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A large cube of volume 64 cubic metre is broken down into small cubes of volume 8 cubic metre each. The percentage increase in the surface area is
Volume of a cube = $$a^3$$
Surface area of a cube = $$6a^2$$.
Side of the large cube = $$64^{1/3}$$ = $$4 m$$.
Surface area of the larger cube = $$6*4^2$$ = $$96 m^2$$.
The large cube is broken down to small cubes of volume $$8 m^3$$.
Number of small cubes = $$\frac{64}{8}$$ = $$8$$.
Side of the small cube = $$8^{1/3}$$ = $$2 m$$.
Surface area of the small cube = $$6*2^2$$ = $$24 m^2$$
Total surface area of the smaller cubes = $$8*24 = 192 m^2$$
%age increase in surface area = $$\frac{192-96}{96}$$
= $$\frac{96}{96}$$
= $$100$$%.
Hence, option B is the right answer.
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