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15 Jan 2017, 08:26
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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% (01:12) correct
38%
(01:25)
wrong
based on 66
sessions
History
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Not Attempted Yet
15 Jan 2017, 08:31
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Is \(q\) > \(r\)?
(1) \(qr\) = 32 --> if q = 32 and r = 1, the answer will be YES but if q = -32 and r = -1, the answer will be NO. Not sufficient.
(2) \(q^2\) > \(r^2\). This implies that |q| > |r|, so q is further from 0 than r is. Not sufficient.
(1)+(2) We can use examples from (1) again. Not sufficient.
Answer: E.
(1) \(qr\) = 32 --> if q = 32 and r = 1, the answer will be YES but if q = -32 and r = -1, the answer will be NO. Not sufficient.
(2) \(q^2\) > \(r^2\). This implies that |q| > |r|, so q is further from 0 than r is. Not sufficient.
(1)+(2) We can use examples from (1) again. Not sufficient.
Answer: E.
15 Jan 2017, 08:35
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Is \(q\) > \(r\)?
(1) \(qr\) = 32
(2) \(q^2\) > \(r^2\)
(1) \(qr\) = 32
(2) \(q^2\) > \(r^2\)
Statement 1: qr = 32
q and r can be = 31,1 or 16,2, or 8,4 Also both of them can be negative as well. Not sufficient.
Statement 2: Clearly not sufficient.
Combined: Just by comparing squares, we can't determine if q>r, since we can have negative values for both.
