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Re: If x and y are positive, is x < 10 < y? [#permalink]
Bunuel
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?
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Re: If x and y are positive, is x < 10 < y? [#permalink]
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honchos
Bunuel
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.
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Re: If x and y are positive, is x < 10 < y? [#permalink]
Bunuel
honchos
Bunuel
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?
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If x and y are positive, is x < 10 < y? [#permalink]
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russ9
Bunuel
honchos
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 D:
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.
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Re: If x and y are positive, is x < 10 < y? [#permalink]
Bunuel
russ9


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.
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Re: If x and y are positive, is x < 10 < y? [#permalink]
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russ9
Bunuel
russ9


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.

Hope it's clear.
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Re: If x and y are positive, is x < 10 < y? [#permalink]
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Hi All,

We're told that X and Y are POSITIVE. We're asked if X < 10 < Y. This is a YES/NO question. We can solve it with a mix of TESTing VALUES and Number Properties.

1) X < Y and (X)(Y) = 100

With the information in Fact 1, we know that X and Y cannot be the same value (X < Y). IF they were the same value, then we would have (10)(10) = 100. Since that's NOT possible though - and both variables are POSITIVE - the only option is for X to DECREASE from 10 and Y to INCREASE from 10. Thus, the answer to the question is ALWAYS YES.
Fact 1 is SUFFICIENT

2) X^2 < 100 < Y^2

Since both variables are POSITIVE, we know that X MUST be less than 10 and that Y MUST be greater than 10. Thus, the answer to the question is ALWAYS YES.
Fact 2 is SUFFICIENT

Final Answer:

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Re: If x and y are positive, is x < 10 < y? [#permalink]
Bunuel
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.

Hello Bunuel, can you please explain why we cannot consider X and Y as +ve fractions? For example - X=100/21 and Y = 21
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Re: If x and y are positive, is x < 10 < y? [#permalink]
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NischalSR
Bunuel
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.

Hello Bunuel, can you please explain why we cannot consider X and Y as +ve fractions? For example - X=100/21 and Y = 21

You can but this values also give the same YES answer to the question: x < 10 < y.
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Re: If x and y are positive, is x < 10 < y? [#permalink]
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Bunuel
If x and y are positive, is x < 10 < y?

(1) x < y and xy = 100
(2) x^2 < 100 < y^2

Similar question from GMAT Club's Test: https://gmatclub.com/forum/is-x-10-y-1- ... 48284.html
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Re: If x and y are positive, is x < 10 < y? [#permalink]
Bunuel
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.
The question does not specify that x and y are integers, it specifies that they are positive, so for the 1st statement, x and y values could be 1/2 and 200. In which case, the answer should be B? Can someone correct my thought process regarding this point.
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Re: If x and y are positive, is x < 10 < y? [#permalink]
Expert Reply
afra94
Bunuel
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.
The question does not specify that x and y are integers, it specifies that they are positive, so for the 1st statement, x and y values could be 1/2 and 200. In which case, the answer should be B? Can someone correct my thought process regarding this point.

Since 1/2 < 10 < 200, doesn't your example, still gives a YES answer to the question?
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Re: If x and y are positive, is x < 10 < y? [#permalink]
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afra94
Bunuel
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.
The question does not specify that x and y are integers, it specifies that they are positive, so for the 1st statement, x and y values could be 1/2 and 200. In which case, the answer should be B? Can someone correct my thought process regarding this point.

Hi afra94,

You are correct that both X and Y COULD be non-integers. However, you still have to answer the question that is ASKED.

In this case, the question is "is X < 10 < Y?"

With your example (X = 1/2 and Y = 200), the answer to the question is still YES (just as it is with any other examples that fit the restrictions in Fact 1). This makes the results in Fact 1 consistent - and thus SUFFICIENT.

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