Correct option is D 125%
Given,
Increment in each edge of a cube = 50%
Let original length of edge of cube = a
Original surface area of cube = 6a2
After increment ,
New length of edge = a + a \(\times\cfrac{50}{100}\) = \(a+\cfrac{a}2\)\(=\cfrac{3a}2\)
New surface area of cube = 6a2 = 6×\(\big(\cfrac{3a}{2}\big)^2\)
= \(6\times\cfrac{9a^2}4=\cfrac{27}{2}a^2\)
Increase in area = \(\cfrac{27}{2}a^2-6a^2=\cfrac{15a^2}2\)
% increase in area = \(\cfrac{increased\,area}{orginal\,area}\times100\)
= \(\cfrac{\frac{15a^2}{2}}{6a^2}\times100\) = 125%
