Question

# 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 to20 10 50 60

Solution

## The correct option is C 50 Since the given volume ratio for five solid cubes is 1:1:8:27:27. The corresponding side ratio of the same cubes is 1:1:2:3:3 and the volume of the big cube is 64k^3, where k is any constant. Hence, its side is 4k.  Percentage by which the sum of surface areas of five cubes exceed the surface area of original cube =100×[6×24k2−6×(4k)2]{6×(4k)2}=[(24−16)16]×100=50

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