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09 Sep 2021, 02:57
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
72% (01:19) correct
28%
(01:09)
wrong
based on 203
sessions
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Date
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If \(x > y > 0\), which of the following are true?
I. \(\frac{x+1}{y+1}\) > \(\frac{x}{y}\)
II. \(\frac{x+1}{y+1}\) = \(\frac{x}{y}\)
III. \(\frac{x+1}{y+1}\) > 1
(A) I only
(B) II only
(C) III only
(D) I and III only
(E) II and III only
I. \(\frac{x+1}{y+1}\) > \(\frac{x}{y}\)
II. \(\frac{x+1}{y+1}\) = \(\frac{x}{y}\)
III. \(\frac{x+1}{y+1}\) > 1
(A) I only
(B) II only
(C) III only
(D) I and III only
(E) II and III only
09 Sep 2021, 17:30
Kudos
Bookmarks
Bunuel
If \(x > y > 0\), which of the following are true?
I. \(\frac{x+1}{y+1}\) > \(\frac{x}{y}\)
II. \(\frac{x+1}{y+1}\) = \(\frac{x}{y}\)
III. \(\frac{x+1}{y+1}\) > 1
(A) I only
(B) II only
(C) III only
(D) I and III only
(E) II and III only
I. \(\frac{x+1}{y+1}\) > \(\frac{x}{y}\)
II. \(\frac{x+1}{y+1}\) = \(\frac{x}{y}\)
III. \(\frac{x+1}{y+1}\) > 1
(A) I only
(B) II only
(C) III only
(D) I and III only
(E) II and III only
----------------ASIDE------------------------------------------------
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
----------------------------------------
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
General Discussion
04 Mar 2022, 06:05
Kudos
Bookmarks
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...
----------------ASIDE------------------------------------------------
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
----------------------------------------
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
BrentGMATPrepNow
Bunuel
If \(x > y > 0\), which of the following are true?
I. \(\frac{x+1}{y+1}\) > \(\frac{x}{y}\)
II. \(\frac{x+1}{y+1}\) = \(\frac{x}{y}\)
III. \(\frac{x+1}{y+1}\) > 1
(A) I only
(B) II only
(C) III only
(D) I and III only
(E) II and III only
I. \(\frac{x+1}{y+1}\) > \(\frac{x}{y}\)
II. \(\frac{x+1}{y+1}\) = \(\frac{x}{y}\)
III. \(\frac{x+1}{y+1}\) > 1
(A) I only
(B) II only
(C) III only
(D) I and III only
(E) II and III only
----------------ASIDE------------------------------------------------
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
----------------------------------------
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
