Is x > y ? (1) |x - y| < |y| (2) |x| > x

Is x > y ?


(1) |x - y| < |y|

The above statement implies that the distance between x and y (|x - y|), on the number line, is less than the distance between y and 0 (|y|):

If y is positive, we can have two cases:

--------------------0---------y---x---
--------------------0-----x---y-------

If y is negative, we can have another two cases:

----------y---x-----0---------------
------x---y---------0----------------

As we can see x may (green cases) or may not (red cases) be greater than y. Not sufficient.

• y cannot be 0 because in this case we'd get |x| < 0, which is not possible: the absolute value of a number cannot be negative.
• It's also possible that x = y AND y ≠ 0, which also gives a NO answer to the question.

(2) |x| > x

The above implies that x is negative. We know nothing about y. Not sufficient.

(1)+(2) From (2) we got that x is negative, so we are left with:

----------y---x-----0---------------
------x---y---------0----------------

And again, x may or may not be greater than y. Not sufficient.


Answer: E.

Hope it helps.

Hi
chetan2u , SaquibHGMATWhiz

I tried but couldn't get a simple solution by squaring (A) both sides for ex.
Can you please share a methodical approach to this question?

Hi happypuppy,

I would prefer to go the way Bunuel has solved this one because we can draw the number line directly from the absolute inequality.

However, we can also do it after doing the square.

\(|x - y|^2 < |y|^2\\
\)
\(x^2 - 2xy + y^2 < y^2 \)
\(x^2 - 2xy < 0\)
\(x(x - 2y) < 0\)

Now if you try to form the number line this can be true in two ways,

Scenario 1

x < 0 and
x - 2y > 0 or x > 2y
So we can easily draw x < 0 here on a number line but we need to understand how we write x > 2y.

If y is less than x (a negative number) here, then y (is negative here) can take any values as 2y is going to be smaller. So in such a case, y can go till - infinity

_y___________x______ 0

If y > x (a negative number) here, then y needs to be negative and y should be greater than x and less than x/2.
This part is a bit tricky and also important to understand The idea here is that y should have such a value so that y falls towards right or x but 2y falls towards left and for that y must be between x and x/2.

So, the second scenario from here looks like
________x______y__(x/2)________0

The point x/2 is given to make sure that you understand exactly where 'y' can be.

You can approach a similar way for scenario 2. x > 0 and x - 2y < 0 or x < 2y.

You should get the possibilities as,

________0____________________x______________y (y can be anywhere towards right of x)
________0______(x/2)___y______x_____________ (y in between x/2 and x)


Once we are done with this, I always suggest considering the zero cases separately. By zero case, I mean that x = y could also be a possibility.
Since I am already getting contradicting answers so I am not really bothered about the x = y scenario as it will not change my answers.


Hope this helps.

happypuppy, I would look at this type of question logically.

(I) What does |x-y|<|y| mean?
If both x and y are of opposite sign then numeric values of x and y will get added and MOD of the sum will surely be greater than y.
Hence, we can say x and y are of same sign.
The next logical conclusion will be that x is between 0 and 2y.

(II) |x|>x
This means x is negative


Combined
We can say that both x and y are negative.
So, two possibilities
a) 2y<x<y<0
b) 2y<y<x<0
Insufficient


E

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