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X and Y. which is true?

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X and Y. which is true? [#permalink]  05 Sep 2009, 10:59
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38% (02:36) correct 61% (01:06) wrong based on 6 sessions
X and Y are positive integers. If 1/X + 1/Y < 2, which of the following must be true?

(A) X+Y>4
(B) X*Y>1
(C) X/Y+Y/X<1
(D) (X-Y)^2>0
(E) None of the above
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Re: X and Y. which is true? [#permalink]  05 Sep 2009, 11:12
whats the OA?is it B?
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Re: X and Y. which is true? [#permalink]  05 Sep 2009, 11:31
ans is b......
since x and y are both +ive int, x*y>1... only exception being x=y=1 as it is not given they are different integers...
however it is given1/x +1/y<2.. this cannot be true if x=y=1.... so one or both have to be > 1
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Re: X and Y. which is true? [#permalink]  05 Sep 2009, 12:20

1/X + 1/Y < 2

The maximum value of 1/X is 1 because if X equals any other number greater than one it will be a fraction. The same is true with 1/Y.

So 1/X and 1/Y will always be less than 2 as long as both X and Y are not both equal to one at the same time.

Another way of putting it is:

X*Y>1
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Re: X and Y. which is true? [#permalink]  06 Sep 2009, 02:02
OA B. Thank you.
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Re: X and Y. which is true? [#permalink]  14 Feb 2011, 08:36
My take on this :

From the equation :
x + y < 2xy

=> xy > (x+y)/2

So if x and y are two different positive integers, taking the two least values as 1 and 2, we have x > 1.5 at least. Hence xy > 1.

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Re: X and Y. which is true? [#permalink]  05 Sep 2011, 14:16
Quote:
X and Y are positive integers. If 1/X + 1/Y < 2, which of the following must be true?

(A) X+Y>4
(B) X*Y>1
(C) X/Y+Y/X<1
(D) (X-Y)^2>0
(E) None of the above

Let X=1,
1+1/Y<2
1/Y<1
1<Y

Y>1 when X=1,
A --> yes and no
B --> yes
C--> yes and no
D--> yes and no

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Re: X and Y. which is true? [#permalink]  05 Sep 2011, 23:20
1
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barakhaiev 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) X*Y>1
(C) X/Y+Y/X<1
(D) (X-Y)^2>0
(E) None of the above

Trying a few values makes us realize that the only relation that holds is (B). But how can we be sure that (B) holds for all acceptable values of X and Y.

1/X + 1/Y < 2 implies (1/X + 1/Y)/2 < 1
A useful property of positive numbers is AM >= GM
Arithmetic Mean >= Geometric Mean

Say, the numbers are 1/X and 1/Y
AM = (1/X + 1/Y)/2
It is given that (1/X + 1/Y)/2 < 1 so we know that AM < 1

GM = \sqrt{\frac{1}{X}*\frac{1}{Y}}

Since GM <= AM,

\sqrt{\frac{1}{X}*\frac{1}{Y}} < 1

\frac{1}{XY} < 1 (Squaring the inequality)

XY > 1 (X and Y are positive so the inequality doesn't change)
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Re: X and Y. which is true? [#permalink]  06 Sep 2011, 02:16
**
Quote:
1/X + 1/Y < 2 implies (1/X + 1/Y)/2 < 1
A useful property of positive numbers is AM >= GM
Arithmetic Mean >= Geometric Mean

Say, the numbers are 1/X and 1/Y
AM = (1/X + 1/Y)/2
It is given that (1/X + 1/Y)/2 < 1 so we know that AM < 1

GM = \sqrt{\frac{1}{X}*\frac{1}{Y}}

Since GM <= AM,

\sqrt{\frac{1}{X}*\frac{1}{Y}} < 1

\frac{1}{XY} < 1 (Squaring the inequality)

XY > 1 (X and Y are positive so the inequality doesn't change)

This is fantastic. A fool-proof method! Thanks!
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Re: X and Y. which is true? [#permalink]  06 Sep 2011, 02:56
VeritasPrepKarishma wrote:
barakhaiev 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) X*Y>1
(C) X/Y+Y/X<1
(D) (X-Y)^2>0
(E) None of the above

Trying a few values makes us realize that the only relation that holds is (B). But how can we be sure that (B) holds for all acceptable values of X and Y.

1/X + 1/Y < 2 implies (1/X + 1/Y)/2 < 1
A useful property of positive numbers is AM >= GM
Arithmetic Mean >= Geometric Mean

Say, the numbers are 1/X and 1/Y
AM = (1/X + 1/Y)/2
It is given that (1/X + 1/Y)/2 < 1 so we know that AM < 1

GM = \sqrt{\frac{1}{X}*\frac{1}{Y}}

Since GM <= AM,

\sqrt{\frac{1}{X}*\frac{1}{Y}} < 1

\frac{1}{XY} < 1 (Squaring the inequality)

XY > 1 (X and Y are positive so the inequality doesn't change)

yup... nice method... thanks Karishma
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Re: X and Y. which is true? [#permalink]  07 Sep 2011, 06:05
Guys, can you guide me how D is not true? coz last time i checked, square of any number is greater than 0. Even if x is less than y, still, it's square would me more than 0..unless, x = y...
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Re: X and Y. which is true? [#permalink]  07 Sep 2011, 07:54
krishnasty wrote:
Guys, can you guide me how D is not true? coz last time i checked, square of any number is greater than 0. Even if x is less than y, still, it's square would me more than 0..unless, x = y...

Given: 1/X + 1/Y < 2
Say X = 2, Y = 2
These values satisfy the inequality: 1/2 + 1/2 < 2

But they do not satisfy (D)
(X-Y)^2>0
(2-2)^2 = 0, not greater than 0
Hence (D) must not be true for all values. There are values that satisfy the inequality but does not satisfy (D)
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Re: X and Y. which is true? [#permalink]  07 Sep 2011, 08:03
Karishma, now i need a confirmation on GMAT questions...
lets say that if two unknowns are given (like X and Y ), can we assume that these two are equals? I thought if we say x and y, they are implicitly different numbers..

VeritasPrepKarishma wrote:
krishnasty wrote:
Guys, can you guide me how D is not true? coz last time i checked, square of any number is greater than 0. Even if x is less than y, still, it's square would me more than 0..unless, x = y...

Given: 1/X + 1/Y < 2
Say X = 2, Y = 2
These values satisfy the inequality: 1/2 + 1/2 < 2

But they do not satisfy (D)
(X-Y)^2>0
(2-2)^2 = 0, not greater than 0
Hence (D) must not be true for all values. There are values that satisfy the inequality but does not satisfy (D)

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Re: X and Y. which is true? [#permalink]  07 Sep 2011, 20:35
krishnasty wrote:
Karishma, now i need a confirmation on GMAT questions...
lets say that if two unknowns are given (like X and Y ), can we assume that these two are equals? I thought if we say x and y, they are implicitly different numbers..

VeritasPrepKarishma wrote:
krishnasty wrote:
Guys, can you guide me how D is not true? coz last time i checked, square of any number is greater than 0. Even if x is less than y, still, it's square would me more than 0..unless, x = y...

Given: 1/X + 1/Y < 2
Say X = 2, Y = 2
These values satisfy the inequality: 1/2 + 1/2 < 2

But they do not satisfy (D)
(X-Y)^2>0
(2-2)^2 = 0, not greater than 0
Hence (D) must not be true for all values. There are values that satisfy the inequality but does not satisfy (D)

Until and unless they mention 'distinct numbers' or 'X not equal to Y', X and Y can be equal. The equality can be a deal breaker/maker sometimes so you have to make sure that you have analyzed its effects too.
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Re: X and Y. which is true? [#permalink]  07 Sep 2011, 21:38
Thanks Karishma for the information.

Quote:
Until and unless they mention 'distinct numbers' or 'X not equal to Y', X and Y can be equal. The equality can be a deal breaker/maker sometimes so you have to make sure that you have analyzed its effects too.

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Re: X and Y. which is true? [#permalink]  26 Dec 2012, 23:51
barakhaiev 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) X*Y>1
(C) X/Y+Y/X<1
(D) (X-Y)^2>0
(E) None of the above

Let x=2
Let y=2

\frac{1}{2} + \frac{1}{2}< 2

A) X+Y=4 OUT!
B) 2*2 > 1 HOLD!
C) 2/2 + 2/2 = 2 < 1 OUT!
D) (2-2)^2 = 0 OUT!

Re: X and Y. which is true?   [#permalink] 26 Dec 2012, 23:51
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