Question 84
A solid metallic cube is melted to form five solid cubes whose volumes are in the ratio 1 : 1 : 8 : 27 : 27. The percentage by which the sum of the surface areas of these five cubes exceeds the surface area of the original cube is nearest to
Solution
Let the volumes of the five cubes be x, x, 8x, 27x and 27x.
Let the sides be a, a, 2a, 3a and 3a. ($$x = a^3$$)
Let the side of the original cube be A.
$$A^3 = x + x + 8x + 27x + 27x$$
$$A = 4a$$
Original surface area = $$96a^2$$
New surface area = $$6(a^2 + a^2 + 4a^2 + 9a^2 + 9a^2)$$ = $$144a^2$$
Percentage increase = $$\frac{144 - 96}{96}*100 =$$ 50%
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