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# If the width of a rectangle is increased by 25% while the length rema

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Math Expert
Joined: 02 Sep 2009
Posts: 52164
If the width of a rectangle is increased by 25% while the length rema  [#permalink]

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04 Feb 2018, 20:54
00:00

Difficulty:

25% (medium)

Question Stats:

68% (01:08) correct 32% (00:51) wrong based on 35 sessions

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If the width of a rectangle is increased by 25% while the length remains constant, the resulting area is what percent of the original area?

(A) 25%

(B) 75%

(C) 125%

(D) 225%

(E) 250%

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Joined: 16 Jan 2018
Posts: 91
Location: New Zealand
Re: If the width of a rectangle is increased by 25% while the length rema  [#permalink]

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04 Feb 2018, 21:15
Initial area = l x b = lb

New width = 25/100 x b + b = b(1 + 25/100) = b x ( 125/100 )

New area = l x b x ( 125/100 ) = lb x (125%) = 125 % of original area.

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examPAL Representative
Joined: 07 Dec 2017
Posts: 867
Re: If the width of a rectangle is increased by 25% while the length rema  [#permalink]

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04 Feb 2018, 23:22
Bunuel wrote:
If the width of a rectangle is increased by 25% while the length remains constant, the resulting area is what percent of the original area?

(A) 25%

(B) 75%

(C) 125%

(D) 225%

(E) 250%

As there are no exact numbers in the question, why not pick some?
This is an Alternative approach.

Say the width is 4 (a number it is easy to increase by 25%) and the length is 1 (a very small number, easy to work with).
Then the width increased by to 5 and the new area is 5*1 = 5.
So the answer is 5/4 = 125%.

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Intern
Joined: 26 Sep 2017
Posts: 31
Re: If the width of a rectangle is increased by 25% while the length rema  [#permalink]

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04 Feb 2018, 23:29
Let the length be 2

Width be 4

Area is 8

Revised width is 5(4*25% +4)
Now revised area is 10(5*2)

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Re: If the width of a rectangle is increased by 25% while the length rema &nbs [#permalink] 04 Feb 2018, 23:29
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