Here is a GMAT DS question from the topic Inequalities. Tests concepts of modulus.
Is |a| > |b|?
1. 1/(a – b) > 1/(b – a)
2. a + b < 0
Correct Answer : Choice C. Both statements together are sufficient.
We need to determine whether |a| is greater than |b|.
The answer to this question will be a conclusive ‘yes’ if |a| > |b|.
The answer will be a conclusive ‘n’ if |a| <= |b|
Let us evaluate statement 1
1/(a – b) > 1/(b – a)
We can rewrite the same inequality as 1/(a – b) > -1(a – b).
If a number is greater than the negative of the number, the number has to be a positive number.
So, we can conclude that a – b > 0 or a > b.
If a > b, |a| may or may not be greater than |b|.
For e.g, a = 4, b = 2. Then |a| > |b|. For positive a and b, when a > b, |a| > |b|.
Let us look at a counter example. a = 2 and b = -10. a > b. But |a| < |b|.
Hence, we cannot conclude from statement 1 whether |a| > |b|.
Statement 1 is NOT sufficient.
Let us evaluate statement 2
a + b < 0
Either both a and b are negative or one of a or b is negative.
If only one of the two numbers is negative, then the magnitude of the negative number is greater than the magnitude of the negative number is greater than the magnitude of the positive number.
For e.g., a = -3 and b = -4. a + b < 0, |a| < |b|
Here is a counter example: a = -4 and b = -3. a + b < 0 and |a| > |b|.
So, statement 2 is NOT sufficient.
Let us combine the two statements.
We know a > b from statement 1 and a + b < 0 from statement 2.
If both a and b are negative, and we know that a > b, then |a| < |b|. Note in negative numbers, lesser the magnitude, greater the number.
If one of ‘a’ or ‘b’ is negative, as a > b, a has to be positive and b has to be negative.
The sum of a + b < 0. Therefore, the magnitude of a has to be lesser than the magnitude of b.
So, we can conclude that |a| < |b|.
Hence, by combining the two statements we can conclude that |a| is not greater than |b|.
So, the two statements taken together are sufficient to answer the question.
Choice C is the correct answer
Here is an alternative explanation for the same.
From statement 1 we know a – b > 0. From statement 2 we know a + b < 0.
So, (a – b)(a + b) < 0
Or a^2 – b^2 < 0 or a^2 < b^2
If a^2 < b^2, we can conclude that |a| < |b|.