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Expert replies
by thp510 » Sun Feb 27, 2011 6:40 am
Is x·|y| > y2?

(1) x > y
(2) y > 0

OA: C

Here's how I tried to rephrase.
Step 1a) First, I lets say Y is positive. So the inequality becomes:
x(y) > (+y)(+y)
Step 1b) Divide out the positive y on both sides and you get
x>y

Now, what if y was negative. Here's how I took the inequality.
x(y) > (-y)(-y)
Step 2a) Divide out the negative y and switch the inequality sign
(x*y)/(-y) > (-y*-y)/(-y)
-x<-y
Step 2b) Now, I divided by -1 again (I hate working with neg variables). When doing so, I switched the inequality once more.
x>y

So no matter what, the DS question is asking, "Is X>Y" ? So what's wrong with my approach other than the obvious numbers I can plug in to prove this is wrong? I initially choose A after doing both scenarios.

[/spoiler]
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Source: — Data Sufficiency |

by manpsingh87 » Sun Feb 27, 2011 6:59 am
thp510 wrote:Is x·|y| > y2?

(1) x > y
(2) y > 0

OA: C

Here's how I tried to rephrase.
Step 1a) First, I lets say Y is positive. So the inequality becomes:
x(y) > (+y)(+y)
Step 1b) Divide out the positive y on both sides and you get
x>y

Now, what if y was negative. Here's how I took the inequality.
x(y) > (-y)(-y)
Step 2a) Divide out the negative y and switch the inequality sign
(x*y)/(-y) > (-y*-y)/(-y)
-x<-y
Step 2b) Now, I divided by -1 again (I hate working with neg variables). When doing so, I switched the inequality once more.
x>y

So no matter what, the DS question is asking, "Is X>Y" ? So what's wrong with my approach other than the obvious numbers I can plug in to prove this is wrong? I initially choose A after doing both scenarios.

[/spoiler]
lxl = x, or -x depending upon whether x>0; or x<0;

now in the step 2a of your post if you assume y to be negative than lyl = -y ; you have made an error there,

also as nothing is mention about x therefore x can be either positive or negative as well; so we have to consider these cases as well.

try to solve it again..!! i'm sure you will figure out the answer..!!!
Last edited by manpsingh87 on Sun Feb 27, 2011 8:17 am, edited 2 times in total.
O Excellence... my search for you is on... you can be far.. but not beyond my reach!
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by maihuna » Sun Feb 27, 2011 7:27 am
1. Is not suff, as if x or y<0, it will be always false, for positive alue either ay happen.

2. Only knowing y is positive will not help any further.

Now both combined : 0<y<x

So we know both sides are positive : since x>y and and y>0, |y| = y so x.y > y.y => x>y which is given as premise. So C
thp510 wrote:Is x·|y| > y2?

(1) x > y
(2) y > 0
[/spoiler]
Charged up again to beat the beast :)
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by Anurag@Gurome » Sun Feb 27, 2011 8:07 pm
Solution:
The question is asking whether x.lyl > y2 or not.
Now y2 = lyl.lyl.
Or we need to know whether x.lyl > lyl.lyl or not?
Since lyl > 0, we divide both sides of the above inequality by lyl without changing the inequality sign.
Or we can say, we need to know whether x > lyl or not.
Let us consider first (1) alone.
It says x > y.
Now, let x = 2 and y = -3. Or lyl = 3.
Here, since 2 > -3, x > y but x < lyl.
Next, let x = 4 and y = 3. Or lyl = 3.
Here, since 4 > 3, x > y and x > lyl.
So, from (1) alone, nothing definite can be said.
Next, let us consider (2) alone.
It says y > 0.
But, we know nothing about x.
Or (2) alone is not sufficient to answer the question.
We next combine both the statements together and check.
On combining, we have that x > y and y > 0.
If y > 0, lyl = y.
Or x > lyl.
So, the answer to the main question is yes.
Or both statements together are sufficient to answer the question.

The correct answer is (C).
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
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