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

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CEO
Joined: 21 Jan 2007
Posts: 2764
Location: New York City
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Kudos [?]: 350 [0], given: 4

X and Y are positive integers. If 1/x + 1/y < 2, which of [#permalink]  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
SVP
Joined: 21 Jul 2006
Posts: 1551
Followers: 8

Kudos [?]: 305 [0], given: 1

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?
Manager
Joined: 08 Nov 2007
Posts: 99
Followers: 1

Kudos [?]: 1 [0], given: 0

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?
SVP
Joined: 21 Jul 2006
Posts: 1551
Followers: 8

Kudos [?]: 305 [0], given: 1

no because the question clearly says that X and Y are positive integers
SVP
Joined: 05 Jul 2006
Posts: 1519
Followers: 5

Kudos [?]: 115 [0], given: 39

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 .

Intern
Joined: 06 Sep 2007
Posts: 8
Followers: 0

Kudos [?]: 0 [0], given: 0

IF X=Y [#permalink]  22 Nov 2007, 20:22
If x and y are same int

is D right?
Manager
Joined: 26 Sep 2007
Posts: 65
Followers: 1

Kudos [?]: 11 [0], given: 5

1/x + 1/y <2>1
Manager
Joined: 26 Sep 2007
Posts: 65
Followers: 1

Kudos [?]: 11 [0], given: 5

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
Intern
Joined: 11 Nov 2007
Posts: 3
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Kudos [?]: 0 [0], given: 0

B is the answer. 1/x + 1/y <2> (x+y) / 2

(x+y)/2 >= 1

so xy > 1
Director
Joined: 09 Aug 2006
Posts: 766
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Kudos [?]: 69 [0], given: 0

Re: 11.35 positive integers [#permalink]  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.
SVP
Joined: 29 Aug 2007
Posts: 2497
Followers: 57

Kudos [?]: 556 [0], given: 19

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.
Intern
Joined: 28 Aug 2007
Posts: 19
Location: London, United Kingdom
Followers: 1

Kudos [?]: 1 [0], given: 0

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.
SVP
Joined: 29 Aug 2007
Posts: 2497
Followers: 57

Kudos [?]: 556 [0], given: 19

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.
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