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M70-17

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Math Expert
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Joined: 02 Sep 2009
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M70-17  [#permalink]

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New post 03 Sep 2018, 04:37
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

100% (01:58) correct 0% (00:00) wrong based on 6 sessions

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Math Expert
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Re M70-17  [#permalink]

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New post 03 Sep 2018, 04:37
Official Solution:

Image
In rectangle ABCD above, \(AE = \frac{1}{3}AD\), \(4×AG = GB\). The area of triangle AEF is what fraction of the area of the area of rectangle FHCI?


A. \(\frac{1}{30}\)
B. \(\frac{1}{16}\)
C. \(\frac{1}{24}\)
D. \(\frac{5}{64}\)
E. \(\frac{1}{32}\)


We’ll go for ALTERNATIVE because all we have is ratio.

We’ll assign easy numbers to AE and AG: AE = 1, AG = 2. This makes AEF’s area \(\frac{1×2}{2}=1\). This means GB = 8 and ED = AD – AE = 3 – 1 = 2. Through symmetry, FI = ED = 2 and FH = GB = 4, so that FHCI = 8 × 2 = 16. Thus, \(\frac{AEF}{FHCI}=\frac{1}{16}\).


Answer: B
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Re: M70-17  [#permalink]

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New post 07 Mar 2019, 06:24
This was helpful . thanks
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Re: M70-17   [#permalink] 07 Mar 2019, 06:24
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