Is a0? (1) a^3-a

Question

Is a>0?

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

Bunuel · Accepted Answer

Is a>0? (1) a^3-a<0 --> a*(a^2-1)<0 Two cases: A. Both multiples are positive: a<0 and a^2-1>0 , (which means a<-1 or a>1 ). As a<0 then the range would be a<-1 . B. Both multiples are negative: a>0 and a^2-1<0 , (which means -10 then the range would be 01 --> |a|>1 --> a<-1 or a>1 . Again two possible answers to the question whether a is positive. Not sufficient. (1)+(2) Intersection of the ranges form (1) and (2) gives unique range a<-1 (From (1): a<-1 OR 01 ). So the answer to the question is a positive is NO. sufficient. Answer: C.

lagomez · Answer

Question:

on statement 1, why couldn't you say:
 a^3-a <0
=a^3 < a
=a^2<1
= -1<a<1

Bunuel · Answer

This is not correct as when you get  a^2<1  from  a^3 < a , you are reducing (dividing) by a. We cannot divide the inequality by the variable (or by any expression) sign of which is unknown.

This is one of the most frequently used catch from GMAT.

NEVER EVER divide inequality by the variable (or expression with variable) with unknown sign.

lagomez · Answer

Yes, good answer. that gets me every time

what if we are told a>0 in the question
then we can divide?

Bunuel · Answer

Not only if we were told that a is positive, but even if we were told that a is negative we could divide. But in this case the question wouldn't make any sense as the question exactly asks if a is positive.

grumpytesttaker · Answer

I wanted to offer my thoughts in case my line of reasoning resonates with someone and helps them better understand. Ok, so we need to know if a>0. given, 1. a^3-a<0 i could factor out a from both terms but that may make it complicated, but I ma told that the left hand side is negative, so whatever the value of a, whether positive or negative, I ought to be able to say that a is bigger than a^3, otherwise a^3-a would not be less than 0. so, a^30 ok, instantly i can see that 1 has to be greater than a^2 for their difference to be positive. so without worrying about what sign mumbojumbo, i can say a^2<1. well, i have learned from mgmat advanced quant book that anytime i see a^2<1, i can simply write it as |a|<1 and that i can write it as -10. Hope my line of reasoning is not a wrong way that still got the right answer, and I hope it at least makes someone follow and understand the solution to this problem.

hfbamafan · Answer

For statement 1 can't you just factor out an a and make it a(a^2-1) where a has to equal zero making it not sufficient?

Edvento · Answer

Hi,

My 2 cents:

Bunuel's explanation is great. Hats off.
To put all confusion to rest note that we cannot divide both sides both sides of the statement (1) by a, as we don't know whether a is positive or negative.

Hence, bunuel's approach is the logical one.

Regards,

Shouvik.

joshhowatt · Answer

I'm confused... Maybe someone can show me where I'm going wrong.
This how I approached (I actually came up with A).

a^3-a<0
a(a^2-1)<0
factors out to a(a-1)(a+1)<0 - so it seems here we have factors that are consecutive, (a-1)(a)(a+1)<0
The only product that would be less than 0 would require all 3 to be negative, so I determined a as negative, which would answer the question "no."

Where am I going wrong?

Bunuel · Answer

The product of three multiples to be negative either all three must be negative or one must be negative and other two must be positive. For more on how to solve such kind of inequalities check: x2-4x-94661.html#p731476 inequalities-trick-91482.html everything-is-less-than-zero-108884.html?hilit=extreme#p868863 xy-plane-71492.html?hilit=solving%20quadratic#p841486 Hope it helps.

joshhowatt · Answer

Exactly...
So if they are consecutive numbers we have a couple of options:

Say the numbers are -1,0,1. This won't work because one is zero, therefore the answer would not be less than 0.

So all three must be negative no?

Bunuel · Answer

We are not told that the  a  is an integer so  a-1 ,  a  and  a+1  are not necessarily consecutive integers.  Please read the solutions above  and plug some numbers from the correct ranges to check whether inequality holds true for them.

For example if  a=\frac{1}{2}  then  a^2-a=\frac{1}{8}-\frac{1}{2}=-\frac{3}{8}<0 .

P.S. The links in my previous post might help you to deal with inequalities.

meghaswaroop · Answer

Statement 1 : a^3-a<0 => a(a^2-1)<0, which can be possible in two cases:
Case 1 : a<0 , a^2 - 1 > 0 Case 2 : a>0 , a^2 - 1 < 0 . Hence, we can't say whether a > 0 .
Statement 2 : 1 - a^2 < 0 => a^2 > 1 , which is again possible in two cases
case 1 : a>0 Case 2: a<0 . Hence, we can't say whether a > 0.

Taking Staement 1 and 2 together : From statement 2 we got to know a^2 > 1 from which using Statement 1 [case 1] we can confirm a<0

Hence Both statements are needed to conclude this.

Temurkhon · Answer

Is a>0? Y/N St1. a^3-a<0 means a(a^2-1)<0 so two options: a>0 & a^2<1 OR a<0 & a^2>1. INSUFF St2. 1-a^2<0 means a^2>1, so can be a<>0. INSUFF St1+St2 means that a<0 & a^2>1. SUFF C

theLucifer · Answer

Is a positive? Statement 1: a^3-a<0 a^30 1>a^2 So -1

CrackverbalGMAT · Answer

Hi kanusha,

Questions like these can be easily solved if you know the process of solving a quadratic inequality. You can go through an article on quadratic inequalities here https://gmatclub.com/forum/inequalities-quadratic-inequalities-231326.html#p1781849

Is a > 0?

Statement 1 : a^3 - a < 0

Simplifying we get
a(a^2 - 1) < 0
a(a - 1)(a + 1) < 0

The critical points here a -1, 0 and +1. Plotting them on the number line and take the regions which are negative.

Attachment:

Number Line 1.png [ 2.88 KiB | Viewed 6572 times ]

So 0 < a < 1 and a < -1

Now a here can either be greater than 0 or less than 0. Insufficient.

Statement 2 : 1 - a^2 > 0

Multiplying throughout by -1 (we always need to keep the variable positive) we get

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

The critical points here are -1 and 1. Plotting them on the number line and take the negative regions

Attachment:

Number Line 2.png [ 2.69 KiB | Viewed 6520 times ]

So -1 < a < 1

Here again a can be greater than 0 or less than 0. Insufficient.

Combining 1 and 2 :

From statement 1 we have 0 < a < 1 and a < -1
From statement 2 we have -1 < a < 1

The only range of a that will satisfy both statements combined is 0 < a < 1. Sufficient.

Attachment:

Number line 3.png [ 4.32 KiB | Viewed 6556 times ]

Answer : C

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

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