# GMAT Quant Questions | GMAT Inequalities Q1

#### Absolute Values | GMAT Sample Questionbank

This GMAT quant practice question is a problem solving question in Inequalities. Concept: Computing solution set to inequalities of absolute values and algebraic expressions. A GMAT 650 level inequalities sample question.

Question 1: If "x" is an integer, which of the following inequalities has (have) a finite range of values of "x" satisfying it (them)?

1. x2 + 5x + 6 > 0
2. |x + 2| > 4
3. 9x - 7 < 3x + 14
4. x2 - 4x + 3 < 0
5. (B) and (D)

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### Explanatory Answer | GMAT Absolute Values

#### Step 1: Find the values of "x" that will satisfy the four inequalities

Choice A: x2 + 5x + 6 > 0

Factorize the given expression: x2 + 5x + 6 > 0 = (x + 2)(x + 3) > 0.
This inequality will hold good when both (x + 2) and (x + 3) are simultaneously positive OR are simultaneously negative.

Possibility 1: Both (x + 2) and (x + 3) are positive.
i.e., x + 2 > 0 AND x + 3 > 0
i.e., x > -2 AND x > -3
Essentially translates to x > -2

Possibility 2 Both (x + 2) and (x + 3) are negative.
i.e., x + 2 < 0 AND x + 3 < 0
i.e., x < -2 AND x < -3
Essentially translates to x < -3

Evaluating both the possibilities, we get the range of values of "x" that satisfy this inequality to be x > -2 or x < -3. i.e., "x" does not lie between -3 and -2.
i.e., x takes values lesser than -3 or greater than -2.
The range of values that x takes is infinite.

Choice B: |x + 2| > 4

|x + 2| > 4 is a modulus function and therefore, has two possibilities

Possiblity 1: x + 2 > 4
i.e., x > 2

Possiblity 2: (x + 2) < -4.
i.e., x < -6
Evaluating the two options together, we get the values of "x" that satisfy the inequality as x > 2 OR x < -6.
i.e., "x" does not lie between -6 and 2.
An infinite range of values.

Choice C: 9x - 7 < 3x + 14
Simplifying, we get 6x < 21 or x < 3.5.
An infinite range of values.

Choice D: x2 - 4x + 3 < 0
Factorizing x2 - 4x + 3 < 0 we get, (x - 3)(x - 1) < 0.
This inequality will hold good when one of the terms (x - 3) or (x - 1) is positive and the other is negative.

Possibility 1: (x -3) is positive and (x - 1) is negative.
i.e., x - 3 > 0 AND x -1 < 0
i.e., x > 3 AND x < 1
Such a number DOES NOT exist. It is an infeasible solution.

Possibility 2: (x - 3) is negative and (x - 1) is positive.
i.e., x - 3 < 0 AND x - 1 > 0
i.e., x < 3 AND x > 1
Essentially translates to 1 < x < 3 Finite range of values for "x".

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