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21 Apr 2017, 06:26
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
69% (02:05) correct
31%
(01:56)
wrong
based on 315
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If 0 < x < y, then which of the following MUST be true?
I) \(\frac{x + 2}{y + 2} > \frac{x}{y}\)
II) \(\frac{x – y}{x} < 0\)
III) \(\frac{2x}{x + y} < 1\)
A) I only
B) I and II only
C) I and III only
D) II and III only
E) I, II and III
*kudos for all correct solutions
I) \(\frac{x + 2}{y + 2} > \frac{x}{y}\)
II) \(\frac{x – y}{x} < 0\)
III) \(\frac{2x}{x + y} < 1\)
A) I only
B) I and II only
C) I and III only
D) II and III only
E) I, II and III
*kudos for all correct solutions
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21 Apr 2017, 06:41
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statement 1: could get simplified as : x+2/y+2 -x/y >0 => 2(y-x)/y*(y+2) >0, considering y>x>0, the inequality does hold good.
statement 2: (x-y)/x <0 is true as the numerator is negative in this case.
statement 3: 2x/(x+y) -1<0 => (x-y)/(x+y) <0, negative numerator as per the question stem will make the inequality get satiated.
option-> E
statement 2: (x-y)/x <0 is true as the numerator is negative in this case.
statement 3: 2x/(x+y) -1<0 => (x-y)/(x+y) <0, negative numerator as per the question stem will make the inequality get satiated.
option-> E
21 Apr 2017, 06:42
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GMATPrepNow
Hi..
Let's see the cases...
I) \(\frac{x + 2}{y + 2} > x/y\)
If x/y <1, and you add same amount to both numerator and denominator, the new fraction will be GREATER than the original value..
Example \(\frac{1}{2}\) will be LESS than \(\frac{1+2}{2+2}\)..
True in all cases even if x and y are fraction..
TRUE
II) \(\frac{x – y}{x} < 0\)
OR \(\frac{x}{x}-\frac{y}{x}=1-\frac{y}{x}\)...
Now y is greater than x, so y/x >1.. and thus 1-y/x will be NEGATIVE..
TRUE
III) \(\frac{2x}{x + y} < 1\)
x<y, so x+x<y+x....2x<y+x..\(\frac{2x}{y+x}\)<1..
TRUE
All three TRUE..
E
