Inequality!!

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adi_800: Legendary Member; Posts: 544; Joined: Thu Oct 08, 2009 9:14 amFollowed by: 2 members; Location: Pune, India; Thanked: 31 times

by adi_800 » Sat Jul 24, 2010 8:12 am

Is a > 0 ?

1. a^3 - a < 0
2. 1 - a^2 > 0

Consider 1st statement => a^3 - a < 0
a(a^2 - 1) < 0 => Two expressions have different signs. One is positive and other is negative and vice-e-versa.

Case I : a > 0 and (a^2-1) < 0
a > 0 and a^2 < 1
a > 0 and |a| < 1
a > 0 and -1 < a < 1
Clubbing above result => 0 < a < 1 => Result of case I

Case II : a < 0 and (a^2-1) > 0
a < 0 and a^2 > 1
a < 0 and |a| > 1
a < 0 and a > 1 or a < -1
Now the book that i m referring to mentions that It is impossible for a to be both negative and greater than 1, so we can focus only on second posibilty : a is both negative and less than -1.

Now why the condition a > 1 was neglected? We could have eliminated the negative scenario
(a < 0) and considered only positive (a > 0) ? whether the reason to eliminate the condition a > 1 was the original equation was not satisfied when a > 1? For e.g. if a = 2, then we can not have below condition being satisfied?
a^3 - a < 0

Btw...OA is C

Please let me know...N yes thanks for being patient in reading all above steps!!

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Source: Beat The GMAT — Data Sufficiency | Sat Jul 24, 2010 8:12 am

akdayal: Senior | Next Rank: 100 Posts; Posts: 52; Joined: Thu Feb 25, 2010 9:08 am; Thanked: 5 times

by akdayal » Sat Jul 24, 2010 8:47 am

@adi_800
just considering case II I hope you have only problem in case II

Case II : a < 0 and (a^2-1) > 0
a < 0 and a^2 > 1
a < 0 and |a| > 1
a < 0 and a > 1 or a < -1
Now the book that i m referring to mentions that It is impossible for a to be both negative and greater than 1, so we can focus only on second posibilty : a is both negative and less than -1.

Now why the condition a > 1 was neglected?

Because to determine final answer you have to take common answer.
Basically at this point you have two choice
1. a < 0 and a > 1
OR
2. a < 0 and a < -1

Now you think which one of above two have common ?
Obviously 2 . and that's why your book refer so.
Hope I cleared your doubt

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adi_800: Legendary Member; Posts: 544; Joined: Thu Oct 08, 2009 9:14 amFollowed by: 2 members; Location: Pune, India; Thanked: 31 times

by adi_800 » Sat Jul 24, 2010 8:55 am

Oh yes..got it..
did not concentrate on the or condition!!

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aloneontheedge: Senior | Next Rank: 100 Posts; Posts: 97; Joined: Thu Jan 28, 2010 2:07 am; Thanked: 9 times

by aloneontheedge » Sat Jul 24, 2010 11:20 am

adi_800 wrote:Is a > 0 ?

1. a^3 - a < 0
2. 1 - a^2 > 0

Consider 1st statement => a^3 - a < 0
a(a^2 - 1) < 0 => Two expressions have different signs. One is positive and other is negative and vice-e-versa.

Case I : a > 0 and (a^2-1) < 0
a > 0 and a^2 < 1
a > 0 and |a| < 1
a > 0 and -1 < a < 1
Clubbing above result => 0 < a < 1 => Result of case I

Case II : a < 0 and (a^2-1) > 0
a < 0 and a^2 > 1
a < 0 and |a| > 1
a < 0 and a > 1 or a < -1
Now the book that i m referring to mentions that It is impossible for a to be both negative and greater than 1, so we can focus only on second posibilty : a is both negative and less than -1.

Now why the condition a > 1 was neglected? We could have eliminated the negative scenario
(a < 0) and considered only positive (a > 0) ? whether the reason to eliminate the condition a > 1 was the original equation was not satisfied when a > 1? For e.g. if a = 2, then we can not have below condition being satisfied?
a^3 - a < 0

Btw...OA is C

Please let me know...N yes thanks for being patient in reading all above steps!!

IS OA correct I'm getting E as the answer

stmnt 1 : a^3 - a < 0
a = 0.1 or a= -2.Not suff

Stmnt 2: 1- a^2 > 0
a = 0.1 or a=-0.1. Not suff

combining a^3 - a < 0
0 < 1- a^2
Adding
a^3 - a + a^2 < 1
a = -2 or 0.1
Hence E

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sat Jul 24, 2010 12:55 pm

aloneontheedge wrote:
adi_800 wrote:Is a > 0 ?

1. a^3 - a < 0
2. 1 - a^2 > 0

Consider 1st statement => a^3 - a < 0
a(a^2 - 1) < 0 => Two expressions have different signs. One is positive and other is negative and vice-e-versa.

Case I : a > 0 and (a^2-1) < 0
a > 0 and a^2 < 1
a > 0 and |a| < 1
a > 0 and -1 < a < 1
Clubbing above result => 0 < a < 1 => Result of case I

Case II : a < 0 and (a^2-1) > 0
a < 0 and a^2 > 1
a < 0 and |a| > 1
a < 0 and a > 1 or a < -1
Now the book that i m referring to mentions that It is impossible for a to be both negative and greater than 1, so we can focus only on second posibilty : a is both negative and less than -1.

Now why the condition a > 1 was neglected? We could have eliminated the negative scenario
(a < 0) and considered only positive (a > 0) ? whether the reason to eliminate the condition a > 1 was the original equation was not satisfied when a > 1? For e.g. if a = 2, then we can not have below condition being satisfied?
a^3 - a < 0

Btw...OA is C

Please let me know...N yes thanks for being patient in reading all above steps!!

IS OA correct I'm getting E as the answer

stmnt 1 : a^3 - a < 0
a = 0.1 or a= -2.Not suff

Stmnt 2: 1- a^2 > 0
a = 0.1 or a=-0.1. Not suff

combining a^3 - a < 0
0 < 1- a^2
Adding
a^3 - a + a^2 < 1
a = -2 or 0.1
Hence E

We can't just add the inequalities as you did. We need to be careful when inequalities contain exponents. A negative solution might force the direction of the inequality to change. Doing a bit of algebra and plugging in numbers is much safer.

Here's how I would approach this problem.

Statement 1:
Rewritten, we get a^3 < a.
a = 1/2 and a = -2 both work. Insufficient.

Statement 2:
Rewritten, we get -a^2 > -1, so a^2 < 1.
a= 1/2 and a= - 1/2 both work. Insufficient.

Statements 1 and 2 together:

Statement 2 tells us that -1 < a < 1. No other values will work.
a=0 doesn't work in statement 1.
A negative fraction won't work in statement 1 because a negative fraction raised to an odd power gets bigger: (-1/2)^3 = -(1/8), and -1/8 > - 1/2.

So to satisfy both statements, 0 < a < 1. Sufficient.

The correct answer is C.

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aloneontheedge: Senior | Next Rank: 100 Posts; Posts: 97; Joined: Thu Jan 28, 2010 2:07 am; Thanked: 9 times

by aloneontheedge » Sat Jul 24, 2010 1:05 pm

GMATGuruNY wrote:
aloneontheedge wrote:
adi_800 wrote:Is a > 0 ?

1. a^3 - a < 0
2. 1 - a^2 > 0

Consider 1st statement => a^3 - a < 0
a(a^2 - 1) < 0 => Two expressions have different signs. One is positive and other is negative and vice-e-versa.

Case I : a > 0 and (a^2-1) < 0
a > 0 and a^2 < 1
a > 0 and |a| < 1
a > 0 and -1 < a < 1
Clubbing above result => 0 < a < 1 => Result of case I

Case II : a < 0 and (a^2-1) > 0
a < 0 and a^2 > 1
a < 0 and |a| > 1
a < 0 and a > 1 or a < -1
Now the book that i m referring to mentions that It is impossible for a to be both negative and greater than 1, so we can focus only on second posibilty : a is both negative and less than -1.

Now why the condition a > 1 was neglected? We could have eliminated the negative scenario
(a < 0) and considered only positive (a > 0) ? whether the reason to eliminate the condition a > 1 was the original equation was not satisfied when a > 1? For e.g. if a = 2, then we can not have below condition being satisfied?
a^3 - a < 0

Btw...OA is C

Please let me know...N yes thanks for being patient in reading all above steps!!

IS OA correct I'm getting E as the answer

stmnt 1 : a^3 - a < 0
a = 0.1 or a= -2.Not suff

Stmnt 2: 1- a^2 > 0
a = 0.1 or a=-0.1. Not suff

combining a^3 - a < 0
0 < 1- a^2
Adding
a^3 - a + a^2 < 1
a = -2 or 0.1
Hence E
We can't just add the inequalities as you did. We need to be careful when inequalities contain exponents. A negative solution might force the direction of the inequality to change. Doing a bit of algebra and plugging in numbers is much safer.

Here's how I would approach this problem.

Statement 1:
Rewritten, we get a^3 < a.
a = 1/2 and a = -2 both work. Insufficient.

Statement 2:
Rewritten, we get -a^2 > -1, so a^2 < 1.
a= 1/2 and a= - 1/2 both work. Insufficient.

Statements 1 and 2 together:

Statement 2 tells us that -1 < a < 1. No other values will work.
a=0 doesn't work in statement 1.
A negative fraction won't work in statement 1 because a negative fraction raised to an odd power gets bigger: (-1/2)^3 = -(1/8), and -1/8 > - 1/2.

So to satisfy both statements, 0 < a < 1. Sufficient.

The correct answer is C.

Thank you...U mean to say whenever we have exponents,we cannot add inequalities as i did before?

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by GMATGuruNY » Sat Jul 24, 2010 1:39 pm

aloneontheedge wrote:
GMATGuruNY wrote:
aloneontheedge wrote:
adi_800 wrote:Is a > 0 ?

1. a^3 - a < 0
2. 1 - a^2 > 0

Consider 1st statement => a^3 - a < 0
a(a^2 - 1) < 0 => Two expressions have different signs. One is positive and other is negative and vice-e-versa.

Case I : a > 0 and (a^2-1) < 0
a > 0 and a^2 < 1
a > 0 and |a| < 1
a > 0 and -1 < a < 1
Clubbing above result => 0 < a < 1 => Result of case I

Case II : a < 0 and (a^2-1) > 0
a < 0 and a^2 > 1
a < 0 and |a| > 1
a < 0 and a > 1 or a < -1
Now the book that i m referring to mentions that It is impossible for a to be both negative and greater than 1, so we can focus only on second posibilty : a is both negative and less than -1.

Now why the condition a > 1 was neglected? We could have eliminated the negative scenario
(a < 0) and considered only positive (a > 0) ? whether the reason to eliminate the condition a > 1 was the original equation was not satisfied when a > 1? For e.g. if a = 2, then we can not have below condition being satisfied?
a^3 - a < 0

Btw...OA is C

Please let me know...N yes thanks for being patient in reading all above steps!!

IS OA correct I'm getting E as the answer

stmnt 1 : a^3 - a < 0
a = 0.1 or a= -2.Not suff

Stmnt 2: 1- a^2 > 0
a = 0.1 or a=-0.1. Not suff

combining a^3 - a < 0
0 < 1- a^2
Adding
a^3 - a + a^2 < 1
a = -2 or 0.1
Hence E
We can't just add the inequalities as you did. We need to be careful when inequalities contain exponents. A negative solution might force the direction of the inequality to change. Doing a bit of algebra and plugging in numbers is much safer.

Here's how I would approach this problem.

Statement 1:
Rewritten, we get a^3 < a.
a = 1/2 and a = -2 both work. Insufficient.

Statement 2:
Rewritten, we get -a^2 > -1, so a^2 < 1.
a= 1/2 and a= - 1/2 both work. Insufficient.

Statements 1 and 2 together:

Statement 2 tells us that -1 < a < 1. No other values will work.
a=0 doesn't work in statement 1.
A negative fraction won't work in statement 1 because a negative fraction raised to an odd power gets bigger: (-1/2)^3 = -(1/8), and -1/8 > - 1/2.

So to satisfy both statements, 0 < a < 1. Sufficient.

The correct answer is C.
Thank you...U mean to say whenever we have exponents,we cannot add inequalities as i did before?

While there might be some instances in which adding would be ok, don't take the risk. There is too much room for error. For example:

Let's say x^2 < 1 and x^3 < 1.
Does x = -2 work in each equality? No, because (-2)^2<1 becomes 4<1, which isn't true.

Adding the two inequalties, we get x^2 + x^3 < 2.
Does x = -2 work here? Yes, because (-2)^2 + (-2)^3 < 2 becomes -4 < 2, which is true.

So adding the inequalities allows for values that are not allowed by the separate inequalities. (The problem is that x^2 < 1 means x<1 AND x>-1. So depending on whether x>0 or x<0, the direction of the inequality has to change, which makes adding problematic.)

So adding inequalities with exponents is very dangerous. Much safer to plug in real numbers.

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rsnaren: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Mon Dec 29, 2008 2:54 am; Location: India

by rsnaren » Mon Jul 26, 2010 6:20 am

GMATGuruNY wrote:
aloneontheedge wrote:
adi_800 wrote:Is a > 0 ?

1. a^3 - a < 0
2. 1 - a^2 > 0

Consider 1st statement => a^3 - a < 0
a(a^2 - 1) < 0 => Two expressions have different signs. One is positive and other is negative and vice-e-versa.

Case I : a > 0 and (a^2-1) < 0
a > 0 and a^2 < 1
a > 0 and |a| < 1
a > 0 and -1 < a < 1
Clubbing above result => 0 < a < 1 => Result of case I

Case II : a < 0 and (a^2-1) > 0
a < 0 and a^2 > 1
a < 0 and |a| > 1
a < 0 and a > 1 or a < -1
Now the book that i m referring to mentions that It is impossible for a to be both negative and greater than 1, so we can focus only on second posibilty : a is both negative and less than -1.

Now why the condition a > 1 was neglected? We could have eliminated the negative scenario
(a < 0) and considered only positive (a > 0) ? whether the reason to eliminate the condition a > 1 was the original equation was not satisfied when a > 1? For e.g. if a = 2, then we can not have below condition being satisfied?
a^3 - a < 0

Btw...OA is C

Please let me know...N yes thanks for being patient in reading all above steps!!

IS OA correct I'm getting E as the answer

stmnt 1 : a^3 - a < 0
a = 0.1 or a= -2.Not suff

Stmnt 2: 1- a^2 > 0
a = 0.1 or a=-0.1. Not suff

combining a^3 - a < 0
0 < 1- a^2
Adding
a^3 - a + a^2 < 1
a = -2 or 0.1
Hence E
We can't just add the inequalities as you did. We need to be careful when inequalities contain exponents. A negative solution might force the direction of the inequality to change. Doing a bit of algebra and plugging in numbers is much safer.

Here's how I would approach this problem.

Statement 1:
Rewritten, we get a^3 < a.
a = 1/2 and a = -2 both work. Insufficient.

Statement 2:
Rewritten, we get -a^2 > -1, so a^2 < 1.
a= 1/2 and a= - 1/2 both work. Insufficient.

Statements 1 and 2 together:

Statement 2 tells us that -1 < a < 1. No other values will work.
a=0 doesn't work in statement 1.
A negative fraction won't work in statement 1 because a negative fraction raised to an odd power gets bigger: (-1/2)^3 = -(1/8), and -1/8 > - 1/2.

So to satisfy both statements, 0 < a < 1. Sufficient.

The correct answer is C.

I understood till the point wherein you explained the cases separately. But i was completely lost and could not figure it out how you managed to bring the two statements together and arrived at 0<a<1?

Could you pls explain in detail?

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by GMATGuruNY » Mon Jul 26, 2010 7:09 am

Here's how I would approach this problem.

Statement 1:
Rewritten, we get a^3 < a.
a = 1/2 and a = -2 both work. Insufficient.

Statement 2:
Rewritten, we get -a^2 > -1, so a^2 < 1.
a= 1/2 and a= - 1/2 both work. Insufficient.

Statements 1 and 2 together:

Statement 2 tells us that -1 < a < 1. No other values will work.
a=0 doesn't work in statement 1.
A negative fraction won't work in statement 1 because a negative fraction raised to an odd power gets bigger: (-1/2)^3 = -(1/8), and -1/8 > - 1/2.

So to satisfy both statements, 0 < a < 1. Sufficient.

The correct answer is C.

I understood till the point wherein you explained the cases separately. But i was completely lost and could not figure it out how you managed to bring the two statements together and arrived at 0<a<1?

Could you pls explain in detail?

To satisfy statement 1, we can use a positive fraction (we used 1/2) or values less than -1 (we used -2).
To satisfy statement 2, we can use a positive fraction (we used 1/2) or a negative fraction (we used -1/2).
To satisfy both statements, we can use ONLY a positive fraction. So 0<a<1.

Does this help?

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rsnaren: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Mon Dec 29, 2008 2:54 am; Location: India

by rsnaren » Mon Jul 26, 2010 10:23 am

GMATGuruNY wrote:Here's how I would approach this problem.

Statement 1:
Rewritten, we get a^3 < a.
a = 1/2 and a = -2 both work. Insufficient.

Statement 2:
Rewritten, we get -a^2 > -1, so a^2 < 1.
a= 1/2 and a= - 1/2 both work. Insufficient.

Statements 1 and 2 together:

Statement 2 tells us that -1 < a < 1. No other values will work.
a=0 doesn't work in statement 1.
A negative fraction won't work in statement 1 because a negative fraction raised to an odd power gets bigger: (-1/2)^3 = -(1/8), and -1/8 > - 1/2.

So to satisfy both statements, 0 < a < 1. Sufficient.

The correct answer is C.

I understood till the point wherein you explained the cases separately. But i was completely lost and could not figure it out how you managed to bring the two statements together and arrived at 0<a<1?

Could you pls explain in detail?
To satisfy statement 1, we can use a positive fraction (we used 1/2) or values less than -1 (we used -2).
To satisfy statement 2, we can use a positive fraction (we used 1/2) or a negative fraction (we used -1/2).
To satisfy both statements, we can use ONLY a positive fraction. So 0<a<1.

Does this help?

Thanks alot. I did understand finally the way you have worked it up.
But i got few doubts, might be silly but i am confused at certain points.

In case 1 : you have re written the equation as a^3<a;
Is it not possible that I can divide both sides by a so that the equation becomes a^2<1 which is similar to that of Case 2 ???

It will be of great help, if you can clear me out. I am literally lost with Inequalities !!!

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by GMATGuruNY » Mon Jul 26, 2010 11:07 am

rsnaren wrote:
GMATGuruNY wrote:Here's how I would approach this problem.

Statement 1:
Rewritten, we get a^3 < a.
a = 1/2 and a = -2 both work. Insufficient.

Statement 2:
Rewritten, we get -a^2 > -1, so a^2 < 1.
a= 1/2 and a= - 1/2 both work. Insufficient.

Statements 1 and 2 together:

Statement 2 tells us that -1 < a < 1. No other values will work.
a=0 doesn't work in statement 1.
A negative fraction won't work in statement 1 because a negative fraction raised to an odd power gets bigger: (-1/2)^3 = -(1/8), and -1/8 > - 1/2.

So to satisfy both statements, 0 < a < 1. Sufficient.

The correct answer is C.

I understood till the point wherein you explained the cases separately. But i was completely lost and could not figure it out how you managed to bring the two statements together and arrived at 0<a<1?

Could you pls explain in detail?
To satisfy statement 1, we can use a positive fraction (we used 1/2) or values less than -1 (we used -2).
To satisfy statement 2, we can use a positive fraction (we used 1/2) or a negative fraction (we used -1/2).
To satisfy both statements, we can use ONLY a positive fraction. So 0<a<1.

Does this help?
Thanks alot. I did understand finally the way you have worked it up.
But i got few doubts, might be silly but i am confused at certain points.

In case 1 : you have re written the equation as a^3<a;
Is it not possible that I can divide both sides by a so that the equation becomes a^2<1 which is similar to that of Case 2 ???

It will be of great help, if you can clear me out. I am literally lost with Inequalities !!!

It's too dangerous to simplify a^3 < a because we don't know whether a is positive or negative, and if a is negative, the direction of the inequality would have to change.

A basic example:

2x < 10
x < 5

-2x < 10
x > -5
When we multiply or divide each side of an inequality by a negative number, we have to change the direction of the inequality.

So looking at a^3<a:

If a>0, then a^2 < 1.
If a<0, then a^2 > 1. (Because when we divide by a negative number, we have to change the direction of the inequality.)

Since we don't know whether a is positive or negative, simplifying is risky.

When a DS question has an inequality that includes a variable, don't multiply or divide each side by the variable unless you know that the variable is definitely positive or definitely negative.

Does this help?

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rsnaren: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Mon Dec 29, 2008 2:54 am; Location: India

by rsnaren » Mon Jul 26, 2010 8:26 pm

GMATGuruNY wrote:
rsnaren wrote:
GMATGuruNY wrote:

When a DS question has an inequality that includes a variable, don't multiply or divide each side by the variable unless you know that the variable is definitely positive or definitely negative.

Does this help?
Yeah, I got it now..!!!!
Thanks for the Help!

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pharmxanthan: Senior | Next Rank: 100 Posts; Posts: 77; Joined: Fri Jan 08, 2010 7:55 am; Thanked: 2 times; GMAT Score:700

by pharmxanthan » Wed Jul 28, 2010 10:23 pm

Can we combine the two statements in the following way?

St A:
a (a^2-1)<0
a (a-1) (a+1)<0
Multiplying both sides by -1, we get
a (1-a) (a+1)>0 ---- I
St B:
(1-a) (1+a)>0
So, equation I becomes:
a>0
Hence answer C.

Is this method right?

Thanks in advance!

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Ian Stewart: GMAT Instructor; Posts: 2623; Joined: Mon Jun 02, 2008 3:17 amFollowed by: 355 members; Location: Montreal; Thanked: 1090 times; GMAT Score:780

by Ian Stewart » Sat Jul 31, 2010 10:40 pm

GMATGuruNY wrote: We can't just add the inequalities as you did. We need to be careful when inequalities contain exponents. A negative solution might force the direction of the inequality to change. Doing a bit of algebra and plugging in numbers is much safer.

I'm not sure what this means. It is *always* perfectly correct to add two inequalities, provided the inequalities face in the same direction (though I'd add, it is *not* correct to subtract one inequality from another). It does not matter whether the inequalities contain exponents, or if there are negative solutions. The one issue with adding inequalities is that you risk losing information. I can provide a very simplistic example as illustration. Suppose you are asked a DS question:

Is x > 2?
1) x > 0
2) x > 2

The answer is clearly B, but what happens if we don't notice that, and decide to use both Statements and add the two inequalities? We find that 2x > 2, or that x > 1. That clearly has to be true if both Statements are true - there's nothing wrong with the math here - but we've lost information. The best information comes from Statement 2 alone, that x > 2, and by adding the two inequalities, we get less information about x. This is the problem with the approach alonetheedge took; while it's fine to add the inequalities in the question in the original post, you're losing information about the value of a when you do so, and the information you lose is critical here.

I can't think of a situation where you'd want to add two inequalities which contain only one unknown. I normally only find it useful to add inequalities on GMAT questions when there are two (or more, I suppose) unknowns, and I can see that by adding the inequalities, some of the unknowns will disappear, and this will lead to an answer to the question (or by adding, I'll get an expression closer to the one the question asks for). So in a question like the following, adding inequalities is a very good approach:

Is x > 3?
1) x + y > 1
2) x - y > 5

Adding the inequalities in the two statements instantly gives us the answer (C). So while adding inequalities can be a useful technique, it's only useful in specific situations. Still, it's perfectly correct, mathematically.

For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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bacchewar_prashant: Senior | Next Rank: 100 Posts; Posts: 39; Joined: Tue Sep 28, 2010 11:34 pm; Thanked: 1 times

by bacchewar_prashant » Fri Nov 12, 2010 6:17 am

Excellent problem. BTW what is the source of problem.

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