a is an odd integer and a ≠ -b. Is a^2/|b+a| > a-b ?

In equation
\(\frac{a^2}{|b+a|} + b-a > 0\),
the \(b+a\) part can be either positive or negetive, so we must consider both to remove the MOD. The two conditions we get, after applying a little algebra, are:
\(\frac{b^2}{b+a} > 0\) --------------------------------E1
and
\(\frac{b^2 - 2a^2}{b+a} > 0\)-------------------------E2
any statement must satisfy both the equations to be considered good.

S1 gives us b=0 (since a is an odd integer), putting that in E1 and E2 above we get:
E1: LHS = 0, RHS = 0. Answers the question.
E2: \(-2a>0 ?\). Does not answer the question.
S1 Not sufficient

S2 gives us a>0, putting that in E1 and E2 above does not answer the question.
S2 Not sufficient

Putting b=0 and a>0 in E1 and E2 above:
E1: LHS = 0, RHS = 0. Answers the question.
E2: \(-2a>0 ?\)(where a is positive int). Answers the question.


Both together are sufficient. C

Thanks Karishma and Stuart. Yes <> means "not equal to". It had a long solution and wanted to know from experts like you if this resembles a GMAT Question. I got it from GMATfix question bank.

Yes, this question could resemble an actual GMAT question. Definitely interesting! Though, it is a higher level question, let us say 49 and above.

Hi I have a question on solving \(a^3>a\). I see that you do it directly, but then if I need to solve it ->
\(a^3>a\) => \(a^3-a>0\) =>\(a(a^2-1)>0\) => thats means either a>0 (i) or \(a^2>1\) -> a>+/- 1 => a>1 or a<-1 (ii)
Final value for a from (i) and (ii) => a<-1 and a>1

@Karishma -Please confirm whether this solution to a is correct by inequality method.

The problem lies in your interpretation as shown by the colored part.
Think ab > 0. When will this happen? When either both a and b are positive or both a and b are negative.
Similarly when ab < 0, we say that either a < 0 and b > 0 or a > 0 and b < 0 i.e. one and only one of a and b is negative.
Only when ab = 0, can we say that either a = 0 or b = 0 or both are 0.

Here is how you will solve the inequality:
\(a^3>a\) => \(a^3-a>0\) =>\(a(a^2-1)>0\)=>\((a-1)a(a + 1)>0\)
For \((a-1)a(a + 1)\) to be positive, either all (a - 1), a and (a + 1) should be positive, or exactly two of them should be negative. Then the entire product will be positive.
To deal with this easily, take them on the number line. Plot 1, 0 and -1 on the number line. Now your number line is divided into 4 sections. Start from the rightmost section. That will always be positive. The next section will be negative, next positive again, next negative again and so on...


The green portions show where the expression will be positive and blue portions show where it will be negative.
So you get x > 1 or -1 < x < 0.
For detailed explanation of why this method works, check out my earlier post:
https://gmatclub.com/forum/inequalities-trick-91482.html#p804990

It is mentioned in the question stem that "a is an odd integer". How can we consider a scenario "either a = 0..." a cannot be 0, right?

Please correct me if I am missing something here?

Please find the solution as attached.

Answer: Option C

Hi,

I don't find in the stem of the question that "a" +ve integer. It is simply an odd integer, which can be either +ve or -ve.

Hey, hopefully I can help a bit...

We are given in the Question Stem that A must be an Odd Integer. This means it can not be a Fraction.

From statement 2: (a)^3 > a

the only values of A that satisfy this inequality are:

-1 < a < 0 ----------> which is not possible given our constraint that A = an Odd INTEGER

or

a > 1 ----------> which is possible


therefore, the Inference we can draw from Statement 2 is the following:

a > 1


combined with statement 1: in which we made the Inference that B must = 0, the Question becomes:

Is: (a)^2 / [a] > a ?



(a)^2 = [a] * [a] -------> the Square of A can be written as the Absolute Value of [A] multiplied by the Absolute Value of [a]


Is: [a] * [a] / [a] > a?

----cancel one of the [a] in the NUM and DEN ------

Is: [a] > a?


statement 2 tells us that A must be greater than > 1

no matter which value greater than 1 that we plug into the question stem, we will always find that:

[a] = a


-C- Together Sufficient


Basically it's that one Constraint that A must be an Integer that rules out the case in statement 2 that A could fall in the Range of:

-1 < a < 0


I hope that helps a little bit? if not, please send any follow up questions and I'll do my best.

All the best!