Advanced Quant Chapter 4, Try it #10

[tomslawsky]

I need help on this one please. i know the answer wont be that complicated, but im stuck.

a^3 - a^2 - 2a <0

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

a<0
a<2
a<-1

I'm golden to this point

However, the answer says
a<-1, 0<a<2

How did we go from a being less than zero and less than two to the final, a is between zero and two. Confused here, please help and thank you.

[mithunsam]

Question is whether a<0?

Statement 1) a^3 < a^2 + 2a

Rewrite this
=> a^3 - a^2 - 2a < 0
=> a(a^2 - a - 2) < 0
=> a(a - 2)(a + 1) < 0
For this statement to be less than 0, one or all of the factors should be -ve. Now, look at factor (a - 2). This will become -ve when a < 2. In other words, a(a - 2)(a + 1) will be less than 0 when a < 2 (omit 0 & -1 as the equation becomes 0).

That means, at least all -ve numbers (other than -1) and 1 could satisfy the statement. So, we have both +ve and -ve solutions. Therefore, we don't know whether a<0. Insufficient.

Statement 2) a^2 > a^3

Rewrite
=>a^2 - a^3 > 0
=>a^2(1-a) > 0
Ignore a^2 as it is always +ve. That means (1-a) > 0

Remember, question didn't state that a is integer. So, a could be any -ve numbers or any value between 0 & 1. Again, we have both +ve and -ve solutions. Insufficient.

Combine Statement 1 & 2. We have the same range in both. Any -ve numbers or value between 0 or 1 could satisfies both statements.

So, correct answer is E

Hope this helps.

[tomslawsky]

thank you for your explanation; it was well written. However, my original thought process has hit a road block. I am trying tom apply what I learned in the EIV guide and I remember that if:

1) a is less than 2

and

2) a is less than 0


then

"Ignore less than 2" because the higher restriction is on less than 0, therefore, a is simply less than 0.


Where am I wrong? Thank you!

[mithunsam]

Actually, a(a - 2)(a + 1) < 0 doesn't mean that a<0 and a<2 and a<-1.

When I say x(x+1)(X+2) < 0, what does that mean? It means, one or all the 3 factor(s) of the equations is/are -ve. Right?

-ve * +ve * +ve < 0 - - -> 1 factor -ve
+ve * -ve * +ve < 0 - - -> 1 factor -ve
+ve * +ve * -ve < 0 - - -> 1 factor -ve
-ve * -ve * -ve < 0 - - -> 3 factors -ve

Similarly, a(a - 2)(a + 1) < 0 means either one of a, a-2 or a+1 is -ve or all the 3 factors are -ve.

To get all the possible values of a, we have to choose the first case (first case covers a bigger range). That is a<0, or, a-2<0, or a+1<0.

This means, equation could be < 0 when a<0, or a<2, or a<-1. We have to pick the higher value a<2 (Remember to remove exceptions like 0).

[tomslawsky]

Now that directly answers my question and I "get it". Thank you very much
Tom

[tim]

glad to hear it!