UPSC CDS I 2019 Math Paper

New volume of balloon = V + 700% of V = 8V
Let r be the initial radius and R be the new radius of the balloon.

R3

r3

=

8V

V

=8
⇒

R

r

=2⇒R=2r
Initial surface area of balloon =4πr2
New surface area of balloon =4π(2r)2=16πr2
Percentage increase in surface area