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by bblast » Sun Aug 07, 2011 3:26 am
d>c>b>a

c is twice as far from a as it is from d
b is twice as far from c as it is from a

(d-b)/(d-a) = ?


2/9
1/3
2/3
7/9
3/2
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by Frankenstein » Sun Aug 07, 2011 3:30 am
Hi,
A--(2/3)---B-----(4/3)-----C---(1)--D
So, d-b = 1+4/3 = 7/3
d-a = 2/3 + 4/3 + 1 = 3
So, ratio is 7/3/3 = 7/9
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by dinaroneo » Sun Aug 07, 2011 3:35 am
D; 7/9
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by Anurag@Gurome » Sun Aug 07, 2011 3:35 am
bblast wrote:d>c>b>a

c is twice as far from a as it is from d
b is twice as far from c as it is from a

(d-b)/(d-a) = ?
We can pick numbers to our choice.
Say, a = 0 and b = 1
Hence, from "b is twice as far from c as it is from a", we have c = (1 + 2*1) = 3
And, from "c is twice as far from a as it is from d", we have d = (3 + 3/2) = 4.5

Hence, required ratio = (d - b)/(d - a) = (4.5 - 1)/(4.5 - 0) = 3.5/4.5 = 7/9

The correct answer is D.
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by ajaykpat » Sun Aug 07, 2011 4:08 am
Here a< b < c< d




Image

for d-b = 1 +4/3 = 7/3

for d-a = 3

d-b/d-a = 7/3*3 = 7/9

hence , ANS= D
[/img]
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by GmatKiss » Sun Aug 07, 2011 6:07 am
IMO:D

7/9

from given,

|___________|______|
2x x
a------b-----c-----d
|_____|___________|
y 2y

hence, we can substitute,

a(0)-----b(1)-----c(3)-----d(4.5)
0 1 2 1.5

subtituting values in (d-b)/(d-a) we get, .7/.9 = 7/9.
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