A reservoir is in the shape of a frustum of a right circular cone. The radii of its circular ends are 4 m and 8 m and its depth is 7 m. How many kiloliter of water (correct up to one decimal place) can it hold?
(Take $$\pi = \frac{22}{7}$$)

We know,

Volume of frustum shaped reservoir = Capacity of the reservoir 

i.e; $$\frac{1}{3}\times\ \pi\ \times\ \left(R^2+r^2+Rr\right)\times\ h$$

given, 

R = 8 m

r = 4 m

h = 7 m

Putting the values, 

= $$\frac{1}{3}\times\ \pi\ \times\ \left(8^2+4^2+8\times\ 4\right)\times\ 7$$

= $$\frac{1}{3}\times\ \frac{22}{7}\ \times\ \left(112\right)\times\ 7$$

=$$\frac{1}{3}\times\ 22\ \times\ 112\ $$

= 821.33 $$m^3$$

Hence, Option A is correct