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# GRE Weekly Challenge #3

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GRE Weekly Challenge #3 [#permalink]

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14 Sep 2011, 21:57
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GMAT Club invites you to test your GRE knowledge for a chance to win! Each week, we will post a new Challenge Problem for you to attempt. If you submit the correct answer, you will be entered into that week’s drawing for a free Manhattan GRE Strategy guide. What are you waiting for? Get out their scrap paper and start solving! Click here to view contest & prize details

This week's question:

 A B |x – 5| – |a – b| |b – a| – |5 – x|

Compare Quantity A and Quantity B using the information given above, and select one of the following answer choices:
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

Please post your answer, along with the explanation, below. Get cracking!

Edit: This challenge is now closed

The winner of this Week's Challenge is... (drum roll).... m2l2. Congratulations! Please send me a pm with your shipping address, and choice of Manhattan GRE Guide

Details of next week's competition will be posted in the Weekly Challenge Master thread... Stay tuned!
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Re: GRE Weekly Challenge #3 [#permalink]

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14 Sep 2011, 23:13
Should be (B)
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Re: GRE Weekly Challenge #3 [#permalink]

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15 Sep 2011, 08:45
Answer B. The difference between B-A= 2|b-a| -2|x-5|>,0 since |b-a|>5 and |x-5|<5.
In details, b>5, a<0, b-a>5; x<10; x-5<5.
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Re: GRE Weekly Challenge #3 [#permalink]

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15 Sep 2011, 11:45
Answer B - Quantity B is greater.‎
‎|x – 5|=|5 – x|;‎
‎|a – b|=|b – a|;‎
As 5<x<10, then 0<|x – 5|<5 and 0<|5 – x|<5;‎
As a<0, b>0, then |a – b|=|b – a|=|a|+|b|;‎
‎|a|>0 and 5<|b|<10, then (|a|+|b|)>5;‎
Therefore, |x – 5| < |a – b|, (|x – 5| – |a – b|) < 0 and ‎
‎(|b – a| – |5 – x|) > 0, which means that‎
‎(|x – 5| – |a – b|) < (|b – a| – |5 – x|) for all a, b, x.‎
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Re: GRE Weekly Challenge #3 [#permalink]

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15 Sep 2011, 13:13
Quantity B is greater.
A=|x-5|-|a-b|
= (x-5)-(-(a-b)

since x>5, so the mod will always open positive.
since a<b, |a-b| will always open negetive.

so A= x-5+a-b ............(i)

B=|b-a|-|5-x|
= (b-a)-(-(5-x)
same logic as above.

so B= b-a+5-x ...............(ii)

so A= -(B) ............... from (i) and (ii)
now,
numerical value of b-a is greater than 5 since 0 and 5 lies between a and b.
numerical value of 5-x is less than 5 since x lies between 5 and 10.
so, B is a positive quantity or A is a negetive quantity.

therefore, B>A
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Re: GRE Weekly Challenge #3 [#permalink]

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15 Sep 2011, 19:53
Ans B

Let |x-5| = |5-x| = c --> 0<c<5
Let |a - b| = |b - a| = d --> 5<d<infinity

A = |x – 5| – |a – b| = (0<c<5) - (5<d<infinity) = -ve
B = |b – a| – |5 – x| = (5<d<infinity) - (0<c<5) = +ve

therefore B is greater
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Re: GRE Weekly Challenge #3 [#permalink]

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15 Sep 2011, 21:55
Ans: D

The intervals is not equal so we can not determine !!
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Re: GRE Weekly Challenge #3 [#permalink]

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16 Sep 2011, 05:23
The correct answer is: B)Quantity B is greater.

Quantity A:

0 < |x-5| < 5

|a-b| > b ==> |a-b| > 5

==>
|x – 5| – |a – b| = (0 < |x-5| < 5) - (|a-b| > 5) <=== This expression is clearly smaller than 0

Quantity B:

|b – a| = b+|a| > 5

|5 – x| < 5 (since 5 < x < 10 )

==>
|b – a| – |5 – x| = ( b+|a| > 5) - (|5 – x| < 5) <=== This expression is clearly greater than 0

So quantity B is greater than quantity A.
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Re: GRE Weekly Challenge #3 [#permalink]

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16 Sep 2011, 10:49
Quantity B

|5-x|-->X varies from 5 to 10 therefore |5-x| varies from 0 to 5
|b-a|--> B varies from 5 to 10 and is greater then X ..and A is less than 0.Therefore |b-a| is either greater then 10 or greater than 5.Which means greater than 5 covers the entire range.
Hence ,|b-a|-|5-x| is always positive...(1)

Quantity A
Since x > 0 , |x-5| is the same range as |5-x| varying from 0 to 5
|a-b| again as |b-a| varies from 5 to 10
Hence . |b-a|-|5-x| is < 0 ....(2)

Therefore Clearly , Quantity B is greater then Quantity A.
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Re: GRE Weekly Challenge #3 [#permalink]

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22 Sep 2011, 22:45
Re: GRE Weekly Challenge #3   [#permalink] 22 Sep 2011, 22:45
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# GRE Weekly Challenge #3

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