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If the width of a rectangle is w, the length is l, the perim

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If the width of a rectangle is w, the length is l, the perim [#permalink] New post 19 Aug 2018, 04:12
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Question Stats:

71% (01:32) correct 28% (01:29) wrong based on 7 sessions
If the width of a rectangle is w, the length is l, the perimeter is p, and w = 2l, what is the area in terms of p?

(A) \(\frac{p^2}{18}\)
(B) \(\frac{p^2}{36}\)
(C) \(\frac{p}{9}\)
(D) \(\frac{p^2}{9}\)
(E) \(\frac{p}{6}\)
[Reveal] Spoiler: OA

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Re: If the width of a rectangle is w, the length is l, the perim [#permalink] New post 21 Aug 2018, 22:46
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We can use Plugging-in technique for this question.
We have given w=2L, so let's plug 2 for L. i.e L=2. so W=4, P=12, A=8.
For option A (12*12)/18 the answer is 8 which is our target answer. Verify for other options too. You might end up with different values for A.
Hope it helps!
Re: If the width of a rectangle is w, the length is l, the perim   [#permalink] 21 Aug 2018, 22:46
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If the width of a rectangle is w, the length is l, the perim

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