by: Kevi
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Difficulty:
35%
(medium)
Question Stats:
63% (01:08) correct
38%
(01:08)
wrong
based on 128
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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
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
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
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.
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Bunuel
Wow, I didn't even catch that. Careless algebra mistake. Thank you, both! Also, thank you for the similar questions!
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