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If w > x > y > z on the number line, y is halfway between x and z, and

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If w > x > y > z on the number line, y is halfway between x and z, and  [#permalink]

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New post 12 Dec 2015, 09:15
2
1
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

71% (01:55) correct 29% (01:50) wrong based on 115 sessions

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If w > x > y > z on the number line, y is halfway between x and z, and x is halfway between w and z, then (y - x)/(y - w) =

A. 1/4
B. 1/3
C. 1/2
D. 3/4
E. 1

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Re: If w > x > y > z on the number line, y is halfway between x and z, and  [#permalink]

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New post 14 Dec 2015, 05:22
1
Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.


If w > x > y > z on the number line, y is halfway between x and z, and x is halfway between w and z, then (y - x)/(y - w) =

A. 1/4
B. 1/3
C. 1/2
D. 3/4
E. 1


Let y-z=t ---> since y is halfway between x and z and x>y we have x-y=t. Moreover x-z=(x-y)+(y-z)=2t. Similarly since x is halfway between w and z, we have w-x=2t. So y-x=-t, y-w=-3t. ---> (y - x)/(y - w) = 1/3.

The answer is (B).
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Re: If w > x > y > z on the number line, y is halfway between x and z, and  [#permalink]

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New post 16 Dec 2015, 19:40
3
Hi All,

This question can be solved by TESTing VALUES.

We're given a few facts about 4 numbers on a number line:
1) w > x > y > z
2) y is halfway between x and z
3) x is halfway between w and z

We're asked for the value of (y - x)/(y - w). I'm going to start by picking a value for z and then working my way 'up' the number line...

IF...
z = 1
y = 2
x = 3
So now y is halfway between x and z. With these values 'locked in', we can't choose w randomly (since x is halfway between w and z).
w = 5

Using those values, (y - x)/(y - w) = (2-3)/(2-5) = (-1)/(-3) = 1/3

Final Answer:

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Re: If w > x > y > z on the number line, y is halfway between x and z, and  [#permalink]

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New post 23 Sep 2018, 15:58
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Re: If w > x > y > z on the number line, y is halfway between x and z, and &nbs [#permalink] 23 Sep 2018, 15:58
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