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# M11-35

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Math Expert
Joined: 02 Sep 2009
Posts: 47983
M11-35  [#permalink]

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16 Sep 2014, 00:45
2
4
00:00

Difficulty:

25% (medium)

Question Stats:

77% (01:04) correct 23% (00:35) wrong based on 116 sessions

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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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16 Sep 2014, 00:45
3
1
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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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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Posts: 47983
Re: M11-35  [#permalink]

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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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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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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.

Posted from my mobile device
Math Expert
Joined: 02 Sep 2009
Posts: 47983
Re: M11-35  [#permalink]

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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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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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Re: M11-35  [#permalink]

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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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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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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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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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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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07 Feb 2018, 07:21
ttaiwo wrote:
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.

If x=y=2, then D is not true.
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Re: M11-35  [#permalink]

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08 Feb 2018, 06:54
1
Aliter

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 &nbs [#permalink] 08 Feb 2018, 06:54
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# M11-35

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