1),
y varies directly as x. y = kx , k ≠ 0
y + 2 kx + 2 = x+ 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) .
y 2 = k ( x − b), k ≠ 0
16 k (9 − b) = 4 k ( 3 − b) 9− b 4 = (2) ÷ (1), 3− b 12 − 4b = 9 − b 3b = 3
By substituting k = 1 and x = 2 into the equation, we have y − 1 = (1)( 2) 3 The two possible pairs of x and y are x = 1 , y = 2 or x = 2 , y = 9 . (or any other reasonable answers)
∴
By substituting x = 9 and y = 4 into the equation, we have 4 2 = k ( 9 − b) ...... (2) 16 = k (9 − b)
By substituting k = 1 and x = 1 into the equation, we have y − 1 = (1)(1) 3
∴
y 2 ∝ ( x − b)
By substituting x = 3 and y = 2 into the equation, we have 2 2 = k (3 − b) ...... (1) 4 = k (3 − b)
When x = 3 ,
∴
∵
∴ (c)
y
2
4 = k (3 − 1) 2k = 4 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 equation, we have
2 = kt 2
= t and y = 2 into the ...... (1)
10
Certificate Mathematics in Action Full Solutions 4B
1 = 3d
By substituting x = t + 1 and y = 8 into the equation, we have ...... (2) 8 = k (t + 1) 2
d = ∴
8 = 2 4=
1 km when the time between the 3
k t2 (t + 1) 2
t 4t 2 = t 2 + 2 t + 1
(2) ÷ (1),
thunder and lightning is only 1 second. 18. (a)
3t 2 − 2t − 1 = 0 t= −
V = 800W 3 When W = 3.5 , V = 800 ( 3.5) 3
2
v ∝ t 16. ∵ v = kt , k ≠ 0 ∴ By substituting t = 2.5 and v = 25 into the equation, we have
(b) Original value = $6400 New value = $[ 2 × 800 (1)]
= $1600
Percentage loss
$( 6400 − 1600 ) × 100 % $6400 = 75 % =
t = 5 .5 ,
v = 10 (5.5) ∴ 17. (a)
= 55 The speed of the body is 55 m/s after it has fallen for 5.5 s. ∵ t ∝ d t = kd , k ≠ 0 ∴ By substituting d = 5 and t = 15 into the equation, we have ∴
(b) (i)
15 = k (5) k = 3 t = 3d
When d = 3 ,
t = 3( 3) ∴
t = 3(8) ∴
(c)
11
19. (a)
The number of kilocalories and the weight of the potato chips are in direct variation.
(b) ∵ E ∝W E = kW , k ≠ 0 ∴ By substituting W = 200 and E = 1000 into the equation, we have ∴ (c)
= 24
It takes 24 seconds for you to hear the thunder from lightning that is 8 km away. When t = 1 ,
1000 = k ( 200 ) k = 5 E = 5W
When W = 70 ,
E = 5(70 )
= 9
It takes 9 seconds for you to hear the thunder from lightning that is 3 km away. (ii) When d = 8 ,
= 34 300 The value of a diamond weighing 3.5 carats is $34 300.
∴
25 = k ( 2.5) k = 10 v = 10 t
When
6400 = k ( 2) 3 k = 800
∴
2 from (1), 2 = k (1) k = 2
∴
V = kW 3 , k ≠ 0
By substituting W = 2 and V = 6400 into the equation, we have
1 or t = 1 3
(b) When t = 1 ,
y = 2x
Let $V be the value of the diamond and W carats be the weight of the diamond. ∵ V ∝W3 ∴
(3t + 1) (t − 1) = 0
∴
The distance apart from the lightning is
k (t + 1) 2
2
1 3
20. (a)
= 350
∴
If Sally has eaten 70 g of potato chips, she has taken 350 kcal.
∵
V ∝ m
∴
V = k 1 m, k 1 ≠ 0
∵
m ∝ t
∴
m = k 2 t, k 2 ≠ 0
10 Variations
V = k1 m = k 1 (k 2 t ) = k1 k 2 t
∴
= k 3 t (where k 3 = k 1 k 2 ≠ 0) ∴
(b)
V varies directly as t.
When x = 2,
V + m = k1 k 2 t + k 2 t
56 2 = 28
y =
= ( k 1 k 2 + k 2 )t = k 4 t (wherek 4 = k 1 k 2 + k 2 ≠ 0)
21. (a)
∴
(V + m ) varies directly as t.
∵
V ∝ (m + t )
∴
V = k 1 (m + t ), where k 1 ≠ 0
∵
m ∝ (t + V )
∴
m = k 2 (t + V ), where k 2 ≠ 0
V = k 1 (m + t )
When x = 7,
56 7 = 8
y =
When x = 8,
56 8 = 7
y =
When x = 28 ,
56 28 = 2
= k 1 {[(k 2 (t + V )] + t}
y =
= 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
When x = 56 ,
k k + k1 V = 1 2 t 1 − k1k 2 V = k 3 t (w h e rek 3 = ∴
56 56 =1
y =
k1k 2 + k1 ≠ 0) 1 − k1k 2
V varies directly as t.
m = k 2 (t + V )
(b)
m = k 2 [t + k 1 ( m + t )]
(b) ∵ ∴
(1 − k 1 k 2 )m = (k 2 + k 1 k 2 )t k + k1k 2 m = 2 t 1 − k1k 2 k + k1k 2 m = k 4 t (w h e rek 4 = 2 ≠ 0) 1 − k1k 2 m varies directly as t.
Exercise 10B (p. 254) Level 1 1.
(a)
∵ ∴
y varies inversely as x.
y =
k , k ≠ 0 x
By substituting x = 4 and y = 14 into the equation, we have
y varies inversely as x.
y =
k ,k ≠ 0 x
By substituting x = 12 and y = 5 into the equation, we have
m = k 2 t + k1k 2 m + k1k 2 t
∴
k 4 k = 56 56 y = x
14 =
k 12 k = 60 5 =
∴
y =
60 x
When x = 2,
60 2 = 30
y =
When y = 15 ,
60 x x = 4
15 =
When x = 6,
60 6 = 10
y =
When y = 6,
12
Certificate Mathematics in Action Full Solutions 4B 60 x x = 10
5.
6 =
(a)
∵ ∴
y varies inversely as 3 x .
y =
3
When y = 4,
By substituting x = 27 and y =
60 x x = 15
4 =
2.
(a)
(i)
equation, we have
∵
y varies inversely as x.
∴
Variation constant
1 = 6
= 30 × 12 = 360
(ii) The equation connecting x and y is
360 . y = x (b) (i)
∵
y varies inversely as x.
∴
Variation constant
= 5 ×5 = 25
(ii) The equation connecting x and y is
25 . y = x 3.
(a)
(b) From the graph, when x = 24 , y = 7.5 4.
(a)
∵
F varies inversely as d 2 .
∴
F =
k , k ≠ 0 d2
By substituting d = 4 and F = equation, we have
1 k = 2 4 4 k = 4 4 F = 2 d
∴
(b) When d = 2 ,
4 22 =1
F =
13
k ,k ≠ 0 x
1 into the 4
∴
3
1 2
k =
1
y = (b) When x =
y =
k 27
3
2 x 1 , 8 1
1 8 1 = 1 2 2 = 1 23
1 into the 6
10 Variations 6.
(a)
( y + 2) varies inversely as x2.
∵
k y+ 2 = 2 ,k ≠ 0 ∴ x 1 By substituting x = and y = 25 into the 3 equation, we have
1.5625y 0 − y 0 × 100% y0
=
(1.5625 − 1) y 0 × 100% y0
= 56.25% ∴
k
25 + 2 =
=
y is increased by 56.25%.
2
1 3 27 = 9k k = 3 3 y+ 2 = 2 x 3 y = 2 − 2 x
∴ i.e.
(b) When y = x 2 ,
3 − 2 x2 x 4 = 3 − 2x 2 x2 =
x 4 + 2x 2 − 3 = 0 ( x 2 − 1)(x 2 + 3) = 0 x 2 = 1 o r x 2 = − 3 (re je cted ) x = ±1 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 ∴ 7.
∵ ∴
The value of y is 1 when y = x 2 .
1 x2 k y = 2 , k ≠ 0 x y ∝
Let x0 and y0 be the original values of x and y respectively. ∴
New value of
x = (1 − 20%) x 0 = 0.8 x 0
y = New value of
∴
k (0.8 x 0 ) 2
k = 1.5625 2 x0 = 1.5625y 0
Percentage change in y
14
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 A k h = , k ≠ 0 A h ∝
By substituting A equation, we have
= 2 and h = 30 into the
k 2 k = 60
30 = ∴
h =
60 A
When h = 15 ,
60 A A = 4
15 =
∴
The base area of the cylinder is 4 cm2 when the height is 15 cm.
Level 2 9.
(a)
∵ ∴
1 x+ a k y = , k ≠ 0 x+ a y ∝
By substituting x = 4 and y = 25 into the equation, we have
k 4+ a k ...... (1) 5 = 4+ a
25 =
By substituting x = 8 and y = 625 into the equation, we have
k 8+ a k ..... (2) 25 = 8+ a
625 =
k 25 = 8+ a 5 k 4+ a 4+ a (2) ÷ (1), 5 = 8+ a 40 + 5a = 4 + a 4a = − 36 a = −9
15
k
(b) From (1), 5 = 4 + ( −9)
∴
k = −25 25 y = − x − 9
10 Variations (c)
When y = 4 ,
25 4 = − x− 9 2 x − 18 = − 25 2x = −7 x = − 10. (a)
7 2
= New pressure
k 0.9V 0
10 k 9 V0 10 = P 9 0 =
Let $V be the value of the flat and A years old be the age of the flat. ∵ ∴
1 A k V = , k ≠ 0 A V ∝
By substituting A = 4 and V = 1 500 000 into the equation, we have
k 4 k = 6 0 0 00 0 0 6 0 0 00 0 0 V = A
1 5 0 00 0 0 = ∴
The age of the flat in 2004 is ( 4 + 4) years old, i.e. 8 years old. When A = 8 ,
6 000 000 8 = 750 000
V =
∴
The value of the flat in 2004 is $750 000.
(b) When V = 2 000 000 ,
6 000 000 A A = 3
2 000 000 =
∴ 11.
The flat was worth $2 000 000 in 1999.
1 P k ∴ V = , k ≠ 0 P k i.e. P = , k ≠ 0 V ∵
V ∝
Let V0 and P0 be the original volume and pressure of the gas respectively. ∴
New volume
= (1 − 10%)V 0 = 0.9V 0
16
Certificate Mathematics in Action Full Solutions 4B ∴
Percentage change in pressure
∴
10 − 1 P 9 0 = × 100% P0 1 = 11 % 9 ∴
12. (a)
The pressure is increased by 11
1 %. 9
1 n k p = , k ≠ 0 n p ∝
∵ ∴
By substituting n = 1600 and p = 2.5 into the equation, we have
k 1600 k = 100 100 p = n
2.5 = ∴
(b) When p = 4 ,
100 n n = 25 4 =
∴ (c)
n = 625 625 bottles are available on the island when the retail price is $4.
When n = 2500 ,
100 2500
p = ∴
13. (a)
∵ ∴
= 2 The retail price is $2 when the number of bottles available on the island is 2500.
1 T k V = , k ≠ 0 T V ∝
By substituting T = 1.8 and V = 265 into the equation, we have
k 1. 8 k = 477
265 = ∴
V =
477 T
When T = 2.5 ,
17
477 2.5 = 190.8
V =
10 P0 − P0 = 9 × 100% P0
The average speed of the helicopter is 190.8 km/h when its travelling time is 2.5 hours.
10 Variations (b) When V = 210 ,
revolutions in travelling 1 km.
477 T T = 2.27 (cor. to 3 sig. fig.)
210 = ∴
The travelling time of the helicopter is 2.27 h when its average speed is 210 km/h.
14. Let n be the number of workers and T be the number of months needed to complete the job.
1 T k ∴ n = , k ≠ 0 T By substituting T = 21 and n = 56 into the ∵
n ∝
equation, we have
k 21 k = 1176 ∴ 1176 n = T When T = 7 , 56 =
1176 7 = 168
n =
∴
15. (a)
168 − 56 = 112 An additional 112 workers has to be employed to complete the same job in 7 months. ∵ ∴
1 d k n = , k ≠ 0 d n ∝
By substituting d = 40 and n = 800 into the equation, we have
k 40 k = 32 000 32 000 n = d
800 = ∴
(b) When d = 80 ,
32 000 80 = 400
n = ∴
(c)
The front wheel of a bicycle with a diameter of 80 cm would make 400 revolutions in travelling 1 km.
When n = 200 ,
32 000 d d = 160
200 = ∴
The diameter of the front wheel of a bicycle is 160 cm if it makes only 200
18
Certificate Mathematics in Action Full Solutions 4B 16. (a)
The length of the rectangle and the width of the rectangle are in inverse variation.
(b) ∵ ∴
1 w k l = , k ≠ 0 w l ∝
By substituting w = 20 and l = 45 into the equation, we have
k 20 k = 900
45 = ∴
l = (c)
900 w
When w = 18 ,
900 18 = 50
l = ∴
The length of the rectangle is 50 cm when its width is 18 cm.
(d) When the rectangle becomes a square, l = w . When l = w ,
900 w = 900
w = w2
w = 30 or w = −30 (rejected) ∴
17. (a)
∵
∴
The length of the rectangle is 30 cm when it becomes a square.
1 1 ( x + y ) ∝ + y x k x + y = ,k ≠ 0 1 1 + x y x+ y =
∴
kxy x+ y
( x + y ) 2 = kxy ( x + y ) 2 ∝ xy
( x + y ) 2 = k x y, k ≠ 0 (b)
x 2 + 2x y + y 2 = k x y x 2 + y 2 = ( k − 2) x y 1 xy = (x2 + 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
19
∴
xy ∝ ( x 2 + y 2 )
10 Variations Exercise 10C (p. 261) Level 1 1.
∴
(a) ∵ 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
k =8
(a)
5.
(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) 1 k = 4 1 2 z = xy 4
∴
t
t =4 t =16
P = 8mn
(b) When m = 4 and n = 6, P =8( 4)( 6) =192 2.
3u 2
(b) When u = 4 and y = 12, 3( 4) 2 12 = t
120 = k (3)( 5)
∴
y=
∵ ∴
u3 v2 ku 3 w = 2 ,k ≠ 0 v
w∝
By substituting u = 6, v = 3 and w = 2 into the equation, we have
2
k (6)3 32 1 k = 12 u3 w= 12v 2 2=
∴
(b) When x = 5 and y = 4, 1 z = (5)( 4) 2 4 = 20
(b) When w = 3u 3 ,
u3 12 v 2 36 v 2 = 1 1 v2 = 36 1 v =± 6 3u 3 =
3.
(a)
∵
a varies directly as b and inversely as c2.
∴
a=
kb ,k ≠ 0 c2
By substituting a = 1, b = 3 and c = 3 into the equation, we have
k (3) 32 k =3 3b a= 2 c 1=
∴
(b) When b = 2 and c = 2,
3( 2) 22 3 = 2
a=
4.
(a)
∵
∴
C ∝ d 2t
∴ C = kd 2 t , 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 = 1 .4 d 0 New value of t = (1 − 20 %) t 0 = 0.8t 0
ku 2 t
New value of C = k (1.4d 0 ) 2 (0.8t 0 ) =1.568 kd 0 2 t 0
,k ≠0
By substituting u = 2, t = 9 and y = 4 into the equation, we have
4=
∵
∴
y varies directly as u2 and inversely as t .
y=
6.
k ( 2) 2 9
=1.568 C 0 ∴
Percentage change in
C =
k =3
1.568 C 0 − C0 C0
×100 %
= 56 .8% ∴
C is increased by 56.8%.
20
Certificate Mathematics in Action Full Solutions 4B
k (4)( 3) 22 k =5 5 pq W = 2 r
15 =
Level 2 7.
(a)
∵ ∴
pq r2 kpq W = 2 ,k ≠0 r W ∝
∴
By substituting p = 4, q = 3, r = 2 and W = 15 into the equation, we have
(b)
When W = 6, p = 3 and r = 5,
5(3) q 52 q =10 6=
8.
T ∝ wn
∵ ∴
T = kwn, k ≠ 0
i.e.
w=
T , 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
∴
New value of w =
2T0 k (3n0 )
=
2 T0 3 kn 0
=
2 w0 3
Percentage change in
2 w0 − w0 w= 3 × 100% w0 1 = −33 % 3
9.
1 3
∴
w is decreased by 33 % .
(a)
∵
V ∝ r 2h
V = kr 2 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 2 h
(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
V = 3(0.5r0 ) 2 ( 2h0 ) New value of
2
= 0.5(3r0 h0 ) = 0.5V0
∴
21
10 Variations
new value of V original value of V 0.5V0 = V0
400 (1200 ) 80 = 6000
The required ratio =
1 2 =1 : 2 =
n=
12. ∵ ∴
10. ∵ ∴
M 1M 2 d2 kM 1M 2 F= ,k ≠ 0 d2
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.
F∝
Let (M1)0, (M2)0, d0 and F0 be the original values of M1, M2, d and F respectively. ∴
V P kV x= ,k ≠0 P x∝
New value of M 1 = 0.5( M 1 )0 New value of M 2 = 0.5( M 2 )0
∴
New value of d = 2d 0
x1 =
kV1 P1
(1)
x2 =
kV2 P2
( 2)
k [0.5( M 1 ) 0 ][ 0.5( M 2 ) 0 ] x1 kV P = 1× 2 (2d 0 ) 2 x2 P1 kV2 k ( M 1 ) 0 ( M 2 ) 0 x1 V P = 0.0625 = 1 × 2 2 ÷ (2), (1) d0 x2 V2 P1 9 9 13 = 0.0625 F0 = ×
New value of F =
x2
∴
13
12
x2 = 12
new value of F ∴ The time required to boil the second kettle of original value of F water is 12 min. 0.0625 F0 = y F0 z∝ 13. ∵ x 1 = ky 16 ∴ z = 1 , where k1 is a non-zero constant … x =1 : 16
The required ratio =
(1)
11.
(a)
∵ ∴
∵
A d kA n= ,k ≠0 d
n∝
∴
By substituting A = 1500, d = 50 and n = 12 000 into the equation, we have
k (1500 ) 50 k = 400 400 A n= d
12 000 =
∴
i.e. (2)
y∝
1 x
k2 , where k2 is a non-zero constant. x k x = 2 , where k2 is a non-zero constant … y y=
By substituting (2) into (1), we have
k1 y k2 y k = 1 ( y2 ) k2
z=
(b) When A = 1200 and d = 80, ∵
k1 and k2 are non-zero constants.
22
Certificate Mathematics in Action Full Solutions 4B
∴
k1 is also a non-zero constant. k2
k1 + 4k 2 = 4
(1) 13 By substituting x = 3 and z = into the 2
k1 , we have k2
k3 =
By letting
4 = k1 + k2 ( 2) 2
equation, we have
z = k3 y 2 , where k3 is a non-zero constant. i.e.
z ∝y
14. ∵ z ∝
∴
z=
13 = k1 + k 2 (3) 2 2 13 k1 + 9k 2 = (2) 2 5 5k 2 = 2 (2) – (1), 1 k2 = 2 1 By substituting k 2 = into (1), we have 2 1 k1 + 4 = 4 2 k1 = 2 1 z = 2 + x2 ∴ 2
2
3
x y2
k1 3 x , where k1 is a non-zero constant … y2 (1)
1 ∵ x∝ y ∴ x= i.e.
k2 , where k2 is a non-zero constant. y
k2 , where k2 is a non-zero constant … (2) x
y=
By substituting (2) into (1), we have
z=
(b) When x = 4, 1 z = 2 + ( 4) 2 2 =10
k1 3 x k2 x
2
1 3
x2 = k1 x × 2 k2 = ∵ ∴
k1 k2
2
7
(x 3 )
k1 and k2 are non-zero constants.
k1 k2
is also a non-zero constant.
2
By letting
z = k3 x
k3 = 7 3
k1 k2
2 , we have
, where k3 is a non-zero constant.
z 3 = k3 x 7 , where k3 is a non-zero constant. i.e. z 3 ∝ x 7
Exercise 10D (p. 270) Level 1 1.
(a)
∵
z is partly constant and partly varies directly as x2.
∴
z = k1 + k 2 x 2 , where k1, k2 ≠ 0
By substituting x = 2 and z = 4 into the equation, we have
23
2.
(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 = 109 324 k 2 = 81 (2) – (1), 1 k2 = 4 By substituting k 2 =
(2)
1 into (1), we have 4
1 3k1 +108 = 28 4 1 k1 = 3 1 1 3 w= s+ s ∴ 3 4
10 Variations
(2) – (1),
(b) When s = 2,
1 1 ( 2) + ( 2)3 3 4 8 = 3
w=
3.
(a)
∵ ∴
k2
1 + k2 = 3 k2 = 2
a = 16 + =
k2 (1)
5.
By substituting y = 4 and z = 7 into the equation, we have
7 = k1 +
k2
(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 + k 2 (300 ) k1 + 300 k2 = 680
(2)
845 = k1 + k 2 (450 )
6 + k2 = 8
k1 + 450 k 2 = 845
k2 = 2 z =6 +
2
(2) – (1),
y
∴
150 k 2 = 165 k 2 = 1.1
k1 + 300 (1.1) = 680 k1 = 350 ∴
E = 350 + 1.1x
(b) When x = 380,
E = 350 +1.1(380 )
a partly varies directly as the square root of b and partly varies inversely as the square of c.
a = k1 b +
k2 c2
, where k1, k2 ≠ 0
By substituting b = 1, c = 1 and a = 3 into the equation, we have
3 = k1 1 +
(2)
By substituting k2 = 1.1 into (1), we have
(b) When y = 25, 2 z =6+ 25 32 = 5 ∵
(1)
By substituting x = 450 and E = 845 into the equation, we have
(2) – (1), k1 = 6 By substituting k1 = 6 into (1), we have
(a)
38 9
∵
4
2k1 + k 2 = 14
4.
2 32
1
k1 + k 2 = 8
∴
2 c2
(b) When b = 16 and c = 3,
By substituting y = 1 and z = 8 into the equation, we have
8 = k1 +
a= b+
∴
, where k1, k2 ≠ 0
y
k1 = 1
By substituting k1 = 1 into (1), we have
z is partly constant and partly varies inversely as the square root of y.
z = k1 +
7 k1 = 7
k2 12
k1 + k 2 = 3
5 into 2
the equation, we have
5 k = k1 4 + 22 2 2 8k1 + k2 = 1 0
6.
= 768
The daily expenditure of hiring a taxi is $768 for travelling a distance of 380 km.
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 + k2 (40)
(1 )
By substituting b = 4, c = 2 and a =
∴
k1 + 40 k 2 = 2000
(1)
By substituting n = 38 and C = 1940 into the equation, we have
1940 = k1 + k 2 (38) k1 + 38 k2 = 1940
(2)
(1) – (2),
(2)
2k 2 = 60 k 2 = 30
24
Certificate Mathematics in Action Full Solutions 4B By substituting k2 = 30 into (2), we have
5 = 5 + k 2 (2) + k3 (2) 2
k1 + 38 (30 ) = 1940
2 k 2 + 4 k3 = 0
k1 = 800
k 2 + 2 k3 = 0
∴ C = 800 + 30n When n = 43,
(2) – (1),
C = 800 + 30 ( 43 ) ∴ 7.
(a)
= 2090
∴
k2 = 2
(b)
By substituting n = 2000 and C = 225 into the equation, we have
( x + 2)(x − 4) = 0 x = − 2 or x = 4
…… (1)
k2 5000 5000 k1 + k2 = 1 050 000
y =5 +2x − x 2 = −( x 2 − 2 x −5) (c)
210 = k1 +
(2)
k1 = 200
By substituting k1 = 200 into (1), we have
2000 (200 ) + k 2 = 450 000 k 2 = 50 000 50 000 n
(b) When n = 8000,
Level 2 8. (a)
= 6 −( x −1) 2 ∴
9. (a)
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 k1 + 400 k 2 = 200 000
150 000 = k1 (1.5) + k 2 (10 ) 2 150 000 = 1.5k1 +100 k 2
The cost of producing a copy of the video game is $206.25.
k1 + ∵
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
200 k 2 = 100 000 3
……
(2) (1) – (2),
5 = k1 + k 2 (0) + k3 (0) 2
1000 k 2 = 100 000 3 k 2 = 300
k1 = 5
By substituting k2 = 300 into (2), we have
By substituting x = –1 and y = 2 into the equation, we have
k1 +
2 = 5 + k2 (−1) + k3 (−1) 2 k 2 − k3 = 3
(1)
By substituting x = 2 and y = 5 into the equation, we have
∴
200 (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
25
……
(1) By substituting l = 1.5, w = 10 and C = 150 000 into the equation, we have
50 000 C = 200 + 8000 = 206 .25
∴
= −( x 2 − 2 x +12 −12 −5) = −( x 2 − 2 x +1) + 6
3000 k1 = 600 000
C = 200 +
When y = –3, − 3 = 5 + 2x − x2 x2 − 2x − 8 = 0
By substituting n = 5000 and C = 210 into the equation, we have
∴
y = 5 +2 x − x2
∴
k C = k1 + 2 , where k1, k2 ≠ 0 n
(2) – (1),
k3 = −1
k 2 − (−1) = 3
C is partly constant and partly varies inversely as n.
k 225 = k1 + 2 2000 2000 k1 + k 2 = 450 000
3k3 = −3
By substituting k3 = –1 into (1), we have
The cost for a class of 43 students is $2090. ∵
(2)
10 Variations ∴
The cost of building a road is $227 500 when the length is 2 km and the width is 15 m.
10. (a)
∵
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 + k 2 (300 000 ) k1 + 300 000 k 2 = 7000
(1)
By substituting A = 400 000 and S = 9000 into the equation, we have
9000 = k1 + k 2 ( 400 000 ) k1 + 400 000 k 2 = 9000 (2) – (1),
(2)
100 000 k2 = 2000 k2 =
By substituting k 2 =
1 50
1 into (1), we have 50
1 k1 + 300 000 = 7000 50 k1 = 1000 1 S = 1000 + A ∴ 50 (b) When A = 350 000,
S = 1000 +
1 (350 000 ) 50
= 8000 ∴ (c)
Her income for that month is $8000.
When S = 20 000,
20 000 = 1000 +
1 A 50
1 A 50 A = 950 000
19 000 =
11.
(a)
∴
The amount of her sales should be $950 000.
∵
C is partly constant and partly varies inversely as n.
∴
C = k1 +
k2 , where k1, k2 ≠ 0 n
By substituting n = 1000 and C = 53 000 into the equation, we have
k2 1000 1000 k1 + k 2 = 53 000 000 …… (1) 53 000 = k1 +
By substituting n = 2000 and C = 58 000 into the equation, we have
k2 2000 2000 k1 + k 2 = 116 000 000 58 000 = k1 +
……
26
Certificate Mathematics in Action Full Solutions 4B
32 000 = k1 (80) + k2 (80) 2
(2)
1000 k1 = 63 000 000
(2) – (1),
k1 = 63 000
By substituting k1 = 63 000 into (1), we have
1000(63 000) + k 2 = 53 000 000 k 2 = −10 000 000 ∴
C = 63 000 −
10 000 000 n
(b) When n = 4000,
C = 63 000 −
10 000 000 4000
= 60 500 ∴ 12. (a)
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)
(2) ÷ (1),
11.3− L0 = 1.75 10.1 − L0 11.3 − L0 = 17.675 − 1.75L0 0.75L0 = 6.375 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
∴ ∵ ∴ (c)
When L = 2(8.5) = 17,
∴
13. (a)
Le = 0.8W L = L0 + Le L = 8.5 + 0.8W
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.
∵
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 k1 + 160k2 = 0
(1)
By substituting x = 80 and P = 32 000 into the equation, we have
27
k1 + 80k2 = 400 80 k 2 = −400 (1) – (2), k 2 = −5
(2)
By substituting k2 = –5 into (1), we have
k1 + 160 ( −5) = 0 k1 = 800 ∴
P = 800x – 5x2
10 Variations (b)
When x = 20, =14 000
∴
y =
P = 800 x −5 x 2 = −5( x 2 −160 x )
(c)
12 x 12 x = 5
2
5=
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 (a) ∵ ∴
4.
y = kx , k ≠ 0 2
∴
y = 4x2
4 , 25
(b) When x = 8 , 138 8= 2y +3
4 = 4x2 25 1 x2 = 25 1 x =± 5 (a) ∵ ∴
16 y + 24 = 138 16 y = 114 57 y= 8
( y +2) ∝
x
y +2 =k
x , k ≠0
5.
(a) ∵ ∴
∴
y +2 = 3 x y =3
10 = 3 x − 2
x = 16
3.
(a) ∵
y varies inversely as x.
k ,k ≠0 x By substituting x = 4 and y = 3 into the ∴
100 = k (5) 2 ( 2) k =2 z = 2x2 y
z = 2( 2) 2 (3) = 24
(b) When y =10 ,
x =4
z = kx 2 y , k ≠ 0
(b) When x = 2 and y = 3 ,
x −2
3 x = 12
z varies jointly as x2 and y.
By substituting x = 5 , y = 2 and z = 100 into the equation, we have
By substituting x = 9 and y = 7 into the equation, we have 7 +2 = k 9 9 = 3k ∴ k =3 i.e.
k 2(10 ) + 3 k =138 138 x = 2 y +3 6=
k =4
(b) When y =
x varies inversely as ( 2 y + 3) . k x= , k ≠0 2 y +3
By substituting x = 6 and y =10 into the equation, we have
By substituting x = 3 and y = 36 into the equation, we have 36 = k (3) 2
2.
(a) ∵ ∴
y varies directly as x2.
∴
12 x
(b) When y = 5 ,
= −5( x 2 −160 x +80 2 −80 2 ) = 32 000 − 5( x −80 )
1.
k 4 k = 12 3=
P =800 ( 20 ) −5( 20 ) 2
6.
(a) ∵ ∴
a varies directly as b and inversely as c2.
a=
kb ,k ≠0 c2
By substituting a = 1 , b = 6 and c = 3 into the equation, we have
y=
equation, we have
28
Certificate Mathematics in Action Full Solutions 4B
(1) − ( 2), 9k1 = 2250
k ( 6) 32 3 k = 2 3b a= 2c 2 1=
∴
k1 = 250 By substituting k1 = 250 into (2), we have
250 + k 2 = 260 k 2 = 10 ∴
(b) When a = 3 and c = 4 , 3b 3= 2( 4) 2
3b = 96 b = 32 7.
(a)
∵
w partly varies directly as varies directly as
x and partly
y .
∴
w = k1 x + k 2
y , where k 1 , k 2 ≠ 0
By substituting x = 4 , y = 9 and w = 13 into the equation, we have
13 = k1 4 + k 2 9 2k1 + 3k2 = 13
...... (1)
By substituting x = 9 , y =16 and w = 18 into the equation, we have
18 = k1 9 + k 2 16 3k1 + 4k2 = 18
...... (2)
(1) × 3 − ( 2) × 2, k 2 = 3 By substituting k 2 = 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 =23
8.
(a)
25
∵ y is partly constant and partly varies inversely as x. ∴
y = k1 +
k2 , where k1 , k 2 ≠ 0 x
By substituting x = 10 and y = 251 into the equation, we have
k2 10 10k1 + k 2 = 2510 ...... (1) 251 = k1 +
By substituting x = 1 and y = 260 into the equation, we have
260 = k1 + k1 + k 2 = 260 29
k2 1
...... (2)
y = 250 +
10 x
10 Variations (b) When x = 5 ,
∴
10 y = 250 + 5 = 252 9.
∵
y ∝
New value of
New value of
∴ 10. ∵ ∴
=1.3k
x0
∴
y is increased by 30%.
1 x2 k y = 2 ,k ≠ 0 x
y∝
New value of
New value of
=
4 y0 9
11. ∵
5 y is decreased by 55 % . 9 z∝
x2 y 2
∴
kx z= ,k ≠0 y
1 .6 z 0 − z 0 ×100 % z0
= 60% z is increased by 60%.
x2 y
∴
z=
kx 2 ,k ≠0 y
Let x0, y0 and z0 be the original values of x, y and z respectively. ∴ New value of x = 2x0 New value of y = 0.5 y0
k (2 x 0 ) 2 z = 0 .5 y 0 New value of
4 y0 − y 0 9 y= × 100% Percentage change in y0
kx 0 2 = 1.6 y 0 = 1.6 z0
z∝
k (1.5 x0 ) 2 4 k 2 9 x0
k (1.2 x0 ) 2 (0.9 y0 )
Percentage change in
= 1.5 x0
=
= 0.9 y0
12. ∵
x = (1 + 50 %) x0
5 = −55 % 9 ∴
∴
= 30 %
y=
∴
New value of
z=
1 .3 y 0 − y 0 × 100 % y0
= 1.2 x0
z=
y = k 1.69 x0
Let x0 and y0 be the original values of x and y respectively. ∴
y = (1 −10 %) y0
= 1.69 x0
Percentage change in
y=
New value of
x = (1 + 69 %) x0
=1.3 y0 ∴
x = (1 + 20 %) x0
x
∴ y = k x , k ≠0 Let x0 and y0 be the original values of x and y respectively. ∴
New value of
kx = 8 0 y0 = 8z 0 z=
∴
Percentage change in
∴
z is increased by 700%.
2
8 z0 − z 0 × 100% z0
= 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 = kd 1 C2 = kd 2
2
...... (1) ...... (2)
Let x0, y0 and z0 be the original values of x, y and z respectively.
30
Certificate Mathematics in Action Full Solutions 4B 2
C1 kd1 = C2 kd2 2
(1) ÷(2),
C1 d1 = C2 d 2 2000 2 = C2 3
2
2
2000 4 = C2 9 C2 = 4500 ∴
31
The value of the larger plate is $4500.
10 Variations 14. Let T be the time taken to drink a bottle of cola and d be the diameter of the straw. ∵ ∴
directly as n. ∴
1 d2 k T = 2 ,k ≠ 0 d T ∝
By substituting n = 150 and C = 3750 into the equation, we have
3750 = k1 + k2 (150) k1 + 150k2 = 3750
Let T0 and d0 be the original values of T and d respectively. ∴ New value of d = 2d 0
7590 = k1 + k 2 (750)
∴
(2) −(1),
0.25T0 − T0 ×100 % T0
= −75 % The time taken is decreased by 75%.
∴
(b) When n = 200 ,
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.
C1 = kT1d1
2
C2 = kT2 d 2
2
17. ∵
k ,k ≠0 x By substituting x = 15 and y = 20 into the y=
equation, we have
...... (2)
k 15 k = 300
20 =
2
∴
2
C1 T d1 = 1 × C2 T2 d2
Take
2
81 ×16 d2 = 64 = 20 .25 d 2 = 4.5 or − 4.5 (rejected)
y Take
y
2
16. (a)
The diameter of the second coin is 4.5 cm. ∵
C is partly constant and partly varies
300 x x =5 , 300 = 5 = 60 x = 10 , 300 = 10 = 30 y=
16 64 = 2 81 d2
∴
The cost of making 200 coats is $4070.
y varies inversely as x.
∴
...... (1)
16 3 4 = × 27 4 d2
32 (200 ) 5
= 4070
∴
C1 kT d = 1 12 C2 kT2 d 2
(1) ÷(2),
600 k2 = 3840
32 5 32 By substituting k 2 = into (1), we have 5 32 k1 +150 = 3750 5 k1 = 2790 32 n ∴ C = 2790 + 5
15. Let $C be the cost of the gold coin, T mm be the thickness of the gold coin and d cm be the diameter of the gold coin. ∵ C ∝ Td 2 ∴
...... (2)
k2 =
Percentage change in
T = ∴
k1 + 750k 2 = 7590
k = 0.25 2 d0 = 0.25T0
...... (1)
By substituting n = 750 and C = 7590 into the equation, we have
k T = ( 2d 0 ) 2 New value of
C = k 1 + k 2 n, where k 1 , k 2 ≠ 0
∴ 18.
The ordered pairs (5, 60) and (10, 30) lie on the graph. (or any other reasonable answers)
(a) ∵
The expenditure ($E) of the party includes buying gifts and food. $400 is spent for the gifts and the expenditure of food is $F.
32
Certificate Mathematics in Action Full Solutions 4B ∴ (b) ∵
∴ (c) ∵ ∴
33
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.
x =
12 000 and it is in inverse x
variation. Each classmate has to pay $40 for the food and the expenditure of food is $F. F = 40 N and it is in direct variation.
10 Variations Level 2 19. (a)
∵
4 3 4 7 (1) ÷(2), 7 3 14 a − 7 2a
y ∝ ( ax − 3)
y = k ( ax −3), k ≠ 0 ∴ By substituting x = 4 and y = 3 into the equation, we have 3 = k ( 4a −3) ...... (1)
By substituting x = 5 and y = 6 into the equation, we have 6 = k (5a −3) ...... (2)
a =2
6 k (5a − 3) = 3 k (4a − 3) 5a − 3 2= (2) ÷(1), 4a − 3 8a − 6 = 5a − 3
(b)
3a = 3 a =1 (b)
3 = k [ 4(1) −3]
(c)
∴ ∵
y = 3( x − 3)
By substituting a = 2 into (1), we have
∴
By substituting a = 1 into (1), we have
k =3 y = 3( x −3)
(c)
4 2x − 1 2x2 − x − 1 = 0 4x =
(2 x + 1)(x − 1) = 0
= 3x −9
x= −
21. (a) ∵ ∴
1 ax − 1 k y= , k ≠0 ax −1
∵
w∝
∴
w=
y∝
4 By substituting x = 2 and y = into the 3
x2 y z3 kx 2 y z3
,k ≠0
= 1 and
k (1) 2 1 13 k = 2 2 =
4 k = 3 a ( 2) −1 4 k ...... (1) = 3 2a −1
equation, we have
1 or x = 1 2
By substituting x =1 , y =1 , z w = 2 into the equation, we have
equation, we have
By substituting x = 4 and y =
4 k = 3 2( 2) −1 k =4 4 y= 2 x −1
When y = 4 x ,
When 2 y = 3 x , y = 2 y −9 y =9
20. (a)
k = 2a −1 k 4a −1 4a −1 = 2a −1 = 12 a − 3 =4
∴
4 into the 7
w=
2x2 y z3
(b) When y = 2 w = 2 x = 2 z , i.e. y = 2 x , w = x and z = x ,
4 k = 7 a ( 4) −1 4 k ...... (2) = 7 4a −1
34
Certificate Mathematics in Action Full Solutions 4B
w= x=
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
2x2 y z3 2x2 2x x3
∴
Total volume of the three metal cubes
x4 = 2x2 2x
= [ k (3)3 + k ( 4)3 + k (5)3 ] cm 3
x2 = 2x 2 x4 = 2x 4 x4 = 8x
= k ( 27 +64 +125 ) cm 3 = 216 k cm 3 Let VL cm3 and xL cm be the volume and the length of the side of the larger cube. ∵
x4 − 8x = 0 x ( x 3 − 8) = 0
∴
x = 0 (rejected) or x 3 = 8 x =2
22. (a)
∵ ∴
VL = kx L
y = k 1 + k 2 x + k 3 x 2 , where
∴ 24. ∵
k1 , k 2 , k 3 ≠ 0 By substituting x = 0 and y = 5 into the equation, we have
5 = k1 + k 2 (0) + k3 (0) 2
∴
k1 = 5 By substituting x = 2 and y =15 into the equation, we have
15 = 5 + k 2 (2) + k3 (2) 2 ....
.. (1) By substituting x = −2 and y =11 into the equation, we have
216 k = kx L 3
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=
Let r0 and h0 be the original values of r and h respectively. ∴
New value of
.... New value of
4k3 = 8 k3 = 2
By substituting k3 = 2 into (1), we have
k 2 + 2( 2) = 5 k2 = 1 ∴ (b)
y = 5 + x +2x2
When y = 6 ,
∴
k′ 0.81h0 k′ h 0
=
10 9
=
10 r0 9
Percentage change in
1 = 11 % 9 ∴
The base radius of the cone is increased by
1 11 %. 9
1 x = − 1 or x = 2 25. ∵
35
= 0.81h0
10 r0 − r0 r= 9 ×100 % r0
6 = 5 + x + 2 x2 2x2 + x − 1 = 0 ( x + 1)(2 x − 1) = 0
h = (1 − 19 %) h0
r=
.. (2) (1) − (2),
3
xL = 216
11 = 5 + k 2 (−2) + k3 ( −2) 2 k2 − 2k3 = −3
3
xL = 6
y is partly constant, partly varies directly as x and partly varies directly as x2.
k 2 + 2 k3 = 5
V = kx 3 , k ≠ 0
E ∝ mv 2
10 Variations E = kmv
∴
2
, 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
E = k ( 2m0 )( 0.8v0 ) 2 New value of
= 1.28 km 0v0
2
= 1.28 E0 ∴
Percentage change in
E= ∴ 26. (a)
1.28 E0 − 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 3 ∴ 1 2 V = r h 3 When V = 540 and h = 27 , k =
1 2 r ( 27 ) 3 = 60
540 = r2
r =
60
= 2 15
36
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 ∴
( 2) − (1), 200 k 2 = 40 k2 =
New value of h = 0.5h0
By substituting k 2 =
1 ( 2r0 ) 2 (0.5h0 ) 3 1 2 = 2 r0 h0 3 = 2V0
V = New value of
∴
V =
∴
Percentage change in
2V0 − V0 ×100 % V0
(b)
(c)
V is increased by 100%.
∵
V =
1 2 r h 3
r =
3V h
∴
∴ (c)
New value of
V = (1 + 36 %) V0
New value of
h = h0
New value of
∴
= 1.36V0
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 2T0 = 30 +
= 1.36 r0 ∴
Percentage change in
1.36 r0 − r0 ×100 % r0
r =
∴
The airtime used in this month is (150 + 2n0 ) minutes.
∵
C is partly constant and partly varies inversely as n.
∴
C = k1 +
= 16 .6% (cor. to 1 d.p.) 27. (a)
∴
r is increased by 16.6%.
∵
T is partly constant and partly varies directly as n.
∴
28. (a)
T = k 1 + k 2 n, where k 1 , k 2 ≠ 0
By substituting n = 400 and T =110 into the equation, we have
110 = k1 + k2 ( 400) k1 + 400k2 = 110
...... (1)
By substituting n = 600 and T = 150 into the equation, we have
150 = k1 + k2 (600) k1 + 600k2 = 150
37
...... (2)
1 n 5 1 T0 = 30 + n0 5
T = 30 +
When T = 2T0 ,
3(1.36V0 ) h0
3V0 = 1.36 h0
The monthly charge is $90 if the airtime used is 300 minutes.
Let $T0 be the mobile phone charge in last month. ∵
r=
1 k1 + 400 = 110 5 k1 = 30 1 T = 30 + n 5
1 T = 30 + (300 ) 5 = 90
Let V0, h0 and r0 be the original values of V, h and r respectively. ∴
1 into (1), we have 5
When n = 300 ,
= 100 % ∴
1 5
k2 , where k1 , k2 ≠ 0 n
By substituting n = 200 and C = 250 into the equation, we have
k2 200 200 k1 + k 2 = 50 000 ...... (1) 250 = k1 +
By substituting n = 400 and C = 200 into the equation, we have
10 Variations
k2 400 400k 1 + k 2 = 80 000 ...... (2) 200 = k 1 +
(2) −(1),
P = 4000 x − 50 x 2 = −50 ( x 2 − 80 x ) (c)
200 k1 = 30 000
= −50 ( x 2 − 80 x + 40 2 ) + 80 000
k1 = 150
= −50 ( x − 40 ) 2 + 80 000
By substituting k1 = 150 into (1), we have
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 $40.
200 (150 ) + k 2 = 50 000 k 2 = 20 000 ∴ (b)
C = 150 +
20 000 n
30. (a)
C = 150 +
20 000 500
By substituting r = equation, we have
k1 + 4k2 = 2
=$( 190 +50 ) =$240
5 = k1 (5) + k 2 (5) 2 k1 + 5k 2 = 1
P is partly varies directly as x and partly varies directly as x2.
By substituting k 2 = −1 into (1), we have
∴
k1 + 4( −1) = 2
By substituting x = 20 and P = 60 000 into the equation, we have
k1 = 6
(b)
When
r=2, C = 6( 2) −( 2) 2 =8
By substituting x = 30 and P = 75 000 into the equation, we have
C = 6r − r 2
75 000 = k1 (30 ) + k2 (30 ) 2 k1 + 30 k2 = 2500
....
= − ( r 2 − 6r ) (c)
= − ( r 2 − 6r + 3 2 − 3 2 )
.. (2)
= − ( r 2 − 6r + 3 2 ) + 9 (2) −(1),
10 k 2 = −500
= − ( 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.
k 2 = −50
By substituting k 2 = −50 into (1), we have
k1 + 20 ( −50 ) = 3000 k1 = 4000 ∴ (b)
C = 6r − r 2
∴
60 000 = k1 (20 ) + k 2 ( 20 ) 2 ....
...... (2)
(2) −(1), k 2 = −1
P = k1 x + k 2 x 2 , where k1 , k 2 ≠ 0
.. (1)
...... (1)
By substituting r = 5 and C = 5 into the equation, we have
The selling price of each rice cooker
k1 + 20 k 2 = 3000
4 and C = 8 into the
8 = k1 (4) + k2 ( 4) 2
25 000 Profit of a rice cooker = $ 500 = $50
∵
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
= 190
29. (a)
∵ ∴
When n = 500 ,
∴
= −50 ( x 2 − 80 x + 40 2 − 40 2 )
P = 4000 x − 50 x 2
When x = 35 ,
P = 4000 (35 ) −50 (35 ) 2 ∴
= 78 750 The total profit is $78 750 if the selling price of each VCD is $35.
31. (a)
∵ ∴
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
38
Certificate Mathematics in Action Full Solutions 4B (b) From the graph, when x = 120 , y = 6000 . y = 6000 By substituting x = 120 and into the equation, we have
6000 = 3000 + k ′(120 ) k ′ = 25
∴ (c)
32. (a)
y = 3000 + 25 x
( x − y) ∝ ( x + y)
∵
3 1 3 1 x − 3 ∝ x + 3 y y
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.
∵
(x2 − y 2 ) ∝ (x2 + y2 )
∴
x 2 − y 2 = k ( x 2 + y 2 ), k ≠ 0
( k +1) y 2 = ( k −1) x 2 k −1 2 y2 = 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
y = k ′x , k ′ ≠ 0 i.e.
y∝x
∵
y∝x
∴
y = k ′′x, where k ′′ ≠ 0
∵
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 1 + k ′′ x− y = = x+ y = = x− y = x+ y
and By letting
39
33. (a)
i.e.
∴
x 2 − y 2 = kx 2 + ky 2
(b)
x − y = k ′′′( x + y ), k ′′′ ≠ 0
k ′′ ≠ 1 .
k ′′′ =
1 − k ′′ 1 + k ′′
, we have
3 1 1 = k x + y3 , k ≠ 0 y3 k + 1 (1 − k ) x 3 = 3 y x3 −
y3 =
k +1 1 × 1 − k x3
y=3 By letting k ′ = 3
∴
k +1 1 × 1− k x k +1 , we have 1−k
y=
k′ , k′ ≠ 0 x
y∝
1 x
10 Variations ∵
3 1 1 = k x + 3 ,k ≠0 3 y y
x3 −
(b)
∴
(5 + 6k ) y = ( 4k − 3) x
1 x 1 x − x 2 + + 2 = y y y
1 x 1 k x + x 2 − + 2 y y y 2 x 1 x − + 2 1 1 y y x + x − = k x 1 y y 2 x + + 2 y y 3. x y − xy + 1 1 x + = k 2 2 y x y + xy + 1 2 k′ 2 k′ x − x + 1 x x = k 2 ′ ′ x 2 k + x k + 1 x x 1 x + y 2 2
k′2 − k ′ + 1 1 x + = k 2 y k′ + k′ + 1
4k − 3 y = x 5 + 6k 4k − 3
′ By letting k = 5 + 6k , we have
y =k ′x, k ′ ≠0
i.e.
Answer: C ∵ x varies directly as y2. ∴ x = ky 2 , k ≠ 0 By substituting y = 2 and x = 20 into the equation, we have 20 = k (2) 2
k =5
∴
1
∴
2.
Answer: A
3 ( x + 1) 2
4( x + 1) 2 = 1 4x 2 + 8x + 3 = 0 ( 2 x + 1)(2 x + 3) = 0 x=− 4.
∴ 5.
1 3 or x = − 2 2
Answer: D ( x + y ) varies inversely as ( x − y ) . ∵ k x +y = , k ≠0 ∴ x −y
x −y =
k , k ≠0 x+y
( x − y ) varies inversely as ( x + y ) .
Answer: B ∵ x varies inversely as y and directly as z2. ∴
x=
kz 2 , where k is a non-zero constant y
xy = k , where k is a non-zero constant z2
=16
y = ±4
3 ( x +1) 2
12=
80 = 5 y 2 y
1 k = 3 (2 +1) 2 1 k = 3 9 k =3
When y =12 ,
x =5 y2 When x = 80 , 2
k , k ≠0 ( x +1) 2
By substituting x = 2 and y = 3 into the equation, we have
y=
Multiple Choice Questions (p. 278) 1.
y=
∴
1 1 = k ′′ x+ , k ′′ ≠ 0 y y
1 1 x − y ∝ x + y
∴
y∝x
Answer: B ∵ y varies inversely as ( x +1) 2 .
k′ 2 − k′ + 1 , we have By letting k ′ = k k′ 2 + k′ + 1 x−
(3 x + 5 y ) ∝ ( 4 x − 6 y ) 3 x + 5 y = k ( 4 x − 6 y ), k ≠ 0
6.
Answer: D z ∝ ( y + 2) and y ∝ ( x −1) ∵
40
Certificate Mathematics in Action Full Solutions 4B ∴
z = k1 ( y + 2) and y = k 2 ( x −1) , where k1 , k 2 ≠ 0
z = k1 ( k2 x − k2 + 2)
∴
= k1k 2 x − k1k 2 + 2k1 7.
Answer: A ∵ y partly varies directly as x and partly varies directly as the cube of x.
y = k1 x + k 2 x 3 , where k1 , k 2 ≠ 0 By substituting x = 3 and y = −21 into the ∴
equation, we have
− 21 = k1 (3) + k 2 (3)3 k1 + 9k 2 = −7
…… (1)
By substituting x = 5 and y = −115 into the equation, we have
∴
(2) – (1),
p. 244
…… 2.
16 k 2 = −16 k 2 = −1
By substituting k 2 = −1 into (1), we have
k1 + 9( −1) = −7 k1 = 2 ∴ 8.
y = 2 x − x3
Answer: A ∵ z varies directly as x2 and inversely as y. ∴
kx 2 , k ≠0 y
z=
Let x0, y0 and z0 be the original values of x, y and z respectively. ∴
New value of
x = (1 + 30 %) x0
New value of
y = (1 + 30 %) y0
= 1.3x0 = 1.3 y0
k (1.3x 0 ) 2 z = 1 .3 y 0 New value of
∴
Percentage increase in
z=
1.3 z0 − z0 × 100 % z0
= 30% 9.
41
kx = 1.3 0 y0 = 1 .3 z 0
Answer: B ∵
2
1 x
Let’s Discuss 1.
(2)
y ∝
10. Answer: D ∵ y is partly constant and partly varies directly as x. y = k1 + k 2 x, where k1 , k 2 ≠ 0 ∴
− 115 = k1 (5) + k 2 (5)3 k1 + 25k 2 = −23
xy = 1 × 315 = 3 × 105 = 5 × 63 = 7 × 45 = 9 × 35 = 315 315 y= x
Yes
Only graph number ③ shows the relation y varies directly as x. It is because the graph of direct variation is a straight line passing through the origin.