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

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Is |A-X| <|X-B|? [#permalink] New post 13 Sep 2013, 03:28
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
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Manager
Manager
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Joined: 29 Aug 2013
Posts: 77
Location: United States
Concentration: Finance, International Business
GMAT 1: 590 Q41 V29
GMAT 2: 540 Q44 V20
GPA: 3.5
WE: Programming (Computer Software)
Followers: 0

Kudos [?]: 19 [0], given: 24

Re: Is |A-X| <|X-B|? [#permalink] New post 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
Manager
avatar
Joined: 29 Aug 2013
Posts: 77
Location: United States
Concentration: Finance, International Business
GMAT 1: 590 Q41 V29
GMAT 2: 540 Q44 V20
GPA: 3.5
WE: Programming (Computer Software)
Followers: 0

Kudos [?]: 19 [0], given: 24

Re: Is |A-X| <|X-B|? [#permalink] New post 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
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Joined: 10 Sep 2013
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Re: Is |A-X| <|X-B|? [#permalink] New post 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.
1 KUDOS received
Manager
Manager
avatar
Joined: 29 Aug 2013
Posts: 77
Location: United States
Concentration: Finance, International Business
GMAT 1: 590 Q41 V29
GMAT 2: 540 Q44 V20
GPA: 3.5
WE: Programming (Computer Software)
Followers: 0

Kudos [?]: 19 [1] , given: 24

Re: Is |A-X| <|X-B|? [#permalink] New post 14 Sep 2013, 22:23
1
This post received
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santoshbs wrote:
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.



Ok.. Let's take an example..

If A<X<Y<B and given that |B-Y| > |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<4<7<20 i.e. |20-7| > |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 < 2X(A-B)

A^2 - B^2 < 2Y (A-B)
i.e. (A-B) (A+B) < 2Y(A-B)
Since A < B, hence A-B will be negative and hence while cancelling the terms, we will have to reverse the sign
i.e. (A+B) > 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..
Re: Is |A-X| <|X-B|?   [#permalink] 14 Sep 2013, 22:23
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