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

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!
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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.
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Jamil1992Mehedi
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.

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

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).
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Bunuel
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

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<y => \(\frac{x}{y}<1\)
MUST BE LESS THAN 1

IMO C
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Bunuel
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

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.
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