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Re M1135 [#permalink]
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15 Sep 2014, 23:45
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Official Solution:If \(x\) and \(y\) are positive integers and \(\frac{1}{x}+\frac{1}{y} \lt 2\), which of the following must be true? A. \(x + y \gt 4\) B. \(xy \gt 1\) C. \(\frac{x}{y} + \frac{y}{x} \lt 1\) D. \((x  y)^2 \gt 0\) E. none of the above Notice that since \(x\) and \(y\) are positive integers then each can be 1, 2, 3, ... Also notice that \(x=y=1\) is not possible since in this case \(\frac{1}{x}+\frac{1}{y}=2\), which violates given condition that \(\frac{1}{x}+\frac{1}{y} \lt 2\). So, at least one out of \(x\) and \(y\) must be more than 1, which makes \(xy\) greater than 1. So, option B is always true. As for the other options: consider \(x=y=2\) to see that neither holds true. Answer: B
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Re: M1135 [#permalink]
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15 Aug 2015, 01:18
Bunuel wrote: Official Solution:
If \(x\) and \(y\) are positive integers and \(\frac{1}{x}+\frac{1}{y} \lt 2\), which of the following must be true?
A. \(x + y \gt 4\) B. \(xy \gt 1\) C. \(\frac{x}{y} + \frac{y}{x} \lt 1\) D. \((x  y)^2 \gt 0\) E. none of the above
Notice that since \(x\) and \(y\) are positive integers then each can be 1, 2, 3, ... Also notice that \(x=y=1\) is not possible since in this case \(\frac{1}{x}+\frac{1}{y}=2\), which violates given condition that \(\frac{1}{x}+\frac{1}{y} \lt 2\). So, at least one out of \(x\) and \(y\) must be more than 1, which makes \(xy\) greater than 1. So, option B is always true. As for the other options: consider \(x=y=2\) to see that neither holds true.
Answer: B Thanks Bunuel, I understand that x=y=1 is not possible but why xy must be greater than 1 , (.5)(.3)=.06 also satisfies the inequality 1/x+1/y <2 and x and y as positive integers.



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Re: M1135 [#permalink]
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17 Aug 2015, 03:02
nancykumar wrote: Bunuel wrote: Official Solution:
If \(x\) and \(y\) are positive integers and \(\frac{1}{x}+\frac{1}{y} \lt 2\), which of the following must be true?
A. \(x + y \gt 4\) B. \(xy \gt 1\) C. \(\frac{x}{y} + \frac{y}{x} \lt 1\) D. \((x  y)^2 \gt 0\) E. none of the above
Notice that since \(x\) and \(y\) are positive integers then each can be 1, 2, 3, ... Also notice that \(x=y=1\) is not possible since in this case \(\frac{1}{x}+\frac{1}{y}=2\), which violates given condition that \(\frac{1}{x}+\frac{1}{y} \lt 2\). So, at least one out of \(x\) and \(y\) must be more than 1, which makes \(xy\) greater than 1. So, option B is always true. As for the other options: consider \(x=y=2\) to see that neither holds true.
Answer: B Thanks Bunuel, I understand that x=y=1 is not possible but why xy must be greater than 1 , (.5)(.3)=.06 also satisfies the inequality 1/x+1/y <2 and x and y as positive integers. First of all, 0.5 and 0.3 are NOT integers. They are positive numbers, not positive integers. Also, 1/0.5 + 1/0.3 = 5.(3) > 2 not < 2.
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Re: M1135 [#permalink]
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07 Oct 2016, 09:02
i didnt quite understand
wouldnt option A always be correct as well?
x + y > 4
so lets say x = 1, y = 4, their sum is 5 which is > 4
now i apply this in the question
1/1+1/4 < 2
5/4 = 1.25 which is less than 2.
so im confused



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Re: M1135 [#permalink]
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07 Oct 2016, 09:13
jonmarrow wrote: i didnt quite understand
wouldnt option A always be correct as well?
x + y > 4
so lets say x = 1, y = 4, their sum is 5 which is > 4
now i apply this in the question
1/1+1/4 < 2
5/4 = 1.25 which is less than 2.
so im confused And what would happen if both the values of x and y is equal to 1. Does the equality hold in such cases? Note that the question is asking for all cases. Posted from my mobile device



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Re: M1135 [#permalink]
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08 Oct 2016, 02:12



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Re: M1135 [#permalink]
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08 Oct 2016, 02:13



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Re: M1135 [#permalink]
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08 Oct 2016, 12:17
Bunuel wrote: shamim2k14 wrote: jonmarrow wrote: i didnt quite understand
wouldnt option A always be correct as well?
x + y > 4
so lets say x = 1, y = 4, their sum is 5 which is > 4
now i apply this in the question
1/1+1/4 < 2
5/4 = 1.25 which is less than 2.
so im confused And what would happen if both the values of x and y is equal to 1. Does the equality hold in such cases? Note that the question is asking for all cases. Posted from my mobile device Please read carefully: Also notice that \(x=y=1\) is not possible since in this case \(\frac{1}{x}+\frac{1}{y}=2\), which violates given condition that \(\frac{1}{x}+\frac{1}{y} \lt 2\).Exactly Bunuel, that was my point too. I was actually stating the same thing that the equality does not hold in that case. Posted from my mobile device



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Re: M1135 [#permalink]
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25 Nov 2016, 22:17
I get the solution but for any number x and y, (x  y)^2 > 0, will always hold true? eg : x = 3 y = 4 > xy = 1 > (xy)^2 = 1 > 0 That would also make option D as correct option.
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Re: M1135 [#permalink]
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25 Nov 2016, 22:41



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Re: M1135 [#permalink]
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12 May 2017, 10:11
Bunuel wrote: If \(x\) and \(y\) are positive integers and \(\frac{1}{x}+\frac{1}{y} \lt 2\), which of the following must be true?
A. \(x + y \gt 4\) B. \(xy \gt 1\) C. \(\frac{x}{y} + \frac{y}{x} \lt 1\) D. \((x  y)^2 \gt 0\) E. none of the above I solved it this way: \(\frac{1}{x}+\frac{1}{y} \lt 2\) \(\frac{x+y}{xy} \lt 2\) \(x+y \lt 2xy\) \(\frac{x+y}{2} \lt xy\) As x=y=1 isn't possible as per given condition. \(1 \lt 1.5 \lt xy\) B.
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Re: M1135 [#permalink]
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07 Feb 2018, 06:03
Hi Bunuel,
When I plug in numbers that comply with the restriction for D...I find it to be always true...how is D not always true?...I would've ended up guessing between D and B in the exam.



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Re: M1135 [#permalink]
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07 Feb 2018, 06:21



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Re: M1135 [#permalink]
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08 Feb 2018, 05:54
Aliter x+y < 2xy put x =y = 1 Least positive integers on LHS 2xy>2 xy>1 QED hence proved.
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