M12-16

Question

If  x  and  y  are positive integers, is   \frac{x}{y}+ \frac{y}{x} \gt 2 ?

(1)  x  does not equal  y 
 
(2)  x  and  y  have no common factor other than 1.

Bunuel · Answer

Official Solution:

If  x  and  y  are positive integers, is   \frac{x}{y}+ \frac{y}{x} \gt 2 ? 
 
Simplify the question:
  
Is  \frac{x}{y}+\frac{y}{x} > 2 ?
 
Is  \frac{x^2+y^2}{xy} > 2 ?
  
Since both variables are positive, we can safely multiply by  xy :
  
Is  x^2+y^2 > 2xy ?
 
Is  x^2-2xy+y^2 > 0 ?
 
Is  (x-y)^2 > 0 ? 
  
Now, if  x  does not equal  y , the answer to this question will be YES, but if  x=y , then the answer will be NO, since in this case  (x-y)^2=0 .
 
(1)  x  does not equal  y . Directly answers the question. Sufficient.
 
(2)  x  and  y  have no common factor other than 1.
 
If  x=y=1 , then the answer is NO, but if  x=1  and  y=2 , then the answer is YES. Not sufficient.

Answer: A

georgetags10 · Answer

Quick question:

Statement 2: would imply that the numbers are consecutive, BUT only if the initial question stem characterized X and Y as two distinct positive integers, correct?

I picked D for this question, upon review I understand why it is not the correct answer. I would like to make sure my understanding of statement two is correct.

Thank you very much.

Bunuel · Answer

No. Even then, x and y have not to be consecutive integers, just co-prime. For example, 1 and any other integer, any two distinct primes (for example 2 and 7), ...

rhnbansal · Answer

I did the square root of (x-y)^2>0 to get x>y. Can we not do that in this case?

Bunuel · Answer

The point is that  \sqrt{x^2}=|x| . So, if you take the square root from  (x-y)^2 \gt 0  you'll get  |x-y| \gt 0 , which is true for all values of x and y but the case when x = y (the same conclusion as we got in the solution).

ProfChaos · Answer

Hello Bunuel At this step, Is (x-y)^2 \gt 0 ? I went further and arrived at Is (x-y) \gt 0 ? --> Is x \gt y ? Statement 1 says that x not equal to y Thus x < y or x > y Hence not sufficient Statement 2 says that x and y donot share any common divisors except 1 So let x be 2, y be 3 xy Hence not sufficient Option E Please tell me what is wrong with my approach. TIA!

Bunuel · Answer

The point is "is  (x-y)^2 \gt 0 ?" after taking the square root gives "is  |x - y| > 0 ?" (recall that  \sqrt{a^2}=|a| ). So,  (x-y)^2 \gt 0  even if x - y < 0, for example, if x - y = -2, then  (x-y)^2=4 \gt 0 . The only case when  (x-y)^2 \gt 0  is not true is when x = y.

Hope its clear.

Magni03 · Answer

Hi Buneul ,

Please correct my understanding here , as sqrt (x) = |x| . Here it would be |x-y|, doesnt it mean that x is not equal to y for x>0 and x<0 . Can we derive this from the stem , and hence S2 will also be sufficient ?

Bunuel · Answer

From the stem we only know that x and y are positive integers. NOTHING more.

(2) says that x and y are co-prime. So, for (2) we know that  x and y are positive integers that are co-prime . The solution on page 1 gives an example: if  x=y=1  then the answer is NO ( \frac{x}{y}+ \frac{y}{x}  is NOT greater than 2), but if  x=1  and  y=2 , then the answer is YES ( \frac{x}{y}+ \frac{y}{x}  IS greater than 2). Not sufficient.

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.

M12-16

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