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04 May 2016, 08:17
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D
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If m > 0, is |m−n| > 0?
(1) n^2 = −64n
(2) m > n
(1) n^2 = −64n
(2) m > n
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04 May 2016, 10:48
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Answer is B
|m-n| can only be >=0. For equality m=n.
n(2)=-64n => n(2)+64n=0 hence n can be zero or 64. so if you put n=0 then m-n>0 but if n=64 and m also =64 then it will be 0 so statment 1 is not sufficient.
Statment 2 says m>n, as m is always positive and n is smaller hence |m-n| always >0 hence statment 2 is enough for answering. So B.
|m-n| can only be >=0. For equality m=n.
n(2)=-64n => n(2)+64n=0 hence n can be zero or 64. so if you put n=0 then m-n>0 but if n=64 and m also =64 then it will be 0 so statment 1 is not sufficient.
Statment 2 says m>n, as m is always positive and n is smaller hence |m-n| always >0 hence statment 2 is enough for answering. So B.
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04 May 2016, 13:22
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I would go with D.
1) n^2 = -64n -> Square of a number cannot be negative, so n is either 0 or -64. As m > 0, |m-n| for both values of n will be greater than 0. Statement 1 is sufficient
2) m> n, |m-n| will always be greater than 0, as long as \(m\neq{n}\). Hence, statement 2 is also sufficient.
1) n^2 = -64n -> Square of a number cannot be negative, so n is either 0 or -64. As m > 0, |m-n| for both values of n will be greater than 0. Statement 1 is sufficient
2) m> n, |m-n| will always be greater than 0, as long as \(m\neq{n}\). Hence, statement 2 is also sufficient.
