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.
A<X<Y<B. Is |A-X| <|X-B|?
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
And since X<Y thus 2X < 2Y
2X 2Y A+B
Hence 2X < A+B
This is what we had to prove..