If every side of a cube is increased by 16\(\frac{2}{3}\)%, then what will be percentage increase in the surface area -

36.11 %

16\(\frac{2}{3}\)% = \(\frac{1}{6}\)

Surface area (S.A.) = = 6s²

ATQ,

             Original   :   New

side   :       6        :      7

S.A.   :     6(6)²    :     6 (7)²

         :      36       :      49

Here,

S.A. increased by = 49 - 36 = 13

Increased percentage = \(\frac{13}{36}\) × 100 = 36.11%