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Stuck at this Inequality :(

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Source: — Data Sufficiency |

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by Brent@GMATPrepNow » Sat Apr 27, 2013 10:02 am
bharti06 wrote:Q
Is mn > 1?
1. m > 1/n
2. m^3 > n^2

can someone please explain how to approach this question and how to think of the right values to try to reach counter examples if any. Many thanks.
Target question: Is mn > 1?

Plugging in numbers is a good strategy when you suspect that a statement is not sufficient. For more on this, see our free video: https://www.gmatprepnow.com/module/gmat- ... cy?id=1102

We want choose a variety of numbers that represent a cross-section of numbers.

Statement 1: m > 1/n
There are several pairs of values that meet this condition. Here are two:
Case a: m = 2 and n = 1, in which case mn > 1
Case b: m = 2 and n = -1, in which case mn < 1
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: m^3 > n^2
There are several pairs of values that meet this condition. Here are two:
Case a: m = 2 and n = 1, in which case mn > 1
Case b: m = 2 and n = -1, in which case mn < 1
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined:
There are still several pairs of values that meet these two conditions. Here are two:
Case a: m = 2 and n = 1, in which case mn > 1
Case b: m = 2 and n = -1, in which case mn < 1
Since we cannot answer the target question with certainty, the combined statements are is NOT SUFFICIENT

Answer = E

Cheers,
Brent
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by Anju@Gurome » Sat Apr 27, 2013 10:05 am
bharti06 wrote:Is mn > 1?

1. m > 1/n
2. m^3 > n^2
As Brent has already posted the plugging numbers approach, I'm posting the algebraic approach.

To determine whether mn > 1 or not, the first thing we need to determine is whether m and are of same sign. If their are of different signs, then their product will be less than zero, i.e. mn < 1

Statement 1: m > 1/n
If n > 0 ---> mn > 1 ---> YES
If n < 0 ---> mn < 1 ---> NO

Not sufficient

Statement 2: n² cannot be negative. So, m³ > n² ≥ 0
So, m³ ≥ 0 ---> m ≥ 0
As we don't know the sign of n, we cannot answer the question.

Not sufficient

1 & 2 Together: We still don't know the sign of n.

Not sufficient

The correct answer is E.
Anju Agarwal
Quant Expert, Gurome

Backup Methods : General guide on plugging, estimation etc.
Wavy Curve Method : Solving complex inequalities in a matter of seconds.

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by srcc25anu » Sat Apr 27, 2013 10:11 am
Is mn > 1?

ST2 compares m^3 and n^2 so lets pick m = 3 and n = (+/-) 4

St1: m > 1/4
if m = 3, n = 4, then 3 > 1/4 and mn = 12 (GREATER THAn 1)BUT
if m = 3, n = -4, then 3 > -1/4 and mn = -12 (LESS THAn 1)
Insufficient

St2: m^3 > n^2
if m = 3, n = 4, then m^3 = 27 > n^2 = 16 and mn = 12 (GREATER THAn 1)
if m = 3, n = -4, then m^3 = 27 > n^2 = 16 and mn = -12 (LESS THAn 1)
Insufficient

Together: Again insufficient
for m = 3, n = 4 mn > 1 but for m = 3, n = -4, mn < 1
Ans E
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by bharti06 » Sat Apr 27, 2013 10:29 am
Thank you so much Brent and Anju!!!
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by barquentine » Sat Apr 27, 2013 4:24 pm
From (1) using m=1/2, n=2 the answer is 'no' but with m=1/2, n=3 the answer is 'yes'. Insufficient.
From (2), it can only be true if either (a) m,n are +ve fractions and m<n, or (b) m >1 and m>=n. There are multiple cases where this is insufficient.
Together, nothing extra is added, therefore answer E.
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by ygdrasil24 » Sun May 05, 2013 11:07 am
My approach:
For mn >1, the [m], [n] should be >1.

1. m>1/n , take m=1/2 and n to be 8 , so 1/2 > 1/8 but still product is not > zero for reason mentioned as above.
2. m3>n2, this holds for all m >0 , n can be any integer, but this again does not say anything about m being greater than 1.

Together also nothin can be deciphered.

Comments welcome :)
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