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Re: If x and y are positive, is x < 10 < y? [#permalink]
01 Oct 2012, 04:18

5

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

SOLUTION

If x and y are positive, is x < 10 < y?

(1) x < y and xy = 100. Since both \(x\) and \(y\) are positive AND \(x < y\), then in order \(xy=100\) to hold true, one multiple must be less than 10 and another greater than 10, thus \(x < 10 < y\). Sufficient.

(2) x^2 < 100 < y^2. Take the square root from all three parts: \(|x|<10<|y|\). Again, since both \(x\) and \(y\) are positive, then it transforms to \(x < 10 < y\). Sufficient.

Each week we'll be posting several questions from The Official Guide for GMAT® Review, 13th Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

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(1) If \(x=y\) and \(xy=100\), then \(x=y=10\). So, if the two positive numbers \(x\) and \(y\) are not equal, one must be smaller than \(10\) and the other one must be greater than \(10\). It is given that \(x<y\), so necessarily \(x<10<y\). Sufficient.

(2) Since we are given that \(x\) and \(y\) are positive, we can take the square root of all the sides in the given inequality and obtain \(x<10<y\). Sufficient.

Answer D. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: If x and y are positive, is x < 10 < y? [#permalink]
04 Oct 2012, 13:37

Expert's post

1

This post was BOOKMARKED

SOLUTION

If x and y are positive, is x < 10 < y?

(1) x < y and xy = 100. Since both \(x\) and \(y\) are positive AND \(x < y\), then in order \(xy=100\) to hold true, one multiple must be less than 10 and another greater than 10, thus \(x < 10 < y\). Sufficient.

(2) x^2 < 100 < y^2. Take the square root from all three parts: \(|x|<10<|y|\). Again, since both \(x\) and \(y\) are positive, then it transforms to \(x < 10 < y\). Sufficient.

Answer: D.

Kudos points given to everyone with correct solution. Let me know if I missed someone. _________________

Re: If x and y are positive, is x < 10 < y? [#permalink]
29 Jan 2013, 07:53

Option 2 is very clear.

Option 1 just drew a number line.

xy=100 Any combination of numbers say 5 X 20 or 4 X 25 etc can give me an 'x' greater than 10 as x should be less than y.

Probably this would be insufficient if x<y would not be given

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_________________

I've failed over and over and over again in my life and that is why I succeed--Michael Jordan Kudos drives a person to better himself every single time. So Pls give it generously Wont give up till i hit a 700+

Re: If x and y are positive, is x < 10 < y? [#permalink]
24 Nov 2013, 22:49

Bunuel wrote:

SOLUTION

If x and y are positive, is x < 10 < y?

(1) x < y and xy = 100. Since both \(x\) and \(y\) are positive AND \(x < y\), then in order \(xy=100\) to hold true, one multiple must be less than 10 and another greater than 10, thus \(x < 10 < y\). Sufficient.

(2) x^2 < 100 < y^2. Take the square root from all three parts: \(|x|<10<|y|\). Again, since both \(x\) and \(y\) are positive, then it transforms to \(x < 10 < y\). Sufficient.

Answer: D.

What If GMAT twist it by not giving that x and y are +ves? _________________

Re: If x and y are positive, is x < 10 < y? [#permalink]
25 Nov 2013, 02:15

Expert's post

honchos wrote:

Bunuel wrote:

SOLUTION

If x and y are positive, is x < 10 < y?

(1) x < y and xy = 100. Since both \(x\) and \(y\) are positive AND \(x < y\), then in order \(xy=100\) to hold true, one multiple must be less than 10 and another greater than 10, thus \(x < 10 < y\). Sufficient.

(2) x^2 < 100 < y^2. Take the square root from all three parts: \(|x|<10<|y|\). Again, since both \(x\) and \(y\) are positive, then it transforms to \(x < 10 < y\). Sufficient.

Answer: D.

What If GMAT twist it by not giving that x and y are +ves?

In this case the answer would be C. _________________

Re: If x and y are positive, is x < 10 < y? [#permalink]
13 Apr 2014, 10:27

Bunuel wrote:

honchos wrote:

Bunuel wrote:

SOLUTION

If x and y are positive, is x < 10 < y?

(1) x < y and xy = 100. Since both \(x\) and \(y\) are positive AND \(x < y\), then in order \(xy=100\) to hold true, one multiple must be less than 10 and another greater than 10, thus \(x < 10 < y\). Sufficient.

(2) x^2 < 100 < y^2. Take the square root from all three parts: \(|x|<10<|y|\). Again, since both \(x\) and \(y\) are positive, then it transforms to \(x < 10 < y\). Sufficient.

Answer: D.

What If GMAT twist it by not giving that x and y are +ves?

In this case the answer would be C.

2 questions:

1) What is +ves? 2) If the stem read that x & y could be positive OR negative, that would mean that only statement one is sufficient. Is that correct?

My reasoning being, the second inequality could become \(x<+-10<y\). Meaning, x would have to be less than -10, lets call it -25 and y would have to be 4 which would yield a "no" for the question. Additionally, if we took the positive squareroot, the question stem would yield a yes. Am I thinking about this the right way?

Re: If x and y are positive, is x < 10 < y? [#permalink]
14 Apr 2014, 00:50

Expert's post

russ9 wrote:

Bunuel wrote:

honchos wrote:

SOLUTION

If x and y are positive, is x < 10 < y?

(1) x < y and xy = 100. Since both \(x\) and \(y\) are positive AND \(x < y\), then in order \(xy=100\) to hold true, one multiple must be less than 10 and another greater than 10, thus \(x < 10 < y\). Sufficient.

(2) x^2 < 100 < y^2. Take the square root from all three parts: \(|x|<10<|y|\). Again, since both \(x\) and \(y\) are positive, then it transforms to \(x < 10 < y\). Sufficient.

Answer: D.

What If GMAT twist it by not giving that x and y are +ves?

In this case the answer would be C.

2 questions:

1) What is +ves? 2) If the stem read that x & y could be positive OR negative, that would mean that only statement one is sufficient. Is that correct?

My reasoning being, the second inequality could become \(x<+-10<y\). Meaning, x would have to be less than -10, lets call it -25 and y would have to be 4 which would yield a "no" for the question. Additionally, if we took the positive squareroot, the question stem would yield a yes. Am I thinking about this the right way?

1. +ve = positive.

2. If we were not told that x and y are positive, then the answer would be C, not A: Is x < 10 < y?

(1) x < y and xy = 100. If x=-20 and y=-5, then the answer is NO but if x=5 and y=20, then the answer is YES. Not sufficient. Notice that from xy = 100 we can deduce that x and y have the same sign.

(2) x^2 < 100 < y^2 --> -10 < x < 10 and |y|>10. So, y can be more than 10 as well as less than -10. Not sufficient.

(1)+(2) Since x < y, then y < -10 is not possible, thus y > 10. So, we have that x < 10 < y. Sufficient.

Re: If x and y are positive, is x < 10 < y? [#permalink]
04 May 2014, 07:21

Bunuel wrote:

russ9 wrote:

2 questions:

1) What is +ves? 2) If the stem read that x & y could be positive OR negative, that would mean that only statement one is sufficient. Is that correct?

My reasoning being, the second inequality could become \(x<+-10<y\). Meaning, x would have to be less than -10, lets call it -25 and y would have to be 4 which would yield a "no" for the question. Additionally, if we took the positive squareroot, the question stem would yield a yes. Am I thinking about this the right way?

1. +ve = positive.

2. If we were not told that x and y are positive, then the answer would be C, not A: Is x < 10 < y?

(1) x < y and xy = 100. If x=-20 and y=-5, then the answer is NO but if x=5 and y=20, then the answer is YES. Not sufficient. Notice that from xy = 100 we can deduce that x and y have the same sign.

(2) x^2 < 100 < y^2 --> -10 < x < 10 and |y|>10. So, y can be more than 10 as well as less than -10. Not sufficient.

(1)+(2) Since x < y, then y < -10 is not possible, thus y > 10. So, we have that x < 10 < y. Sufficient.

Answer: C.

Hope it's clear.

Hi Bunuel,

I see how you can prove that 1 is NOT sufficient although I'm having a hard time with #2. The equality reads x^2 < 100 < y^2. Doesn't that yield that x < +/- 10? Wouldn't that make it is x<10 or -x>10? I can tell that my signs are off but if I follow the math, they seem fine. What am I missing here? Assuming that this part of the problem is resolved(as I can see the light at the end of the tunnel), I still don't see how you get insufficient.

Assuming that the correct inequalities are -10<x<10 and 10<y<-10, are you saying that since the final inequality can be y<-10<x or x<10<y, therefore insufficient? But wouldn't we say that only x<10<y pertains to the main equation and therefore sufficient?

Re: If x and y are positive, is x < 10 < y? [#permalink]
04 May 2014, 07:30

Expert's post

russ9 wrote:

Bunuel wrote:

russ9 wrote:

2 questions:

1) What is +ves? 2) If the stem read that x & y could be positive OR negative, that would mean that only statement one is sufficient. Is that correct?

My reasoning being, the second inequality could become \(x<+-10<y\). Meaning, x would have to be less than -10, lets call it -25 and y would have to be 4 which would yield a "no" for the question. Additionally, if we took the positive squareroot, the question stem would yield a yes. Am I thinking about this the right way?

1. +ve = positive.

2. If we were not told that x and y are positive, then the answer would be C, not A: Is x < 10 < y?

(1) x < y and xy = 100. If x=-20 and y=-5, then the answer is NO but if x=5 and y=20, then the answer is YES. Not sufficient. Notice that from xy = 100 we can deduce that x and y have the same sign.

(2) x^2 < 100 < y^2 --> -10 < x < 10 and |y|>10. So, y can be more than 10 as well as less than -10. Not sufficient.

(1)+(2) Since x < y, then y < -10 is not possible, thus y > 10. So, we have that x < 10 < y. Sufficient.

Answer: C.

Hope it's clear.

Hi Bunuel,

I see how you can prove that 1 is NOT sufficient although I'm having a hard time with #2. The equality reads x^2 < 100 < y^2. Doesn't that yield that x < +/- 10? Wouldn't that make it is x<10 or -x>10? I can tell that my signs are off but if I follow the math, they seem fine. What am I missing here? Assuming that this part of the problem is resolved(as I can see the light at the end of the tunnel), I still don't see how you get insufficient.

Assuming that the correct inequalities are -10<x<10 and 10<y<-10, are you saying that since the final inequality can be y<-10<x or x<10<y, therefore insufficient? But wouldn't we say that only x<10<y pertains to the main equation and therefore sufficient?

Hope my question is clear.

\(x^2 < 100\) means that \(|x| < 10\) --> \(-10 < x < 10\) (so x IS less than 10). \(y^2>100\) means that \(|y| > 10\) --> \(y< -10\) or \(y>10\) (so y may be less as well as greater than 10).

For example, if \(x=0\) and \(y=100\), then YES \(x < 10 < y\) but if \(x=0\) and \(y=-100\), then \(x < 10 < y\) dose not hold true.

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