Inequalities advanced question MGMAT

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Inequalities advanced question MGMAT

by waggrave » Tue Aug 23, 2011 6:13 am
Is a > 0 ?

(1) a^3 - a < 0
(2) 1 - aÂ² > 0

My answer :

(1)

a^3 - a = a(a+1)(a-1) < 0
Hence a < 0 cause none of them can equal to 0 since the product is inferior to 0, and the 3 of them have to be inferior to 0. SUFFICIENT

It makes sens doesn't it ?

(2) is useless,

My answer is A.
In the mgmat book on inequalities, they do it another way and their answer is C. What's wrong with mine ?

Thanks

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by samyukta » Tue Aug 23, 2011 6:27 am
St 1 :

a ( a^2 -1) < 0

so a <0 or ( a^2 -1) <0 . Insufficient.

St 2: 1 - aÂ² > 0

which is nothing but (a^2 -1 )<0 ( Just multiply -1 in the prompt of st 2)

Still we are not sure of a as just a^2 <1, where in which a can be -1<0<1. Insufficent.

Combining St1 & st 2, we have (a^2 -1 )<0, so a must be greater than 0 for the st 1 to hold true.

Pick C.

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by kmittal82 » Tue Aug 23, 2011 6:29 am
Hence a < 0 cause none of them can equal to 0 since the product is inferior to 0, and the 3 of them have to be inferior to 0.
What if a = 0.1? Then you have 0.1 x 1.1 x -0.9 < 0 . This means this statement has a solution for a > 0 as well as a < 0.

If the question had stated a is an integer, then your explanation would have been good.

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by GMATGuruNY » Tue Aug 23, 2011 7:02 am
waggrave wrote:Is a > 0 ?

(1) a^3 - a < 0
(2) 1 - aÂ² > 0

My answer :

(1)

a^3 - a = a(a+1)(a-1) < 0
Hence a < 0 cause none of them can equal to 0 since the product is inferior to 0, and the 3 of them have to be inferior to 0. SUFFICIENT

It makes sens doesn't it ?

(2) is useless,

My answer is A.
In the mgmat book on inequalities, they do it another way and their answer is C. What's wrong with mine ?

Thanks
One way to evaluate the statements is to determine the critical points. I posted an explanation of this approach here:

https://www.beatthegmat.com/inequality-c ... 89518.html
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by saketk » Wed Aug 24, 2011 9:54 pm
waggrave wrote:Is a > 0 ?

(1) a^3 - a < 0
(2) 1 - aÂ² > 0

My answer :

(1)

a^3 - a = a(a+1)(a-1) < 0
Hence a < 0 cause none of them can equal to 0 since the product is inferior to 0, and the 3 of them have to be inferior to 0. SUFFICIENT

It makes sens doesn't it ?

(2) is useless,

My answer is A.
In the mgmat book on inequalities, they do it another way and their answer is C. What's wrong with mine ?

Thanks
HI-- Since it is not given that X is an integer, you need to keep in mind that the number can be a fraction as well. Let's take an example -- X=1/2
in this case, although your number is positive the answer will be less than zero..

(1/8) - (1/2) <0
Also if we take a negative value -- say for example (-2) then also we will get the resultant less than 0. hence the first condition is also NOT sufficient by itself.

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by navami » Thu Aug 25, 2011 3:03 am
(1) a^3 - a < 0
(2) 1 - aÂ² > 0

Lets consider the first case (1) a^3 - a < 0
a^3 < a --->(A)
here a could be +ve/ -ve Integer fraction or zero
now lets assume a is +ve Integer. Equation (A) will not be satisfied.
now lets assume a is -ve Integer. Equation (A) will be satisfied.
now lets assume a is +ve fraction. Equation (A) will be satisfied.
now lets assume a is -ve fraction. Equation (A) will not be satisfied.

only combining option (2) we can come out with a ans
This time no looking back!!!
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by GMATGuruNY » Thu Aug 25, 2011 7:44 pm
waggrave wrote:Is a > 0 ?

(1) a^3 - a < 0
(2) 1 - aÂ² > 0
I received a PM requesting that I show how this DS can be solved by determining the critical points.

Statement 1: aÂ³ - a < 0.
a(a+1)(a-1) < 0.
The critical points are a=0. a=-1, a=1.
These are the only values where a(a+1)(a-1) = 0.
When a is any other value, a(a+1)(a-1) < 0 or a(a+1)(a-1) > 0.
To determine the range of a, test one value to the left and right of each critical point.

Plug a < -1 into aÂ³ - a < 0:
Let a = -2.
(-2)Â³ - (-2) <0.
-6 < 0.
This works.

Plug -1 < a < 0 into aÂ³ - a < 0:
Let a = -1/2.
(-1/2)Â³ -(-1/2) < 0.
3/8 < 0.
Doesn't work.

Plug 0 < a < 1 into aÂ³ - a < 0:
Let a = 1/2.
(1/2)Â³ - 1/2 < 0.
-3/8 < 0.
This works.

Plug a > 1 into aÂ³ - a < 0:
Let a = 2
(2)Â³ - 2 < 0.
6 < 0.
Doesn't work.

Two ranges work in statement 1:
a < -1.
0 < a < 1.
Since a can be negative or positive, insufficient.

Statement 2: 1 - aÂ² > 0
1 - aÂ² > 0
(1+a)(1-a) > 0.
The critical points are a = -1 and a = 1.
These are the only values where 1 - aÂ² = 0.
When a is any other value, 1 - aÂ² < 0 or 1 - aÂ² > 0.
To determine the range of a, test one value to the left and right of each critical point.

Plug a < -1 into 1 - aÂ² > 0:
Let a = -2.
1 - (-2)Â² > 0.
-3 > 0.
Doesn't work.

Plug -1 < a < 1 into 1 - aÂ² > 0:
Let a = 0.
1 - 0Â² > 0.
1 > 0.
This works.

Plug a > 1 into 1 - aÂ² > 0:
Let a = 2.
1 - 2Â² > 0.
-3 > 0.
Doesn't work.

The only range that works is -1 < a < 1.
Since a can be negative or positive, insufficient.

Statements 1 and 2 combined:
Ranges that satisfy statement 1: 0 < a < 1 or a < -1.
Range that satisfies statement 2: -1 < a < 1.
The only range that satisfies both statements is 0 < a < 1.
Since a must be positive, sufficient.

The correct answer is C.
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by waggrave » Sat Aug 27, 2011 7:52 am
Thank you all for your answers.

What level is this question to your mind ?

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