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Ratio of distance on the number line

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Ratio of distance on the number line

by Vemuri » Fri May 29, 2009 9:37 pm
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
D. 7/20
E. 3/5

OA: C. Please explain your answers.
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by DanaJ » Fri May 29, 2009 10:34 pm
The distance from a to b is actually |b - a| - the absolute value of numbers actually is defined as a distance!

For the sake of brevity, we'll assume that all the numbers here are positive. In this case, |b - a| = b - a. If we were to consider that some numbers are negative, establishing whether |b - a| is b - a or -(b - a) would take up too much time.

So you know that:

1. The distance between a and b will be |b - a| = b - a.
The distance between d and b will be |d - b| = d - b.
b - a = (d - b)/4
4b - 4a = d - b
5b - 4a = d


2. c - a = 3(d - c)
c - a = 3d - 3c
4c - a = 3d
(4c - a)/3 = d

Combine the last things to get: (4c - a)/3 = 5b - 4a, which means that 4c - a = 15b - 12a. Subtract 3a from each side to get 4c - 4a = 15b - 15a
4c - 4a = 15b - 15a
4(c - a) = 15(b - a)
(b - a)/(c - a) = 4/15

Edit: I realized that I mixed up the second stmt.
Last edited by DanaJ on Fri May 29, 2009 11:03 pm, edited 1 time in total.
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by sureshbala » Fri May 29, 2009 10:37 pm
Let the distance from a to b, b to c and c to d be x, y and z respectively.

Given 4x = y + z

and x + y = 3z

We need to find x/(x+y)

Adding the above two equations we get , 5x = 4z i.e. x = 4z/5

Now x/(x+y) = (4z/5)/3z = 4/15
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