If height of a circular cone is decreased by 10% and its radius is increased by 10%, then what will be the change in its volume?

Let's assume the initial radius and height of the cone are R and H respectively.

the volume of cone = $$\frac{1}{3}\times\ \pi\ \times\ \ R^2\times\ H$$

= $$\frac{1}{3}\ \pi\ R^2H$$    Eq.(i)

If height of a circular cone is decreased by 10% and its radius is increased by 10%.

the volume of cone after change in its height and radius =

= $$\frac{1}{3}\times\ \pi\ \times\ \ \left(1.1R\right)^2\times\ 0.9H$$

= $$\frac{1}{3}\times\ \pi\ \times1.21R^2\times\ 0.9H$$

= $$\frac{\pi\ }{3}\times1.089R^2H$$    Eq.(ii)

percentage change = $$\frac{\left(Eq.\left(ii\right)-Eq.\left(i\right)\right)}{Eq.\left(i\right)}\times100$$

= $$\frac{\frac{\pi\ }{3}\times1.089R^2H-\frac{1}{3}\ \pi\ R^2H}{\frac{1}{3}\ \pi\ R^2H}\times100$$

= $$\frac{\left(1.089-1\right)}{1}\times100$$

= $$0.089\times100$$

= 8.9% increase