Data Sufficiency

a^3-a<0 ==> a(a^2-1)<0
1-a^2>0 ==> a^2-1<0
since from the 2nd stm we know that a^2-1<0 then a must be >0

answ is C

IMO - E
(1) a^3-a<0 => a^2<1
if a = 1/2 so 1/4<1 => 1/2>0
if a = -1/2 so 1/4<1 => -1/2<0
=> ambivalent Insufficient

(2) 1-a^2>0 => a^2<1 same as (1) => ambivalent => Insufficient

(1+2) IS too.
=> E

Oops..
I have made a mistake dividing inequality by an unknown: a^3<a \:a => a^2<1 - NOT CORRECT(sign of a is unknown)
Shame on me....

The answer is C.

ANSWER: C

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

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.

C!

Good question. Answer is C.

Statement 1: a(a^2-1)<0 Not Sufficient

Statement 2: 1-a^2>0
=> a^2-1<0 i.e. a^2-1 is -ve Not Sufficient

Combining the 2 statements. a(-ve term)<0.
This means that a has to be positive. C wins.

Try drawing the graphs for both of them. This method is quick and easy.

Is a > 0?

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


Let's check both the options one by one.

Option 1 -
a^3 - a = a(a^2 - 1)

Let's consider a is 0.5,
=> 0.5(0.25 -1) = 0.5 * (-0.75) < 0

Thus a =0.5 satisfies this equation.
Any value of a >=1 will not satisfy this equation. Thus, if a is positive number than it has to be
0<a<1.

But a = -2 will also satisfy this equation,
-2 * (4-1) <0

Thus option 1 is not sufficient to answer this question.

a<-1 & 0<a<1 will satisfy this equation.

Now let's consider option 2 alone to answer the question.

Option 2-
1 - a^2 > 0

a= -0.5 will satisfy this equation
a = 0.5 will satisfy this equation.

but any value of a>1 or a<-1 will not satisfy this equation.

Thus -1<a<1 will satisfy option 2.
Thus, option 2 alone is also not sufficient to answer the question.

However using both the options will solve the question.

Combining both the options we can conlcude that a will be in the range of 0<a<1.
Hence C.

1. a^3 - a <0
a^3 < a
it means that either a is negative or a is a positive fraction
ex: a = -2 so, a^3 = -8
a = 1/2 so, a^3 = 1/8
INSUFFICIENT

2. 1 - a^2 >0
=> a^2 < 1
-1 <a<1
so, a could be a negative fraction, zero or positive fraction
INSUFFICIENT

Using both (1) and (2), the only common value that a can take by satisfying both (1) and (2) is a positive fraction!

Ans: [C]

checking for Statements 1 and 2 together:
a^3-a<0
multiply both sides by -1
a-a^3>0
a(1-a^2)>0

no from statement 2 we know that 1-a^2>0 i.e. is +ve.
therefore "a" must be +ve or a>0
Ans: C

C the answer.

The best way to solve such problems is to draw them on the number line as shown in the diagram. Draw the first equation on the top of the number line and second equation at the bottom.

Looking at the picture. we can see that a>1 only when we combine the answer of both the equations.

GMATGuruNY wrote: 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 completely missed that! I don't believe it..