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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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Bunuel
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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] New post   11 Nov 2019, 02:43
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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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D
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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] New post   11 Nov 2019, 06:18
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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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harshi17
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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] 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] New post   01 Mar 2020, 03:36
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harshi17 wrote:
i did not understand the explanation above, can anyone simplify it or give an alternative method?


d-a = ad
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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Re: Let d > c > b > a. If c is twice as far from a as it is from d, and b [#permalink] New post   17 Jun 2020, 11:39
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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] New post   20 Jun 2020, 15:27
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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

Answer: D
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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] New post   20 Jun 2020, 20:44
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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 --> 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
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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] New post   13 Aug 2020, 07:26
Assuming all values will give solution asap, but trick is which values to choose and how.
I always try to assume the value on right side on the last as we can adjust it accordingly to Q .
In this question also, if we assume values of a,b,c which comes later in Q then problem is easy to solve ,

b is twice as far from a than C,
Let B = 2 , A =1, c=4
so c-b =2 * b-a
Now C is twice as far from A as from D, Now
c-a= 3 , D need to be half of C-A , i.e 1.5 : D= C+1.5 =4+1.5 =5.5
Solving we have
5.5-2 /5.5-1
3.5/4.5
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 d, and b [#permalink] New post   16 Jun 2023, 10:33
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