Variations)

Certificate Mathematics in Action Full Solutions 4B

10 Variations • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •

Activity

4.

(a)

∴

Activity 10.1 (p. 263) 1.

(a)

∵

$25x

(b) $30y 2.

∴

8 : 6 = 16 : x 8 16 = 6 x 8 x = 96 x = 12

∵

( x + 1) : ( x − 1) = 3 : 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.

(c)

Follow-up Exercise

∴

p. 238

1.

(a)

Cost rate

21 / cup 6 = $3.5 / cup =$

1.

20 km 20 (b) Travel rate h 60 = 60 km/h

2.

$24 30 dozen 12 = $9.6 /dozen

(a)

30 cm = 50 cm 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.

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

(a)

∵ ∴

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 x (b) When = 6, y = 3× 6 = 18

=

Cost rate

x +1 3 = x −1 2 2 x + 2 = 3x − 3 x=5

p. 242

=

(c)

x:4 = 5:2 x 5 = 4 2 2 x = 20 x = 10

2.

∵ 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 x = 32, When z = 3 × 32 = 96

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

∴ Percentage change in y 1.3 y0 − y0 = × 100% y0 =

3.

∵

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

Percentage change in

= 21%

2.

∴

y is increased by 21%.

(a)

∵

C ∝ r2

p. 251 1.

y∝ x

(a)

y = k x, k ≠ 0 By substituting x = 16 and y = 8 into the equation, we have ∴

∴

∵

20 = k (6) 2 20 k= 36 5 = ∴ 9 5 C = r2 9 C = 45, (b) (i) When 5 45 = r 2 9 2 ∴ r = 81 r =9 ∴ The radius of the hemisphere is 9 cm. r = 0.18 × 100 = 18, (ii) When 5 C = (18) 2 9 = 180 ∴ The cost of painting a hemisphere is $180.

∴ y = kx 2 , k ≠ 0 Let x0 and y0 be the original values of x and y respectively. ∴ New value of x = (1 + 10%) x0 = 1.1x0

∴

= 30% y is increased by 30%.

C = kr 2 , k ≠ 0 By substituting r = 6 and C = 20 into the equation, we have

y ∝ x2

New value of

∴

∴

p. 243 1.

(a)

(1.3 − 1) y0 × 100% y0

8 = k ( 16 ) 8 = 4k k=2

k 1 k =3 3 y= x x = 3, 3=

∴

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 New value of y

y varies inversely as x. k y= ,k ≠0 ∴ x By substituting x = 1 and y = 3 into the equation, we have ∵

(b) When

3 3 =1

y=

= 2 1.69 x 0 = 1.3( 2 x 0 ) = 1.3 y 0

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

∴

When

1 k = 3 1 1 k= 3 1 z= 3x z = 2,

1 3x 6x = 1 1 x= 6 2=

3.

y2 varies inversely as x. k 2 ∴ y = ,k ≠0 x By substituting x = 4 and y = 3 into the equation, we have ∵

k 4 k = 36 36 y2 = x y = 5, 32 =

∴

When

36 x 25 x = 36 36 x= 25 52 =

p. 253 1.

(a)

1 x2 k y= 2,k ≠0 ∴ x By substituting x = 2 and y = 4 into the equation, we have ∵

y∝

k 22 k = 16 16 y= 2 x 4=

∴

(b) Let x0 and y0 be the original values of x and y respectively. ∴ New value of x = (1 + 25%) x0 = 1.25 x0 = New value of y

3

16 (1.25 x0 ) 2

 16  = 0.64 2   x0  = 0.64 y0

10 Variations ∴ Percentage change in y 0.64 y0 − y0 = × 100% y0 =

= −36% y is decreased by 36%.

∴ 2.

y∝

∵

(0.64 − 1) y0 × 100% y0

y=

1 x k

,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 x0 ∴

y= New value of

∴

3.

0.64 x0

 k   = 1.25  x   0  = 1.25 y0

Percentage change in y 1.25 y0 − y0 = × 100% y0 =

∴

k

(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 10 k = 300 300 T= n n = 12,

30 = ∴

When

300 12 = 25 The food can last for 25 days if 2 more people go ashore on the island. T=

∴

4

Certificate Mathematics in Action Full Solutions 4B p. 261 1.

(a) (b) (c)

2.

(a)

∴ z = kx 2 y 3 kd s= t kb3 a2 = c

∴

z varies jointly as x2 and

∴

z = kx 2 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 z = 2x

2

(a)

2.

(a)

y = k1 x + k 2 x 2

(b)

a = k1 +

(c)

A = k1B 2 +

(a)

∵

y

(b) When x = 3 and z = 2(3) 2 16 = 72 3.

y = 16

,

z∝ z=

New value of

y = (1 + 50%) y0 = 1.5 y0 z =

New value of

k (0.75 x 0 ) 1.5 y 0

kx = 0.5  0   y0  = 0.5 z 0

B

z partly varies directly as x and partly varies directly as y. ∴ 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 4 = k1 (2) + k2 (1)

2k1 + 3 = 4 1 2 1 z = x + 3y 2

k1 = ∴

y=2 (b) When x = 3 and , 1 z = (3) + 3( 2) 2 15 = 2

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 = 0.75 x0 ∴

k2

4k1 + 3k2 = 11 …… (2) ( 2) − (1) × 2, k 2 = 3 By substituting k2 = 3 into (1), we have

k (2) 12 k=2 ∴ 2y x= 2 z y=3 (b) When and z = 2 , 2(3) x= 2 2 3 = 2

∵

k2 c2

2k1 + k2 = 4 …… (1) By substituting x = 4 , y = 3 and z = 11 into the equation, we have 11 = k1 ( 4) + k 2 (3)

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 ∵

4=

4.

= −50% z is decreased by 50%.

p. 269

∵

∴

Percentage change in z 0.5z 0 − z 0 = × 100% z0

3.

(a)

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 + k 2 = 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

5

10 Variations By substituting k1 = 4 into (1), we have

1 x 4 x = 12 3=

4 + k2 = 2 k 2 = −2 ∴

P = 4x −

When y = 6, 1 6= x 4 x = 24

2 x

(b) When x = 4 , 2 P = 4( 4) − 4 31 = 2 4.

(a)

∵

When x = 40, 1 y = (40) 4 = 10 When x = 60, 1 y = (60) 4 = 15

C is partly constant and partly varies directly as n. C = k 1 + k 2 n, where k 1 , k 2 ≠ 0

∴ By substituting n = 10 and C = 1750 into the equation, we have 1750 = k1 + k 2 (10)

(b) ∵ ∴

k1 + 10k2 = 1750

…… (1) By substituting n = 25 and C = 3625 into the equation, we have 3625 = k1 + k 2 ( 25)

By substituting x = 3 and y = 6.6 into the equation, we have 6.6 = k × 3 k = 2.2 ∴ y = 2.2 x y When = 2.2, 2.2 = 2.2 x x =1

k1 + 25k 2 = 3625 …… (2) ( 2) − (1), 15k 2 = 1875 k 2 = 125 By substituting k2 = 125 into (1), we have k1 + 10(125) = 1750 k1 = 500

∴

When x = 6, y = 2.2(6) = 13.2

C = 500 + 125n

(b) When n = 20 , C = 500 + 125( 20) = 3000 The cost per head

When y = 17.6, 17.6 = 2.2 x x = 8

3000 20 = $150 =$

Exercise

When y = 19.8, 19.8 = 2.2 x x=9 2.

(a)

(i)

Exercise 10A (p. 245) Level 1 1.

(a)

∵ ∴

y varies directly as x. 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 When y = 3,

y varies directly as x. y = kx , k ≠ 0

∵

y varies directly as x.

∴

20 30 Variation constant 2 = 3 =

(ii) The equation connecting x and y is 2 y= x. 3 (b) (i)

∵ ∴

y varies directly as x. −20 Variation constant = 10 = −2

(ii) The equation connecting x and y is y = −2 x .

6

Certificate Mathematics in Action Full Solutions 4B 3.

(b) When V = 32 , 1 32 = d 3 2 3 d = 64 ∴ d =4

(a)

6.

(a)

∵

y∝ x

∴

y = k x, k ≠ 0

By substituting x = 16 and y = 2

2 into the 3

equation, we have 2 2 = k 16 3 8 = k ( 4) 3 2 ∴ k= 3 2 y= x 3 (b) When x = 9 , 2 y= 9 3 2 = (3) 3 =2

(b) From the graph, when x = 8 , y = 12 4.

(a)

∵

A varies directly as l 2.

∴

A = kl 2 , k ≠ 0

7.

∴

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 (a)

∵

V varies directly as d 3 .

∴

V = kd 3 , k ≠ 0

By substituting d = 5 and V = 62 equation, we have 1 62 = k (5)3 2 125 = k (125) 2 1 ∴ k= 2 1 3 V = d 2

7

( y − 1) ∝ 3 x

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

5.

∵

y −1 7 −1 = 32 3 27 64 y2 − 1 6 = 3 4 8 = y2 − 1 y2 = 9

1 into the 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

10 Variations

New value of

y = k (0.25 x 0 ) 2 = 0.0625kx 0

2

= 0.0625 y 0 ∴

Percentage change in y 0.0625 y 0 − y 0 = × 100% y0 =

∴

(0.0625 − 1) y 0 × 100% y0

= −93.75% y is decreased by 93.75%.

8

Certificate Mathematics in Action Full Solutions 4B 9.

∵ ∴

S ∝ t S = kt , k ≠ 0

3 × 60 = 45 and S = 45 into 4 the equation, we have By substituting t =

45 = k ( 45) k =1 ∴ S = t When t = 25 , S = 25 ∴ The train travels 25 km in 25 minutes. 10. ∵ ∴

C ∝ x2 C = kx 2 , k ≠ 0

Level 2 13. (a)

∴

C = 6x When x = 3 ,

(c)

14. (a)

2

∵

( y − 1) ∝ x 3

∴

y − 1 = kx 3 , k ≠ 0

By substituting k = 1 and x = 1 into the equation, we have y − 1 = (1)(1) 3 y = 2 By substituting k = 1 and x = 2 into the equation, we have y − 1 = (1)( 2) 3 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) 12. ∵ ∴ y + x +

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 = 3x , 3x = 5 x − 2 2x = 2 x = 1 ∵

y 2 ∝ ( x − b)

∴

y 2 = k ( x − b), k ≠ 0

By substituting x = 3 and y = 2 into the equation, we have 2 2 = k ( 3 − b) ...... (1) 4 = k ( 3 − b) By substituting x = 9 and y = 4 into the equation, we have 4 2 = k (9 − b) ...... (2) 16 = k (9 − b)

C = 6(3) 2 = 54 ∴ The cost of painting a cube of side 3 cm is $54. 11.

( y + 2) ∝ x y + 2 = kx , k ≠ 0

By substituting x = 3 and y = 13 into the equation, we have 13 + 2 = k (3) k = 5 ∴ y + 2 = 5x y = 5x − 2 i.e. (b) When x = 2 , y = 5( 2) − 2 = 8

By substituting x = 4 and C = 96 into the equation, we have 96 = k ( 4) 2 k = 6

∵ ∴

k ( 9 − b) 16 = 4 k (3 − b) 9 −b 4 = (2) ÷ (1), 3− b 12 − 4b = 9 − b 3b = 3 b =1

(b) From (1), ∴ (c)

y2

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 = t and y = 2 into the equation, we have ...... (1) 2 = kt 2

9

10 Variations By substituting x = t + 1 and y = 8 into the equation, we have ...... (2) 8 = k (t + 1) 2 k (t + 1) 2 8 = 2 kt 2 (t + 1) 2 4 = t2 2 2 4t = t + 2t + 1

(2) ÷ (1),

∴

18. (a)

or t = 1

(b) When t = 1 , 2 from (1), 2 = k (1) k = 2 ∴ y = 2x 2

16. ∵ 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)

∵ t ∝ d ∴ t = kd , k ≠ 0 By substituting d = 5 and t = 15 into the equation, we have 15 = k (5) k = 3 ∴ t = 3d 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

∴

V = 800(3.5) 3 = 34 300 The value of a diamond weighing 3.5 carats is $34 300.

(b) Original value = $6400 New value = $[2 × 800(1)] = $1600 $(6400 − 1600) × 100% Percentage loss = $6400 = 75% 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 1000 = k ( 200) k = 5 ∴ E = 5W (c)

(b) (i)

(c)

V = kW 3 , k ≠ 0

By substituting W = 2 and V = 6400 into the equation, we have 6400 = k ( 2) 3 k = 800 ∴ V = 800W 3 When W = 3.5 ,

3t − 2t − 1 = 0 (3t + 1)( t − 1) = 0 1 3

Let $V be the value of the diamond and W carats be the weight of the diamond. ∵ V ∝ W3 ∴

2

t = −

1 3 km when the time between the thunder and lightning is only 1 second. The distance apart from the lightning is

20. (a)

When W = 70 , E = 5(70) = 350 ∴ If Sally has eaten 70 g of potato chips, she has taken 350 kcal. ∵ ∴ ∵ ∴

V V m m

∝ m = k 1 m, k 1 ≠ 0 ∝ t = k 2 t, k 2 ≠ 0

V = k1 m = k 1 (k 2 t ) = k1 k 2 t = k 3 t (where k 3 = k 1 k 2 ≠ 0) ∴

V varies directly as t.

10

Certificate Mathematics in Action Full Solutions 4B

(b)

When x = 28, 56 y = 28 = 2

V + m = k1 k 2 t + k 2 t = ( k 1 k 2 + k 2 )t = k 4 t (where k 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

When x = 56, 56 y = 56 = 1

m ∝ (t + V ) m = k 2 (t + V ), where k 2 ≠ 0 V = k 1 (m + t )

(b) ∵

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 y = x When x = 2,

= 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

V varies directly as t.

60 2 = 30

m = k 2 (t + V )

(b)

y =

m = k 2 [t + k 1 ( m + t )] m = k 2 t + k1 k 2 m + k1k 2 t

When y = 15, 60 15 = x x = 4

(1 − k 1 k 2 )m = ( k 2 + k 1 k 2 ) t k + k1 k 2 m =  2  1 − k1 k 2

t  

m = k 4 t (where k 4 = ∴

k 2 + k1 k 2 ≠ 0) 1 − k1 k 2

When x = 6, 60 y = 6 = 10

m varies directly as t.

Exercise 10B (p. 254) Level 1 1.

(a)

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 y = x When x = 2,

When y = 6, 60 6 = x x = 10

∵

56 2 = 28

When y = 4, 60 4 = x x = 15 2.

(a)

(i)

∵ ∴

y =

When x = 7, 56 y = 7 = 8 When x = 8, 56 y = 8 = 7

11

y varies inversely as x. = 30 × 12 Variation constant = 360

(ii) The equation connecting x and y is 360 y = . x (b) (i)

∵ ∴

y varies inversely as x. = 5×5 Variation constant = 25

(ii) The equation connecting x and y is 25 y = . x

10 Variations 3.

(a)

6.

(a)

(b) From the graph, when x = 24 , y = 7.5 4.

(a)

∵ ∴

F varies inversely as d 2 . k F = 2 , k ≠ 0 d

(b) When y = x 2 ,

x4

x 4 + 2x 2 − 3 = 0 (x

∵ ∴

equation, we have 1 k = 3 6 27 1 k = ∴ 2 1 y = 3 2 x 1 , 8 1 y = 1 3 2 8 1 = 1 2   2 = 1

(b) When x =

− 1)( x 2 + 3) = 0

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 .

∴

y varies inversely as 3 x . k y = 3 , k ≠ 0 x

By substituting x = 27 and y =

2

x 2 = 1 or x 2 = −3 (rejected) x = ±1 By substituting x = 1 into y = x 2 , we have

(b) When d = 2 , 4 F = 2 2 =1 (a)

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 ∴ 4 F = 2 d

5.

( 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 k 25 + 2 = 2 1    3 27 = 9k k = 3 3 y + 2 = 2 ∴ x 3 y = 2 − 2 i.e. x ∵

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 = 0.8 x 0 ∵

y ∝

y = New value of

∴

 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%.

12

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 , k ≠ 0 ∴ h = A By substituting A = 2 and h = 30 into the equation, we have k 2 k = 60 60 h = A

30 = ∴

When h = 15 , 60 15 = A A = 4 ∴

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 By substituting x = 4 and y = 25 into the equation, we have ∵

y ∝

k 4 + a ...... (1) k 5 = 4 + a By substituting x = 8 and y = 625 into the equation, we have k 625 = 8+ a k ..... (2) 25 = 8+ a 25 =

k 25 8 + a = 5 k 4 + a 4 + a (2) ÷ (1), 5 = 8+ a 40 + 5a = 4 + a 4a = −36 a = −9 k (b) From (1), 5 = 4 + ( −9) k = −25 25 y = − ∴ x − 9

13

10 Variations (c)

10. (a)

When y = 4 , 25 4 = − x − 9 2 x − 18 = −25 2 x = −7 7 x = − 2 Let $V be the value of the flat and A years old be the age of the flat. 1 ∵ V ∝ A k , k ≠ 0 ∴ V = A By substituting A = 4 and V = 1 500 000 into the equation, we have k 1 500 000 = 4 k = 6 000 000 ∴ 6 000 000 V = 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 ∴

The value of the flat in 2004 is $750 000.

(b) When V = 2 000 000 , 6 000 000 2 000 000 = A A = 3 ∴ 11.

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 = 0.9V 0 ∵

V ∝

= New pressure

k 0.9V 0

10  k    9  V0  10 = P 9 0 =

14

Certificate Mathematics in Action Full Solutions 4B ∴

Percentage change in pressure 10 P0 − P0 = 9 × 100% P0  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 ∴ 100 p = n ∵

p ∝

(b) When p = 4 , 100 4 = n n = 25 n = 625 ∴ 625 bottles are available on the island when the retail price is $4. (c)

13. (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 ∵

V ∝

k 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. 265 =

15

10 Variations (b) 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. 14. Let n be the number of workers and T be the number of months needed to complete the job. 1 ∵ n ∝ T k , k ≠ 0 ∴ n = T By substituting T = 21 and n = 56 into the equation, we have k 21 k = 1176 ∴ 1176 n = T When T = 7 ,

16. (a)

The length of the rectangle and the width of the rectangle are in inverse variation. 1 (b) ∵ l ∝ w k , k ≠ 0 ∴ l = w By substituting w = 20 and l = 45 into the equation, we have

∴

(c)

56 =

1176 7 = 168 168 − 56 = 112 An additional 112 workers has to be employed to complete the same job in 7 months. n =

∴

15. (a)

1 d k , k ≠ 0 ∴ n = 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.

k 20 k = 900 900 l = w

45 =

When w = 18 , 900 l = 18 = 50 ∴ 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 = 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)

∵

1 1 ( x + y ) ∝  +  y x k ,k ≠ 0 1 1 + x y kxy x + y = x + y

x + y = ∴

∴

( x + y ) 2 = kxy ( x + y ) 2 ∝ xy ( x + y ) 2 = kxy , k ≠ 0

(b)

x

2

+ 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 ∴

xy ∝ ( x 2 + y 2 )

16

Certificate Mathematics in Action Full Solutions 4B Exercise 10C (p. 261) Level 1 1.

(a)

12 =

∵ 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

(a)

5.

∵ 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

(a)

4.

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 (b) When b = 2 and c = 2, 3( 2) a= 2 2 3 = 2 (a)

(a)

∵

∵

2

y varies directly as u and inversely as ku

,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 ∴

y=

2

(b) When u = 4 and y = 12,

17

u3 v2 ku 3 ∴ 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∝

(b) When w = 3u 3 , u3 3u 3 = 12v 2 36v 2 = 1 1 v2 = 36 1 v=± 6

(b) When x = 5 and y = 4, 1 z = (5)(4) 2 4 = 20 3.

t

t =4 t = 16

(b) When m = 4 and n = 6, P = 8(4)(6) = 192 2.

3(4) 2

6.

∵

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 = 1.4d 0

New value of t = (1 − 20%)t 0 = 0.8t 0 New value of C = k (1.4d 0 ) 2 (0.8t 0 ) = 1.568kd 0 2 t 0 = 1.568C0

t .

1.568C 0 − C0

∴

Percentage change in C =

∴

= 56.8% C is increased by 56.8%.

C0

× 100%

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 ∵

W ∝

10 Variations k (4)(3) 22 k =5 5 pq W = 2 r

15 =

∴ (b)

8.

M 1M 2 d2 kM 1M 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 ∴ New value of M 2 = 0.5( M 2 ) 0

10. ∵

When W = 6, p = 3 and r = 5, 5(3)q 6= 52 q = 10

New value of d = 2d 0 New value of F =

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

∴

∴

1 w is decreased by 33 % . 3

(a)

∵

1 16 = 1 : 16 =

11.

(a)

V ∝ r 2h

2 ∴ V = kr 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 new value of V original value of V 0.5V0 = V0

The required ratio = ∴

= 0.0625 F0 new value of F The required ratio = original value of F 0.0625 F0 = F0

   

2 w0 − w0 w= 3 × 100% Percentage change in w0

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  

1 = −33 % 3

9.

F∝

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.

12. ∵

x∝

1 2 = 1: 2 =

18

Certificate Mathematics in Action Full Solutions 4B

∴

kV1 P1

 (1)

kV x2 = 2 P2

 (2)

x1 =

z=

 k2     x 

2

1

= k1 x 3 ×

x1 kV1 P2 = × x2 P1 kV2

x2 2 k2 7

x1 V1 P2 (1) ÷ (2), x = V × P 2 2 1

= ∵

9 9 13 = × x2 13 12

∴

x2 = 12 ∴

k1 3 x

The time required to boil the second kettle of water is 12 min.

k1 (x 3 ) 2 k2

k1 and k2 are non-zero constants. k1 2 is also a non-zero constant. k2

By letting k 3 =

k1 2 , we have k2

7

y 13. ∵ z ∝ x ky ∴ z = 1 , where k1 is a non-zero constant … (1) x 1 ∵ y∝ x k2 ∴ y= , where k2 is a non-zero constant. x k2 i.e. x = , where k2 is a non-zero constant … (2) y By substituting (2) into (1), we have ky z= 1 k2 y k1 2 = (y ) k2 ∵ ∴

z = k3 x 3 , 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 4 = k1 + k 2 ( 2) 2 k1 + 4k2 = 4

By substituting x = 3 and z = equation, we have 13 = k1 + k 2 (3) 2 2 13 k1 + 9k 2 = 2 5 5k 2 = 2 (2) – (1), 1 k2 = 2

k1 and k2 are non-zero constants. k1 is also a non-zero constant. k2

k1 By letting k3 = , we have k2 z = k3 y 2 , where k3 is a non-zero constant. i.e.

 (1)

z ∝ y2

By substituting k2 = 14. ∵ z ∝ ∴ z=

3

x y2

∴

1 ∵ x∝ y

1 into (1), we have 2

z = 2+

1 2 x 2

(b) When x = 4, 1 z = 2 + ( 4) 2 2 = 10

k2 ∴ x= , where k2 is a non-zero constant. y i.e. y =

19

 (2)

1 k1 + 4  = 4 2 k1 = 2

k1 3 x , where k1 is a non-zero constant … (1) y2

k2 , where k2 is a non-zero constant … (2) x By substituting (2) into (1), we have

13 into the 2

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 ∵

10 Variations square of c.

56 = k1 (6) + k 2 (6) 3

k2 , where k1, k2 ≠ 0 c2 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)

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 + 1728k 2 = 436 3k1 + 432k 2 = 109 324k2 = 81 (2) – (1), 1 k2 = 4 By substituting k2 =

 (2)

By substituting b = 4, c = 2 and a = equation, we have 5 k = k1 4 + 22 2 2 8k1 + k 2 = 10

1 into (1), we have 4

∵ ∴

z is partly constant and partly varies inversely as the square root of y. k z = k1 + 2 , where k1, k2 ≠ 0 y

∴

5.

4.

(a)

∵

(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 + 450k2 = 845

k1 = 350 ∴

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.

y

a partly varies directly as the square root of b and partly varies inversely as the

 (2)

(2) – (1), 150k 2 = 165 k 2 = 1.1 By substituting k2 = 1.1 into (1), we have k1 + 300(1.1) = 680

2

(b) When y = 25, 2 z = 6+ 25 32 = 5

2 c2

k1 + 300k 2 = 680  (1) By substituting x = 450 and E = 845 into the equation, we have 845 = k1 + k2 (450)

k2 = 2 z = 6+

a= b+

(b) When b = 16 and c = 3, 2 a = 16 + 2 3 38 = 9

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

∴

 (2)

k2 = 2

(b) When s = 2, 1 1 w = (2) + (2)3 3 4 8 = 3 (a)

5 into the 2

(2) – (1), 7 k1 = 7 k1 = 1 By substituting k1 = 1 into (1), we have 1 + k2 = 3

1 3k1 + 108  = 28 4 1 k1 = 3 1 1 3 ∴ w= s+ s 3 4

3.

a = k1 b +

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

20

Certificate Mathematics in Action Full Solutions 4B equation, we have 2000 = k1 + k 2 ( 40)

2 = 5 + k2 (−1) + k3 (−1) 2 k 2 − k3 = 3  (1) By substituting x = 2 and y = 5 into the equation, we have 5 = 5 + k 2 ( 2) + k3 (2) 2

k1 + 40k 2 = 2000  (1) By substituting n = 38 and C = 1940 into the equation, we have 1940 = k1 + k 2 (38) k1 + 38k2 = 1940

2 k 2 + 4 k3 = 0

 (2)

k 2 + 2 k3 = 0

(1) – (2), 2k 2 = 60 k 2 = 30

(2) – (1),

7.

(a)

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 2000k1 + k 2 = 450 000 …… (1)

(2) – (1), 3000k1 = 600 000 k1 = 200 By substituting k1 = 200 into (1), we have 2000(200) + k2 = 450 000 k2 = 50 000 ∴

C = 200 +

50 000 n

(b) When n = 8000, 50 000 C = 200 + 8000 = 206.25 ∴ The cost of producing a copy of the video game is $206.25. Level 2 8. (a)

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

21

k2 = 2 y = 5 + 2x − x2

∴ (b)

When y = –3, − 3 = 5 + 2x − x2 x2 − 2x − 8 = 0 ( x + 2)( x − 4) = 0 x = − 2 or x = 4

∵

By substituting n = 5000 and C = 210 into the equation, we have k 210 = k1 + 2 5000 5000k1 + k2 = 1 050 000  (2)

3k3 = −3

k3 = −1 By substituting k3 = –1 into (1), we have k2 − (−1) = 3

By substituting k2 = 30 into (2), we have k1 + 38(30) = 1940 k1 = 800 ∴ C = 800 + 30n When n = 43, C = 800 + 30( 43) = 2090 ∴ The cost for a class of 43 students is $2090.

 (2)

y = 5 + 2x − x2 = −( x 2 − 2 x − 5) (c)

= −( x 2 − 2 x + 12 − 12 − 5) = −( x 2 − 2 x + 1) + 6 = 6 − ( x − 1) 2 The maximum value of y is 6.

∴ 9. (a)

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 + 400k 2 = 200 000 …… (1) By substituting l = 1.5, w = 10 and C = 150 000 into the equation, we have 150 000 = k1 (1.5) + k 2 (10) 2 150 000 = 1.5k1 + 100k 2 200 …… (2) k 2 = 100 000 3 1000 (1) – (2), 3 k 2 = 100 000 k 2 = 300 By substituting k2 = 300 into (2), we have 200 k1 + (300) = 100 000 3 k1 = 80 000 ∴ C = 80 000l + 300w2 k1 +

(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

10 Variations 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 000k2 = 7000  (1) By substituting A = 400 000 and S = 9000 into the equation, we have 9000 = k1 + k 2 ( 400 000) k1 + 400 000k2 = 9000 (2) – (1),

 (2)

100 000k 2 = 2000

1 50 1 By substituting k2 = into (1), we have 50  1  k1 + 300 000  = 7000  50  k1 = 1000 k2 =

∴

S = 1000 +

1 A 50

(b) When A = 350 000, 1 S = 1000 + (350 000) 50 = 8000 ∴ Her income for that month is $8000. (c)

When S = 20 000, 20 000 = 1000 +

1 A 50

1 A 50 A = 950 000 The amount of her sales should be $950 000. 19 000 =

∴

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 1000k1 + k 2 = 53 000 000 …… (1) ∵

By substituting n = 2000 and C = 58 000 into the equation, we have k 58 000 = k1 + 2 2000 2000k1 + k 2 = 116 000 000 …… (2) (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

22

Certificate Mathematics in Action Full Solutions 4B

∴

C = 63 000 −

(b)

10 000 000 n

(b) When n = 4000, C = 63 000 −

10 000 000 4000

P = 800 x − 5 x 2

= 60 500 ∴ 12. (a)

11.3 − L0 = k (3.5)

 (2)

11.3 − L0 ( 2) ÷ (1), = 1.75 10.1 − L0

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.

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 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.

(b) When y =

2.

∴

23

P = 800x – 5x2

( 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 y+2=3 x

i.e.

k1 + 160k2 = 0  (1) By substituting x = 80 and P = 32 000 into the equation, we have 32 000 = k1 (80) + k 2 (80) 2

k1 = 800

(a) ∵ ∴

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 =3 x −2

(b) When y = 10 , 10 = 3 x − 2 3 x = 12

 (2)

(1) – (2), 80k2 = −400 k2 = −5 By substituting k2 = –5 into (1), we have k1 + 160( −5) = 0

4 , 25

4 = 4x2 25 1 x2 = 25 1 x=± 5

∵

k1 + 80k 2 = 400

y = kx 2 , k ≠ 0

By substituting x = 3 and y = 36 into the equation, we have 36 = k (3) 2 k=4 ∴ y = 4x2

L0 = 8.5

13. (a)

(a) ∵ y varies directly as x2. ∴

0.75 L0 = 6.375

(c)

= −5( x 2 − 160 x + 802 − 802 ) = 32 000 − 5( x − 80) 2

11.3 − L0 = 17.675 − 1.75 L0

∴

= −5( x 2 − 160 x)

(c)

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)

When x = 20, P = 800(20) − 5( 20) 2 = 14 000

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 b = 32

k 4 k = 12 12 y= x 3=

∴

(b) When y = 5 , 12 5= x 12 x= 5 4.

7.

(a)

∴

By substituting x = 5 , y = 2 and z = 100 into the equation, we have 100 = k (5) 2 (2) k=2 ∴ 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

w = k 1 x + k 2 y , where k 1 , k 2 ≠ 0

3k1 + 4k 2 = 18

...... (2)

(1) × 3 − (2) × 2, k 2 = 3

By substituting k2 = 3 into (1), we have 2k1 + 3(3) = 13 k1 = 2 ∴ (b)

w = 2 x +3 y

When x = 16 and y = 25 , w = 2 16 + 3 25 = 23

8. z = kx 2 y, k ≠ 0

y.

2k1 + 3k 2 = 13 ...... (1) By substituting x = 9 , y = 16 and w = 18 into the equation, we have 18 = k1 9 + k 2 16

(b) When x = 8 , 138 8= 2y + 3 16 y + 24 = 138 16 y = 114 57 y= 8

∴

x and partly

By substituting x = 4 , y = 9 and w = 13 into the equation, we have 13 = k1 4 + k 2 9

(a) ∵ x varies inversely as ( 2 y + 3) . k ,k≠0 ∴ x= 2y + 3

(a) ∵ z varies jointly as x2 and y.

w partly varies directly as varies directly as

By substituting x = 6 and y = 10 into the equation, we have k 6= 2(10) + 3 k = 138 ∴ 138 x= 2y + 3

5.

∵

(a)

∵ y is partly constant and partly varies inversely as x. k ∴ y = k1 + 2 , where k1 , k 2 ≠ 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 + k 2 = 260 (1) − (2), 9k1 = 2250 k1 = 250 By substituting k1 = 250 into (2), we have 250 + k2 = 260 k2 = 10 ∴

y = 250 +

10 x

24

Certificate Mathematics in Action Full Solutions 4B (b) When x = 5 , y = 250 +

z=

10 5

New value of

 kx 2  = 1.6 0   y   0  = 1.6 z0

= 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 y0 1.3 y0 − y0 × 100% y0

= 30% ∴ y is increased by 30%.

y=

k (1.5 x0 ) 2

=

4  k  9  x0 2 

=

4 y0 9

5 = −55 % 9 5 ∴ y is decreased by 55 % . 9 z∝ z=

New value of

25

12. ∵

z∝

y = (1 − 10%) y0 = 0.9 y0

x2 y

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 z=

z = New value of

k (2 x 0 ) 2 0.5 y 0

 kx = 8 0  y0 = 8z0

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 = kd 2

2

...... (2) 2

C1 kd = 12 C2 kd 2

x2 y

kx 2 ,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 = 1.2 x0 ∴

= 60% ∴ z is increased by 60%.

∴

4 y0 − y0 9 y = × 100% ∴ Percentage change in y0

11. ∵

1.6 z0 − z0 × 100% z0

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 = 1.5 x0

New value of

∴ Percentage change in z =

∴

∴ Percentage change in y =

k (1.2 x0 ) 2 (0.9 y0 )

(1) ÷(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 + k2 (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),

32 5 32 By substituting k 2 = into (1), we have 5  32  k1 + 150  = 3750  5  k1 = 2790 32 n ∴ C = 2790 + 5

Let T0 and d0 be the original values of T and d respectively. ∴ New value of d = 2d 0 T = New value of

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

...... (2) 2

C1 kT d = 1 12 C2 kT2 d 2 C1 T1  d1  = ×  C2 T2  d 2  16 3  4  (1) ÷(2), 27 = 4 ×  d   2 16 64 = 2 81 d2

2

2

81 × 16 64 = 20.25 d 2 = 4.5 or − 4.5 (rejected) 2

d2 =

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 y= 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) 18.

(a) ∵

∴ (b) ∵

∴ The diameter of the second coin is 4.5 cm. ∴ 16. (a)

∵

C is partly constant and partly varies directly as n. C = k 1 + 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)

32 (200) 5

= 4070 ∴ The cost of making 200 coats is $4070.

= −75% ∴ The time taken is decreased by 75%.

∴

k2 =

(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)

600k2 = 3840

(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 = 40 N and it is in direct variation.

k1 + 150k2 = 3750

...... (1) By substituting n = 750 and C = 7590 into the equation, we have

26

Certificate Mathematics in Action Full Solutions 4B Level 2 19. (a)

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)

(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 , 4 2x − 1 2 x2 − x − 1 = 0 ( 2 x + 1)( x − 1) = 0 1 x=− or x = 1 2

k (5a − 3) 6 = 3 k ( 4a − 3) 5a − 3 2 = (2) ÷(1), 4a − 3 8a − 6 = 5a − 3 3a = 3 a =1 (b)

(c)

4x =

By substituting a = 1 into (1), we have 3 = k [4(1) − 3] k =3 y = 3( x − 3) ∴

21. (a)

∵

w∝

∴

w=

∵ ∴

1 y∝ ax − 1 k y= ,k ≠0 ax − 1

By substituting x = 2 and y =

equation, we have 4 k = 7 a ( 4) − 1 4 k ...... (2) = 7 4a − 1 4 k 3 = 2a − 1 4 k 7 4a − 1 (1) ÷(2), 7 4a − 1 = 3 2a − 1 14a − 7 = 12a − 3 2a = 4 a=2

27

kx 2 y z3

,k≠0

4 into the 3

w=

2x2 y z3

(b) When y = 2w = 2 x = 2 z , i.e. y = 2 x , w = x and z = x , 2x2 y w= z3 x=

equation, we have 4 k = 3 a ( 2) − 1 4 k ...... (1) = 3 2a − 1 By substituting x = 4 and y =

z3

By substituting x = 1 , y = 1 , z = 1 and w = 2 into the equation, we have k (1) 2 1 2 = 13 k = 2

y = 3( x − 3) = 3x − 9 When 2 y = 3 x , y = 2y − 9 y=9 ∵

∴ 20. (a)

x2 y

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

10 Variations 5 = k1 + k 2 (0) + k3 (0) 2 k1 = 5 By substituting x = 2 and y = 15 into the equation, we have 15 = 5 + k2 ( 2) + k3 (2) 2 k 2 + 2 k3 = 5

Let r0 and h0 be the original values of r and h respectively. h = (1 − 19%) h0 ∴ New value of = 0.81h0 r=

...... New value of

(1) By substituting x = −2 and y = 11 into the equation, we have 11 = 5 + k2 (−2) + k3 (−2) 2 k2 − 2k3 = −3 (1) − (2),

...... (2)

4k3 = 8

∴

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 (27 + 64 + 125) cm3

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

3

xL = 216 xL = 6

24. ∵

∴

=

10 r0 9

10 r0 − r0 r= 9 × 100% Percentage change in r0 1 = 11 % 9 The base radius of the cone is increased by 1 11 %. 9 E ∝ mv 2 E = kmv 2 , k ≠ 0

New value of v = (1 − 20%)v0 = 0.8v0 New value of

E = k ( 2m0 )(0.8v0 ) 2 = 1.28 km0v0

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

= 1.28 E0 ∴ ∴

= 216k cm3

∴

10  k ′    9  h0 

V = kx 3 , k ≠ 0

Total volume of the three metal cubes = [k (3)3 + k ( 4)3 + k (5)3 ] cm3

∴

=

Let m0, v0 and E0 be the original values of m, v and E respectively. ∴ New value of m = 2m0

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 ∴

k′ 0.81h0

26. (a)

Percentage change in E =

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 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

28

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

V = New value of

∴

2V0 − V0 × 100% Percentage change in V = V 0

∴

V is increased by 100%.

∵

V =

(c)

= 100%

(c)

1 2 r h 3

h = h0

r= New value of

3(1.36V0 ) h0

 3V0 = 1.36   h0 

∴    

= 1.36 r0

28. (a)

1.36 r0 − r0 × 100% r0

∴

Percentage change in r =

∴

r is increased by 16.6%.

∵

T is partly constant and partly varies directly as n. T = k 1 + k 2 n, where k 1 , k 2 ≠ 0

= 16.6% (cor. to 1 d.p.)

k1 + 400k 2 = 110

...... (1) By substituting n = 600 and T = 150 into the equation, we have 150 = k1 + k2 (600) k1 + 600k 2 = 150

By substituting k1 = 150 into (1), we have 200(150) + k 2 = 50 000

( 2) − (1), 200k 2 = 40

29

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 ∵

(2) − (1), 200k1 = 30 000 k1 = 150

...... (2)

k 2 = 20 000

1 5

1 By substituting k 2 = into (1), we have 5 1   k1 + 400  = 110 5 k1 = 30

The airtime used in this month is (150 + 2n0 ) minutes.

By substituting n = 400 and C = 200 into the equation, we have k 200 = k1 + 2 400 400k 1 + k 2 = 80 000 ...... (2)

∴ By substituting n = 400 and T = 110 into the equation, we have 110 = k1 + k 2 (400)

k2 =

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

= 1.36V0

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

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 ∴

T = 30 +

∴ (b)

C = 150 +

20 000 n

When n = 500 , C = 150 + = 190

20 000 500

10 Variations (2) − (1), k2 = −1 By substituting k 2 = −1 into (1), we have k1 + 4( −1) = 2

25 000 Profit of a rice cooker = $ 500 = $50 ∴ The selling price of each rice cooker = $(190 + 50) = $240 29. (a)

∵ ∴

P is partly varies directly as x and partly varies directly as x2. P = k1 x + k 2 x 2 , where k1 , k 2 ≠ 0

By substituting x = 20 and P = 60 000 into the equation, we have 60 000 = k1 ( 20) + k2 ( 20) 2 k1 + 20k 2 = 3000

k1 = 6 ∴ (b)

(2) − (1),

= −( 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.

By substituting k2 = −50 into (1), we have k1 + 20(−50) = 3000 k1 = 4000 ∴ (b)

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.

From the graph, when x = 120 , y = 6000 . y = 6000 x = 120 By substituting and 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.

= −50( x 2 − 80 x) = −50( x 2 − 80 x + 40 2 − 402 ) = −50( x 2 − 80 x + 40 2 ) + 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 + k2 r 2 , where k1 , k 2 ≠ 0

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 ∵

(b)

P = 4000 x − 50 x 2 (c)

= − ( r 2 − 6r + 3 2 − 3 2 ) = − ( r 2 − 6r + 3 2 ) + 9

31. (a)

k 2 = −50

= − ( r 2 − 6r )

(c)

...... (2)

10k 2 = −500

When r = 2 , C = 6(2) − (2) 2 =8 C = 6r − r 2

...... (1) By substituting x = 30 and P = 75 000 into the equation, we have 75 000 = k1 (30) + k2 (30) 2 k1 + 30k 2 = 2500

32. (a)

∵

( x2 − y 2 ) ∝ (x2 + y 2 )

∴

x 2 − y 2 = k ( x 2 + y 2 ), k ≠ 0 x 2 − y 2 = kx 2 + ky 2 ( k + 1) y 2 = ( k − 1) x 2 k −1 2 y2 = x k +1

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 = k 1 (5) + k 2 (5) 2 k 1 + 5k 2 = 1

...... (2)

C = 6r − r 2

y= ∵ ∴

k −1 x k +1

k≠0 k −1 ≠0 k +1

30

Certificate Mathematics in Action Full Solutions 4B k −1 , we have k +1 y = k ′x, k ′ ≠ 0 y∝x

By letting k ′ = i.e. (b)

∵ ∴

3 (b) x −

 1 1  = k  x 3 + 3 , k ≠ 0 3 y y    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  2 x 1  y y x + + 2  y y  

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 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 = x + y

33. (a)

i.e.

( 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 (1 − k ) x3 = 3 y k +1 1 3 y = × 1 − k x3 x3 −

y=3 By letting k ′ = 3

 x 2 y 2 − 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   k ′2 − 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− ∴

1.

k +1 1 × 1− k x

k +1 , we have 1− k

k′ , k′ ≠ 0 x 1 y∝ x

Answer: C ∵ x varies directly as y2. ∴ x = ky 2 , k ≠ 0 y=2 By substituting and x = 20 into the equation, we have 20 = k ( 2) 2 k =5 ∴ x = 5y2 When x = 80 , 80 = 5 y 2 y 2 = 16 y =±4

2.

Answer: A ∵ (3 x + 5 y ) ∝ ( 4 x − 6 y ) 3 x + 5 y = k (4 x − 6 y ), k ≠ 0 ∴ (5 + 6k ) y = ( 4k − 3) x 4k − 3  y =  x  5 + 6k  4k − 3 k′ = By letting 5 + 6k , we have i.e.

31

 1  1  x −  ∝  x +  y  y 

Multiple Choice Questions (p. 278)

y= ∴

 1 1 = k ′′ x + , k ′′ ≠ 0 y y 

y = k ′x, k ′ ≠ 0 y∝x

10 Variations 3.

Answer: B ∵ y varies inversely as ( x + 1) 2 . k y= , k≠0 ∴ ( x + 1) 2

By substituting x = 5 and y = −115 into the equation, we have

1 y= By substituting x = 2 and 3 into the equation, we have 1 k = 3 ( 2 + 1) 2 1 k = 3 9 ∴ k =3 3 y= ( x + 1) 2 When y = 12 , 3 12 = ( x + 1) 2

(2) – (1), 16k 2 = −16 k 2 = −1

− 115 = k1 (5) + k 2 (5)3 k1 + 25k 2 = −23

…… (2)

By substituting k 2 = −1 into (1), we have k1 + 9(−1) = −7 k1 = 2 ∴ 8.

4( x + 1) 2 = 1 4x + 8x + 3 = 0 ( 2 x + 1)(2 x + 3) = 0 2

y = 2 x − x3

Answer: A ∵ z varies directly as x2 and inversely as y. kx 2 ,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 New value of

1 3 x=− or x = − 2 2

y = (1 + 30%) y0 = 1.3 y0 z =

4.

Answer: D ∵ ( x + y ) varies inversely as ( x − y ) . k x+ y = , k≠0 ∴ x− y x− y = ∴

5.

New value of

k , k≠0 x+ y

∴

Answer: D ∵ z ∝ ( y + 2) and y ∝ ( x − 1) ∴ z = k1 ( y + 2) and y = k 2 ( x − 1) , where k1 , k 2 ≠ 0

z = k1 ( k2 x − k2 + 2) = k1k2 x − k1k2 + 2k1 7.

  

1.3z0 − z0 × 100% z0

9.

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

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

2

= 30%

xy =k , where k is a non-zero constant z2 6.

 kx = 1.3 0  y0 = 1.3 z 0

Percentage increase in z =

( x − y ) varies inversely as ( x + y ) .

Answer: B ∵ x varies inversely as y and directly as z2. kz 2 x= ∴ , where k is a non-zero constant y

k (1.3x 0 ) 2 1.3 y 0

p. 244 1.

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.

2.

Yes

…… (1)

32