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10

138

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Key Concepts and Formulae

Variations

10A

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10.2 Direct Variation

If x and y are in direct variation, then y kx= , where k is a non-zero constant.

x 9 11 13 15

y 4 12 20 28 36

x 8 12 20 28

y 1.6 6.4 8 9.6 12.8

1. Given that y varies directly as x, complete the tables below.

(a)

(b)

2. If A varies directly as l2 and A = 100 when l = 5, find

(a) an equation connecting A and l,

(b) the value of A when l = 7 .

Solution

(a) Q A varies directly as l2. (b) When l = 7 ,

∴ A k= ( ), k ≠ 0 A ==

( )( )

)(By substituting l = ( ) and

A = ( ) into the equation,

we have

1 3 5 7

44 52 60

3.2 4.8

4 16 24 32

11.2

100 5

4

4

2

2

=

=

=

k

k

A

( )

l∴

l2

5

100

4

28

( )7 2

139

10 Variations

3. If V varies directly as r3 and V = 48when r = 4, find

(a) an equation connecting V and r,

(b) the value of r when V = 6.

Solution

(a)

(b)

4. If y x∝ and y = 21 when x = 81,find

(a) an equation connecting x and y,

(b) the value of y when x = 16.

Solution

(a)

(b)

Q y varies directly as x .

∴ y k x= , k ≠ 0

By substituting x = 81 and y = 21

into the equation, we have

21 81

7

3

=

=

k

k

∴ y x= 7

3

When x = 16 ,

y =

=

7

3

28

3

16

When V = 6,

6

8

2

3

4

24

3

3

3

=

=

=

=

r

r

r∴

Q V varies directly as r 3.

∴ V kr= 3 , k ≠ 0

By substituting r = 4 and V = 48


48 4 3

3

3

4

3

4

=

=

=

k

k

V r

( )

∴

140

Number and AlgebraNumber and Algebra

5. If ( )y + 1 varies directly as x3 andy = 14 when x = 27, find the value of

y when x = 64 without finding thevariation constant.

Solution

6. It is given that y x∝ 2 . If x is increasedby 10%, find the percentage change in y.

Solution

Q ( )y + 1 varies directly as x3 .

∴

y k x

y k x

1 13

2 23

1

1

+ =

+ =

……(1)

……(2)

From (1), we have

k

y

x= +1

13

1

From (2), we have

k

y

x= +2

23

1

∴

y

x

y

x

y

y

y

1

13

2

23

1 1

15

3

14 1

27

1

643 3

1 4

20

19

+ +=

=

+ = ×

=

=

+ +

∴

Q y x∝ 2

∴ y kx= 2, k ≠ 0

Let x0 and y0 be the original values of x

and y respectively.

∴ New value of x

= +

=

( %)

.

1 10

1 1

0

0

x

x

New value of y

=

=

=

k x

kx

y

( . )

.

.

1 1

1 21

1 21

02

02

0

∴ Percentage change in y

= ×

= ×

=

−

−

1 21

1 21 1

0 0

0

0

0

100

100

21

.

( . )

%

%

%

y y

y

y

y

∴ y is increased by 21%.

141

10 Variations

7. It is given that y + 1 varies directly as x.If y = 11 when x = 4, find


(b) the value of y when x = 7,

(c) the value of x when y x= 5 .

Solution

(a)

(b)

(c)

8. If y x∝ 2 and y = 24 when x t= ,y = 54 when x t= + 1, find

(a) the values of t,

(b) an equation connecting x and ywhen t is positive.

Solution

(a)

(b)

Q ( )y x+ ∝1

∴ y kx+ =1 , k ≠ 0

By substituting x = 4 and y = 11


11 1 4

3

+ =

=

k

k

( )

∴

y x

y x

+ =

= −

1 3

3 1

When x = 7,

y = −

=

3 7 1

20

( )

When y x= 5 ,

5 3 1

2 1

1

2

x x

x

x

= −

= −

= −

Q y x∝ 2

∴ y kx= 2, k ≠ 0

By substituting x t= , y = 24 into

the equation, we have

24 2= kt ……(1)

By substituting x t= + 1, y = 54


54 1 2= +k t( ) ……(2)

( )( )

:21

54

24

1

9

4

2 1

2

5

2

2

2

2

9 4 8 4

5 8 4 0

5 2 2 0

2 2

2

=

=

= + +

− − =

+ − =

= −

+

+ +

k t

kt

t t

t

t t t

t t

t t

t t

( )

( )( )

= 2∴ or

By substituting t = 2 into (1), we

have

24 2

6

6

2

2

=

=

=

k

k

y x

( )

∴

142


9. If y 2 varies directly as ( )x b+ and y = 6 when x = 13, y = 9 when x = 463

4, find

(a) the value of b,

(b) the relation between x and y,

(c) the value of x when y = 16 .

Solution

(a)

(b)

(c)

Q y x b2 ∝ +( )

∴ y k x b2 = +( ), k ≠ 0

By substituting y = 6 and x = 13 into the equation, we have

36 13= +k b( ) ……(1)

By substituting y = 9 and x = 46 3

4 into the equation, we have

81 46 3

4= +

k b ……(2)

( )

( ):2

1

81

36

1874

13

1053 81 1683 36

45 630

14

=

+ = +

=

=

+

+

k b

k b

b b

b

b

( )

By substituting b = 14 into (1), we have

36 13 14

14

4

3

4

32

= +

=

= +

k

k

y x

( )

( )∴

By substituting y = 16 into y x2 4

314= +( ) , we have

16 14

192 14

178

2 4

3= +

= +

=

( )x

x

x

∴ x = 178 when y = 16.

143

10 Variations

11. If v varies directly as ( )m t− , and mvaries directly as ( )t v+ , show that

(a) v varies directly as t,

(b) m varies directly as t.

Solution

(a)

(b)

10. The value (in $) of a piece of jade variesdirectly as the cube of its volume (incm3).

(a) If the value of a piece of jade ofvolume 2 cm3 is $6720, find thevalue of a piece of jade of volume3.5 cm3.

(b) If the piece of jade of volume 2 cm3

is broken into two parts of equalvolume, what would be thepercentage loss in its value?

Solution

(a)

(b)

Let $P and V cm3 be the value and

volume of the jade respectively.

P kV= 3 , k ≠ 0

By substituting P = 6720, V = 2


6720 2

840

3=

=

k

k

( )

∴ P V= 840 3

When V = 3 5. ,

P =

=

840 3 5

36 015

3( . )

∴ The value of the jade of volume

3.5 cm3 is $36 015.

The value of the two broken pieces

of jade with equal volume 1 cm3

= × ×

=

2 840 1

1680

3( )

$

The percentage loss in the value

= ×

=

−6720 1680

6720100

75

%

%

v k m t= −1( ), k1 0≠ ……(1)

m k t v= +2( ) , k2 0≠ ……(2)

By substituting (2) into (1),

v k k t v t

k k t k k v k t

= + −

= + −1 2

1 2 1 2 1

{[ ( )] }

v k k k k t

v tk k

k k

( ) ( )

( )

1 11 2 1 2

1 2

1 2

1

1

− = −

=−

−

Assume k k1 2 1≠ .

v t∝

∴ v varies directly as t.


m k t k m t

k t k k m k k t

m k k k k t

m tk k

k k

= + −

= + −

− = −

= −−

2 1

2 1 2 1 2

1 2 2 11 1

2 1

1 2

1

1

[ ( )]

( ) ( )

( )

Assume k k1 2 1≠ .

m t∝

∴ m varies directly as t.

10

144

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Variations

10B

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10.3 Inverse Variation

If x and y are in inverse variation, then yk

x= , where k is a non-zero constant.

x 100 50 30 8 4

y 5 20 25 662

3

x 3 15 25 60

y 225 561

422

1

27

1

24

1

2

1. Given that y varies inversely as x, complete the tables below.

(a)

(b)

2. If K varies inversely as v2 and K = 7

3 when v = 3, find

(a) an equation connecting K and v, (b) the value of K when v = 7 .

Solution

(a) Q K varies inversely as v2. (b) When v = 7,

∴ Kk=

( ), k ≠ 0 K =

=

(

(

)

)

By substituting K = 7

3 and v = 3


(

(

()

)

)

=

=

=

k

k

K∴

200 40 15

10 125 250 33 1

3

2 8 20 100

150 30 18

v 2

7

3 32

21

21v 2

21

72

21

49

37

=

145

10 Variations

4. If (W − 2) varies inversely as x3 and

W = 5

2 when x = 4, find

(a) an equation connecting x and W,

(b) the value of W when x = 1

2.

Solution

(a)

(b)

3. If y varies inversely as x and y = 78when x = 36, find


(b) the value of y when x = 1

64.

Solution

(a)

(b)

Q y varies inversely as x .

∴ y k

x= , k ≠ 0

By substituting y = 78 and x = 36


78

468

36

468

=

=

=

k

x

k

y∴

Q ( )W − 2 varies inversely as x 3.

∴ ( )W k

x− =2

3, k ≠ 0

By substituting W = 5

2 and x = 4


5

2 4

1

2 64

2

32

3− =

=

=

k

k

k

∴ W

x− =2 32

3

When x = 1

2,

W

W

− =

=

2

258

32

1

2

3

∴

When x = 1

64,

y =

=

468

1

64

3744

146


5. It is given that y varies inversely as x2. Ifx is decreased by 50%, find thepercentage change in y.

Solution

6. For a prism of a given volume, theheight of the prism varies inversely asthe base area. The height of the prism is45 cm when the base area is 2 cm2, whatis the height of prism when the base areais 9 cm2?

SolutionQ y varies inversely as x 2.

∴ y k

x=

2, k ≠ 0

Let x0 and y0 be the original value of x

and y respectively.

∴ New value of x

= −

=

( %)

.

1 50

0 5

0

0

x

x

New value of y

=

=

=

k

x

k

x

y

( . )0 5 02

0

24

4 0

∴ Percentage change in y

= ×

= ×

=

−4

3

0 0

0

0

0

100

100

300

y y

y

y

y

%

%

%

∴ y is increased by 300%.

Let h cm and A cm2 be the height and the

base area of the prism respectively.

Q h varies inversely as A.

∴ h

k

A= , k ≠ 0

By substituting h = 45 and A = 2 into

the equation, we have

45

90

2

90

=

=

=

k

A

k

h∴

When A = 9,

h =

=

90

9

10

∴ The height of the prism is 10 cm when

the base area is 9 cm2.

147

10 Variations

7. If y2 varies inversely as ( )x a+ and y = 5 when x = 6, y = 10 when x = 0, find

(a) the value of a, (b) the relation between x and y,

(c) the value of x when y = 4.

Solution

(a)

(b)

Q y 2 varies inversely as (x + a).

∴ y k

x a2 =

+, k ≠ 0 ……(1)

By substituting y = 5 and x = 6 into (1), we have

5

25 6

2

6=

= ++k

a

k a( ) ……(2)

By substituting y = 10 and x = 0 into (1), we have

100

100

0=

=+k

a

k a ……(3)

From (2) and (3),

25 6 100

6 4

3 6

2

( )+ =

+ =

=

=

a a

a a

a

a∴

Q y k

x2

2=

+

By substituting x = 6 and y = 5 into the above equation, we have

5

200

2

2

6 2

200

2

=

=

=

+

+

k

x

k

y∴

148


(c)

8. A job can be completed in 12 months by58 workers. If the same job has to becompleted in 8 months, how manyworkers should be added?

Solution

9. If 1 1

x y+

varies inversely as ( )x y+ ,

show that

(a) ( )x y xy+ ∝2 ,

(b) x y xy2 2+ ∝ .

Solution

(a)

(b)

Let x be number of workers added.

58 12 58 8

58 87

29

× = +

+ =

=

( )x

x

x

∴ 29 workers should be added in order

to complete the job in 8 months.

Q

1 1

x y+

varies inversely as

( )x y+ .

∴

1 1

x y

k

x y+ =

+, k ≠ 0

x y

xy

k

x y

x y kxy

x y xy

++

=

+ =

+ ∝

( )

( )

2

2∴

By using the result of (a),

( )

( )

x y kxy

x xy y kxy

x y k xy

+ =

+ + =

+ = −

2

2 2

2 2

2

2

Assume k ≠ 2.

∴ x y xy2 2+ ∝

When y = 4,

4

2 12 5

10 5

2 200

2=

+ =

=

+x

x

x

.

.∴

10

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Variations

10C

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10.4 Joint Variation

(a) If z varies jointly as x and y, then z kxy= , where k is a non-zero constant.

(b) If z varies directly as x and inversely as y, then zkx

y= , where k is a non-

zero constant.

1. If Q varies jointly as x and y, and Q = 120 when x = 3 and y = 8, find

(a) an equation connecting x, y and Q,

(b) the value of Q when x = 5 and y = 12 .

Solution

(a) Q Q varies jointly as x and y.

∴ Q k= ( )( ) , k ≠ 0

By substituting x = ( ), y = ( ) and Q = ( ) into the equation, we have

( ) ( )( )

( )

( )

===

k

k

Q∴

(b) When x = 5 and y = 12 ,

Q ==

( ( (

(

) ) )

)

x y

3 8 120

120 3 8

5

5xy

5 5 12

300

150


2. If K varies jointly as m and v2, andK = 352 when m = 11 and v = 8,find

(a) an equation connecting K, m and v,

(b) the value of K when m = 5 andv = 13.

Solution

(a)

(b)

3. If P varies directly as u2 and inversely asv, and P = 832 when u = 16 andv = 4, find

(a) an equation connecting P, u and v,

(b) the value of v when P = 637 andu = 28 .

Solution

(a)

(b)

Q K varies jointly as m and v 2.

∴ K k mv= 2 , k ≠ 0

By substituting K = 352, m = 11

and v = 8 into the equation, we

have

352 11 8 2

2

1

2

1

2

=

=

=

K

K

K mv

( )( )

∴

When m = 5 and v = 13,

K =

=

1

25 13

422 5

2( )( )

.

Q P varies directly as u 2 and

inversely as v.

∴ P ku

v=

2

, k ≠ 0

By substituting P = 832 , u = 16

and v = 4 into the equation, we

have

832

13

16

4

13

2

2

=

=

=

k

k

P u

v∴

When P = 637 and u = 28,

637

16

13 28 2

=

=

( )

v

v∴

151

10 Variations

4. It is given that wu

v∝

3, and w = 6

when u = 4 and v = 3. Find

(a) an equation connecting u, v and w,

(b) the value of u when wv

= 93.

Solution

(a)

(b)

5. The cost ($C) of making a cylindricalrod varies jointly as the square of itsbase radius (r cm) and its height (h cm).Find the percentage change in C if r isincreased by 20% and h is decreased by10%.

Solution

Q w u

v∝

3

∴ w k

u

v=

3, k ≠ 0

By substituting w = 6, u = 4 and

v = 3 into the equation, we have

6

81

4

3

81

3

3

=

=

=

k

k

w u

v∴

When w

v= 9

3,

9 81

1

9

1

81

3 3v

u

v

u

u

=

=

=∴

Let $C0, h0 cm and r0 cm be the original

cost, height and base radius of the

cylindrical rod.

C kr h0 02

0= , k ≠ 0

New value of r

= +

=

( %)

.

1 20

1 2

0

0

r

r

New value of h

= −

=

( %)

.

1 10

0 9

0

0

h

h

New value of C

=

=

=

k r h

kr h

C

( . ) ( . )

. ( )

.

1 2 0 9

1 296

1 296

02

0

02

0

0

The percentage change in C

= ×

=

−1 296 0 0

0

100

29 6

.%

. %

C C

C

∴ C is increased by 29.6%

152


6. The attractive force (F) between twounlike charges varies directly as theproduct of their charges (q1 and q2) andinversely as the square of their distance(d). If q1 and q2 are both doubled and d ishalved, find the percentage change in F.

Solution

7. It is given that z varies directly as ywhen x is constant and varies inverselyas x2 when y is constant. Show that

z y∝ 3 if yx

∝ 1.

Solution

Let F0 be the original value of F.

Q F varies directly as the product of q1 and

q2, and inversely as the square of d.

∴ F

kq q

d0

1 2

2= , k ≠ 0

The new value of F

=

=

=

=

k q q

kq q

kq q

d

d

d

F

( )( )2 2

4

16

1 2

2

1 2

1 2

2

2

4

2

16 0

Percentage change in F

= ×

=

−16 0 0

0

100

1500

F F

F%

%

∴ F is increased by 1500 %.

Q z varies directly as y when x is

constant and varies inversely as x 2

when y is constant.

∴ z varies jointly as y and

12x

i.e. z

k y

x= 1

2, k1 0≠ ……(1)

Q y

x∝ 1

∴ y

k

x= 2 , k2 0≠

i.e. xk

y= 2 ……(2)


zk y

k

y

k y

k

=

=

1

2

2

13

2

2

∴ z y∝ 3

10

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Variations

10D

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10.5 Partial Variation

(a) If z partly varies directly as x and partly varies directly as y, thenz k x k y= +1 2 , where k1 and k2 are two non-zero constants.

(b) If z is partly constant and partly varies directly as x, then z k k x= +1 2 ,where k1 and k2 are two non-zero constants.

(c) If z partly varies directly as x and partly varies inversely as y, then

z k xk

y= +1

2 , where k1 and k2 are two non-zero constants.

1. If w is partly constant and partly varies directly as x, and w = 9 when x = 4, w = 27 whenx = 16, find

(a) an equation connecting x and w,

(b) the value of w when x = 8.

Solution

(a) Q w is partly constant and partly varies directly as x.

∴ w k k= +1 2 ( ), where k1, k2 0≠

By substituting x = 4 and w = 9 into the equation, we have

9 4

4 9

1 2

1 2

= +

+ =

k k

k k

( )

……(1)

x

154


By substituting x = 16 and w = 27 into the equation, we have

(b) When x = 8,

w = +

=

( ( ( ) ) )

( )

(2) – (1),

12 182

23

2

k

k

=

=

By substituting k2

3

2= into (1), we have

k

k

1

1

4 9

3

3

2+

=

=

∴ w x= +3 3

2

3

15

8 3

2

27 16

16 27

1 2

1 2

= +

+ =

k k

k k

( )

……(2)

155

10 Variations

2. It is given that z partly varies directly as x and partly varies directly as y, and z = 5 whenx = 8 and y = 1, z = 20 when x = 16 and y = 7.

(a) Express z in terms of x and y.

(b) Find the value of z when x = 4, y = 7

4.

Solution

(a)

(b)

Q z partly varies directly as x and partly varies directly as y.

∴ z k x k y= +1 2 , where k1, k2 0≠

By substituting z = 5, x = 8 and y = 1 into the equation, we have

5 8 1

8 5

1 2

1 2

= +

+ =

k k

k k

( ) ( )

……(1)

By substituting z = 20, x = 16 and y = 7 into the equation, we have

20 16 7

16 7 20

1 2

1 2

= +

+ =

k k

k k

( ) ( )

……(2)

(2) – (1) x 2,

5 10

2

2

2

k

k

=

=

By substituting k2 2= into (1), we have

8 2 51

13

8

k

k

+ =

=

∴ z x y= +3

82

When x = 4 and y = 7

4,

z = +

=

3

87

44 2

5

( )

156


3. Given that w partly varies directly as x and partly varies directly as x2, and w = 58 whenx = 2, w = 126 when x = 3, find

(a) an equation connecting w and x,

(b) the value of w when x = 4.

Solution

(a)

(b)

Q w partly varies directly as x and partly varies directly as x 2.

∴ w k x k x= +1 22 , where k1, k2 0≠

By substituting w = 58 and x = 2 into equation, we have

58 2 2

2 29

1 22

1 2

= +

+ =

k k

k k

( ) ( )

……(1)

By substituting w = 126 and x = 3 into the equation, we have

126 3 3

3 42

1 22

1 2

= +

+ =

k k

k k

( ) ( )

……(2)

(2) – (1),

k2 13=


k

k

1

1

2 13 29

3

+ =

=

( )

∴ w x x= +3 13 2

When x = 4,

w = +

=

3 4 13 4

220

2( ) ( )

157

10 Variations

4. If y partly varies directly as x and partly varies inversely as x, and y = −2 when x = 2,y = 8 when x = 4, find


(b) the value of y when x = 32 .

Solution

(a)

(b)

Q y partly varies directly as x and partly varies inversely as x.

∴ y k x

k

x= +1

2 , where k1, k2 0≠

By substituting y = −2 and x = 2 into the equation, we have

− = +

+ = −

2 2

2 2

1

1

2

2

2

2

k

k

k

k

( )

……(1)

By substituting y = 8 and x = 4 into the equation, we have

8 4

4 8

1

1

2

2

4

4

= +

+ =

k

k

k

k

( )

……(2)

(2) – (1) × 2,

− =

= −

3

4 2

2

12

16

k

k

By substituting k2 16= − into (1), we have

2 2

3

1

1

16

2k

k

+ = −

=

−( )

∴ y x

x= −3 16

When x = 32,

y = −

=

3 32

95 5

16

32( )

.

158


5. The expenditure ($E) of holding a birthday party is partly constant and partly varies directly asthe number of guests (n). For 50 guests, the expenditure is $1450, and for 70 guests, theexpenditure is $1790. Find the expenditure per head when 100 guests join the party.

Solution

Q E is partly constant and partly varies directly as n.

∴ E k k n= +1 2 , where k1, k2 ≠ 0

By substituting E = 1450 and n = 50 into the equation, we have

1450 501 2= +k k ……(1)

By substituting E = 1790 and n = 70 into the equation, we have

1790 701 2= +k k ……(2)

(2) – (1),

340 20

17

2

2

=

=

k

k


1450 50 17

600

1

1

= +

=

k

k

( )

∴ E n= +600 17

The expenditure per head for 100 guests

=

=

+$

$

( )600 17 100

100

23

159

10 Variations

6. The cost ($C) of building a road varies partly as the length ( lm) and partly as the square of thelength of the road. If building a road of 100 m long costs $520 000, and building a road of 50 mlong costs $135 000.

(a) Express C in terms of l.

(b) If the cost of building a road is $2 040 000, what is its length?

Solution

(a)

(b)

Q The cost of building a road varies partly as the length and partly as the square of the

length

∴ C k k= +1 22l l , k1, k2 0≠

By substituting C = 520 000 and l = 100 into the equation, we have

520 000 100 100

100 5200

1 22

1 2

= +

+ =

k k

k k

( ) ( )

……(1)

By substituting C = 135 000 and l = 50 into the equation, we have

135 000 50 50

50 2700

1 22

1 2

= +

+ =

k k

k k

( ) ( )

……(2)

(1) – (2),

50 2500

50

2

2

k

k

=

=


k1 5200 100 50

200

=

=

– ( )

∴ C = +200 50 2l l

When C = 2 040 000 ,

2 040 000 200 50

4 40 800 0

200

2

2

or 204 (rejected)

= +

+ − =

= = −

l l

l l

l l

∴ The length of the road is 200 m.

10

160

Name :

Date :

Mark :

10E

Variations

Multiple Choice Questions

1. If y varies directly as x2 and y = 350when x = 10, find y when x = 5.

A. 50.5

B. 65.5

C. 87.5

D. 89

2. If ( ) ( )2 32 2x y x y+ ∝ − , then

A. y x∝ .

B. yx

∝ 1.

C. y x∝ 2 .

D. yx

∝ 12.

3. If w varies inversely as y and directly asx3, then

A.wy

x 3 is a constant.

B. wxy2 is a constant.

C.wx

y

3

is a constant.

D.w x

y

2 3

is a constant.

4. It is given that x y∝ 2 . If x is increasedby 44%, find the percentage increase in y.

A. 20% B. 32%

C. 40% D. 44%

5. It is given that zx

y∝

2. If x is increased

by 20% and y is decreased by 20%, findthe percentage increase in z.

A. 20% B. 42.5%

C. 60% D. 87.5%

6. If x xy y

+ ∝ −1 1, then

A. x y∝ .

B. xy

3 1∝ .

C. yx

∝ 1.

D. yx

2 1∝ .

7. The following table shows two quantitiesx and y and some of their correspondingvalues.

Which of the following is the correctrelation between x and y?

A. x y∝

B. xy

∝ 1

C. xy

∝ 1

D. xx

2 1∝

x 2 11

2

1

3y 1 4 16 36

x 2 11

2

1

3y 1 4 16 36

C

C

A

A

D

C

B

161

10 Variations

8. It is given that xy

2 1∝ . If y is decreased

by 75%, then x will be

A. increased by 100%.

B. increased by 200%.

C. decreased by 100%.

D. decreased by 200%.

9. It is given that y partly varies directly as

x2 and partly varies directly as 1

x. When

x = 1, y = 2 and when x = 2,

y = 9

2. The relation between x and y is

A. xy

+ =12.

B. y xx

= +2 1.

C. x yy

= +1 2 .

D. y xx

= +1

23 2 .

10. z partly varies directly as x2 and partlyvaries directly as y. If x is decreased by20% and y is decreased by 36%, then z is

A. decreased by 20%.

B. decreased by 36%.

C. increased by 10%.

D. increased by 20%.

11. If w partly varies directly as x and

partly varies directly as y , andw = 13 when x = 4 and y = 9 ,w = 18 when x = 9 and y = 16 , findw when x = 16 and y = 25.

A. 20 B. 21

C. 22 D. 23

12. It is given that z varies directly as x2 andinversely as y. If x is doubled and y ishalved, then z is

A. increased by 200%.

B. increased by 400%.

C. increased by 700%.

D. unchanged.

13. If y varies directly as x, which of thefollowing is/are true?

I. y x2 21∝ +( )

II. ( ) ( )2 2x y x y+ ∝ −III. x y xy2 2+ ∝

A. I only

B. II only

C. I and II only

D. II and III only

14. It is given that ( )x − 1 varies inverselyas the square root of y. If y t= whenx = 3 5. and y t= + 1 when x = 3,find t.

A.1

9

B.2

9

C.5

9

D.16

9

A

B

B

D

C

B

D

162


15. It is given that zx

y∝

2

. If both x and y

are increased by 10%, find the percentagechange in z.

A. Increased by 10%

B. Decreased by 10%

C. Increased by 21%

D. Decreased by 15%

16. It is given that y ax∝ +( )3 . If y = 69when x = 4, and y = 84 when x = 5,find the value of a.

A. 1

B. 2

C. 5

D. 8

17. If x varies directly as y, which of thefollowing is/are false?

I. ( ) ( )x y x y+ ∝ −2 3

II. ( )x y xy− ∝2 22

III. yx

∝ 1

A. I only

B. I and II only

C. II and III only

D. I, II and III

18. Which of the following graphs showsthat y is partly constant and partly variesdirectly as x?

A.

B.

C.

D.

x

y

0

x

y

0

x

y

01

y

0 x1

A

C

D

A

163

10 Variations

19. If y z2 3∝ and zx

∝ 1, which of the

following is/are true?

I. xz

∝ 12

II. yx

2 12

∝

III. xyx

∝ 1

A. I only

B. II only

C. III only

D. I and III only

20. If 1 1 1

x y y x−

∝

−, which of the

following is/are true?

I. x y xy2 2+ ∝II. ( ) ( )x y x y+ ∝ +2 2 2

III. ( ) ( )x y x y+ ∝ −

A. I only

B. II only

C. I and II only

D. I, II and III

C

C

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