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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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Joined: 12 Dec 2016
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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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06 Jan 2018, 22:18
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45% (medium)

Question Stats:

69% (02:47) correct 31% (02:09) wrong based on 52 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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06 Jan 2018, 22:40
1
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 [ 20.6 KiB | Viewed 1002 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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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
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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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08 Apr 2019, 15:15
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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] 08 Apr 2019, 15:15
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