M11-35

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

Bunuel · Accepted 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} < 2 . Therefore, we can conclude that at least one of  x  and  y  must be greater than  1 , and hence  xy>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

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

Mohit1994 · Answer

Why not A could be the ans ??? if y=1 and for any value of x >3 say- 4,5,6,7,8......... and so on then the resulting equation asked in the question will be satisfied as sum of 1+ reciprocal of any value of X would be always less than 2.

Bunuel · Answer

Please note that the question asks which of the following MUST be true, not what COULD be true. While option A, x + y > 4, could be true for specific values of x and y - say, for instance, if x = y = 10, it isn't necessarily always true. For example, consider x = 1 and y = 2. Only option B, xy > 1, is always true, as we concluded earlier, at least one of x or y must be greater than 1, and thus, xy > 1.

Does this make sense?

Mohit1994 · Answer

To be very honest, i didn't get, u put any positive integer value in x & y subject to condition satisfied for option A. You'll always get sum less than 2.
Even in worst case if u consider y=1 than reciprocal value of any integer X will always be less than 1 and sum of both the value would always be less than 2.

(worst scinario) case 1- y=1 and x=4 ( to satisfy option A equation- x+y > 4)

then ans would be 1/4+ 1/1= 1.25 which is less than 2

case 2- x=y=10 as u replied 
then ans would be 1/10 +1/10= 0.2 (less than 2)

case 3- y= 1 and x=10000
then ans would be 1/10000 +1/1= 1.0001 (less than 2)

case 4- x= 2 and y=3 
then ans would be 1/2+ 1/3= 0.833 (less than 2)

u try with any integer value for x & y subject to option A condition satisfied you will always get sum less than 2.

Bunuel · Answer

It's the other way around.

Given:  x  and  y  are positive integers and  \frac{1}{x}+\frac{1}{y} \lt 2 .
Question: Which of the following must be true?

Option A proposes:  x + y \gt 4 . Must this option always be true? No. For example, if x = 1 and y = 2, (please note that these values satisfy the given conditions that  x  and  y  are positive integers and  \frac{1}{x}+\frac{1}{y} \lt 2 ), then x + y > 4 is NOT true. Hence, option A must not always be true.

Hope it helps.

BottomJee · Answer

I think this is a  high-quality  question and I agree with explanation.

siri_99 · Answer

Simple explanation: 
 \frac{1}{x} + \frac{1}{y} = 2  => x + y < 2xy
Given x and y are positive integers, and hence must be greater than 0. The minimum possible value for either of them is 1. Which means, 2xy must be greater than 2
2xy > 1 + 1
Hence, xy > 1

Spectre26 · Answer

Concept:
AM >= GM >= HM (Harmonic Mean)

Note:
HM =  \frac{2}{(1/a + 1/b)} 
HP (Harmonic Progression):  \frac{1}{a}, \frac{1}{b}, \frac{1}{c}  .... are in HP, if  a, b, c  ... are in AP

Now coming to the question:

AM >= GM >= HM
 \frac{(x+y)}{2 }  >=  \sqrt{xy}  >=  \frac{2}{(1/x + 1/y)}

Given in question:
 1/x + 1/y  <  2 
 \frac{1}{(1/x+1/y)}  >  1/2 
 \frac{2}{(1/x+1/y)}  >  1

Since GM >= HM,
  \sqrt{xy}  >=  \frac{2}{(1/x + 1/y)}  >  1 
 xy  >  1 , Matches Option B

Correct Option: B

Shabaz18gmat · Answer

I don’t quite agree with the solution. Hello  Bunuel ,

If you consider x and y as different positive integers, provided x is not equal to y, then option D will be a must true, because whatever difference you get for x-y, the square of it will be always greater than zero.

Bunuel · Answer

The square of any integer is always greater than   or equal to 0  . So, for example, if x = y = 2, then (x - y)^2 = 0, and D is not necessarily true. Also, note that unless explicitly stated otherwise, different variables can represent the same value.

Shabaz18gmat · Answer

In the case of x=y, then xy>1 will also fail because if you take x=y=1, then you won't get xy>1, and moreover x cannot be equal to y because if x=y, then it will become:

1/x + 1/x = 2 
2/x = 2
x = 1 = y,

and if x=y=1, then 1/x + 1/y = 2

but in the question it is strictly mentioned as 1/x + 1/y < 2.

I think, you should remove one of the 2 options from the question

Bunuel · Answer

You are missing a point. We are asked to find which of the following must be true given the condition  \frac{1}{x}+\frac{1}{y} \lt 2 . If x = y = 1, then this condition is violated. I suggest reviewing the solution carefully again.

agrasan · Answer

I like the solution - it’s helpful.

M11-35

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