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Where am I going wrong - Inequalities

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Where am I going wrong - Inequalities

by imhimanshu » Thu Aug 18, 2011 6:52 am
Hi Experts,
Can you please verify where I am going wrong.

If m <> 0, is m^3> m^2

1) m>0
2) m^2>m

my approach -

Since it is given that m^3> m^2

and m^2 > 0 (always). So, it is safe to divide by m^2
Hence, the question becomes is m>1

1. Insufficient as m can lie b/w 0-1
2. Here I am going wrong.

Given - m^2>m
My take -
m^2-m > 0
m(m-1)>0
so, m>0 or m>1
So,m must be greater than 1.
But as per solution, It is incorrect. I know, this can be solved by plugging the values. However, I usually avoid them because I usually miss one or the other scenario. I would appreciate if someone can help me in catching the error.

Thanks
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Source: — Data Sufficiency |
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by kmittal82 » Thu Aug 18, 2011 6:58 am
imhimanshu wrote:Hi Experts,
m(m-1)>0
so, m>0 or m>1
So,m must be greater than 1.
You missed m < 1 as a solution as well. For a product of 2 numbers to be greater than 0, they are either both positive (the case you considered), or they are both negative (the case you missed). Hope that helps
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by Anurag@Gurome » Thu Aug 18, 2011 7:06 am
imhimanshu wrote:Hi Experts,
Can you please verify where I am going wrong.

If m <> 0, is m^3> m^2

1) m>0
2) m^2>m

my approach -

Since it is given that m^3> m^2

and m^2 > 0 (always). So, it is safe to divide by m^2
Hence, the question becomes is m>1

1. Insufficient as m can lie b/w 0-1
2. Here I am going wrong.

Given - m^2>m
My take -
m^2-m > 0
m(m-1)>0
so, m>0 or m>1
So,m must be greater than 1.
But as per solution, It is incorrect. I know, this can be solved by plugging the values. However, I usually avoid them because I usually miss one or the other scenario. I would appreciate if someone can help me in catching the error.

Thanks
m(m - 1) > 0 implies any of the following two cases,
  • 1. m and (m - 1) both positive --> m > 1
    2. m and (m - 1) both negative ---> m < 0
Hence, statement 2 is not sufficient.

Taking both statements together: As m is positive from statement 1, m must be greater than 1 from statement 2. SUFFICIENT

The correct answer is C.
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by GMATGuruNY » Thu Aug 18, 2011 7:11 am
imhimanshu wrote: Given - m^2>m
My take -
m^2-m > 0
m(m-1)>0
so, m>0 or m>1
So,m must be greater than 1.
The reasoning in red is incorrect.

Given that m(m-1)>0:
The critical points are m=0 and m=1.
These are the only values for m that will make the left hand side equal 0.
When m equals any other value, the left hand side will be less than or greater than 0.
To determine the range of m, test one value to the left and right of each critical point.

m<0:
Plug m=-1 into m(m-1)>0:
-1*(-1-1) > 0
-1 * -2 > 0
2 > 0.
This works.
m<0 IS part of the range.

0<m<1:
Plug m= 1/2 into m(m-1)>0:
(1/2)*((1/2)-1) > 0
1/2 * -1/2 > 0
-1/4 > 0.
Doesn't work.
0<m<1 is NOT part of the range.

m>1:
Plug m= 2 into m(m-1)>0:
2*(2-1) > 0
2 * 1 > 0
2 > 0.
This works.
m>1 IS part of the range.

Thus, statement 2 indicates that m<0 OR that m>1.
Insufficient.

When the two statements are combined, the only range for m that satisfies both statements is m>1.
Sufficient.

The correct answer is C.
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by imhimanshu » Thu Aug 18, 2011 7:28 am
Thanks Experts for catching the error. I hope not to repeat the same again.
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