If x and y are positive integers such that x

Question

If x and y are positive integers such that x < y, which of the following expressions must be less than 1?

I.  \sqrt{\frac{y}{x}}

II.  \frac{x^2 - 100}{y^2 - 100}

III.  \frac{x}{y}

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) II and III only

DavidTutorexamPAL · Answer

As there are specific rules that govern if a fraction is larger or smaller than 1, we'll use them.
This is a Precise approach.

A fraction is smaller than 1 if its numerator is smaller than its denominator.
That is,  \frac{smaller}{larger}<1 . Since x<y and both are positive, then  \frac{x}{y}<1 .
III is true so (A), (B), (D) are eliminated.
So all we need to know is if II is true. Since x < y and both are positive integers then x^2 < y^2 meaning that x^2 - 100 < y^2 - 100.
So, if both are positive then  \frac{x^2 - 100}{y^2 - 100}  is smaller than 1.
But what happens if both numerator and denominator are negative? In this case, then the numberator is a 'smaller number' = 'larger negative' and the denominator a 'larger number' = 'smaller negative' and once we cancel out the minus sign we get a fraction whose numerator is larger than its denominator, meaning that it is larger than 1.

(C) s our answer.

qwertybd · Answer

The question states that x and y are positive integers, hence our answer will be (E)

DavidTutorexamPAL · Answer

Try x = 2 and y = 3.
Then:
x^2 - 100 = -96
y^2 - 100 = -91.

Their quotient is 96/91 which is larger than 1.

I think the problem came from lack of clarity in my answer: 'both are negative' doesn't refer to x and y but to the expressions (x^2 - 100) and (y^2 - 100)
I edited the original answer for clarity, thanks!

Jamil1992Mehedi · Answer

Hi,  Bunuel  sir. we expect your clarification for answer of the above question.
Why the answer shows the option E. 
Where, it is easily found that x=2, y=3, so (x^2 - 100)/(y^2 -100) >1.
So ans should be C only
Then based on which cogent logic it should be the answer option E.

Bunuel · Answer

Yes, the answer is C. Edited the OA. Thank you.

exc4libur · Answer

given: 0<x<y positive integers
I.  \sqrt{\frac{y}{x}} : sqrt of a number greater than 1 is not less than 1, out.
II.  \frac{x^2 - 100}{y^2 - 100}=\frac{1 - 100}{81 - 100}=-99/-19>1  out.
III.  \frac{x}{y}<1

Answer (C).

Kinshook · Answer

Given: x and y are positive integers such that x < y, Asked: Which of the following expressions must be less than 1? I. \sqrt{\frac{y}{x}} \frac{y}{x}>1 \sqrt{\frac{y}{x}}>1 NOT LESS THAN 1 II. \frac{x^2 - 100}{y^2 - 100} If x^2 - 100>0 & y^2 - 100>0 => LESS THAN 1 If x^2 - 100<0 & y^2 - 100<0 => NOT LESS THAN 1 NOT NECESSARILY LESS THAN 1 III. \frac{x}{y} x \frac{x}{y}<1 MUST BE LESS THAN 1 IMO C

amarsinha · Answer

1. y/x will always be greater than 1 as x<y. So sqrt(y/x) will always be greater than 1. 
2. It can be both greater than 1 and less than 1. 
 If x, y <10. x = 8, y=9 , the expression becomes -36/-19 = 36/19 which is greater than 1. 
 But if x,y >10, x = 12, y=13 , the expression becomes 44/69 which is less than 1. 
3. x/y will always be less than 1. as x is less than y.

bumpbot · Answer

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If x and y are positive integers such that x < y, which of the followi

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