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

---

2. 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

---

3. 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

---

4. 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

---

5. 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

---

6. 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

---