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

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If x and y are positive, is x < 10 < y?  [#permalink]

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01 Oct 2012, 05:17
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If x and y are positive, is x < 10 < y?

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

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Question: 52
Page: 279
Difficulty: 600

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

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01 Oct 2012, 05:18
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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.

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

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01 Oct 2012, 08:19
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Bunuel wrote:
The Official Guide for GMAT® Review, 13th Edition - Quantitative Questions Project

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

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

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Question: 52
Page: 279
Difficulty: 600

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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.

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

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04 Oct 2012, 14:37
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.

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

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29 Jan 2013, 08:53
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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Re: If x and y are positive, is x < 10 < y?  [#permalink]

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24 Nov 2013, 23:49
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.

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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25 Nov 2013, 03:15
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.

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]

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13 Apr 2014, 11:27
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.

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

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14 Apr 2014, 01:50
1
1
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.

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.

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

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04 May 2014, 08:21
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.

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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04 May 2014, 08:30
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.

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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23 Jun 2014, 18:42
I get the entire process, but I selected E, because when I took the square roots I thought that left me with two options
x < -10 < y and,
x < 10 < y

Because it is telling me that x and y are positive, then I no longer take into account a -10 as a possibility when I take the square root of 100?
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Re: If x and y are positive, is x < 10 < y?  [#permalink]

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24 Jun 2014, 06:05
carolinanmd wrote:
I get the entire process, but I selected E, because when I took the square roots I thought that left me with two options
x < -10 < y and,
x < 10 < y

Because it is telling me that x and y are positive, then I no longer take into account a -10 as a possibility when I take the square root of 100?

Since x and y are positive then x^2 < 100 < y^2 implies that x < 10 < y. How can it be x < -10?

Sorry don't understand what your question is.
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Re: If x and y are positive, is x < 10 < y?  [#permalink]

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19 Apr 2016, 14:16
statement 1 gives information that as product xy = 100 and x<y , x is less than 10 and y is greater than 10. sufficient
statement 2 gives information that |x|<10<|y| but as it is given x<y clearly sufficient
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Re: If x and y are positive, is x < 10 < y?  [#permalink]

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22 May 2016, 15:30
Here we are told both x and y are positive, IS x<10<y?

Before I begin, I say, x could be 9, or .5, or 7.5(if x < 10 < y is true)
y could be 10.01 or 100, or 24.5 (if x < 10 < y is true)

So, here we go

1) x<y and xy = 100
since x is less than y, and both x and y are positive, we know that x must be less than 10, and y more then 10 (sufficient)
2) x^2 < 100 < y^2
since x and y are both positive, this is sufficient

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

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28 Jun 2017, 09:24
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.

Hi Bunuel ,

As per statement 1, why can't we consider two possibilities of x=1, y =100 & the other usual values of x=10,y=10 ?
Both the sets follow x<y & xy=100.

We can, similarly, use this set of values for statement 2 as well. Considering these, shouldn't the answer be E. ?
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Re: If x and y are positive, is x < 10 < y?  [#permalink]

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28 Jun 2017, 09:30
TaN1213 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.

Hi Bunuel ,

As per statement 1, why can't we consider two possibilities of x=1, y =100 & the other usual values of x=10,y=10 ?
Both the sets follow x<y & xy=100.

We can, similarly, use this set of values for statement 2 as well. Considering these, shouldn't the answer be E. ?

I understood where I went Wrong. Apology for the question.

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

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14 Dec 2017, 16:09
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

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

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09 Jan 2019, 13:34
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.

Hello!
Thank you for the solution!
We can also look into fractions that are positive as well as it doesnt mention anywhere that it has to be integers.
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Re: If x and y are positive, is x < 10 < y?  [#permalink]

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25 Nov 2019, 09:39
1. x < y & xy = 100

So this can be 2 X 50 or 4 X 25 or 5 X 20. Thus x = 2,4,5 & y = 20,25,50. SUFFICIENT.

2. x^2 < 100 < y^2 => x^2 < 10^2 < y^2 => x < 10 < y. SUFFICIENT.

Re: If x and y are positive, is x < 10 < y?   [#permalink] 25 Nov 2019, 09:39

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