BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Inequality!!

Expert replies

by LalaB » Fri Jul 22, 2011 1:55 am
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
Join the discussion

by olegpoi » Tue Aug 30, 2011 10:56 am
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
Join the discussion

by olegpoi » Tue Aug 30, 2011 11:04 am
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.
Join the discussion

by Deependra1 » Sun Sep 04, 2011 7:23 am
ANSWER: C
Join the discussion

by prashant misra » Fri Sep 09, 2011 12:41 am
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
Join the discussion

by GMATGuruNY » Fri Sep 09, 2011 4:04 am
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.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Join the discussion

by parul9 » Sat Oct 01, 2011 10:30 pm
Join the discussion

by immaculatesahai » Mon Nov 14, 2011 1:48 am
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.
Join the discussion

by spartacus1412 » Sun Apr 01, 2012 11:27 pm
Try drawing the graphs for both of them. This method is quick and easy.
Join the discussion

by Shubhu@MBA » Thu Jul 26, 2012 10:49 am
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.
Join the discussion

by mparakala » Tue Oct 30, 2012 8:10 am
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]
Join the discussion

by [email protected] » Sun Nov 04, 2012 4:19 am
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
Join the discussion

by rajeshsinghgmat » Thu Jan 24, 2013 4:11 pm
C the answer.
Join the discussion

by sohailabbas4 » Thu Apr 25, 2013 1:01 pm
Image

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.
Join the discussion

by petrifiedbutstanding » Wed May 29, 2013 12:08 am
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..
Join the discussion