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Re: If mn ≠ 0, is m > n? (1) 1/m < 1/n (2) m^2 > n^2 [#permalink]

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22 May 2017, 08:46

1

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statement 1st will boils down to (n-m)/mn <0 => (n-m) and mn are of opposite sign => not sufficient and statement 2nd implies( m^2 -n^2) >0 => (m+n)(m-n) >0 and that isn't sufficient and taking them together won't lead to any unique solution either -> option ->E

Statement 1: 1/m < 1/n This statement doesn't FEEL sufficient, so I'll TEST some values. There are several values of m and n that satisfy statement 1. Here are two: Case a: m = 2 and n = 1. In this case m > n Case b: m = -3 and n = 1. In this case m < n Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: m² > n² Before I start choosing numbers to test, I'll see if I can REUSE my numbers from statement 1. Yes I can! Those same values satisfy the conditions in statement 2. Case a: m = 2 and n = 1. In this case m > n Case b: m = -3 and n = 1. In this case m < n Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined IMPORTANT: Notice that I was able to use the same counter-examples to show that each statement ALONE is not sufficient. So, the same counter-examples will satisfy the two statements COMBINED. In other words, Case a: m = 2 and n = 1. In this case m > n Case b: m = -3 and n = 1. In this case m < n Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT

All hail Bunuel !! Please help us with a logical (algebraic) solution.

If mn ≠ 0, is m > n?

(1) 1/m < 1/n.

Two cases: If m and n have the same sign (so if both are negative, or both are positive), then when cross multiplying we'll get n < m. Answer: YES. If m and n have different signs signs (so if n is positive and m is negative), then when cross multiplying we'll get n > m. Answer: NO. Not sufficient.

(2) m^2 > n^2. Take the square root from both sides: |m| > |n|. This implies that m is further from 0 than n is. Clearly insufficient to say which is greater.

(1)+(2) The second statement still allows m and n to have the same sign (m = 2 and n = 1) as well as m and n to have different signs (m = -2 and n = 1) thus we can still have two different answers: n < m and n > m. Not sufficient.

Algebraic approach.. Given that neither of m and n is 0, they can be negative, positive..

Let's see the statements.. 1) \(\frac{1}{m}<\frac{1}{n}.....\frac{1}{n}-\frac{1}{m}.....\frac{m-n}{mn}>0\) So two cases.. mn>0... m-n>0, m>n..... That is if both n and m are of SAME sign, m>n mn<0.. m-n<0, m<n... That is if both are of different sign, m<n.. Insufficient

2) \(m^2>n^2.....m^2-n^2>0......(m-n)(m+n)>0\).. Two cases.. m+n>0...m-n>0, m>n... That is if both are POSITIVE, or one is ATLEAST positive, m>n m+n<0... m-n<0.. So if both are NEGATIVE m<n... Again different possibilities Insufficient

Combined.. Find the common in both statements.. A) when both are NEGATIVE, m<n B) when both are of different sign, m>n Again insufficient

E

Just for info

Ofcourse when both are positive ans is different from each statement, so not possible.. It seems Bunuel has already replied in a crisp manner, this is on same line slightly in detail
_________________

If mn ≠ 0, is m > n? (1) 1/m < 1/n (2) m^2 > n^2 [#permalink]

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14 Aug 2017, 05:06

Great question.

The bottom line is => If you don't know the signs of the inequality then do not change the sides. i.e => If 1/m<1/n then it does not mean n<m unless we know the signs for the involved numbers.

As for this question -> Here is an easy way out Taking two examples => (100,1) and (-100,1) => Push the E option. _________________

I'm happy to respond. :-) This is brilliant question!

Statement #1: \(\frac{1}{m} < \frac{1}{n}\) I think this is the trickier of the two statements.

Here positive/negative sign is crucial. Case I: if m = 5 and n = 2, then: \(\frac{1}{5} < \frac{1}{2}\) and m > n Case II: if m = -5 and n = 2, then: \(-\frac{1}{5} < \frac{1}{2}\) but m < n

Two different choices consistent with the statement produce two different answers. Thus, this statement, alone and by itself, is insufficient.

Statement #2: \(m^2> n^2\)[/quote] Clearly, positive/negative signs make a difference here. We can use the same two choices. Case I: if m = 5 and n = 2, then: \(5^2> 2^2\) and and m > n Case II: if m = -5 and n = 2, then: \((-5)^2> 2^2\) but m < n

Again, two different choices consistent with the statement produce two different answers. Thus, this statement, alone and by itself, is insufficient.

Combining the statements produces no additional restraints, and both pairs still can be used with the combination. Thus, everything is insufficient.

OA = (E)

A truly wonderful question!

Does all this make sense? Mike :-)
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