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CEO
Joined: 21 Jan 2007
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X and Y are positive integers. If 1/x + 1/y < 2, which of [#permalink]
 22 Nov 2007, 05:34
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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SVP
Joined: 21 Jul 2006
Posts: 1553
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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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Manager
Joined: 08 Nov 2007
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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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SVP
Joined: 21 Jul 2006
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no because the question clearly says that X and Y are positive integers
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SVP
Joined: 05 Jul 2006
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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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Intern
Joined: 06 Sep 2007
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If x and y are same int
is D right?
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Manager
Joined: 26 Sep 2007
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Answer is B (XY>1)
1/x + 1/y <2>1
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Manager
Joined: 26 Sep 2007
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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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Intern
Joined: 11 Nov 2007
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B is the answer. 1/x + 1/y <2> (x+y) / 2
(x+y)/2 >= 1
so xy > 1
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Director
Joined: 09 Aug 2006
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Re: 11.35 positive integers [#permalink]
22 Nov 2007, 22:54
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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CEO
Joined: 29 Aug 2007
Posts: 2530
Followers: 41
Kudos [?]:
357
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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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Intern
Joined: 28 Aug 2007
Posts: 19
Location: London, United Kingdom
Followers: 1
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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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CEO
Joined: 29 Aug 2007
Posts: 2530
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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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