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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 given \(\frac{1}{x} +\frac{1}{y}<2.\). this cannot be true if x=y=1.... so one or both have to be > 1
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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) 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

Answer: B
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"The best day of your life is the one on which you decide your life is your own. No apologies or excuses. No one to lean on, rely on, or blame. The gift is yours - it is an amazing journey - and you alone are responsible for the quality of it. This is the day your life really begins." - Bob Moawab

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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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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Appreciation in KUDOS please! Knewton Free Test 10/03 - 710 (49/37) Princeton Free Test 10/08 - 610 (44/31) Kaplan Test 1- 10/10 - 630 Veritas Prep- 10/11 - 630 (42/37) MGMAT 1 - 10/12 - 680 (45/34)

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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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)

_________________

Appreciation in KUDOS please! Knewton Free Test 10/03 - 710 (49/37) Princeton Free Test 10/08 - 610 (44/31) Kaplan Test 1- 10/10 - 630 Veritas Prep- 10/11 - 630 (42/37) MGMAT 1 - 10/12 - 680 (45/34)

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.
_________________

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.

_________________

Appreciation in KUDOS please! Knewton Free Test 10/03 - 710 (49/37) Princeton Free Test 10/08 - 610 (44/31) Kaplan Test 1- 10/10 - 630 Veritas Prep- 10/11 - 630 (42/37) MGMAT 1 - 10/12 - 680 (45/34)

Re: x and y are positive integers. If 1/x + 1/y < 2, which of [#permalink]

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12 Feb 2016, 18:06

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) xy>1 (C) x/y + y/x < 1 (D) (x - y)^2 > 0 (E) None of the above

I thought it is some kind of trap here.. we can rewrite the original as: x+y<2xy

A - x=2, y=2 -> x+y is not greater than 4, yet 1/2 + 1/2 < 2. so A is out. B - if x and y are both positive integers, xy>1 all the times - looks good. C - x/y +y/x <1 or x^2 + y^2 < xy - which will never be true, if x and y are positive integers. D - x^2 + y^2 > 2xy - suppose x=2 and y=2. 4+4 = 8. 2*2*2=8. 8=8, it's not an inequality. E - since B works, e is out.

X and Y are positive integers. If 1/x + 1/y < 2, which of the followin [#permalink]

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24 Nov 2017, 16:51

jedit 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

\(\frac{1}{x} + \frac{1}{y} < 2\)

Let x = 1 and y = 2

\(\frac{1}{1} + \frac{1}{2} = \frac{3}{2}\)

\(\frac{3}{2} < 2\) -- Those numbers for x and y work

MUST be true?

A. X+Y>4 - NO 1 + 2 = 3, which is not greater than 4

B. XY>1 YES Because x and y are positive integers, there is only one way XY would NOT be greater than 1: if both x and y = 1. Then XY = 1. But x = y = 1 violates the prompt: their reciprocals summed must be less than 2; in that case, they equal 2. This choice must be true.

C. X/Y + Y/X < 1 - NO \(\frac{1}{2} + \frac{2}{1}=\frac{5}{2}\) \(\frac{5}{2}\) is not less than 1

D. (x-y)^2 > 0 NO For this option, let x=y=2. \((2-2)^2 = 0^2 = 0\) 0 is not greater than 0

E. none - NO - One of the answers, B, must be true.