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On a number line, a<b<c<d. The distance from a to b is 1⁄4 of the

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On a number line, a<b<c<d. The distance from a to b is 1⁄4 of the [#permalink]

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New post 06 Jan 2018, 22:18
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Difficulty:

  55% (hard)

Question Stats:

65% (04:28) correct 35% (02:34) wrong based on 43 sessions

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On a number line, \(a < b < c < d\). The distance from \(a\) to \(b\) is \(\frac{1}{4}\) of the distance from \(b\) to \(d\). The distance from \(a\) to \(c\) is \(3\) times the distance from \(c\) to \(d\). What is the value of \(\frac{(b-a)}{(c-a)}\)?


A. \(\frac{3}{20}\)

B \(\frac{1}{5}\)

C. \(\frac{4}{15}\)

D. \(\frac{7}{20}\)

E. \(\frac{3}{5}\)

Official explanation: Let's try to pick numbers that will be easy to work with. Let's begin by choosing a simple value for a, say a = 0. If a = 0, we'll need to pick additional values that make our calculations simple.
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Re: On a number line, a<b<c<d. The distance from a to b is 1⁄4 of the [#permalink]

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New post 06 Jan 2018, 22:40
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chesstitans wrote:
On a number line, a < b < c < d. The distance from a to b is 1⁄4 of the distance from b to d. The distance from a to c is 3 times the distance from c to d. What is the value of (b - a) ⁄ (c - a)?


A. 3/ 20
B 1⁄5
C. 4⁄15 [4][/15]
D. 7⁄20
E. 3⁄5

There is a same topic that has been locked. There is also a note that requests a re-post for discussion. I still have no idea how to have the proper format for fraction.

Official explanation: Let's try to pick numbers that will be easy to work with. Let's begin by choosing a simple value for a, say a = 0. If a = 0, we'll need to pick additional values that make our calculations simple.


Refer to below image

Attachment:
number line.jpg
number line.jpg [ 20.6 KiB | Viewed 613 times ]



Let the distance between b & d be \(4x\) so distance between a & b will be \(x\)

similarly if the distance between c & d is \(y\) then distance between c & a will be \(3y\)

We need ratio of distance between a & b and c & a i.e \(\frac{x}{3y}\)

it is evident that \(5x=4y\)

\(=>\frac{x}{y}=\frac{4}{5}\)

hence \(\frac{x}{3y}=\frac{4}{15}\)

so our ratio is \(\frac{4}{15}\)

Option C
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Re: On a number line, a<b<c<d. The distance from a to b is 1⁄4 of the [#permalink]

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New post 07 Jan 2018, 06:45
a<b<c<d
Let dist(b,d) =x
dist (a,b) = x/4

dist (c,d)=y; dist (a,c)=3*y

I need to find the ratio (x/4)/(3*y)
x+x/4 = y+3*y
=> x/4 = 4y/5

Resubstituting in the equation gives 4/15
Re: On a number line, a<b<c<d. The distance from a to b is 1⁄4 of the   [#permalink] 07 Jan 2018, 06:45
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