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

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Math Expert
Joined: 02 Sep 2009
Posts: 50000
If x and y are positive integers such that x < y, which of the followi  [#permalink]

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29 Mar 2018, 00:21
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45% (medium)

Question Stats:

67% (01:44) correct 33% (01:15) wrong based on 100 sessions

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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

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Posts: 715
Re: If x and y are positive integers such that x < y, which of the followi  [#permalink]

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Updated on: 29 Mar 2018, 10:03
Bunuel wrote:
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

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.

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Originally posted by DavidTutorexamPAL on 29 Mar 2018, 00:49.
Last edited by DavidTutorexamPAL on 29 Mar 2018, 10:03, edited 1 time in total.
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Joined: 02 Jul 2017
Posts: 2
Re: If x and y are positive integers such that x < y, which of the followi  [#permalink]

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29 Mar 2018, 09:56
DavidTutorexamPAL wrote:
Bunuel wrote:
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

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 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.

The question states that x and y are positive integers, hence our answer will be (E)
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Joined: 07 Dec 2017
Posts: 715
Re: If x and y are positive integers such that x < y, which of the followi  [#permalink]

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29 Mar 2018, 10:01
qwertybd wrote:
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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Posts: 21
GPA: 3.35
Re: If x and y are positive integers such that x < y, which of the followi  [#permalink]

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10 Apr 2018, 22:50
1
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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Joined: 02 Sep 2009
Posts: 50000
Re: If x and y are positive integers such that x < y, which of the followi  [#permalink]

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10 Apr 2018, 23:41
Jamil1992Mehedi wrote:
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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Re: If x and y are positive integers such that x < y, which of the followi &nbs [#permalink] 10 Apr 2018, 23:41
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