Problem Solving

Hi All,

We're told that the width, depth, AND length of a rectangular box are each decreased by 50%. We're asked by what percent the volume of the box decreases. This question can be solved rather easily by TESTing VALUES. There's also a great 'concept shortcut' that you can use to answer this question without too much math.

If you reduce ANY of the three dimensions by 50%, then the total volume will decrease by 50%. By decreasing EACH dimension by 50%, the decrease becomes greater and greater. When each dimension is HALF of what it used to be, then the total volume will be (1/2)(1/2)(1/2) = 1/8 of what it was originally. 1/8 is clearly less than 1/4, so the correct answer must be greater than 75%.

Final Answer: E

GMAT assassins aren't born, they're made,
Rich

AAPL wrote:If the width, depth, and length of a rectangle box were each decreased by 50%, by how many percent would the volume of the box decrease?

A. 12.5%
B. 25%
C. 50%
D. 75%
E. 87.5%

We can let L, W, and H be the length, width, and depth of the original rectangular box. Thus, the original volume is L x W x H = LWH, and the new volume is 0.5L x 0.5W x 0.5H = 0.125LWH

The percentage change is: (0.125LWH - LWH)/LWH x 100% = -0.875 x 100% = -87.5%, i.e., a 87.5% decrease.

Alternate Solution:

Assume that the original dimensions of the box were 2 x 2 x 2, with a volume of 8. Reduce each dimension by 50% to get new dimensions of 1 x 1 x 1, with a volume of 1. The percentage decrease will be (old - new)/old x 100% = (8 - 1)/8 x 100% = 7/8 x 100% = 87.5%.

Answer: E