If the width, depth and length of a rectangle box were each decreased

Method \(1\):

Let the length, width and depth of rectangle box be \(= l, b, h\) respectively.

Volume \(= l * b * h = lbh\)

Length, width, depth of the rectangle box is decreased by \(50\)%.

Therefore, New length, width and depth of rectangle box \(= \frac{1}{2}l, \frac{1}{2}b, \frac{1}{2}h = (0.5)l, (0.5)b, (0.5)h\)

New Volume \(= (0.5)l* (0.5)b* (0.5)h = (0.125)lbh\)

Difference in Volume \(= lbh - (0.125)lbh = lbh(1-0.125) = (0.875)lbh\)

Required Percentage \(=\) (Difference \(/\) Original Volume) \(* 100 = \frac{(0.875)lbh}{lbh} * 100 = 0.875 * 100 = 87.5\)%

Answer (E)...

Method \(2\):

Let the length (\(l\)), width (\(b\)) and depth (\(h\)) of rectangular box be \(= 4, 2, 2\) respectively.

Volume \(= l*b*h = 4*2*2 = 16\)

Length, width, depth of the rectangle box is decreased by \(50\)%.

Therefore, New length, width and depth of rectangle box be \(= 2, 1, 1\) respectively.

New Volume \(= l*b*h = 2*1*1 = 2\)

Difference in Volume \(= 16 - 2 = 14\)

Required Percentage \(=\) (Difference \(/\) Original Volume) \(* 100 = \frac{14}{16} * 100 = 87.5\)%

Answer (E)...

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