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X and Y are positive integers. If 1/x + 1/y < 2, which of

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CEO
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X and Y are positive integers. If 1/x + 1/y < 2, which of [#permalink] New post 22 Nov 2007, 04:34
X and Y are positive integers. If 1/x + 1/y < 2, which of the following must be true?

A. X+Y>4
B. XY>1
C. X/Y + Y/X < 1
D. (x-y)^2 > 0
E. none


how can i solve this quickly? not even sure how to start the approach
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 [#permalink] New post 22 Nov 2007, 07:00
this is what I think:

when looking at 1/x + 1/y < 2, i know for a fact that both X and Y can't be 1. because if they are, then 1/1 + 1/1 = 2 which doesn't agree with the question. The only way that the 2 fractions can be smaller is to have either one of the variables bigger than 1 or both variables bigger than 1. Therefore, for sure at least that XY must be greater than 1, therefore my answer is B.


what's the OA?
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 [#permalink] New post 22 Nov 2007, 07:16
tarek99 wrote:
this is what I think:

when looking at 1/x + 1/y < 2, i know for a fact that both X and Y can't be 1. because if they are, then 1/1 + 1/1 = 2 which doesn't agree with the question. The only way that the 2 fractions can be smaller is to have either one of the variables bigger than 1 or both variables bigger than 1. Therefore, for sure at least that XY must be greater than 1, therefore my answer is B.


what's the OA?


Couldnt you have one as -1 and the other as or even 0.5?
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 [#permalink] New post 22 Nov 2007, 08:11
no because the question clearly says that X and Y are positive integers
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 [#permalink] New post 22 Nov 2007, 10:42
X and Y are positive integers. If 1/x + 1/y < 2, which of the following must be true?

A. X+Y>4
B. XY>1
C. X/Y + Y/X < 1
D. (x-y)^2 > 0
E. none

NO NEED TO SOLVE , (X-Y)^2 IS ALWAYS +VE IE > 0 .

MY ANSWER IS D
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IF X=Y [#permalink] New post 22 Nov 2007, 20:22
If x and y are same int

is D right?
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 [#permalink] New post 22 Nov 2007, 20:59
Answer is B (XY>1)

1/x + 1/y <2>1
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 [#permalink] New post 22 Nov 2007, 21:00
Answer is B (XY>1)

Both x and y are positive integers, so minimum positive value for one of the variable can be 1 and minimum value for other variable will be greater than 1 in order to satisfy the equation.

So, XY>1
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 [#permalink] New post 22 Nov 2007, 21:07
B is the answer. 1/x + 1/y <2> (x+y) / 2

(x+y)/2 >= 1

so xy > 1
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Re: 11.35 positive integers [#permalink] New post 22 Nov 2007, 21:54
bmwhype2 wrote:
X and Y are positive integers. If 1/x + 1/y <2>4
B. XY>1
C. X/Y + Y/X <1> 0
E. none


how can i solve this quickly? not even sure how to start the approach


Getting B.

1/x + 1/y < 2 = (x + y)/xy < 2

Since x and y are +ve integers (given), x & y cannot both be equal to 1. Since one of them has to be greater than 1 (lets say y), then the product of xy has to be greater than 1 as well.
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 [#permalink] New post 23 Nov 2007, 00:34
bmwhype2 wrote:
X and Y are positive integers. If 1/x + 1/y < 2, which of the following must be true?

A. X+Y>4
B. XY>1
C. X/Y + Y/X < 1
D. (x-y)^2 > 0
E. none


seems the question is poorly structures: both B and D are correct since x and y are +ve integers.
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 [#permalink] New post 23 Nov 2007, 05:53
D isn't always true. x could be 2 and y could be 2

Then D would = 0.

B is only option that is always true.
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 [#permalink] New post 23 Nov 2007, 15:00
mantymooney wrote:
D isn't always true. x could be 2 and y could be 2

Then D would = 0.

B is only option that is always true.


agree. you are correct.
  [#permalink] 23 Nov 2007, 15:00
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