[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..

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