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

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

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22 May 2017, 08:35
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If mn ≠ 0, is m > n?

(1) 1/m < 1/n
(2) m^2 > n^2
[Reveal] Spoiler: OA

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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
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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

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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, 09:04
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hazelnut wrote:
If mn ≠ 0, is m > n?

(1) 1/m < 1/n
(2) m² > n²

Target question: Is m > n?

Given: mn ≠ 0

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

Aside: For more on this idea of plugging in values when a statement doesn't feel sufficient, read my article: http://www.gmatprepnow.com/articles/dat ... lug-values

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

[Reveal] Spoiler:
E

Cheers,
Brent
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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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12 Aug 2017, 06:19
Hi Brent,

in statement one 1)

case a ) if M=2 and N=1 . then how come M > N ? could you please clarify

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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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14 Aug 2017, 04:22
hazelnut wrote:
If mn ≠ 0, is m > n?

(1) 1/m < 1/n
(2) m^2 > n^2

Calling the master of Algebra!

All hail Bunuel !!
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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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14 Aug 2017, 04:49
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rekhabishop wrote:
hazelnut wrote:
If mn ≠ 0, is m > n?

(1) 1/m < 1/n
(2) m^2 > n^2

Calling the master of Algebra!

All hail Bunuel !!

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.

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

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14 Aug 2017, 05:01
hazelnut wrote:
If mn ≠ 0, is m > n?

(1) 1/m < 1/n
(2) m^2 > n^2

Hi..

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
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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.

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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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14 Aug 2017, 05:07
hazelnut wrote:
If mn ≠ 0, is m > n?

(1) 1/m < 1/n
(2) m^2 > n^2

Number plugging was faster IMO!

1) $$\frac{1}{m}$$ < $$\frac{1}{n}$$
m = 2, n = 1 => Yes
m = -2, n = 1 => No
Insufficient.

2) $$m^2$$ > $$n^2$$
m = 2, n = 1 => Yes
m = -2, n = 1 => No
Insufficient.

1+2)
Nothing new. Insufficient.
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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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14 Aug 2017, 05:19
Clearly its E

1) 1/m<1/n
(m-n)/mn >0
numerator and denominator both have to be simultaneously positive or negative to be the statement to be true. A,D ruled out

2) m^2>n^2
m<-n and m>n and therefore B is out

combine them

put m=-2 n=1 and also m=2 and n=1
both the condition will satisfy

clearly we cant say if m>n or not. C is out

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If mn not equal to 0, is m>n? [#permalink]

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17 Aug 2017, 13:59
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If $$mn\neq{0}$$ is m>n?

1. $$\frac{1}{m} < \frac{1}{n}$$
2. $$m^2> n^2$$
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Re: If mn not equal to 0, is m>n? [#permalink]

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17 Aug 2017, 16:53
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Gnpth wrote:
If $$mn\neq{0}$$ is m>n?

1. $$\frac{1}{m} < \frac{1}{n}$$
2. $$m^2> n^2$$

Dear Gnpth,

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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Re: If mn not equal to 0, is m>n?   [#permalink] 17 Aug 2017, 16:53
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