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is a>0 Data Sufficiency

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is a>0 Data Sufficiency

by xiamsamx » Sat Oct 10, 2015 11:43 am
Is a>0?

(1) a^3 - a < 0
(2) 1 - a^2 > 0


I know that you can solve theoretically but I wanted to see where I went wrong when I tried to solve algebraically.

Here is what I did:

1) a^3 - a < 0 can be rewritten as a*(a^2 - 1) < 0. I know one solution is a <0.
Looking at (a^2 - 1) < 0, I then rewrote that as (a+1)(a-1) < 0.
My second set of solutions then is a < -1 and a < 1.
Statement 1 is insufficient.

2) 1 - a^2 > 0 can be rewritten as (1+a)(1-a) >0. Your solutions are then a > -1 and a < 1.
Statement 2 is also insufficient.

Looking at both statements together, you still get a range of -1 < a < 1, so I picked answer choice E.

The correct answer is C.
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by theCEO » Sat Oct 10, 2015 1:03 pm
xiamsamx wrote:Is a>0?

(1) a^3 - a < 0
(2) 1 - a^2 > 0


I know that you can solve theoretically but I wanted to see where I went wrong when I tried to solve algebraically.

Here is what I did:

1) a^3 - a < 0 can be rewritten as a*(a^2 - 1) < 0. I know one solution is a <0.
Looking at (a^2 - 1) < 0, I then rewrote that as (a+1)(a-1) < 0.
My second set of solutions then is a < -1 and a < 1.
Statement 1 is insufficient.

2) 1 - a^2 > 0 can be rewritten as (1+a)(1-a) >0. Your solutions are then a > -1 and a < 1.
Statement 2 is also insufficient.

Looking at both statements together, you still get a range of -1 < a < 1, so I picked answer choice E.

The correct answer is C.
1) a^3 - a < 0 can be rewritten as a*(a^2 - 1) < 0. I know one solution is a <0.
Looking at (a^2 - 1) < 0, I then rewrote that as (a+1)(a-1) < 0.
My second set of solutions then is a < -1 and a < 1.
two solutions: 0<a<1 or a<-1
Statement 1 is insufficient.

2) 1 - a^2 > 0 can be rewritten as (1+a)(1-a) >0. Your solutions are then a > -1 and a < 1.
two solutions a<-1 or a<1
Statement 2 is also insufficient.

combination
(0<a<1 or a<-1) and (a<-1 or a<1)
a<-1 is common in both

therefore we can answer the question: is a>0
ans = c
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by GMATGuruNY » Sat Oct 10, 2015 1:09 pm
xiamsamx wrote:Is a>0?

(1) a^3 - a < 0
(2) 1 - a^2 > 0
Statement 1: a³ < a.
a³ < a.
a³ - a < 0.
a(a+1)(a-1) < 0.
The critical points are a=0. a=-1, a=1.
These are the only values for a 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 a³ < a:
Let a = -2.
(-2)³ < 2.
-8 < -2.
This works.

Plug -1 < a < 0 into a³ < a:
Let a = -1/2.
(-1/2)³ < -1/2
-1/8 < - 1/2.
Doesn't work.

Plug 0 < a < 1 into a³ < a:
Let a = 1/2.
(1/2)³ < 1/2
1/8 < 1/2.
This works.

Plug a > 1 into a³ < a:
Let a = 2
(2)³ < 2
8 < 2.
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 - a² > 0
1 - a² > 0
(1+a)(1-a) > 0.
The critical points are a = -1 and a = 1.
These are the only values for a where 1 - a² = 0.
When a is any other value, 1 - a² < 0 or 1 - a² > 0.
To determine the range of a, test one value to the left and right of each critical point.

Plug a < -1 into 1 - a² > 0:
Let a = -2.
1 - (-2)² > 0.
-3 > 0.
Doesn't work.

Plug -1 < a < 1 into 1 - a² > 0:
Let a = 0.
1 - 0² > 0.
1 > 0.
This works.

Plug a > 1 into 1 - a² > 0:
Let a = 2.
1 - 2² > 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 ceilidh.erickson » Mon Oct 19, 2015 8:00 am
Here's an easier approach:

The target question is: is a positive?

(1) Rewrite as a³ < a Then ask yourself: where on the number line do we have numbers whose cube is less than the number itself?

Image

Here, a³ < a only when a < -1 or when 0 < a < 1. So, a could be positive or negative. Insufficient.

(2) Rewrite 1 - a² > 0 as 1 > a² . Again, ask yourself: where on the number line is this true? What kinds of squares are less than 1?

Image

This will be true for any positive or negative fraction, and for 0. Thus, where -1 < a < 1. Since this range includes positives and negatives, this is insufficient.

(1) and (2) combined:
Statement 1 tells us that a < -1 or 0 < a < 1
Statement 2 tells us that -1 < a < 1

If we combine these statements, then a must be in the range 0 < a < 1 and therefore a must be positive. Sufficient.

The answer is C.
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by ceilidh.erickson » Mon Oct 19, 2015 8:02 am
I have often found that my students have an easier time visualizing inequalities on a number line than doing algebra. For more practice with this concept, see:
https://www.beatthegmat.com/x-x-x-which- ... tml#583371
https://www.beatthegmat.com/is-x-5-x-4-t ... tml#690059
https://www.beatthegmat.com/x-0-t279441.html#730063
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by Max@Math Revolution » Wed Oct 21, 2015 11:15 am
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is a>0?

(1) a^3 - a < 0
(2) 1 - a^2 > 0

If the range of the question includes that of the condition, the condition is sufficient.
There is one variable (a), and so we need one equation, whereas 2 equations are given from the 2 conditions.

from condition 1, a(a-1)(a+1)<0 ==> a<-1, 0<a<1, but this is not included in the range of the question, so this condition is insufficient.
From condition 2, a^2-1<0, (a-1)(a+1)<0 ==> -1<a<1, and with the same reason this is also a insufficient condition.
But if we combine the 2 conditions, 0<a<1 is what we get, and this is included in the range of the question, so the conditions combined make a sufficient condition, and the answer is (C).

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