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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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Joined: 02 Sep 2009
Posts: 65785
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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11 Nov 2019, 01:43
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55% (hard)

Question Stats:

67% (02:55) correct 33% (02:49) wrong based on 180 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 $$\frac{(d - b)}{(d - a) }= ?$$

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

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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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11 Nov 2019, 05:18
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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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Joined: 19 Aug 2019
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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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25 Nov 2019, 10:07
i did not understand the explanation above, can anyone simplify it or give an alternative method?
Intern
Joined: 14 Jul 2019
Posts: 34
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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01 Mar 2020, 02:36
harshi17 wrote:
i did not understand the explanation above, can anyone simplify it or give an alternative method?

d-b = bd

ad = ab + bc + cd
bd = bc + cd

So to get the rate of (bd)/(ac), we have to somehow convert this ratio to a common unit. Let's choose bc.

"c is twice as far from a as it is from d":
ac = 2*cd
ab + bc = 2*cd (1)

"b is twice as far from c as it is from a"
bc = 2*ab
ab = 1/2*bc (2)

Combine (1) and (2) we have bc*(1+1/2) = 2*cd
3/2*bc = 2*cd
cd = 3/4*bc

(d-b)/(d-a) = (bc + cd)/(ab + bc + cd) = (bc +3/4*bc)/(1/2*bc + bc + 3/4*bc) = (1+3/4)/(1/2+1+3/4) = (7/4)/(9/4) = 7/9

IMO D
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Joined: 03 Dec 2017
Posts: 2
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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17 Jun 2020, 10:39
I found assigning values to each point A,B,C and D to be an easier way to solve the question.

As per the question we know that the AC = 2CD and CB=2AB

Let the value of A = 1 and B = 2, so distance between AB= 2-1 =1.
This gives us CB to be 2 units i.e C=4 (4-2=2).
Similarly, since AC = 3 units CD= 1.5 units
which gives us D=5.5 (4+1.5)

Now, using these values we can solve the equation:
(5.5-2)/(5.5-1) = 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 d, and b  [#permalink]

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20 Jun 2020, 14:27
1
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

Solution:

We can let a = 0 and b = 1. So c is 3 (since b is 2 units from c, which is twice as far as it is from a), and d is 4.5 (since c is 3 units from a, which is twice as far as it is from d).

Substituting these values into the expression, we have:

(4.5 - 1)/(4.5 - 0) = 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 d, and b  [#permalink]

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20 Jun 2020, 19:44
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 --> correct: a <b<c<d, let's say distance from a to b = 2p. Given "b is twice as far from c as it is from a", so distance from b to c = 2*2p=4p => distance from a to c = distance from a to b + distance from b to c = 2p+4p=6p. Given "c is twice as far from a as it is from d" so distance from c to d = (distance from a to c)/2 =6p/2=3p. $$\frac{(d - b)}{(d - a) }= \frac{ (distance-from-b-to-c + distance-from-c-to-d)}{(distance-from-a-to-b+distance-from-b-to-c+distance-from-c-to-d)} = \frac{ (4p+3p)}{(2p+4p+3p)} = \frac{ 7p}{9p}= \frac{ 7}{9}$$
E. 3/2
Re: Let d > c > b > a. If c is twice as far from a as it is from d, and b   [#permalink] 20 Jun 2020, 19:44