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# Is |A-X| <|X-B|?

Author Message
Intern
Joined: 10 Sep 2013
Posts: 7

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13 Sep 2013, 03:28
1
Found this question in GMAT Club Math book. The question has only one portion of DS to illustrate a concept.
I am unable to understand how to solve this.

Problem:
A<X<Y<B. Is |A-X| <|X-B|?
1) |Y-A|<|B-Y|

Rgds
SBS

--== Message from the GMAT Club Team ==--

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Manager
Joined: 29 Aug 2013
Posts: 74
Location: United States
GMAT 1: 590 Q41 V29
GMAT 2: 540 Q44 V20
GPA: 3.5
WE: Programming (Computer Software)

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13 Sep 2013, 03:41
santoshbs wrote:
Found this question in GMAT Club Math book. The question has only one portion of DS to illustrate a concept.
I am unable to understand how to solve this.

Problem:
A<X<Y<B. Is |A-X| <|X-B|?
1) |Y-A|<|B-Y|

Rgds
SBS

Logically if you think in terms of distances then,
Question is asking you if the distance between A and X is smaller than distance between X and B.

Since A<X<Y<B the number line looks like this
_______________________________
A----------------X-------------Y-------------- B

Now, If it is given to you that distance between B and Y is larger than the distance between Y and A
then Distance YB > Distance XA + Distance XY

and Distance XY + Distance YB = Distance XB

therefore Distance XB - Distance XY > Distance XA + Distance XY
Distance XB > Distance XA + 2 (Distance XY)

If Distance XB is greater than XA + 2 XY then Distance XB is obviously greater than Distance XA

i.e. |A-X| < |X-B|

Hope it helps.
Manager
Joined: 29 Aug 2013
Posts: 74
Location: United States
GMAT 1: 590 Q41 V29
GMAT 2: 540 Q44 V20
GPA: 3.5
WE: Programming (Computer Software)

### Show Tags

13 Sep 2013, 03:53
santoshbs wrote:
Found this question in GMAT Club Math book. The question has only one portion of DS to illustrate a concept.
I am unable to understand how to solve this.

Problem:
A<X<Y<B. Is |A-X| <|X-B|?
1) |Y-A|<|B-Y|

Rgds
SBS

Alternatively, we can also square the terms of LHS and RHS since both are Absolute values
in that case,
we have to prove that A^2 + X^2 - 2A*X < X^2 + B^2 - 2X*B which is A^2 - B^2 < 2X(A-B) ----- (1)

Now similarly with given statement,
Y^2 + A^2 -2A*Y < B^2 + Y^2 - 2B*Y
Again, A^2 - B^2 < 2Y (A-B) [Y^2 terms get cancelled on both the sides] ------ (2)

Also given is that X<Y, so if Again, A^2 - B^2 < 2Y (A-B) then also A^2 - B^2 < 2X(A-B)

Hence we have proved. Thus sufficient. Let me know if its not clear
Intern
Joined: 10 Sep 2013
Posts: 7

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14 Sep 2013, 10:25
shameekv wrote:
santoshbs wrote:
Found this question in GMAT Club Math book. The question has only one portion of DS to illustrate a concept.
I am unable to understand how to solve this.

Problem:
A<X<Y<B. Is |A-X| <|X-B|?
1) |Y-A|<|B-Y|

Rgds
SBS

Alternatively, we can also square the terms of LHS and RHS since both are Absolute values
in that case,
we have to prove that A^2 + X^2 - 2A*X < X^2 + B^2 - 2X*B which is A^2 - B^2 < 2X(A-B) ----- (1)

Now similarly with given statement,
Y^2 + A^2 -2A*Y < B^2 + Y^2 - 2B*Y
Again, A^2 - B^2 < 2Y (A-B) [Y^2 terms get cancelled on both the sides] ------ (2)

Also given is that X<Y, so if Again, A^2 - B^2 < 2Y (A-B) then also A^2 - B^2 < 2X(A-B)

Hence we have proved. Thus sufficient. Let me know if its not clear

Still not clear conceptually.
Manager
Joined: 29 Aug 2013
Posts: 74
Location: United States
GMAT 1: 590 Q41 V29
GMAT 2: 540 Q44 V20
GPA: 3.5
WE: Programming (Computer Software)

### Show Tags

14 Sep 2013, 22:23
1
[quote="santoshbs"][quote="shameekv"][quote="santoshbs"]Found this question in GMAT Club Math book. The question has only one portion of DS to illustrate a concept.
I am unable to understand how to solve this.

Problem:
A |Y-A| i.e. the distance between Y and B is greater than distance between Y and A

then we can assume numbers to be 1 |7-1| i.e. 13 > 6
Now if we add anything to 13 and subtract anything to 6
i.e. 13 + Distance|X-Y| > 6 - Distance|X-Y|

which is nothing but Distance |Y-B| + Distance|X-Y| = Distance |X-B|
and Distance |Y-A| - Distance |X-Y| = Distance |A-X|

This will be |B-X| > |A-X|
i.e. 13 + 3 > 6 - 3 i.e. 16 > 3

For 2nd method,

If there are absolute values on both the sides, you can always square both the terms. Since anything squared will always be positive. So there is no discrepancy whether the terms will be negative or positive.

As suggested in the previous posts, open up the terms you will get
AND A^2 - B^2 2Y

Similarly for the other term you will get
A+B > 2X

Now imagine a number line

Given that X<Y

X will be to the left of Y on number line.

and we have just got that A+B is to the right of 2Y,

So number line will look something like this

______________________
2Y A+B

And since X<Y thus 2X < 2Y

Therfore

____________________
2X 2Y A+B

Hence 2X < A+B

This is what we had to prove..

--== Message from the GMAT Club Team ==--

THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION.
This discussion does not meet community quality standards. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
Non-Human User
Joined: 09 Sep 2013
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07 Jul 2018, 00:00
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Re: Is |A-X| <|X-B|? &nbs [#permalink] 07 Jul 2018, 00:00
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