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Is x > y^2? (1) x > y+5 (2) x^2-y^2 = 0 [#permalink]

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12 Jul 2010, 17:53

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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

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.

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.

Mike
_________________

Mike McGarry Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

Re: Is x > y^2? (1) x > y+5 (2) x^2-y^2 = 0 [#permalink]

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07 Dec 2014, 02:18

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 \(|x|=|y|\)

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is x > y^2?

(1) x > y+5

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

-> In the original condition, there are 2 variables(x, y), which should match with the number of equations. So, you need 2 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer. In 1) & 2), it is x-y>5 for 1). In case of 2), x=y, x=-y. When x=y, it is 0>5 from x-y>5, which is impossible and becomes x=-y. Thus, x>y^2? --> -y>y^2? --> 0>y^2+y? --> 0>y(y+1)? --> -1<y<0?. In 1), -y>y+5, -5>2y, -5/2>y, -2.5>y, which is no and sufficient. Therefore, the answer is C.

During the exam, it is better to choose C since there are 2 variables.

-> For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
_________________

Re: Is x > y^2? (1) x > y+5 (2) x^2-y^2 = 0 [#permalink]

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23 Nov 2014, 23:03

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Re: Is x > y^2? (1) x > y+5 (2) x^2-y^2 = 0 [#permalink]

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07 Dec 2014, 01:29

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?

Re: Is x > y^2? (1) x > y+5 (2) x^2-y^2 = 0 [#permalink]

Show Tags

11 Dec 2015, 01:22

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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Re: Is x > y^2? (1) x > y+5 (2) x^2-y^2 = 0 [#permalink]

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18 Jul 2017, 12:19

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.

Bunuel, the question is: IS x>y^2? the rephrase is: is x greater or equal to 0? So, if u combine 1 and 2, then u get x=-y, right? Now, if y=10, then x=-10---->No if y=-10, then x=10--->Yes. So, how C make sense? Thank you__
_________________

“The heights by great men reached and kept were not attained in sudden flight but, they while their companions slept, they were toiling upwards in the night.” ― Henry Wadsworth Longfellow

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.

Bunuel, the question is: IS x>y^2? the rephrase is: is x greater or equal to 0? So, if u combine 1 and 2, then u get x=-y, right? Now, if y=10, then x=-10---->No if y=-10, then x=10--->Yes. So, how C make sense? Thank you__

y = 10 and x = -10 is not possible. These values does not satisfy the first statement. See, when combining we got that \(y<-\frac{5}{2}<0\).
_________________

Re: Is x > y^2? (1) x > y+5 (2) x^2-y^2 = 0 [#permalink]

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18 Jul 2017, 12:34

Bunuel wrote:

iMyself wrote:

Bunuel wrote:

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.

Bunuel, the question is: IS x>y^2? the rephrase is: is x greater or equal to 0? So, if u combine 1 and 2, then u get x=-y, right? Now, if y=10, then x=-10---->No if y=-10, then x=10--->Yes. So, how C make sense? Thank you__

y = 10 and x = -10 is not possible. These values does not satisfy the first statement. See, when combining we got that \(y<-\frac{5}{2}<0\).

Thank you Bunuel. God be with you!
_________________

“The heights by great men reached and kept were not attained in sudden flight but, they while their companions slept, they were toiling upwards in the night.” ― Henry Wadsworth Longfellow

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

Statement 2: X² - y² = 0 I.e. IxI = IyI I.e. for integer values of x and y X < y² For 0<x<1 x> y² Not Sufficient

Combining Now we can only take values of x out of the range 0 and 1

For all values of x and y now, X is not greater than y^2 hence

Sufficient

C

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