Advanced Quant Problem

[me.parashar]

Hi Ron,

I'm trying to solve following problem from advanced quant and can't understand the answer explanation. could you please help me.

If yz ≠ 0, is 0 < y < 1?
(1) y < 1 / y
(2) y = z square

Thanks.

[jlucero]

If you're having trouble on this problem, I'm going to assume that it comes mostly from the first statement. When it comes to inequalities, there are two basic rules: 1- treat them like equations but 2- be careful of negative values.

Statement (1) says y < 1/y and it's tempting to simply move the y over to the other side and say, y^2 < 1. However, we'd be ignoring the possibility that y could be positive or negative. Technically, we need to solve this equation twice, once where y is positive and once where y is negative:

(A) IF y is positive:
y < 1/y
y^2 < 1
y < 1
Check: test out a number that is both positive and less than 1: 1/2
1/2 < 1/(1/2) TRUE

(B) IF y is negative:
y < 1/y
y^2 > 1 (gotta flip the sign here)
(-y) > 1 (remember when you take the square root of this value, the variable y is going to be negative, so -y will be a positive value.
y < -1 (flip the sign again)
Check: test out a number that is both negative and less than -1: -3
-3 < -1/3 TRUE

Therefore, statement 1 is insufficient because we can algebraically prove that y could be between 0 and 1, but it also could be any number less than -1.

Statement 2 is insufficient by itself since we don't know anything about the value of y, but it does tell us that y must be a positive number, since z^2 must be a positive number.

Therefore, combining statement 1 & 2, we know that y must be positive and therefore between 0 and 1.

C is the correct answer.

[me.parashar]

Thanks.

[jlucero]

Sure thing.