If x and y are positive integers and 1/x + 1/y

Question

If  x  and  y  are positive integers and  \frac{1}{x}+\frac{1}{y} \lt 2 , which of the following must be true?

A.  x + y \gt 4  
B.  xy \gt 1  
C.  \frac{x}{y} + \frac{y}{x} \lt 1  
D.  (x - y)^2 \gt 0  
E. none of the above

M11-35

chetan2u · Accepted Answer

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

Bunuel · Answer

Official Solution:

If  x  and  y  are positive integers and  \frac{1}{x}+\frac{1}{y} \lt 2 , which of the following must be true?

A.  x + y \gt 4  
B.  xy \gt 1  
C.  \frac{x}{y} + \frac{y}{x} \lt 1  
D.  (x - y)^2 \gt 0  
E. none of the above

Given that  x  and  y  are positive integers, it follows that they can take on the values  1, 2, 3, \ldots . However, if  x=y=1 , then  \frac{1}{x}+\frac{1}{y}=2 , which violates the condition that  \frac{1}{x}+\frac{1}{y}1 . This means that option B is always true.
 
We can demonstrate that options A, C, and D are not always true by providing counterexamples. For instance, if we set  x=y=2 , then we find that none of the options holds true.

Answer: B

h2polo · Answer

Answer is B:

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

subhashghosh · Answer

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.

Answer B.

gmatopoeia · Answer

Let X=1, 1+1/Y<2 1/Y<1 11 when X=1, A --> yes and no B --> yes C--> yes and no D--> yes and no Answer: B

KarishmaB · Answer

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)

krishnasty · Answer

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...

KarishmaB · Answer

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)

krishnasty · Answer

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..

KarishmaB · Answer

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.

krishnasty · Answer

Thanks Karishma for the information.

mbaiseasy · Answer

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!

Answer: B

AkshdeepS · Answer

My Solution:

Given : x and y are positive integers

Stem: 1/x+1/y<2---x+y<2xy (as x and y are positive we can cross multiply)

So A) x+y>4 Try x=1 & y = 2 (Not true)

B) xy>1 Try x=1 and y = 2 (Always true) ) [Note (x=1 and y =1 is not possible values because with these values stem doesn't holds true]

This is our answer as not more then one correct answer choice is possible but we can try all choices for more clarity:

C) x/y+y/x<1 Try x=1 and y = 2 (Not true)

D) (x-y)^2 Try x=2 and y=2 (Not true)

E) Can never be true

Answer is B

mvictor · Answer

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.

generis · Answer

\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. Answer B

ablatt4 · Answer

A) 1/2+1/1<2 but 1+2 is not >4 B) x+y <2xy --> 0<2xy-x-y IF xy<1 then one of x or y is less than .5 which would make 1/x +1/y >2 so xy must be >1 C)1/2 +2/1 = 2 1/2 which is greater than 2 D) if x is greater than y this can work if y is greater than x this can work, but this can work with so many solutions that it doesn't solve anything. B

If x and y are positive integers and 1/x + 1/y < 2, which of the

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