If x > y > 0, which of the following are true?

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Key property: If \(\frac{a}{b}\) is a positive fraction, then \(\frac{a + k}{b + k}\) gets closer and closer to 1 as the value of k increases.

Some examples:

EXAMPLE #1: \(\frac{1}{5}=0.2 \)

Add 1 to numerator and denominator to get: \(\frac{1+1}{5+1}=\frac{2}{7}≈0.29 \)

Add 7 to numerator and denominator to get: \(\frac{1+7}{5+7}=\frac{8}{12}≈0.67 \)

Add 30 to numerator and denominator to get: \(\frac{1+30}{5+30}=\frac{31}{35}≈0.89 \)

Add 400 to numerator and denominator to get: \(\frac{1+400}{5+400}=\frac{401}{405}≈0.99 \)

So, when \(\frac{a}{b}\) is less than 1, \(\frac{a}{b} < \frac{a + k}{b + k} <1\) for all positive values of k


EXAMPLE #2: \(\frac{8}{2}=4 \)

Add 1 to numerator and denominator to get: \(\frac{8+1}{2+1}=\frac{9}{3}=3 \)

Add 6 to numerator and denominator to get: \(\frac{8+6}{2+6}=\frac{14}{8}=1.75 \)

Add 100 to numerator and denominator to get: \(\frac{8+100}{2+100}=\frac{108}{102}≈1.06 \)

So, when \(\frac{a}{b}\) is greater than 1, \(1<\frac{a + k}{b + k}<\frac{a}{b} \) for all positive values of k

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Now onto the question!!!!

Since we're told \(x > y > 0\), we know that \(\frac{x}{y}\) is greater than 1, which means \(1<\frac{x + k}{y + k} < \frac{x}{y} \) (for all positive values of k)

So statement III is true.

Answer: C
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Hi Brent. I am confused here. We've learned that the results increases when we add the same number to the nominator and denominator which is why I really thought that the answer was D but here it is suddenly not the case...

I'm not sure where you learned that, but it's not true.
Sometimes, the value of the fraction increases when we add the same number to the numerator and denominator, and sometimes the fraction decreases when we add the same number to the numerator and denominator.
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Hi Brent. Thank you for the clarification. I thought it was the takeaway from of the lessons but the key property was that the fraction gets closer to 1 as the value of k increases. But here since we know that both x and y are positive and that the value of k is +1 why isn't the answer D?
Since x and y are positive it seems that that adding+1 to both the numerator and denominator will result in a higher result than the initial fraction

Here are the two rules with regard to adding the same value to numerator and denominator:

Rule #1: When \(0< \frac{a}{b} < 1\),then \(\frac{a}{b} < \frac{a + k}{b + k} <1\) for all positive values of k

Rule #2: When \(\frac{a}{b} > 1\), then \(1<\frac{a + k}{b + k}<\frac{a}{b} \) for all positive values of k

Since we're told that \(0 < y < x\) (i.e., x is greater than y), then we know that \(\frac{x}{y} > 1\), which means Rule #2 applies, and Rule #1 does not apply.

Does that help?
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Hi Brent. Yes that helps. Thank you!

Hi BrentGMATPrepNow, if we are told \(x > y > 0\), not quite sure how do you arrive at \(\frac{x}{y}\) is greater than 1? and not 0< x/y <1 ?
Thanks Brent

If x > y > 0, then x is greater than y, and both x and y are positive.
Since x is greater than y, it could be the case that x = 3 and y = 2, in which case x/y = 3/2, and 3/2 is greater than 1.
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Get it thanks Brent BrentGMATPrepNow