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M11-35

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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

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Re M11-35 [#permalink]

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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: M11-35 [#permalink]

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New post 15 Aug 2015, 02:18
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: M11-35 [#permalink]

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New post 17 Aug 2015, 04:02
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: M11-35 [#permalink]

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New post 07 Oct 2016, 10:02
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: M11-35 [#permalink]

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New post 07 Oct 2016, 10:13
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.

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Re: M11-35 [#permalink]

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New post 08 Oct 2016, 03:12
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


The question asks: which of the following must be true? If x = 1 and y = 3, x + y = 4, so A is NOT always true.
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Re: M11-35 [#permalink]

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New post 08 Oct 2016, 03:13
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 \(\frac{1}{x}+\frac{1}{y} \lt 2\).
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M11-35 [#permalink]

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New post 08 Oct 2016, 13:17
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.

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Re: M11-35 [#permalink]

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New post 25 Nov 2016, 23:17
I get the solution but for any number x and y, (x - y)^2 > 0, will always hold true?

eg : x = 3 y = 4 --> x-y = -1 --> (x-y)^2 = 1 > 0

That would also make option D as correct option.
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Re: M11-35 [#permalink]

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New post 25 Nov 2016, 23:41
sidoknowia wrote:
I get the solution but for any number x and y, (x - y)^2 > 0, will always hold true?

eg : x = 3 y = 4 --> x-y = -1 --> (x-y)^2 = 1 > 0

That would also make option D as correct option.


This is explained couple of times in the discussion above as well as in the official solution.

The question asks: which of the following must be true not could be true. If x=y=2, then D is not true!
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Re: M11-35 [#permalink]

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New post 12 May 2017, 11:11
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: M11-35 [#permalink]

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New post 07 Feb 2018, 07:03
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: M11-35 [#permalink]

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New post 07 Feb 2018, 07:21
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Re: M11-35 [#permalink]

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New post 08 Feb 2018, 06:54
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x+y < 2xy
put x =y = 1 Least positive integers
on LHS
2xy>2
xy>1

QED hence proved.
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Re: M11-35   [#permalink] 08 Feb 2018, 06:54
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