SOLUTION: If the surface area of one cube is 44% larger than the surface area of a second
cube, what is the closest whole percent by which the volume of the first cube
exceeds the volume o
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Question 257234: If the surface area of one cube is 44% larger than the surface area of a second
cube, what is the closest whole percent by which the volume of the first cube
exceeds the volume of the second cube?
Answer by drk(1908) (Show Source): You can put this solution on YOUR website!
Think of surface area of first cube as 100.
The surface area of second cube is 144.
ratio of surface areas is 100/144
This means that ratio of sides is 10/12 or 5/6
The ratio of volumes becomes 1000/1728 or 125/216
So, 1728/1000 = 1.728 = 172.8% or about 173%.
The first cube volume exceeds the second cube volume by 73%.
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