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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]

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05 May 2011, 07:37
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66% (03:35) correct 34% (02:57) wrong based on 274 sessions

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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]

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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

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Re: If d > c > b > a [#permalink]

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05 May 2011, 08:06
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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
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Re: If d > c > b > a [#permalink]

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05 May 2011, 16:34
1
KUDOS
<----2x----><-x-->
a---b------c------d

<-y><--2y->

3y = 2x

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

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Re: If d > c > b > a [#permalink]

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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

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

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01 Jan 2014, 06:13
2
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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

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]

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06 May 2014, 15:38
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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]

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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]

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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

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

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

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30 Nov 2014, 11:44
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

When i first wrote two equations 2(d-c) = c-a & 2(b-a) = b-c and tried to solve, i realized its taking a lot of time...hence shifted to plug-in method

Let a=1 b=2 ...hence c = 4 d=5.5 from the given realtionship.

So d-b / d-a = 3.5/4.5 = 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]

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03 Dec 2014, 04:49
jlgdr wrote:
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

Hope it helps!
Cheers!

J

Hello,
could you explain these steps?

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

Replace y and one will get 7/9

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

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17 Jan 2015, 08:11
Spidy001 wrote:
<----2x----><-x-->
a---b------c------d

<-y><--2y->

3y = 2x

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

awesome !!
Kudos !!
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Lucky

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

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17 Jan 2015, 11:44
Ans D

take distance between c and d as x=>distance between b and c is 2x
take distance between a and b as y=>distance between a and c is 2y

with this we will get 3y=2x

BD = X+4X/3 = 7x/3

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

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13 Jan 2016, 08:22
Number line says that ratio is 1>x>2/3

only option is 7/9

D
Re: Let d > c > b > a. If c is twice as far from a as it is from   [#permalink] 13 Jan 2016, 08:22
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