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# X and Y are positive integers. If \frac{1}{X} + \frac{1}{Y}

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X and Y are positive integers. If \frac{1}{X} + \frac{1}{Y} [#permalink]  02 Apr 2008, 14:48
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$$X$$ and $$Y$$ are positive integers. If $$\frac{1}{X} + \frac{1}{Y} \lt 2$$ , which of the following must be true?

* $$X + Y \gt 4$$

* $$XY \gt 1$$

* $$\frac{X}{Y} + \frac{Y}{X} \lt 1$$

* $$(X - Y)^2 \gt 0$$

* none of the above
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Re: PS: Inequality [#permalink]  02 Apr 2008, 18:28
I would say II...can explain if correct...OA?
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Re: PS: Inequality [#permalink]  02 Apr 2008, 18:32
neelesh wrote:
$$X$$ and $$Y$$ are positive integers. If $$\frac{1}{X} + \frac{1}{Y} \lt 2$$ , which of the following must be true?

* $$X + Y \gt 4$$

* $$XY \gt 1$$

* $$\frac{X}{Y} + \frac{Y}{X} \lt 1$$

* $$(X - Y)^2 \gt 0$$

* none of the above

1/x +1/y < 2 <--> (x+y)/xy <2
Since x>0. y> 0 --> x+y <2xy ---> xy>(x+y)/2
x, y are intergers--> x,y>=1 --> x+y >=2 --> (x+y)/2 >=1

---> xy>1--> B
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Re: PS: Inequality [#permalink]  02 Apr 2008, 19:52
i get B as well..remember x and y are positive integers
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Re: PS: Inequality [#permalink]  02 Apr 2008, 20:13
this can be illustarted well with an example

1/x+1/y<2 resolves to x+y/xy < 2

x y x+y xy
1 1 2 1 -> equation doesnt hold true
1 2 3 2 -> equation holds true
2 2 4 4 -> equation holds true
2 3 5 6 -> equation holds true
and so on...

for all the values other than first, the equation holds true. Hence the answer is xy > 1. (B)
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Re: PS: Inequality [#permalink]  04 Apr 2008, 23:28
neelesh wrote:
$$X$$ and $$Y$$ are positive integers. If $$\frac{1}{X} + \frac{1}{Y} \lt 2$$ , which of the following must be true?

* $$X + Y \gt 4$$

* $$XY \gt 1$$

* $$\frac{X}{Y} + \frac{Y}{X} \lt 1$$

* $$(X - Y)^2 \gt 0$$

* none of the above

for 1/x + 1/y to be 2, x=1 and y =1
for 1/x + 1/y less than 2, at least one of x and y should be greater than 1.
so xy will always be greater than 1
hence 'B'

Last edited by vshaunak@gmail.com on 05 Apr 2008, 16:27, edited 1 time in total.
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Re: PS: Inequality [#permalink]  05 Apr 2008, 13:40
Expert's post
pmenon wrote:
why would d be incorrect ?

for example, x=2 and y=2 --> (2-2)^2>0 - is not true but must be true.
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Re: PS: Inequality   [#permalink] 05 Apr 2008, 13:40
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