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Inequalities Problem - Experts Please

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Inequalities Problem - Experts Please

by fk27 » Sun Jul 03, 2011 5:21 pm
Source: GMATHacks

If m and n are negative, is m/n less than 1?

1) mn<1
2)m-n>n

I'm having trouble following the explanation in the manual and was looking for a little feedback from the Experts. I seem to be having problems with tackling DS questions with variables and inequalities..... Is the best approach to plug in #'s and see what works? If anyone can outline a methodical approach to dealing with DS questions with variables it would be greatly appreciated.
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by amit2k9 » Mon Jul 04, 2011 1:03 am
for variables we have to check for : 1 +|-ive integers 2. fractions 3. equating such as x=y 4. zeros
5. 1

a mn<1 means m and n can be -0.5 each or -1 and -0.6 each. not sufficient.

b m>2n means m = -0.5 n = -0.4 giving m/n > 1 and m = -1 n = -2 giving m/n < 1. not sufficient.

a+b

m= -0.5 n = -0.4 m/n > 1 similarly m= -0.3 and n= -0.5. m/n < 1.

hence,E is sufficient.
Last edited by amit2k9 on Tue Jul 05, 2011 10:17 pm, edited 1 time in total.
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by top_business_2011 » Mon Jul 04, 2011 1:40 am
I am not an expert but noting that no one has responded yet, I wanted you to consider my approach till the experts come with better approaches.

For problems involving variables and especially inequalities, one useful approach is the graphic method.

First, start by simplifying the question: Is m/n <1 can be simplified as (m/n) - 1 <0
= m-n <0
= m<n
So the question is: Is m less than n? Here draw a line for this inequality. Bear in mind that m and n are both less than 0; so while drawing lines, consider m<0 and n<0. That is, consider only quadrant III.

Stat. 1: mn<1
= mn-1<0. Draw the graph of this inequality. Observe the area covered by this inequality along with the graph developed in the question stem; you shall see that for some part, m<n for other parts m>n. Insufficient.
stat. 2: m-n>n
This can be simplified as: m-2n>0. Similarly, draw a graph. And you shall see that this statement alone is insufficient.

Both: looking at the intersection of the graphs drawn in statement 1 and statement 2, one cannot still unequivocally answer the question.
Therefore, IMO the answer is E.

But what is the OA?
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by Frankenstein » Mon Jul 04, 2011 2:30 am
fk27 wrote:Source: GMATHacks

If m and n are negative, is m/n less than 1?

1) mn<1
2)m-n>n
Hi,
From(1):
if m = -0.5, n = -0.4, m/n= 1.25 > 1
if m = -0.4, n = -0.5, m/n = 0.8 < 1
Not sufficient
From(2):
m-n > n
=> m > 2n.
As n is negative m/n < 2
Not sufficient
Both (1)&(2):
if m = -0.5, n = -0.4, m/n= 1.25 > 1
if m = -0.4, n = -0.5, m/n = 0.8 < 1
Not sufficient

Hence, E
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by fk27 » Mon Jul 04, 2011 8:08 am
Hi guys, yes, OA is E. Nice explanation Frankenstein, thanks. I just had a quick question. Do you have a methodical approach for handling these questions where you have to pick numbers? It seems like it can get quite time consuming to 1)choose integers of both +ve and -ve 2) choose fractions of +ve and -ve 3) choose 0 when testing out the inequalities.
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by Frankenstein » Mon Jul 04, 2011 8:20 am
Hi,
You don't need to disprove something with counter example, if it is generic enough to perceive.
For instance statement (2), we get m/n < 2. So, we don't need to pick values of m and n.
Coming to statement (1) mn<1, if we think a bit we get a feeling that we cannot strictly restrict m/n to a finite range.
If we are able to see this, it is fine no need to picking counter examples. If not it is always better to pick counter examples.
Generally, number properties differ in the following ranges
n<-1
-1<n<0
0<n<1
n>1.
So, depending on the question we choose various values.This is something we get by intuition or sometimes by practicing similar questions.

P.S.:I am bad at generalizing things. So, whatever makes sense for you, follow it.
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by GMATGuruNY » Tue Jul 05, 2011 12:35 pm
fk27 wrote:Source: GMATHacks

If m and n are negative, is m/n less than 1?

1) mn<1
2)m-n>n

I'm having trouble following the explanation in the manual and was looking for a little feedback from the Experts. I seem to be having problems with tackling DS questions with variables and inequalities..... Is the best approach to plug in #'s and see what works? If anyone can outline a methodical approach to dealing with DS questions with variables it would be greatly appreciated.
I received a PM asking me to comment.
Use algebra where possible.
Plug in numbers if you don't see a clear algebraic approach.

Is m/n < 1?
Try to rephrase the question:
Since n<0, if we multiply each side of the inequality by n, the direction of the inequality must change:
Question rephrased: Is m > n?

Statement 1: mn < 1.
No way to determine whether m > n.
Insufficient.

Statement 2: m-n > n.
m > 2n.
No way to determine whether m > n.
Insufficient.

Statements 1 and 2: mn < 1 and m > 2n.
Here I would plug in values.
To make the process efficient, try to determine what the question is testing.
mn < 1 makes me think of fractions:
m = -1/3 and n = -1/2 satisfies both statements because (-1/3)(-1/2) < 1 and -1/3 > 2(-1/2).
m = -1/2 and n = -1/3 satisfies both statements because (-1/2)(-1/3) < 1 and -1/2 > 2(-1/3).
Since in the first case m > n and in the second case m < n, insufficient.

The correct answer is E.
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by amit2k9 » Tue Jul 05, 2011 10:07 pm
r^2+2rs+s^2 = r^2+16 + s^2 + 16

thus for,

a r=8 we get after solving s=2

b s=2 we get after solving r=8.

hence D it is.
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