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27 Oct 2016, 05:04
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
50% (01:41) correct
50%
(01:29)
wrong
based on 248
sessions
History
Date
Time
Result
Not Attempted Yet
27 Oct 2016, 08:09
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Bunuel
With these types of problems, one must always consider the sign of the variables before you multiply them. If y>0, then the question is asking if \(x>3y\). If y<0, then the question is asking if \(x<3y\).
(1) 3x > 9y. Divide by 3 since it's always positive. \(x>3y\). We don't know if y is positive or negative. Insufficient.
(2) y > 3y. This tells us that y<0. We still can't say anything about the relationship between x and y. Obviously insufficient.
Taking them both together, we know that y<0 and thus that the question is asking if \(x<3y\). Statement 1 says that \(x>3y\). Thus \(x/y <3\). Sufficient.
Answer: C
03 Nov 2016, 03:08
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OE is E
1) x>3y
8>-6 (3*-2) or 8>6 (3*2). Not sufficient
2) y>3y
y is negative. But not sufficient
1)+2) - x=8, 8>-ve y
x=-2 > -ve y (say 3y=-6).
Nice question. Thanks.
1) x>3y
8>-6 (3*-2) or 8>6 (3*2). Not sufficient
2) y>3y
y is negative. But not sufficient
1)+2) - x=8, 8>-ve y
x=-2 > -ve y (say 3y=-6).
Nice question. Thanks.
