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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 |
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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).
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