Question

The radius of cone is increased by 25% and height is decreased by 20%, then percentage change in volume is​

Answer 1

Assuming initial radius, height and thus, volume to be r, h, and 1/3 πr^2 h = V.

Now, new radius:

r′ = r + 25/100 × r = r(1 + 1/4) = 5/4 r

New height:

h′ = h - 20/100 h = h(1 - 1/5) = 4/5 h

So, V′ = 1/3 × π(5/4 r)^2 × (4/5) h

=> V′ = 1/3 πr^2 h × (5/4)

=> V′ = 5/4 V

Now, percentage in change in volume:

Change = (V′ - V)/V × 100%

=> ∆ (in %) = V(5/4 - 1)/(V) × 100%

=> ∆ (in %) = (1/4)100%

=> ∆ (in %) = + 25%.

So, change was of + 25%.

(∆ means change.)