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)?

- x
^{2}+ 5x + 6 > 0 - |x + 2| > 4
- 9x - 7 < 3x + 14
- x
^{2}- 4x + 3 < 0 - (B) and (D)

@ INR

**Choice A**: x^{2} + 5x + 6 > 0

Factorize the given expression: x^{2} + 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:** x^{2} - 4x + 3 < 0

Factorizing x^{2} - 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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