Data Sufficiency

Statement 1:
a^3 - a < 0 => a(a+1)(a-1) < 0 -- alone NOT sufficient
Statement 2:
1 - a^2 > 0 => a^2 - 1 < 0 => (a-1)(a+1) < 0 -- alone not sufficient

If we put the value of second equation in first statement,
we have to make 'a' POSITIVE to get the result of a(a+1)(a-1) as NEGATIVE, given that (a-1)(a+1) is NEGATIVE from II statement.
Hence answer is C.

Hope it helps.

Statement 1: a^3 - a < 0
Or, a(a+1)(a-1) < 0

Case I : a>0, a+1 >0, a-1<0 Or, a>0,a>-1,a<1 => since a >0 as well as a > -1, a value is ambiguous=> data insufficient.


------------I-----I-----I--------
-1 0 +1
Case II : a>0, a+1 <0, a-1>0 Or, a>0,a <-1,a>1 => impossible solution since, a <-1 & a >1
Case III: a<0, a+1 >0, a-1>0 Or, a<0,a >-1,a>1 => impossible solution since, a <0 & a > 1

Statement 2: 1-a^2 > 0
Or, (1+a)(1-a) > 0
Case I : 1+a > 0,1-a > 0 Or, a > -1, a < 1 i.e. -1<a<1=> ambiguity in value of a i.e. positive or negative
Case II: 1+a < 0, 1-a < 0 Or, a < -1, a >1 i.e. -1<a<1=> ambiguity in value of a i.e. positive or negative


------------I-----I-----I--------
-1 0 +1

Combining Statement I & II, 0<a<1 & -1<a<1 gives correct nature of a.

Hence, correct option is (C). [/img]

Here is a demonstration of why combining the inequalities could miss the true answer:

Is a > 0 ?

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

Therefore a^3 - a < a^2 - 1

So a^3 - a^2 - a + 1 < 0

Factorising gives:

(a+1)(a-1)(a-1) < 0

Roots = -1 and 1

so 3 regions are to be looked at:

A) a < -1 so choose a = -2 (example)
B) -1 < a < 1 so choose a = 0
C) a > 1 so choose a = 2

Using (A) a = -2:
1. (-2)^3 - (-2) = -6 < 0
2. 1 - (-2)^2 = -3 IS NOT > 0
Hence a is not < -1

Using (B) a = 0:
1. (0)^3 - (0) = 0 IS NOT < 0
2. 1 - (0)^2 = 1 IS > 0
Hence, combining A and B, a is not < 1

Using (C) a = 2:
1. (2)^3 - (2) = 6 IS NOT < 0
2. 1 - (2)^2 = -3 IS NOT > 0
Hence a is not > 1

Therefore there would appear to be no answer!

However, as a cubic inequality must have an answer, the contradiction indicates that
combining the 2 given inequalities has limited our vision.

This leaves fractional values to test between -1 < a < 1

Consider:
D) a = -0.5
E) a = 0.5

Using (D) a = -0.5:
1. (-0.5)^3 - (-0.5) = 3/8 IS NOT < 0
2. 1 - (-0.5)^2 = 0.75 > 0
Hence a is not < -0.5

Using (E) a = 0.5:
1. (0.5)^3 - (0.5) = -3/8 < 0
2. 1 - (0.5)^2 = 0.75 > 0
Hence, combining 0 < a < 1

Therefore, yes a > 0.

i personally found the line no approach faster and accurate

I.(a^3)-a<0
a((a^2)-1)<0
a(a+1)(a-1)<0. so, the value of a will be less than zero for a<-1 and 0<a<1

II. 1-(a^2)>0 for -1<a<1

so my answer is E----

since 0<a<1 is common in part I & II, so answer will be 'C' (I thank you DavidG@VeritasPrep Sir for your guidance..)

I.(a^3)-a<0
a((a^2)-1)<0
a(a+1)(a-1)<0. so, the value of a will be less than zero for a<-1 and 0<a<1

II. 1-(a^2)>0 for -1<a<1

so my answer is E

Remember that when you're testing together, only values that satisfy both statements can be used. For the first statement, you found that:
a<-1 or 0<a<1

For the second statement, you found that -1<a<1

If a is less than -1, we will violate statement 2, so you can't consider those values. For example, -2 is not between -1 and 1.

If a is between -1 and 0, we'll violate statement 1, so we can't use those values. For example, (-1/2) is not less than -1 or between 0 and 1.

The only values that will satisfy both statements at the same time will fall between 0 and 1. (The previous poster's number line is a nice visual demonstration of this.) Because we know that 0 < a < 1, we'll know that a must be positive here, so C is correct.

GMATGuruNY wrote:

prashant misra wrote:i thought the answer to be D but i mistook the question as it also includes real values.gmat guru i understood the statements individually but how come -1<x<1

When we evaluate the two statements together, we may consider only values that satisfy both statements.

Two ranges satisfy statement 1: 0 < a < 1 or a < -1.
One range satisfies statement 2: -1 < a < 1.

In statement 1 it is possible that a < -1, but statement 2 requires that a > -1.
Since a number cannot be both less than -1 and greater than -1 at the same time, there are no negative values that satisfy both statements.

If a is not negative, then statement 1 requires that 0 < a.
Both statements require that a < 1.
Thus, the only range that satisfies both statements is 0 < a < 1.
Thus, we know that a > 0.
Sufficient.

The correct answer is C.

Aren't the ranges for which statement 1 is satisfied -1<a<0 and a>1? Same solution of the data sufficiency question, but a different outcome for a that is negative

anshuman09 wrote:Statement 1:
a^3 - a < 0 => a(a+1)(a-1) < 0 -- alone NOT sufficient
Statement 2:
1 - a^2 > 0 => a^2 - 1 < 0 => (a-1)(a+1) < 0 -- alone not sufficient

If we put the value of second equation in first statement,
we have to make 'a' POSITIVE to get the result of a(a+1)(a-1) as NEGATIVE, given that (a-1)(a+1) is NEGATIVE from II statement.
Hence answer is C.

Hope it helps.

Great, it is the simplest way to solve this question.

Only 1 is sufficient to answer this

GMATGuruNY wrote:

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?

I am lost completely in this example.... please help to understand how you found С
How did you get 1/2 and -2?
Sorry for the silly question but I did not get it... Please help ))\

kalini wrote:[I am lost completely in this example.... please help to understand how you found С
How did you get 1/2 and -2?
Sorry for the silly question but I did not get it... Please help ))\

Is a > 0 ?

(1) a^3 - a < 0
(2) 1 - a² > 0

Determine the CRITICAL POINTS.

Statement 1: a³ - 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 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
2³ < 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 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:
Let a = -2.
1 > (-2)².
1 > 4.
Doesn't work.

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

Plug a > 1 into 1 > a²:
Let a = 2.
1 > 2².
1 > 4.
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.