# An Official Question on Absolute Values Now that we have discussed some important absolute value properties, let’s look at how they can help us in solving official questions.

Knowing these basic properties can help us quickly analyze the question and arrive at the answer without getting stuck in analyzing different ranges, a cumbersome procedure.

First we will look at a GMAT Prep question.

Question 1: Is |m – n| > |m| – |n|?

Statement 1: n < m
Statement 2: mn < 0

Solution 1:

Recall the property number 2

Property  2: For all real x and y, |x – y| >= |x| – |y|

|x – y| = |x| – |y| when (1) x and y have the same sign and x has greater (or equal) absolute value than y (2) y is 0

|x – y| > |x| – |y| in all other cases

So if m and n have the same sign with |m| >= |n|, equality will hold.

Also, if n is 0, equality will hold.

If we can prove that both these conditions are not met, then we can say that |m – n| is definitely greater than |m| – |n|.

Statement 1: n < m

We have no idea about the signs of m and n. Are they same? Are they opposite? We don’t know. Also n may or may not be 0. Hence we don’t know whether the equality will hold or the inequality. Statement 1 alone is not sufficient to answer the question.

Statement 2: mn < 0

Since mn is negative, it means one of m and n is positive and the other is negative. This also implies that n is definitely not 0. So we know that m and n do not have the same sign and n is not 0. So under no condition will the equality hold.  Hence |m – n| is definitely greater than |m| – |n|. Statement 2 alone is sufficient to answer the question.

Let’s look at one more question now.

Question 2: If xyz ? 0, is x(y + z) >= 0?

Statement 1: |y + z| = |y| + |z|
Statement 2: |x + y| = |x| + |y|

Solution 2:  xyz ? 0 implies that all, x, y and z, are non zero numbers.

Question: Is x(y + z) >= 0?

If we can prove that x(y + z) is not negative that is x and (y+z) do not have opposite signs, we can say that x(y + z) is positive or 0.

Looking at the statements given, let’s review our property number 1:

Property 1: For all real x and y, |x + y| <= |x| + |y|

|x + y| = |x| + |y| when (1) x and y have the same sign (2) at least one of x and y is 0.

|x + y| < |x| + |y| when (1) x and y have opposite signs

The two statements give us equalities which means that the relevant part of the property is this:
|x + y| = |x| + |y| when (1) x and y have the same sign (2) at least one of x and y is 0.

We are also given in the question stem that x, y and z are not 0. Hence, given |x + y| = |x| + |y|, we can infer that x and y have the same sign.

Statement 1: |y + z| = |y| + |z|

This implies that y and z have the same signs. But we have no information about the sign of x hence this statement alone is not sufficient.

Statement 2: |x + y| = |x| + |y|

This implies that x and y have the same signs. But we have no information about the sign of z hence this statement alone is not sufficient.

Using both statements together, we know that x, y and z have the same sign. Whatever is the sign of y and z, the same will be the sign of (y+z). Hence x and (y+z) have the same sign. This implies that x(y + z) cannot be negative.

Hence we can answer our question with a definite ‘yes’.