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niteshwaghray
GMAT 1: 700 Q45 V40
Posts: 59
23 May 2017, 00:40
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
63% (01:24) correct
37%
(02:01)
wrong
based on 179
sessions
History
Date
Time
Result
Not Attempted Yet
What is the value of \(\frac{(y^2–x^2)}{x^2y^2}\) ?
(1) \(\frac{x}{x+y}=\frac{1}{2}\)
(2) \(\frac{y}{x+y}=\frac{1}{2}\)
(1) \(\frac{x}{x+y}=\frac{1}{2}\)
(2) \(\frac{y}{x+y}=\frac{1}{2}\)
23 May 2017, 00:57
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What is the value of \(\frac{(y^2–x^2)}{x^2y^2}\) ?
(1) \(\frac{x}{x+y}=\frac{1}{2}\)
2x = x + y;
x = y.
Thus, \(\frac{(y^2–x^2)}{x^2y^2}=\frac{(x^2–x^2)}{x^2x^2}=\frac{0}{x^4}=0\). Sufficient.
(2) \(\frac{y}{x+y}=\frac{1}{2}\)
The same here: 2y= x + y;
x = y.
Thus, \(\frac{(y^2–x^2)}{x^2y^2}=\frac{(x^2–x^2)}{x^2x^2}=\frac{0}{x^4}=0\). Sufficient.
Answer: D.
(1) \(\frac{x}{x+y}=\frac{1}{2}\)
2x = x + y;
x = y.
Thus, \(\frac{(y^2–x^2)}{x^2y^2}=\frac{(x^2–x^2)}{x^2x^2}=\frac{0}{x^4}=0\). Sufficient.
(2) \(\frac{y}{x+y}=\frac{1}{2}\)
The same here: 2y= x + y;
x = y.
Thus, \(\frac{(y^2–x^2)}{x^2y^2}=\frac{(x^2–x^2)}{x^2x^2}=\frac{0}{x^4}=0\). Sufficient.
Answer: D.
10 Jul 2019, 22:19
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niteshwaghray
S1:
\(\frac{x}{x+y} = \frac{1}{2}\)
\(=> \frac{x+y}{x} = 2\)
\(=> 1 + \frac{y}{x} = 2\)
=> x =y
\(=> \frac{(y^2–x^2)}{x^2y^2} = 0\)
SUFFICIENT
S2:
\(\frac{y}{x+y}=\frac{1}{2}\)
\(=> \frac{x+y}{y} = 2\)
\(=> \frac{x}{y} =1\)
=> x =y
\(=> \frac{(y^2–x^2)}{x^2y^2} = 0\)
SUFFICIENT
IMO D
