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10 Jul 2013, 19:35
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B
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% (01:08) correct
38%
(01:08)
wrong
based on 128
sessions
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m and n are both integers, and m ≠ n, does n = 0?
(1) m = 7
(2) mn = n^2
(1) m = 7
(2) mn = n^2
Got this question from a Kaplan advanced quant library. I don't understand why statement 2 is sufficient. I understand that m cannot be equal to n, however, how is n forced to be equaled to be 0? I haven't found a question similar to this in the OG yet. If you have similar questions, please post! Thanks
10 Jul 2013, 21:14
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m and n are both integers, and m (does not equal) n, does n = 0?
1) m = 7
So \(n\neq{7}\), not sufficient.
2) mn = n^2
\(mn-n^2=0\), \(n(m-n)=0\), this gives us two possibilities: or \(n=0\) or \(m-n=0\) (or both). But since we know that \(n\neq{m}\) the second case \(n-m=0\) or \(n=m\) is not possible, so the only possibility that remains is \(n=0\).
Sufficient
1) m = 7
So \(n\neq{7}\), not sufficient.
2) mn = n^2
\(mn-n^2=0\), \(n(m-n)=0\), this gives us two possibilities: or \(n=0\) or \(m-n=0\) (or both). But since we know that \(n\neq{m}\) the second case \(n-m=0\) or \(n=m\) is not possible, so the only possibility that remains is \(n=0\).
Sufficient
10 Jul 2013, 22:15
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DelSingh
m and n are both integers, and m#n, does n = 0?
(1) m = 7. Clearly insufficient to get the value of n.
(2) mn = n^2 --> \(mn-n^2=0\) --> \(n(m-n)=0\) --> \(n=0\) or \(m-n=0\) (\(m=n\)). Since given that \(m\neq{0}\), then only the first case is left, thus \(n=0\). Sufficient.
Answer: B.
Similar question to practice: if-xy-0-does-xy-y-3-1-xy-3-2-y-105766.html
Hope it helps.
