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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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Re: If x and y are positive, is x < 10 < y? [#permalink]
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Re: If x and y are positive, is x < 10 < y? [#permalink]
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Bunuel wrote: The Official Guide for GMAT® Review, 13th Edition  Quantitative Questions ProjectIf x and y are positive, is x < 10 < y? (1) x < y and xy = 100 (2) x^2 < 100 < y^2 Practice Questions Question: 52 Page: 279 Difficulty: 600 GMAT Club is introducing a new project: The Official Guide for GMAT® Review, 13th Edition  Quantitative Questions ProjectEach 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. We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation. Thank you! (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.
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Re: If x and y are positive, is x < 10 < y? [#permalink]
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01 Oct 2012, 20:09
Answer is D. (Not providing any explanation coz Eva has covered the same important points.
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Re: If x and y are positive, is x < 10 < y? [#permalink]
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02 Oct 2012, 20:32
Bunuel wrote: If x and y are positive, is x < 10 < y?
(1) x < y and xy = 100 (2) x^2 < 100 < y^2
I think the answer is since (1) X = 1/10 and Y = 1000 and X = 1/100 and Y = 10,000 yield the same answer where X < Y and xy = 100, insufficient (2) same as above. X = (1/10)^2 and y = 1000^2 and X = (1/100)^2 and Y = 10,000^2 satisfy the equation x^2 < 100 < y^2 Together the 2 statements do not yield any new info, hence Nevermind, the question is not asking for the value of x and y. The above posters are correct



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Re: If x and y are positive, is x < 10 < y? [#permalink]
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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.
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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25 Nov 2013, 03:15



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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.
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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Re: If x and y are positive, is x < 10 < y? [#permalink]
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14 Apr 2014, 01:50
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. Answer: C. 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. 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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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. 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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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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Re: If x and y are positive, is x < 10 < y? [#permalink]
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19 Apr 2016, 14:16
correct answer is option D 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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