10
Variations
Name :
10A
10.2
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Date :
Mark :
Direct Variation
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Key Concepts and Formulae
If x and y are in direct variation, then y = kx , where k is a non-zero constant.
1.
Given that y varies directly as x, complete the tables below.
(a)
(b)
2.
x
1
3
5
7
9
11
13
15
y
4
12
20
28
36
44
52
60
x
4
8
12
16
20
24
28
32
y
1.6
3.2
4.8
6.4
8
9.6
11.2
12.8
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.
∴
A = k ( l2 ) , k ≠ 0
By substituting l = ( 5 ) and
A = ( 100
we have
100 = k (5)2
k = 4
∴
138
A = 4l 2
) into the equation,
(b) When l =
7,
A = ( 4 )(( 7)2 )
= ( 28 )
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10
3.
If V varies directly as r3 and V = 48
when r = 4 , find
4.
If y ∝
find
Variations
x and y = 21 when x = 81 ,
(a) an equation connecting V and r,
(a) an equation connecting x and y,
(b) the value of r when V = 6 .
(b) the value of y when x = 16 .
Solution
Solution
(a) Q V varies directly as r 3.
(a) Q y varies directly as x .
∴ V = kr 3 , k ≠ 0
∴ y = k x, k ≠ 0
By substituting r = 4 and V = 48
By substituting x = 81 and y = 21
into the equation, we have
into the equation, we have
48 = k (4)3
3
k =
4
∴
V =
3 3
r
4
(b) When V = 6 ,
6 =
r3 =
3 3
r
4
24
3
21 = k 81
k =
∴
7
3
y =
7
3
x
(b) When x = 16 ,
y =
=
7
3
16
28
3
= 8
∴
r = 2
139
Number and Algebra
5.
If ( y + 1) varies directly as 3 x and
y = 14 when x = 27 , find the value of
y when x = 64 without finding the
variation constant.
Solution
Q (y + 1) varies directly as
∴
y 1 + 1 = k x1
……(1)
y2 + 1 = k 3 x2
……(2)
3
3
6.
It is given that y ∝ x 2 . If x is increased
by 10%, find the percentage change in y.
Solution
Q
y ∝ x2
∴
y = kx 2 , k ≠ 0
x.
Let x0 and y0 be the original values of x
and y respectively.
∴ New value of x = (1 + 10%)x 0
From (1), we have
y1 + 1
k =
3
= 1.1x 0
New value of y = k (1.1x 0 )2
x1
= 1.21kx 0
From (2), we have
k =
∴
3
y1 + 1
3
x1
14 + 1
3
∴ Percentage change in y
x2
=
=
27
y +1=
y2 + 1
3
y + 1
3
140
=
1.21y 0 − y 0
y0
x2
64
15
×4
3
= 20
∴
= 1.21y 0
y2 + 1
y = 19
=
(1.21 − 1)y 0
y0
× 100%
× 100%
= 21%
∴ y is increased by 21%.
2
10
7.
It is given that y + 1 varies directly as x.
If y = 11 when x = 4 , find
8.
Variations
If y ∝ x 2 and y = 24 when x = t ,
y = 54 when x = t + 1, find
(a) an equation connecting x and y,
(a) the values of t,
(b) the value of y when x = 7 ,
(b) an equation connecting x and y
when t is positive.
(c) the value of x when y = 5x .
Solution
Solution
(a) Q (y + 1) ∝ x
∴
y + 1 = kx , k ≠ 0
By substituting x = 4 and y = 11
into the equation, we have
(a) Q y ∝ x 2
∴
By substituting x = t , y = 24 into
the equation, we have
24 = kt 2
11 + 1 = k (4)
∴
y = kx 2 , k ≠ 0
……(1)
k = 3
By substituting x = t + 1, y = 54
y + 1 = 3x
into the equation, we have
54 = k (t + 1)2
y = 3x − 1
(b) When x = 7,
……(2)
(2)
:
(1)
54
k(t + 1)
=
24
kt 2
y = 3(7) − 1
= 20
9
t
=
4
2
2
+ 2t + 1
t
2
9t 2 = 4t 2 + 8t + 4
5t 2 − 8t − 4 = 0
(5t + 2)(t − 2) = 0
t = −
∴
(c) When y = 5x ,
5x = 3x − 1
1
2
5
or t = 2
(b) By substituting t = 2 into (1), we
have
2x = −1
x = −
2
24 = k (2)2
k = 6
∴
y = 6x 2
141
Number and Algebra
9.
3
4
If y 2 varies directly as ( x + b) and y = 6 when x = 13, y = 9 when x = 46 , find
(a) the value of b,
(b) the relation between x and y,
(c) the value of x when y = 16 .
Solution
(a) Q y 2 ∝ (x + b)
y 2 = k (x + b), k ≠ 0
∴
By substituting y = 6 and x = 13 into the equation, we have
36 = k (13 + b)
……(1)
By substituting y = 9 and x = 46
3
into the equation, we have
4
81 = k  46 3 + b  ……(2)


4
(2)
:
(1)
81
=
36
k
187

 4 + b
k (13 + b)
1053 + 81b = 1683 + 36b
45b = 630
b = 14
(b) By substituting b = 14 into (1), we have
36 = k (13 + 14)
k =
∴
y2 =
4
3
4
(x + 14)
3
(c) By substituting y = 16 into y 2 =
162 =
4
(x + 14)
3
192 = x + 14
x = 178
∴ x = 178 when y = 16.
142
4
(x + 14) , we have
3
10
10. The value (in $) of a piece of jade varies
directly as the cube of its volume (in
cm3).
(a) If the value of a piece of jade of
volume 2 cm3 is $6720, find the
value of a piece of jade of volume
3.5 cm3.
(b) If the piece of jade of volume 2 cm3
is broken into two parts of equal
volume, what would be the
percentage loss in its value?
Variations
11. If v varies directly as (m − t ) , and m
varies directly as (t + v ) , show that
(a) v varies directly as t,
(b) m varies directly as t.
Solution
(a) v = k1(m − t ) , k1 ≠ 0
m = k 2 (t + v ) , k 2 ≠ 0
……(1)
……(2)
By substituting (2) into (1),
v = k1{[k 2 (t + v )] − t }
= k1k 2t + k1k 2v − k1t
Solution
(a) Let $P and V cm3 be the value and
v (1 − k1k 2 ) = k1(k 2 − 1)t
v =
volume of the jade respectively.
P = kV , k ≠ 0
3
k1(k 2 − 1)
1 − k1k 2
Assume k1k 2 ≠ 1.
By substituting P = 6720 , V = 2
v ∝t
into the equation, we have
∴ v varies directly as t.
6720 = k (2)
t
3
k = 840
∴ P = 840V 3
When V = 3.5 ,
(b) By substituting (1) into (2),
P = 840(3.5)3
m = k 2 [t + k1(m − t )]
= 36 015
∴ The value of the jade of volume
3.5 cm3 is $36 015.
= k 2t + k1k 2m − k1k 2t
m(1 − k1k 2 ) = k 2 (1 − k1)t
m =
(b) The value of the two broken pieces
of jade with equal volume 1 cm
= 2 × 840 × (1)
3
3
k 2 (1 − k1)
1 − k1k 2
t
Assume k1k 2 ≠ 1.
m ∝t
∴ m varies directly as t.
= $1680
The percentage loss in the value
=
6720 − 1680
× 100%
6720
= 75%
143
10
Variations
Name :
10B
10.3
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Date :
Mark :
Inverse Variation
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Key Concepts and Formulae
If x and y are in inverse variation, then y =
1.
Given that y varies inversely as x, complete the tables below.
(a)
(b)
2.
k
, where k is a non-zero constant.
x
x
200
100
50
40
y
5
10
20
25
x
2
3
8
15
y
225
150
56
If K varies inversely as v2 and K =
1
4
30
30
33
1
3
4
125
250
25
60
100
18
7
66
20
22
8
15
1
2
2
3
(b) the value of K when v = 7 .
Solution
∴
K varies inversely as v2.
K =
k
, k ≠ 0
( v2 )
By substituting K =
(
7
3
) =
k
( 32 )
k = ( 21 )
∴
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21 
K = 
 v2 
(b) When v = 7 ,
K =
7
and v = 3
3
into the equation, we have
4
7
when v = 3, find
3
(a) an equation connecting K and v,
(a) Q
1
2
( 21 )
( 72 )
=
21
49
=
3
7
1
2
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10
3.
If y varies inversely as
when x = 36 , find
x and y = 78
If (W − 2) varies inversely as x3 and
W=
(a) an equation connecting x and y,
(b) the value of y when x =
4.
1
.
64
Variations
5
when x = 4, find
2
(a) an equation connecting x and W,
(b) the value of W when x =
1
.
2
Solution
Solution
(a) Q y varies inversely as x .
(a) Q (W − 2) varies inversely as x 3.
k
∴ y =
x
, k ≠ 0
By substituting y = 78 and x = 36
5
k
−2 = 3
2
4
36
1
k
=
2
64
468
x
k = 32
∴ W −2 =
(b) When x =
1
,
64
(b)
When x =
468
32
x
64
∴
3
1
,
2
W −2 =
1
= 3744
5
and x = 4
2
into the equation, we have
k = 468
y =
, k ≠ 0
k
78 =
y =
x3
By substituting W =
into the equation, we have
∴
k
∴ (W − 2) =
32
 1
 2
3
W = 258
145
Number and Algebra
5.
It is given that y varies inversely as x2. If
x is decreased by 50%, find the
percentage change in y.
Solution
Q y varies inversely as x 2.
∴
y =
k
x2
6.
For a prism of a given volume, the
height of the prism varies inversely as
the base area. The height of the prism is
45 cm when the base area is 2 cm2, what
is the height of prism when the base area
is 9 cm2?
Solution
, k ≠ 0
Let h cm and A cm2 be the height and the
Let x0 and y0 be the original value of x
and y respectively.
∴ New value of x = (1 − 50%)x 0
= 0.5x 0
New value of y =
k
(0.5x 0 )
2
 k 
= 4 2 
 x0 
base area of the prism respectively.
Q h varies inversely as A.
∴ h =
By substituting h = 45 and A = 2 into
the equation, we have
45 =
= 4y 0
∴ Percentage change in y
=
=
4y 0 − y 0
y0
3y 0
y0
= 300%
∴ y is increased by 300%.
k
2
k = 90
∴
h =
× 100%
× 100%
k
, k ≠ 0
A
90
A
When A = 9 ,
h =
90
9
= 10
∴ The height of the prism is 10 cm when
the base area is 9 cm2.
146
10
7.
Variations
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) Q y 2 varies inversely as (x + a).
∴ y2 =
k
, k ≠ 0
x + a
……(1)
By substituting y = 5 and x = 6 into (1), we have
52 =
k
6 + a
k = 25(6 + a)
……(2)
By substituting y = 10 and x = 0 into (1), we have
100 =
k
0 + a
k = 100a
……(3)
From (2) and (3),
25(6 + a) = 100a
6 + a = 4a
3a = 6
∴
a = 2
(b) Q y 2 =
k
x + 2
By substituting x = 6 and y = 5 into the above equation, we have
52 =
k
6 + 2
k = 200
∴
y2 =
200
x + 2
147
Number and Algebra
(c) When y = 4 ,
42 =
200
x + 2
x + 2 = 12.5
x = 10.5
∴
9.
1
1
If  +  varies inversely as ( x + y ) ,
x
y
show that
(a) ( x + y )2 ∝ xy ,
(b) x 2 + y 2 ∝ xy .
Solution
(a) Q  1 +
1
varies inversely as
x
y
(x + y ) .
8.
A job can be completed in 12 months by
58 workers. If the same job has to be
completed in 8 months, how many
workers should be added?
∴
x + y
k
=
xy
x + y
Solution
Let x be number of workers added.
1
1
k
, k ≠ 0
+
=
x
y
x + y
(x + y )2 = kxy
∴ (x + y )2 ∝ xy
58 × 12 = (58 + x )8
58 + x = 87
x = 29
∴ 29 workers should be added in order
to complete the job in 8 months.
(b) By using the result of (a),
(x + y )2 = kxy
x 2 + 2xy + y 2 = kxy
x 2 + y 2 = (k − 2)xy
Assume k ≠ 2 .
∴
148
x 2 + y 2 ∝ xy
10
Variations
Name :
10C
10.4
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Date :
Mark :
Joint Variation
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Key Concepts and Formulae
(a) If z varies jointly as x and y, then z = kxy , where k is a non-zero constant.
kx
(b) If z varies directly as x and inversely as y, then z =
, where k is a nony
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 ( x )( y ) , k ≠ 0
By substituting x = ( 3 ) , y = ( 8 ) and Q = ( 120 ) into the equation, we have
( 120
∴
) = k ( 3 )( 8 )
k = ( 5 )
Q = ( 5xy )
(b) When x = 5 and y = 12 ,
Q = ( 5 )( 5 )( 12 )
= ( 300 )
149
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Number and Algebra
2.
If K varies jointly as m and v2, and
K = 352 when m = 11 and v = 8 ,
find
3.
If P varies directly as u2 and inversely as
v, and P = 832 when u = 16 and
v = 4 , find
(a) an equation connecting K, m and v,
(a) an equation connecting P, u and v,
(b) the value of K when m = 5 and
v = 13 .
(b) the value of v when P = 637 and
u = 28 .
Solution
Solution
(a) Q K varies jointly as m and v 2.
(a) Q P varies directly as u 2 and
∴ K = k mv 2 , k ≠ 0
By substituting K = 352 , m = 11
and v = 8 into the equation, we
have
352 = K (11)(8)2
∴
K =
1
2
K =
1
mv 2
2
inversely as v.
∴ P =
By substituting P = 832 , u = 16
and v = 4 into the equation, we
have
832 = k
K =
150
P =
2
13u
v
(b) When P = 637 and u = 28 ,
1
(5)(13)2
2
= 422.5
162
4
k = 13
∴
(b) When m = 5 and v = 13 ,
ku 2
, k ≠ 0
v
637 =
∴
13(28)2
v
v = 16
10
4.
It is given that w ∝
u
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 w =
9
.
v3
Solution
5.
Variations
The cost ($C) of making a cylindrical
rod varies jointly as the square of its
base radius (r cm) and its height (h cm).
Find the percentage change in C if r is
increased by 20% and h is decreased by
10%.
Solution
Let $C0, h0 cm and r0 cm be the original
(a) Q w ∝
cost, height and base radius of the
u
v3
cylindrical rod.
u
∴ w = k
v3
2
, k ≠ 0
C 0 = kr0 h0 , k ≠ 0
New value of r = (1 + 20%)r0
By substituting w = 6 , u = 4 and
v = 3 into the equation, we have
= 1.2r0
New value of h = (1 − 10%)h0
4
6 = k
= 0.9h0
33
New value of C = k (1.2r0 )2 (0.9h0 )
k = 81
2
= 1.296(kr0 h0 )
81 u
∴ w =
3
v
= 1.296C 0
The percentage change in C
=
(b) When w =
9
v
∴
3
=
9
v3
,
81 u
v
u =
1
9
u =
1
81
1.296C 0 − C 0
C0
× 100%
= 29.6%
∴ C is increased by 29.6%
3
151
Number and Algebra
6.
The attractive force (F) between two
unlike charges varies directly as the
product of their charges (q1 and q2) and
inversely as the square of their distance
(d). If q1 and q2 are both doubled and d is
halved, find the percentage change in F.
Solution
7.
It is given that z varies directly as y
when x is constant and varies inversely
as x2 when y is constant. Show that
z ∝ y 3 if y ∝
Solution
Q z varies directly as y when x is
Let F0 be the original value of F.
constant and varies inversely as x 2
Q F varies directly as the product of q1 and
when y is constant.
q2, and inversely as the square of d.
∴ F0 =
kq1q 2
d
2
, k ≠ 0
The new value of F =
=
=
∴ z varies jointly as y and
i.e. z =
d
 2
2
4kq1q 2
d
2
16kq1q 2
Q
y ∝
1
x
∴
y =
k2
i.e. x =
k2
F0
, k1 ≠ 0
x
, k2 ≠ 0
……(2)
y
z =
× 100%
=
= 1500%
∴ F is increased by 1500 %.
k1y
 k2 
y
k1y 3
k2
∴
z ∝ y3
2
2
1
x2
……(1)
By substituting (2) into (1),
Percentage change in F
16F0 − F0
x2
d2
= 16F0
=
k1y
k (2q1)(2q 2 )
4
152
1
.
x
10
Variations
Name :
10D
10.5
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Date :
Mark :
Partial Variation
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Key Concepts and Formulae
(a) If z partly varies directly as x and partly varies directly as y, then
z = k1 x + k2 y , where k1 and k2 are two non-zero constants.
(b) If z is partly constant and partly varies directly as x, then z = k1 + k2 x ,
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 = k1 x +
1.
k2
y
, where k1 and k2 are two non-zero constants.
If w is partly constant and partly varies directly as x, and w = 9 when x = 4 , w = 27 when
x = 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 = k1 + k2 ( x ), where k1, k2 ≠ 0
By substituting x = 4 and w = 9 into the equation, we have
9 = k1 + k 2 (4)
k1 + 4k 2 = 9
……(1)
153
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Number and Algebra
By substituting x = 16 and w = 27 into the equation, we have
27 = k1 + k 2 (16)
k1 + 16k 2 = 27
……(2)
(2) – (1),
12k 2 = 18
k2 =
3
2
By substituting k 2 =
3
2
into (1), we have
k 1 + 4 3  = 9
 2
k1 = 3
∴ w = 3+
3
x
2
(b) When x = 8 ,
w = ( 3 )+(
= ( 15 )
154
3
2
)( 8 )
10
2.
Variations
It is given that z partly varies directly as x and partly varies directly as y, and z = 5 when
x = 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) Q z partly varies directly as x and partly varies directly as y.
∴ z = k1x + k 2 y , where k1, k 2 ≠ 0
By substituting z = 5 , x = 8 and y = 1 into the equation, we have
5 = k1(8) + k 2 (1)
8k1 + k 2 = 5
……(1)
By substituting z = 20 , x = 16 and y = 7 into the equation, we have
20 = k1(16) + k 2 (7)
16k1 + 7k 2 = 20
……(2)
(2) – (1) x 2,
5k 2 = 10
k2 = 2
By substituting k 2 = 2 into (1), we have
8k1 + 2 = 5
k1 =
z =
∴
3
8
3
x + 2y
8
(b) When x = 4 and y =
z =
7
,
4
(4) + 2 7 
 4
8
3
= 5
155
Number and Algebra
3.
Given that w partly varies directly as x and partly varies directly as x2, and w = 58 when
x = 2 , w = 126 when x = 3 , find
(a) an equation connecting w and x,
(b) the value of w when x = 4 .
Solution
(a) Q w partly varies directly as x and partly varies directly as x 2.
∴ w = k1x + k 2 x 2 , where k1, k 2 ≠ 0
By substituting w = 58 and x = 2 into equation, we have
58 = k1(2) + k 2 (2)2
k1 + 2k 2 = 29
……(1)
By substituting w = 126 and x = 3 into the equation, we have
126 = k1(3) + k 2 (3)2
k1 + 3k 2 = 42
……(2)
(2) – (1),
k 2 = 13
By substituting k 2 = 13 into (1), we have
k1 + 2(13) = 29
k1 = 3
∴ w = 3x + 13x 2
(b) When x = 4 ,
w = 3(4) + 13(4)2
= 220
156
10
4.
Variations
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
(a) an equation connecting x and y,
(b) the value of y when x = 32 .
Solution
(a) Q y partly varies directly as x and partly varies inversely as x.
k2
∴ y = k1x +
x
, where k1, k 2 ≠ 0
By substituting y = −2 and x = 2 into the equation, we have
−2 = k1(2) +
2k1 +
k2
= −2
2
k2
2
……(1)
By substituting y = 8 and x = 4 into the equation, we have
8 = k1(4) +
4k1 +
k2
= 8
4
k2
4
……(2)
(2) – (1) × 2,
−
3
k 2 = 12
4
k 2 = −16
By substituting k 2 = −16 into (1), we have
2k1 +
(−16)
2
= −2
k1 = 3
∴
y = 3x −
16
x
(b) When x = 32 ,
y = 3(32) −
16
32
= 95.5
157
Number and Algebra
5.
The expenditure ($E) of holding a birthday party is partly constant and partly varies directly as
the number of guests (n). For 50 guests, the expenditure is $1450, and for 70 guests, the
expenditure 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 = k1 + k 2n , where k1, k2 ≠ 0
By substituting E = 1450 and n = 50 into the equation, we have
1450 = k1 + 50k 2
……(1)
By substituting E = 1790 and n = 70 into the equation, we have
1790 = k1 + 70k 2
……(2)
(2) – (1),
340 = 20k 2
k 2 = 17
By substituting k 2 = 17 into (1), we have
1450 = k1 + 50(17)
k1 = 600
∴ E = 600 + 17n
The expenditure per head for 100 guests = $
600 + 17(100)
100
= $23
158
10
6.
Variations
The cost ($C) of building a road varies partly as the length ( l m) and partly as the square of the
length of the road. If building a road of 100 m long costs $520 000, and building a road of 50 m
long 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) Q The cost of building a road varies partly as the length and partly as the square of the
length
∴ C = k1l + k 2 l 2 , k1, k 2 ≠ 0
By substituting C = 520 000 and l = 100 into the equation, we have
520 000 = k1(100) + k 2 (100)2
k1 + 100k 2 = 5200
……(1)
By substituting C = 135 000 and l = 50 into the equation, we have
135 000 = k1(50) + k 2 (50)2
k1 + 50k 2 = 2700
……(2)
(1) – (2),
50k 2 = 2500
k 2 = 50
By substituting k 2 = 50 into (1), we have
k1 = 5200 – 100(50)
= 200
∴ C = 200l + 50l 2
(b) When C = 2 040 000 ,
2 040 000 = 200l + 50l 2
l 2 + 4l − 40 800 = 0
l = 200 or l = −204 (rejected)
∴ The length of the road is 200 m.
159
10
Variations
Name :
10E
Date :
Mark :
Multiple Choice Questions
1.
If y varies directly as x2 and y = 350
when x = 10 , find y when x = 5 .
3.
4.
160
It is given that z ∝
x
. If x is increased
y2
A. 50.5
by 20% and y is decreased by 20%, find
the percentage increase in z.
B.
65.5
A. 20%
B.
C.
87.5
C.
D. 87.5%
D. 89
2.
5.
60%
42.5%
6.
If (2 x 2 + y ) ∝ ( x 2 − 3 y ) , then
If x +
1
1
∝ x − , then
y
y
A.
y ∝ x.
A.
x ∝ y.
B.
1
y ∝ .
x
B.
x3 ∝
C.
y ∝ x2 .
C.
y ∝
D.
y ∝
D.
y2 ∝
1
.
x2
C
If w varies inversely as y and directly as
x3, then
7.
1
.
y
1
.
x
1
.
x
C
The following table shows two quantities
x and y and some of their corresponding
values.
A.
wy
is a constant.
x3
B.
wxy2 is a constant.
x
2
1
1
3
C.
wx 3
is a constant.
y
1
2
y
1
4
16
36
D.
w2 x 3
is a constant.
y
A
It is given that x ∝ y 2 . If x is increased
by 44%, find the percentage increase in y.
A. 20%
B.
32%
C.
D.
44%
40%
D
C
A
Which of the following is the correct
relation between x and y?
A.
x ∝
y
B.
x ∝
1
C.
x ∝
D.
x2 ∝
y
1
y
1
x
B
10
8.
It is given that x 2 ∝
1
. If y is decreased
y
x and
B.
increased by 200%.
partly varies directly as y , and
w = 13 when x = 4 and y = 9 ,
w = 18 when x = 9 and y = 16 , find
w when x = 16 and y = 25.
C.
decreased by 100%.
A. 20
B.
C.
D. 23
by 75%, then x will be
A. increased by 100%.
D. decreased by 200%.
9.
11. If w partly varies directly as
Variations
A
It is given that y partly varies directly as
x2 and partly varies directly as
1
. When
x
x = 1, y = 2 and when x = 2 ,
9
y = . The relation between x and y is
2
1
A. x +
= 2.
y
1
.
x
B.
y = x2 +
C.
x =
1
+ y2 .
y
D.
y =
1
+ 3x 2 .
2x
B
C.
increased by 10%.
D. increased by 20%.
12. It is given that z varies directly as x2 and
inversely as y. If x is doubled and y is
halved, then z is
A. increased by 200%.
B.
increased by 400%.
C.
increased by 700%.
D. unchanged.
I.
y 2 ∝ ( x + 1)2
II.
(2 x + y ) ∝ (2 x − y )
C
III. x 2 + y 2 ∝ xy
A. I only
B.
II only
C.
I and II only
D. II and III only
A. decreased by 20%.
decreased by 36%.
D
13. If y varies directly as x, which of the
following is/are true?
10. z partly varies directly as x2 and partly
varies directly as y. If x is decreased by
20% and y is decreased by 36%, then z is
B.
22
21
B
B
14. It is given that ( x − 1) varies inversely
as the square root of y. If y = t when
x = 3.5 and y = t + 1 when x = 3 ,
find t.
A.
1
9
B.
2
9
C.
5
9
D.
16
9
D
161
Number and Algebra
15. It is given that z ∝
x2
. If both x and y
y
are increased by 10%, find the percentage
change in z.
18. Which of the following graphs shows
that y is partly constant and partly varies
directly as x?
y
A.
A. Increased by 10%
B.
Decreased by 10%
C.
Increased by 21%
D. Decreased by 15%
A
16. It is given that y ∝ ( ax + 3) . If y = 69
when x = 4 , and y = 84 when x = 5 ,
find the value of a.
x
0
y
B.
A. 1
B.
2
C.
5
x
D. 8
17. If x varies directly as y, which of the
following is/are false?
I.
( x + y )2 ∝ ( x − y )3
II.
( x − y )2 ∝ 2 xy 2
III. y ∝
0
C
1
x
I and II only
C.
II and III only
D. I, II and III
y
D.
D
0
162
1
x
0
A. I only
B.
y
C.
1
x
A
10
19. If y 2 ∝ z 3 and z ∝
Variations
1
, which of the
x
following is/are true?
1
z2
I.
x ∝
II.
y2 ∝
1
x2
III. xy ∝
1
x
A. I only
B.
II only
C.
III only
D. I and III only
C
1
1
1
20. If  −  ∝
, which of the
x
y
y − x
following is/are true?
I.
x 2 + y 2 ∝ xy
II.
( x + y )2 ∝ ( x 2 + y 2 )
III. ( x + y ) ∝ ( x − y )
A. I only
B.
II only
C.
I and II only
D. I, II and III
C
163