Variation – Exercise 3.6.2 – Class IX

  1. If x varies as y and if x = 6 when y = 3, find x when y = 10.

Solution:

Given x varies as y and if x = 6 we have y = 3

x α y

x=6                 y=3

x=?                  y=10

x=ky

6=k.3 ⇒k = 2

x = 2 x y = 2 x 10 = 20


  1. If p varies as q and if p = 5 when q = 10, find when p = 20.

Solution:

Given p varies as q and if p = 5 when q = 10

p = kq

5 = kx10

k = 5/10 = 1/2

If p = 20 then q=?

p = kq

20 = 1/2 x q

q = 20 x 2 = 40


  1. If y varies directly as √x and y = 24 when x = 3, find y when x = 16

Solution:
Given  y varies directly as √x and y = 24 when x = 3

Such that, y = k√x

24 = k√3

k = 24/√3 = 24/√3 x √3/√3 = 24x√3/3 = 8√3

If x = 16 then  y = ?

y=k√x

y = 8√3 x √16 = 8√3×4 = 32√3


  1. Given that the volume of a sphere varies as the cube of its radius and its volume is 179.7 cm3 when radius is 3.5, find the volume when radius is 1.75 cm

Solution:

Given, the volume of a sphere varies as the cube of its radius.

Volume of sphere = (radius)3

Volume=179.7cm3                               Radius=3.5 cm

Volume=?                                           radius=1.75cm

volume of sphere α r3

volume = k. (radius)3

179.7 = k(3.5)3

k =  179.7/3.5×3.5×3.5 = 1797/3.5×3.5x.3.5 = 1797/428.75 = 4.1912

k = 4.1912

volume = k. (radius)3

=4.1912 x (1.75)3

= 22.46cm3


  1. The distance through which a body falls from rest varies as square of time it takes to fall that distance. It is known that the body falls 64 cm in 2 seconds. How far does that body fall in 6 seconds?

Solution:

h1 α t2

h = 64cm                    t = 2sec

h = ?                            t = 6 sec

h1 α t2

h1 =k. t2

64 = k. (2)2

k = 64/4 = 16

If t = 6sec.

h1 =k. t2

h1 = 16 x 62

= 16 x 36 = 576 cm


  1. the area of an isosceles: right angled triangle varies directly as the square of the length of its leg. If the area is 18 cm2 when the angle of its leg is 6 cm, find (i) the relation between area and length

(ii) area of triangle when length of its leg is 5 cm.

Solution:

A of triangle α b2

A = 18 cm2

b = 6 cm.

(i) the relation between area and length

A of triangle α b2

A = kb2

18 = k.62

k = 18/36 = 1/2

(ii) area of triangle when length of its leg is 5 cm.

A = kb2

A = 1/2 x (5)2 = 25/2 = 12.5 cm2


 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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