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## Comment on

Inequalities and Absolute Value## Hello Brent,

For following question, when |Z|<1, Z will be <1. So Z^2 has to be less than 1. Answer choice A has Z^2<=1. How can Z^2 can be equal to 1 and the right option. Please advise me.

http://greprepclub.com/forum/if-z-1-which-of-the-following-statements-must-be-true-2146.html

Thanks :)

## Hi Vinodhini,

Hi Vinodhini,

Question link: http://greprepclub.com/forum/if-z-1-which-of-the-following-statements-mu...

The notation z² ≤ 1 doesn't imply that z² could equal one; it means z² is less than OR equal to 1.

Likewise, the inequality |3| ≥ -2 is 100%. However, it doesn't mean that |3| could ever equal -2.

Does that help?

## Hello Brent,

It is still not clear to me how can that inequality be true. "z² is less than OR equal to 1." From this are we saying that the option is true when any of the condition (less than or equal) succeeds ? i.e when z^2 is less than 1 or when z^2 = 1 ? Please advise me

Thanks :)

## Hi Vinodhini,

Hi Vinodhini,

I believe you are interpreting the word "or" as one might interpret the word "and."

Consider the following statement: Ottawa is the capital of Canada OR Ottawa is the President of the United States.

Even though Ottawa is NOT the President of the United States, the statement is still 100% correct, because the word "or" does not imply that both parts of the sentence are true.

Likewise, it's perfectly acceptable to write: 3 ≤ 8

In other words, it is true that 3 is less than or equal to 8.

Does that help?

## Hello Brent,

So in that case, for the question in doubt why options B and C are not correct answers ? Option B says Z^2 is less than OR equal to Z. When we take Z as 0, Z^2 = Z. (0^2 = 0) and option C (0^3=0).

Please help me.

Thanks :)

## Great question.

Great question.

The keyword here is MUST.

For statement A, it MUST be true that, for EVERY possible value of z, it is the case that z² ≤ 1. In other words, for EVERY possible value of z, it is the case that EITHER z² < 1 OR z² = 0

For statement B, it is NOT necessarily true that, for EVERY possible value of z, it is the case that z² ≤ z. For example, one possible value of z is -1/2. However, it is NOT the case that (-1/2)² ≤ -1/2

For statement C, it is NOT necessarily true that, for EVERY possible value of z, it is the case that z³ ≤ z. For example, one possible value of z is -1/2. However, it is NOT the case that (-1/2)³ ≤ -1/2

Does that help?

## Hello Brent,

Yes, it is absolutely clear now. Thanks a lot :-)

## |x|>|y| and x+y>O

Quantity A: y

Quantity B: x

Question from Manhattan 5 lb. Pg 450

## Here's my full solution:

Here's my full solution: https://greprepclub.com/forum/x-y-and-x-y-10335.html#p30840

Cheers,

Brent

## ∣x2+3x−1∣<3,

would you please solve this equation.

## Let's first examine |k| < 3

Let's first examine |k| < 3

This means -3 < k < 3

So, we can write: -3 < x² + 3x − 1 < 3

Let's examine -3 < x² + 3x − 1 and x² + 3x − 1 < 3 separately.

Start with: -3 < x² + 3x − 1

Turn in an EQUATION: x² + 3x − 1 = -3

Set equal to zero: x² + 3x + 2 = 0

Factor: (x + 2)(x + 1) = 0

Solve, x = -2 and x = -1

These to values of x satisfy the EQUATION x² + 3x − 1 = -3

Start with: x² + 3x − 1 < 3

Make an EQUATION: x² + 3x − 1 = 3

Set equal to zero: x² + 3x − 4 = 0

Factor: (x + 4)(x - 1) = 0

Solve, x = -4 and x = 1

These to values of x satisfy the EQUATION x² + 3x − 1 = 3

The values x = -2, x = -1, x = -4 and x = 1 divide the number line into 5 separate regions:

1) x < -4

2) -4 < x < -2

3) -2 < x < -1

4) -1 < x < 1

5) 1 < x

Now test a value of x from each REGION to see if it satisfies the original inequality, ∣x² + 3x − 1∣ < 3

When we do so, we find that:

1) x < -4: values in this region do NOT satisfy ∣x² + 3x − 1∣ < 3

2) -4 < x < -2: values in this region DO satisfy ∣x² + 3x − 1∣ < 3

3) -2 < x < -1: values in this region do NOT satisfy ∣x² + 3x − 1∣ < 3

4) -1 < x < 1: values in this region DO satisfy ∣x² + 3x − 1∣ < 3

5) 1 < x: values in this region do NOT satisfy ∣x² + 3x − 1∣ < 3

Solutions: -4 < x < -2 and -1 < x < 1

Cheers,

Brent

## Why x</-2, while it should be

## I'm not entirely sure what

I'm not entirely sure what you're referring to.

Are you referring 2:30 in the video when we analyze |x| ≥ 2?

If so, here's one way to show that part of the solution cannot be x ≥ -2

IF it were true that the solution is x ≥ -2, then one possible solution would be x = 0 (since x = 0 meets the condition that x ≥ -2).

However, we can quickly see that x = 0 is not a solution to the original inequality |x| ≥ 2

Does that help?

Cheers,

Brent

## I found this question on GMAT

So i set it up like this M +X + Y + Z = 80

X + Y + Z < 20 ( I wanted to check if this a correct statement I can say from the problem) Since it says that all the other students scored below average. When I do this I got 61 the right answer. Please let me know if that is the right approach. Thanks!

## Question link: https:/

Question link: https://gmatclub.com/forum/mary-and-three-other-students-took-a-math-tes...

Tricky!!!!

Your conclusion that X + Y + Z < 20 is perfect. Nice work, Ravin!