**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

**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

**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

**Given that the volume of a sphere varies as the cube of its radius and its volume is 179.7 cm**^{3}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=179.7cm^{3} Radius=3.5 cm

Volume=? radius=1.75cm

volume of sphere α r^{3}

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}

= 22.46cm^{3}

**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:

h_{1} α t^{2}

h = 64cm t = 2sec

h = ? t = 6 sec

h_{1} α t^{2}

h_{1} =k. t^{2}

64 = k. (2)^{2}

k = ^{64}/_{4} = 16

If t = 6sec.

h_{1} =k. t^{2}

h_{1} = 16 x 6^{2}

= 16 x 36 = 576 cm

**the area of an isosceles: right angled triangle varies directly as the square of the length of its leg. If the area is 18 cm**^{2 }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 b^{2}

A = 18 cm^{2}

b = 6 cm.

(i) the relation between area and length

A of triangle α b^{2}

A = kb^{2}

18 = k.6^{2}

k = ^{18}/_{36} = ^{1}/_{2}

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

A = kb^{2}

A = ^{1}/_{2} x (5)^{2} = ^{25}/_{2} = 12.5 cm^{2}

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