CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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 to


  1. 20

  2. 10

  3. 50

  4. 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×[6×24k26×(4k)2]{6×(4k)2}=[(2416)16]×100=50

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
View More


People also searched for
View More



footer-image