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

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Let d > c > b > a. If c is twice as far from a as it is from [#permalink] New post 05 May 2011, 07:37
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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 (d - b)/(d - a) = ?

A. 2/9
B. 1/3
C. 2/3
D. 7/9
E. 3/2
[Reveal] Spoiler: OA

Last edited by Bunuel on 19 Nov 2013, 02:05, edited 1 time in total.
Renamed the topic and edited the question.
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Re: If d > c > b > a [#permalink] New post 05 May 2011, 07:44
Draw a number line and label with x and y. You would have:

a b c d
----- ---- ----
y 2y x
------------
2x

Before calculating recognize it would be more effcient to express x or y in terms of the other. here 2x = 3y so x =3/2 y.

Now you can calculate d -b and d -a

d - b= x +2y = 7/2 y

d - a = x + 2x = 3x = 9/2y

Dividing both you get: 7/9

Answer D
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Re: If d > c > b > a [#permalink] New post 05 May 2011, 08:06
Better to draw a number line,
from 1st statement --> c-a = 2x and d-c = x; d-a = 3x
from 2nd statement --> b-a = y and c-b = 2y i.e. c-a = y+2y = 3y

From the 2 underlined eqs; 2x = 3y

d - b = x + 2y (d-c = x and c-b = 2y)
d -a = 3x

d - b/d-a = x + 2y/3x

substituting the value, = 7x/9x = 7/9
Answer is D.
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Re: If d > c > b > a [#permalink] New post 05 May 2011, 16:34
<----2x----><-x-->
a---b------c------d

<-y><--2y->

3y = 2x

(d-b)/(d-a) = (3x-y)/(3x) = 7/9

Answer is D.
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Re: If d > c > b > a [#permalink] New post 11 May 2011, 07:06
I tried algebra, took me some time to figure out.

c - a = 2(d-c)

=> 3c = 2d + a

c-b = 2(b-a)

=> c + 2a = 3b

d - a = d - (3c - 2d) = 3d - 3c

d - b = d - (c+2a)/3 = (3d - c - 2a)/3

= [3d - c - 2(3c - 2d)]/3

= (7d - 7c)/3

=> (d-b)/(d-a) = 7/3/3 = 7/9

Answer - D
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Re: Let d > c > b > a. If c is twice as far from a as it is from [#permalink] New post 01 Jan 2014, 06:13
kannn 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 (d - b)/(d - a) = ?

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


OK, this may get a bit complicated I suggest giving variables 'x' and 'y' to each and then solve. We don't even need smart numbers for this one

So we have

a---(y)--b--(2y)---c---(x)----d
-----------(2x)-------

Now we have that 3y = 2x

So, y = 2x/3

d-b/d-a = 3x-y/3x which will translate to 1 - y/3x

Replace y and one will get 7/9

Answer is thus D

Hope it helps!
Cheers!

J :)
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Re: Let d > c > b > a. If c is twice as far from a as it is from [#permalink] New post 06 May 2014, 15:38
For me this problem turns easy if we pick up some numbers. For example, lets start assuming that a = 0 and c = 6

If c is twice as far from a as it is from d, then d should be 9.
a------c---d
0 6 9

Then we have that b is twice as far from c as it is from a, so b must be 2
a--b----c---d
0 2 6 9

Then just replace: (9-2)/9=7/9
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Re: Let d > c > b > a. If c is twice as far from a as it is from [#permalink] New post 08 Jul 2014, 06:59
Let
a=0
d=3x
c=2x
b=2x/3

solve as per the relation = 7/9
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Re: Let d > c > b > a. If c is twice as far from a as it is from [#permalink] New post 10 Jul 2014, 09:32
Nayimoni wrote:
Draw a number line and label with x and y. You would have:

a b c d
----- ---- ----
y 2y x
------------
2x


Before calculating recognize it would be more effcient to express x or y in terms of the other. here 2x = 3y so x =3/2 y.

Now you can calculate d -b and d -a

d - b= x +2y = 7/2 y

d - a = x + 2x = 3x = 9/2y

Dividing both you get: 7/9

Answer D


Hi
I could not understand highlighted part...Can you pls elaborate ???

Thanks
Re: Let d > c > b > a. If c is twice as far from a as it is from   [#permalink] 10 Jul 2014, 09:32
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