The way I did it was this:

Statement 1: A^3-A<0. Unpack into (A-1)A(A+1)<0 . From here we see that A can be any negative number (neg^3 still neg.) and the statement still holds true OR A can be positive(such as A can be 1) and the statement still holds true. Insufficient.

Statement 2: 1-A^2>0 move A^2 to the right side, and it reads 1>A^2. We now can deduce -1<A<1. Insufficient.

Statements combined: Start plugging in some smart numbers into statement 1. I started with a neg fraction. That yields a positive number, so we know A CANNOT be negative. Just to be sure I plugged in a positive fraction. A is negative. SUFFICIENT.

I always say.....PITA (Plug in the answer choices) in these sort of inequalities.

if you get two answers ...eliminate

combined together:if u r getting a constant answer indicating the 'truth'...C

It is a LOT complicated doing the inequality shifting.You get points for the correct answer not the process here!

MY approach...


i) a^3 - a < 0

=>a (a^2-1) <0
=> a <0 and a^2 > 1 or a>0 and a^2 <1
==> range of a is ..... (-infinity)-------(-1) and 0------(+1) => Insufficient


ii) a^2 -1 >0
=> -1 <a <=1
clearly insufficient


Combine both
Range of a will be 0<a<1 ===> Sufficient

hi:
this is how I came to the conclusion:

Statement #1: a^3 - a < 0

consider that a no can be 0, greater than 1, smaller than -1, between 0 and 1, between 0 and -1

assume a=0, a >1, a <-1, a>-1, a <1

a=0 ==> doesn't work becasue 0-0==0
a>1 ==> doesn;t work becasue cube of a positive no greater than 1 is greater than the no (2^3- 2 = 6)
a<-1 ==> works because cube of a negative no smaller than 1 is smaller than the no (-2^3 - (-2)) == -6) reverse logic to above
a >-1 ==> doesn't work because cube of a fraction below 1 is smaller in magnitude but retains the sign ( (-1/2) ^3 -(-1/2)) ==> -1/8 + 1/2 == greater than 0)
a <1 ==> works because it is reverse of above statement ((1/2)^3 - (1/2) == 1/8 - 1/2= negative)

So a <1 or a < -1

Statement #2: 1 - a^2 > 0

this means that a^2 < 1

assume a=0, a >1, a <-1, a>-1, a <1,
a==0 ==> works
a>1 ==> doesn;t work because square of a positive no greater than 1 is bigger than 1.
a<-1 doesn't work because sign of a aq is always +ve and the sq of a no greater than 1 is always bigger than 1.
a>-1 ==> works because sign of a swquare is always +ve and the sq of a no lesser than 1 is always lesser (-1/4 * -1/4) = 1/16
a<1 = same logic as above. square of a sub-1 no is alwyas less than 1 (1/2 * 1/2) = 1/4

so -1 <a < 1
only way we can answer the question (a >0?) conclusively is by combining the two conditions:
a is positive and < 1

Experts, kindly tell whether the following approach is correct because as with pharmxanthan, i too used the same approach

Can we combine the two statements in the following way?

St A:
a (a^2-1)<0
a (a-1) (a+1)<0
Multiplying both sides by -1, we get
a (1-a) (a+1)>0 ---- I
St B:
(1-a) (1+a)>0
So, equation I becomes:
a>0
Hence answer C.

Is this method right?

Thanks in advance!


Thanks

Quick approach.

a>0?

stmt1,

a^3 - a < 0

a=0.5,a^3-a<0--> a>0

a=-2,a^3-a<0-->a<0

Not suff

stmt2,

1 - a^2 > 0

a=-0.5,1 - a^2 > 0 -->a<0

a=0.5,1 - a^2 > 0 -->a>0

Not suff

Combining 1 and 2,

1 - a^2 > 0 or a^2-1<0

a^3 - a < 0

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

We know that a^2-1<0.So a must be greater than zero so that the above equation satisfied.

Pick C

Is a > 0 ?

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


Statement - 1

If we take a = 1.
1-1 < 0

If we take a = -1.
-1+1<0

If we take a = -2.
-8+2<0
-6<0

If we take a = 2
8-2<0
6<0

So we can conclude that a is not a>0, it can be less than 0 also.


Statement-2

If we take a=1.
1-1>0

If we take a=1.
1-1>0

If we take a=-2.
1-4>0
-3>0

If we take a=2.
1-4>0
-3>0

I am confused, the conditions doesn't seem to match the answer. Statement-2 is not true, whether a > 0 or a < 0 ???

What am I doing???

Great Problem Good Solution

Ok. i definitely need to clear this one up.

Statement 1:

Result which i got from solving it : 0<a <1

Statement 2:

Result which i got from solving it : -1 < a< 1

Q Asks is a>0.

So cant we say from Statement 1 that yes a>0. (assuming my solutions are correct)

Gr8 question.

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?

ankur.agrawal wrote:Ok. i definitely need to clear this one up.

Statement 1:

Result which i got from solving it : 0<a <1

Statement 2:

Result which i got from solving it : -1 < a< 1

Q Asks is a>0.

So cant we say from Statement 1 that yes a>0. (assuming my solutions are correct)

Gr8 question.

Not really Ankur..

St 1 gives : a (a^2 -1) < 0

So we can have a < 0 or (a^2 -1) <0

St 2 : 1 - a^2 > 0

Multiply by -1 on both sides.

a^2 -1 < 0

Here a can be a positive number or a negative number. Insufficient.

Combining both the sts,:

a (a^2 -1) < 0

We know that negative * positive is negative number
. Frm st 2 wkt, (a^2 -1) is negative . So "a" has to be positive to satisfy st 1.

Thus a>0.

Dont bother abt the range .

OK Here it Goes. I thought it a little differently. More GMAT style:

1. a^3 - a < 0 ; a (a^2-1) <0 ; two terms must have Opp signs <0. Don't know which one is + & which is -. NS

2. 1 - a^2 > 0------> a^2<1 ---> Taking square root on both sides------> modulus a <1 ; ---> -1<a<1. NS

Now Looking at two together we note :

We can rewrite the 2nd statement as : a^2-1 <0 . So looking at statement 1 now, we know (a^2 -1) is Negative.

So in order for the 1st statement to be true i.e : a (a^2-1) <0 ; a has to be + i.e >0 becoz we know from 2nd statement that (a^2 -1) is Negative.

SO 'C'

I hope i am absolutely clear.javascript:emoticon(':D')

adi_800 wrote:Is a > 0 ?

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

Consider 1st statement => a^3 - a < 0
a(a^2 - 1) < 0 => Two expressions have different signs. One is positive and other is negative and vice-e-versa.

Case I : a > 0 and (a^2-1) < 0
a > 0 and a^2 < 1
a > 0 and |a| < 1
a > 0 and -1 < a < 1
Clubbing above result => 0 < a < 1 => Result of case I

Case II : a < 0 and (a^2-1) > 0
a < 0 and a^2 > 1
a < 0 and |a| > 1
a < 0 and a > 1 or a < -1
Now the book that i m referring to mentions that It is impossible for a to be both negative and greater than 1, so we can focus only on second posibilty : a is both negative and less than -1.

Now why the condition a > 1 was neglected? We could have eliminated the negative scenario
(a < 0) and considered only positive (a > 0) ? whether the reason to eliminate the condition a > 1 was the original equation was not satisfied when a > 1? For e.g. if a = 2, then we can not have below condition being satisfied?
a^3 - a < 0

Btw...OA is C

Please let me know...N yes thanks for being patient in reading all above steps!!

Thanks buddy dat was a really good insight.

I have posted a different solution for this question. Have a look .

gmatmachoman wrote:

ankur.agrawal wrote:Ok. i definitely need to clear this one up.

Statement 1:

Result which i got from solving it : 0<a <1

Statement 2:

Result which i got from solving it : -1 < a< 1

Q Asks is a>0.

So cant we say from Statement 1 that yes a>0. (assuming my solutions are correct)

Gr8 question.

Not really Ankur..

St 1 gives : a (a^2 -1) < 0

So we can have a < 0 or (a^2 -1) <0

St 2 : 1 - a^2 > 0

Multiply by -1 on both sides.

a^2 -1 < 0

Here a can be a positive number or a negative number. Insufficient.

Combining both the sts,:

a (a^2 -1) < 0

We know that negative * positive is negative number
. Frm st 2 wkt, (a^2 -1) is negative . So "a" has to be positive to satisfy st 1.

Thus a>0.

Dont bother abt the range .

Statement 1: a^3-a<0; a(a^2-1)<0
Now two cases arise, either a<0 & (a^2-1)>0 or a>0 & (a^2-1)<0
Now, it is insufficient to determine the nature of a

Statement 2: 1-a^2>0; 1>a^2; -1<a<1
Again, a's nature cannot be determined

Combining both: Statement 1: a(a^2-1)<0 and Statement 2: 1-a^2>0 which is a^2-1<0
Now as (a^2-1) is negative and substituting this in statement 1 we get a*(-ve)<0; that means a has to be positive.
Thus, a>0
Answer C

I got the 2 inequalities but i did not get the answer as C. Instead i got the answer as E.
Can anybody please explain the following statements:

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.

[email protected] wrote:I got the 2 inequalities but i did not get the answer as C. Instead i got the answer as E.
Can anybody please explain the following statements:

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.

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 both negative and 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.