# GMAT Practice Question - Inequalities

Concept: Finite and infinite range of values.

This GMAT quant practice question is a problem solving question from Inequalities. Concept: Computing solution set to inequalities.

#### Question: 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)

Video explanation will be added soon

We have to find values of "x" that will satisfy the four inequalities given in the answer choices and identify the choice(s) for which the range of values of x satisfying it is finite.

##### Choice A: x2 + 5x + 6 > 0

Factorize the given equation: 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".

Choice D is the correct answer.

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