An edge of a cube is increased by 10percent. Find the percentage by which surface area is increased. Please reply fast........

Let edge initially = a
After increasing by 10%, edge = a + (10% of a) = a + (10/100 x a) = a+(a/10) = 11a/10

Initial Surface area = 6(a)^2
Final Surface area = 6(11a/10)^2
Increase in Surface area = Final Surface Area - Initial Surface Area = 6(11a/10)^2 - 6(a)^2
= 6 [(11a/10)^2 - (a)^2] = 6[121a^2/100 - a^2] = 6[21a^2 / 100]

So, percentage increase in surface area =
(Increase in surface area / original surface area)*100
= { 6[21a^2 / 100] / 6a^2 }* 100
= (6 * 21a^2 / 6a^2 * 100)*100
= 21 %