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Inequality Doubts - please help

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Inequality Doubts - please help

by bharti06 » Sat Apr 27, 2013 9:43 am
Q1. Is m^3 > m^2?
1. m > 0
2. m < 1

My question here is whether or not we can simplify this further - m^3 > m^2? by taking everything to LHS and making RHS 0 as in -
m^3 - m^2 >0
m^2 (m-1) > 0
now either m^2 > 0 or m - 1 > 0

this way the new question would be - if m > 0 or m > 1? is that ok?


Q2
Is 1/m > 1/n?
1. m < 2 < n < 6
2. 0 < mn < 4

my question here is can we convert 1/m > 1/n into m < n? can this be our new simplified question?


Many thanks for your help and time.
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by Anju@Gurome » Sat Apr 27, 2013 9:57 am
bharti06 wrote:My question here is whether or not we can simplify this further - m^3 > m^2? by taking everything to LHS and making RHS 0 as in -
m^3 - m^2 >0
m^2 (m-1) > 0
now either m^2 > 0 or m - 1 > 0
this way the new question would be - if m > 0 or m > 1? is that ok?
Yes, we can take everything to LHS and proceed like that.
But your conclusion is wrong.

If m²(m - 1) > 0, then either {m² > 0 and (m - 1) > 0} or {m² < 0 and (m - 1) < 0}
Now, the second case is not possible as for any real m, m² cannot be negative.
So, m² > 0 and (m - 1) > 0
--> m > 1

SO, the question is asking "Is m > 1?"
bharti06 wrote:my question here is can we convert 1/m > 1/n into m < n? can this be our new simplified question?
No, as we don't know whether m and n are positive or negative.
As signs of inequality changes depending upon the sign of the variable we are multiplying.

To determine whether 1/m > 1/n or not, we need to know the relative magnitudes of m and n, and the signs of m and n.
Statement 1: We know, m < n and n is positive but we don't whether m is positive or not. So, we can't answer the question. For example, Consider the following two cases,
  • m = 1, n = 3 ---> 1/m > 1/n
    m = -1, n = 3 ---> 1/m < 1/n
Not sufficient

Statement 2: This means either m and n are both positive or they are both negative. But we don't know whether m > n or not. For example, Consider the following two cases,
  • m = 1, n = 3 ---> 1/m > 1/n
    m = -1, n = -3 ---> 1/m < 1/n
Not sufficient

1 & 2 Together: Now, we know that m < n and both m and n are positive.
Hence, 1/m > 1/n

Sufficient

The correct answer is C.
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by bharti06 » Sat Apr 27, 2013 10:18 am
Anju@Gurome wrote:
bharti06 wrote:My question here is whether or not we can simplify this further - m^3 > m^2? by taking everything to LHS and making RHS 0 as in -
m^3 - m^2 >0
m^2 (m-1) > 0
now either m^2 > 0 or m - 1 > 0
this way the new question would be - if m > 0 or m > 1? is that ok?
Yes, we can take everything to LHS and proceed like that.
But your conclusion is wrong.

If m²(m - 1) > 0, then either {m² > 0 and (m - 1) > 0} or {m² < 0 and (m - 1) < 0}
Now, the second case is not possible as for any real m, m² cannot be negative.
So, m² > 0 and (m - 1) > 0
--> m > 1

SO, the question is asking "Is m > 1?"
bharti06 wrote:my question here is can we convert 1/m > 1/n into m < n? can this be our new simplified question?
No, as we don't know whether m and n are positive or negative.
As signs of inequality changes depending upon the sign of the variable we are multiplying.

To determine whether 1/m > 1/n or not, we need to know the relative magnitudes of m and n, and the signs of m and n.
Statement 1: We know, m < n and n is positive but we don't whether m is positive or not. So, we can't answer the question. For example, Consider the following two cases,
  • m = 1, n = 3 ---> 1/m > 1/n
    m = -1, n = 3 ---> 1/m < 1/n
Not sufficient

Statement 2: This means either m and n are both positive or they are both negative. But we don't know whether m > n or not. For example, Consider the following two cases,
  • m = 1, n = 3 ---> 1/m > 1/n
    m = -1, n = -3 ---> 1/m < 1/n
Not sufficient

1 & 2 Together: Now, we know that m < n and both m and n are positive.
Hence, 1/m > 1/n

Sufficient

The correct answer is C.
Anju, Thanks a ton for the detailed explanation to the second question. I got that.

But I still have some doubt in the first one -
here I am just copy pasting the part of your explanation which I have doubts in -

You wrote that If m²(m - 1) > 0, then either {m² > 0 and (m - 1) > 0} or {m² < 0 and (m - 1) < 0}

I am not able to understand the part I highlighted in red - how did you reach the second condition - ok may be m² < 0 because m has even power and we need to take it once positive and once negative...but how come you changed the inequality sign for the part of (m-1) too?

Then you said -
So, m² > 0 and (m - 1) > 0
--> m > 1 (and you combined both conditions into one)

should we not use OR (instead of and) here...I mean either m > 0 or (m - 1) > 0 and if any of the statements is able to provide sufficiency for any of the two conditions that should be sufficient...am i wrong here?

Thanks so much!
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by Anju@Gurome » Sat Apr 27, 2013 10:28 am
bharti06 wrote:...You wrote that If m²(m - 1) > 0, then either {m² > 0 and (m - 1) > 0} or {m² < 0 and (m - 1) < 0}

I am not able to understand the part I highlighted in red - how did you reach the second condition - ok may be m² < 0 because m has even power and we need to take it once positive and once negative...but how come you changed the inequality sign for the part of (m-1) too?
m²(m - 1) > 0 means the product of m² and (m - 1) is positive.
Product of two numbers is positive only when either both of the numbers are positive or both of the numbers are negative.

In this case either both m² and (m - 1) are positive or both m² and (m - 1) are negative
Which in mathematical language means either {m² > 0 and (m - 1) > 0} or {m² < 0 and (m - 1) < 0}

I hope this clears your confusion about AND and OR.
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by bharti06 » Sat Apr 27, 2013 10:32 am
Anju@Gurome wrote:
bharti06 wrote:...You wrote that If m²(m - 1) > 0, then either {m² > 0 and (m - 1) > 0} or {m² < 0 and (m - 1) < 0}

I am not able to understand the part I highlighted in red - how did you reach the second condition - ok may be m² < 0 because m has even power and we need to take it once positive and once negative...but how come you changed the inequality sign for the part of (m-1) too?
m²(m - 1) > 0 means the product of m² and (m - 1) is positive.
Product of two numbers is positive only when either both of the numbers are positive or both of the numbers are negative.

In this case either both m² and (m - 1) are positive or both m² and (m - 1) are negative
Which in mathematical language means either {m² > 0 and (m - 1) > 0} or {m² < 0 and (m - 1) < 0}

I hope this clears your confusion about AND and OR.
oh yeah...I got it now. Thank you sooo much Anju!!! I highly appreciate your help.
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by bharti06 » Sat Apr 27, 2013 10:43 am
Anju@Gurome wrote:
bharti06 wrote:My question here is whether or not we can simplify this further - m^3 > m^2? by taking everything to LHS and making RHS 0 as in -
m^3 - m^2 >0
m^2 (m-1) > 0
now either m^2 > 0 or m - 1 > 0
this way the new question would be - if m > 0 or m > 1? is that ok?
Yes, we can take everything to LHS and proceed like that.
But your conclusion is wrong.

If m²(m - 1) > 0, then either {m² > 0 and (m - 1) > 0} or {m² < 0 and (m - 1) < 0}
Now, the second case is not possible as for any real m, m² cannot be negative.
So, m² > 0 and (m - 1) > 0
--> m > 1

SO, the question is asking "Is m > 1?"
Can you also please confirm as the new question now is if m > 1?
So is statement 2 sufficient because it says that m < 1 which means we have a clear answer which is No.

Actually the answer for this question given is C...I am just wondering as per our simplified question the answer should be B....no?
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by Anju@Gurome » Sat Apr 27, 2013 10:47 am
bharti06 wrote:Can you also please confirm as the new question now is if m > 1?
So is statement 2 sufficient because it says that m < 1 which means we have a clear answer which is No.

Actually the answer for this question given is C...I am just wondering as per our simplified question the answer should be B....no?
Yes, the correct answer should be B.

There is no value of m such that m < 1 and m³ > m²
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by bharti06 » Sat Apr 27, 2013 10:59 am
Anju@Gurome wrote:
bharti06 wrote:Can you also please confirm as the new question now is if m > 1?
So is statement 2 sufficient because it says that m < 1 which means we have a clear answer which is No.

Actually the answer for this question given is C...I am just wondering as per our simplified question the answer should be B....no?
Yes, the correct answer should be B.

There is no value of m such that m < 1 and m³ > m²
you seem to be right. But I just checked the source and it says that B is not sufficient because m could equal zero too.....though the source in question here is not very reliable...so would rely on your judgement here...please confirm...thanks.
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by Anju@Gurome » Sat Apr 27, 2013 11:02 am
bharti06 wrote:...it says that B is not sufficient because m could equal zero too.....though the source in question here is not very reliable...so would rely on your judgement here...please confirm...thanks.
If 0 < m < 1, then m³ < m² ---> Answer to the original question is no.
If m = 0, then m³ = m² ---> Answer to the original question is no.
If m < 0, then m³ < 0 and m² > 0 ---> m³ < m² ---> Answer to the original question is no.

So, if m < 1, answer to the original question is always NO.
Hence, statement 2 is sufficient.
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by bharti06 » Sat Apr 27, 2013 11:07 am
Anju@Gurome wrote:
bharti06 wrote:...it says that B is not sufficient because m could equal zero too.....though the source in question here is not very reliable...so would rely on your judgement here...please confirm...thanks.
If 0 < m < 1, then m³ < m² ---> Answer to the original question is no.
If m = 0, then m³ = m² ---> Answer to the original question is no.
If m < 0, then m³ < 0 and m² > 0 ---> m³ < m² ---> Answer to the original question is no.

So, if m < 1, answer to the original question is always NO.
Hence, statement 2 is sufficient.
Yeah that makes complete sense. Appreciate your help!!! Thanks a ton!!!
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by GMATGuruNY » Sun Apr 28, 2013 2:33 am
bharti06 wrote:Q1. Is m^3 > m^2?
1. m > 0
2. m < 1
Is m³ > m²?
For it to be true that m³ > m², m ≠ 0, implying that m²>0.
Thus, we can safely divide each side of the question stem by m², knowing that the direction of the inequality will not change:
m³/m² > m²/m²
m > 1.

Question stem rephrased: Is m > 1?

Statement 1: m>0
It's possible that m>1 or that m<1.
INSUFFICIENT.

Statement 2: m<1
SUFFICIENT.

The correct answer is B.
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by bharti06 » Sun Apr 28, 2013 4:19 am
GMATGuruNY wrote:
bharti06 wrote:Q1. Is m^3 > m^2?
1. m > 0
2. m < 1
Is m³ > m²?
For it to be true that m³ > m², m ≠ 0, implying that m²>0.
Thus, we can safely divide each side of the question stem by m², knowing that the direction of the inequality will not change:
m³/m² > m²/m²
m > 1.

Question stem rephrased: Is m > 1?

Statement 1: m>0
It's possible that m>1 or that m<1.
INSUFFICIENT.

Statement 2: m<1
SUFFICIENT.

The correct answer is B.
Wow Mitch! You made the rephrasing super easy by dividing with m^2. That will surely save lot of time as we dont even need to try different values. Thanks so much for this technique.
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