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LakerFan24
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If the width, depth and length of a rectangle box were each decreased 30 Aug 2017, 16:04
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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%
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E
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If the width, depth and length of a rectangle box were each decreased 30 Aug 2017, 19:27
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LakerFan24
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%
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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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If the width, depth and length of a rectangle box were each decreased 30 Aug 2017, 19:54
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LakerFan24
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%
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Assume values.

Let the original box have these dimensions [a cube is a special kind of rectangular box]

L = 2
W = 2
H = 2

Its volume is length * width * height = 8

Decrease each length by half, and new dimensions are

L = 1
W = 1
H = 1

The new volume is 1 * 1 * 1 = 1

By what percent did the volume decrease?

\(\frac{change}{original}\) * 100 =

\(\frac{(8-1)}{8}\) * 100 =

\(\frac{7}{8}\) = 87.5 * 100 is 87.5 percent

You can also use multipliers, as sashiim20 has done above. See below for a very helpful post by a GMAT Club expert.*

ANSWER E

*McGarry, Mike. Scale Factors on the GMAT: Percent Increases and Decreases. https://magoosh.com/gmat/2012/scale-factors-on-the-gmat-percent-increases-and-decreases/. August 31,2012.
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If the width, depth and length of a rectangle box were each decreased 02 Sep 2017, 07:24
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LakerFan24
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%
Show more

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 decrease is: (0.125LWH - LWH)/LWH = -0.875 x 100 = -87.5%

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
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If the width, depth and length of a rectangle box were each decreased 11 Sep 2017, 04:00
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L=10
B=10
H=10

V=10X10X10=1000

50% reduction

L=5
B=5
H=5

Volume is 5X5X5=125

\(\frac{125}{(1000)}\)*100=12.5 ie new volume is 12.5 % of old volume or 100-12.5=87.5% reduction in volume
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If the width, depth and length of a rectangle box were each decreased 19 Sep 2017, 10:05
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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%

since we're not given hard #s in the question, i thought immediately to pick #s.
> original box dimensions: 2, 2, 2. v=8
> new box dimensions: 1, 1, 1. v=1
> % change = \(\frac{new-old}{old}\) = \(\frac{1-8}{8}\) = \(\frac{-7}{8}\)
> key: multiply numerator & denominator by 12.5 to get 100 = (-7)*(12.5) = 87.5%

Kudos please if you find this helpful :)
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If the width, depth and length of a rectangle box were each decreased 07 Feb 2023, 11:02
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