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Let d > c > b > a. If c is twice as far from a as it is from d, and b

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Math Expert
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Joined: 02 Sep 2009
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Let d > c > b > a. If c is twice as far from a as it is from d, and b  [#permalink]

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New post 11 Nov 2019, 02:43
2
1
5
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

54% (02:48) correct 46% (02:47) wrong based on 46 sessions

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GMAT 1: 780 Q51 V45
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Let d > c > b > a. If c is twice as far from a as it is from d, and b  [#permalink]

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New post 11 Nov 2019, 06:18
1
1
We can label the lengths and turn this into an algebra question. Let A = b - a, B = c - b, C = d - c. Then "c is twice as far from a as it is from d" means A + B is twice the value of C. Then the corresponding equality would be (A+B) = 2C. "b is twice as far from c as it is from a" means 2A = B. We can put everything in terms of A, B = 2A, and 2C = A + B = A + 2A = 3A.
Finally, the fraction is asking what is (B + C) / (A + B + C), we can double everything to make it easier to plug in 2C = 3A.

(2B + 2C) / (2A + 2B + 2C) = (4A + 3A) / (2A + 4A + 3A) = 7 / 9.

Ans: D

Alternatively, we could label the lengths since we only care about ratios. C = 1, A + B = 2. Then knowing b is closer to a, we split A + B into three equal lengths, A is 2/3 and B is 4/3. Next we plug in to get (B + C) / (A + B + C) = 7/3 / (3) = 7/9.

Bunuel wrote:
Let \(d > c > b > a\). If c is twice as far from a as it is from d, and b is twice as far from c as it is from a, then \(\frac{(d - b)}{(d - a) }= ?\)

A. 2/9
B. 1/3
C. 2/3
D. 7/9
E. 3/2


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Re: Let d > c > b > a. If c is twice as far from a as it is from d, and b  [#permalink]

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New post 25 Nov 2019, 11:07
i did not understand the explanation above, can anyone simplify it or give an alternative method?
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Re: Let d > c > b > a. If c is twice as far from a as it is from d, and b   [#permalink] 25 Nov 2019, 11:07
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