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Is x > y^2? (1) x > y+5 (2) x^2-y^2 = 0 [#permalink]
12 Jul 2010, 16:53

1

This post received KUDOS

2

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

40% (02:34) correct
60% (01:35) wrong based on 217 sessions

Is x > y^2?

(1) x > y+5

(2) x^2-y^2 = 0

Hello,

I was wondering if someone can help with providing a detailed explanation as to how they arrived at correct answer . The explanation on the test (GMAT Club Test m2#19) review is a bit brief. Thanks

Re: Is x > y^2? (1) x > y+5 (2) x^2-y^2 = 0 [#permalink]
12 Jul 2010, 18:39

3

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

tonebeeze wrote:

Hello,

I was wondering if someone can help with providing a detailed explanation as to how they arrived at (c). The explanation on the test review is a bit brief. Thanks

Is x>y^2?

(1) x>y+5

(2) x^2-y^2 = 0

Is \(x>y^2\)?

(1) \(x>y+5\) --> \(x-y>5\). Clearly insufficient, for example: if \(x=1\) and \(y=-10\) then the answer is NO, but if \(x=10\) and \(y=1\) then the answer is YES. Two different answers, hence not sufficient.

(2) \(x^2-y^2=0\) --> \((x-y)(x+y)=0\) --> so either \(x-y=0\) or \(x+y=0\). Also insufficient: if \(x=1\) and \(y=1\), then answer is NO, buy if \(x=\frac{1}{2}\) and \(y=\frac{1}{2}\), then the answer is YES. Two different answers, hence not sufficient.

(1)+(2) As from (1) \(x-y>5\neq{0}\), then from (2) must be true that \(x+y=0\) --> so \(x=-y\) --> substitute \(x\) in (1) --> \(-y-y>5\) --> \(y<-\frac{5}{2}<0\), as \(x=-y\), then \(x>\frac{5}{2}>0\), so \(y^2\) (or which is the same \(x^2\)) will always be more than \(x\), thus the answer to the question "Is \(x>y^2\)" is NO. Sufficient.

To elaborate more as \(x=-y>0\), the only chance for \(x>y^2\) to hold true (or which is the same for \(x>x^2\) to hold true) would be if \(x\) is fraction (\(0<x<1\)). For example if \(x=\frac{1}{2}\) and \(y=-\frac{1}{2}\) then \(x=\frac{1}{2}>y^2=\frac{1}{4}\). But the fact that \(x>\frac{5}{2}>0\) rules out this option.

Re: Is x > y^2? (1) x > y+5 (2) x^2-y^2 = 0 [#permalink]
08 Jan 2012, 21:58

2

This post received KUDOS

Expert's post

Hi, there! I'm happy to help with this.

The question: is x > y^2

Statement #1: x > y + 5

This doesn't necessarily tell us anything. If y = 1, and x = 7, then x > y^2, but if y = -6 and x = 0, then x < y^2. But itself, Statement #1 is not sufficient.

Statement #2: x^2 - y^2 = 0

This means that x^2 = y^2, which means that x = ±y. Same absolute value, but both could be positive, both could be negative, or either one could be positive and the other negative. We know that y^2 will be positive, but the x can be positive or negative, so by itself, Statement #2 is insufficient.

Combined Now, we know that x^2 - y^2 = 0 ---> x = ±y, AND we know that x > y + 5. This leads immediate to a few conclusions (a) x is positive and y is negative --- that's the only way they could have the same absolute value, but with x bigger than y + 5 (b) x and y must have an absolute value greater than 2.5, so that the different between positive x and negative y is more than 5

So we are comparing a positive number x, greater than 2.5, to the square of the negative number with the same absolute value. Of course, x^2 and y^2 are equal, so the question really boils down to: given that x > 2.5, is x > x^2?

For all x greater than one, the square of x is greater than x. That's because, squaring is multiplying a number by itself, and when you multiply anything by a number greater than one, it gets bigger.

Thus, if x > 2.5, when we square it, it will get bigger. Therefore, x^2 = y^2 > x for all values of x > 2.5.

Thus, combined, the statements are sufficient together. Answer = C

Does that make sense? Please let me know if you have any questions.

Re: Is x > y^2? (1) x > y+5 (2) x^2-y^2 = 0 [#permalink]
07 Dec 2014, 01:18

1

This post received KUDOS

usre123 wrote:

Hello, could someone please remove the highlighted part from the original post? (also from my post now, I suppose). Also, I just wanted to know, if we could also write \(x^2-y^2=0\) as \(x^2=y^2\) which is the same as \(|x|=|y|\). Just asking because I've become slightly comfortable with solving with absolute values. So is this ok?

yes, you can use \(x^2-y^2=0\) as \(x^2=y^2\) or \(|x|=|y|\)

Re: Is x > y^2? (1) x > y+5 (2) x^2-y^2 = 0 [#permalink]
23 Nov 2014, 22:03

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Re: Is x > y^2? (1) x > y+5 (2) x^2-y^2 = 0 [#permalink]
07 Dec 2014, 00:29

Bunuel wrote:

tonebeeze wrote:

Hello,

I was wondering if someone can help with providing a detailed explanation as to how they arrived at (c). The explanation on the test review is a bit brief. Thanks

Is x>y^2?

(1) x>y+5

(2) x^2-y^2 = 0

Is \(x>y^2\)?

(1) \(x>y+5\) --> \(x-y>5\). Clearly insufficient, for example: if \(x=1\) and \(y=-10\) then the answer is NO, but if \(x=10\) and \(y=1\) then the answer is YES. Two different answers, hence not sufficient.

(2) \(x^2-y^2=0\) --> \((x-y)(x+y)=0\) --> so either \(x-y=0\) or \(x+y=0\). Also insufficient: if \(x=1\) and \(y=1\), then answer is NO, buy if \(x=\frac{1}{2}\) and \(y=\frac{1}{2}\), then the answer is YES. Two different answers, hence not sufficient.

(1)+(2) As from (1) \(x-y>5\neq{0}\), then from (2) must be true that \(x+y=0\) --> so \(x=-y\) --> substitute \(x\) in (1) --> \(-y-y>5\) --> \(y<-\frac{5}{2}<0\), as \(x=-y\), then \(x>\frac{5}{2}>0\), so \(y^2\) (or which is the same \(x^2\)) will always be more than \(x\), thus the answer to the question "Is \(x>y^2\)" is NO. Sufficient.

To elaborate more as \(x=-y>0\), the only chance for \(x>y^2\) to hold true (or which is the same for \(x>x^2\) to hold true) would be if \(x\) is fraction (\(0<x<1\)). For example if \(x=\frac{1}{2}\) and \(y=-\frac{1}{2}\) then \(x=\frac{1}{2}>y^2=\frac{1}{4}\). But the fact that \(x>\frac{5}{2}>0\) rules out this option.

Answer: C.

Hope it's clear.

Hello, could someone please remove the highlighted part from the original post? (also from my post now, I suppose). Also, I just wanted to know, if we could also write \(x^2-y^2=0\) as \(x^2=y^2\) which is the same as \(|x|=|y|\). Just asking because I've become slightly comfortable with solving with absolute values. So is this ok?

Back to hometown after a short trip to New Delhi for my visa appointment. Whoever tells you that the toughest part gets over once you get an admit is...