This GMAT quant practice question is a problem solving question from Inequalities. Concept: Finding solution set to an algebraic inequality. Level of difficulty: Easy to moderate. GMAT 650 level inequalities sample question.

Question 2: What range of values of 'x' will satisfy the inequality 15x - \\frac{2}{x}\\) > 1?

- x > 0.4
- x < \\frac{1}{3}\\)
- -\\frac{1}{3}\\) < x < 0.4, x > \\frac{15}{2}\\)
- -\\frac{1}{3}\\) < x < 0, x > \\frac{2}{5}\\)
- x < -\\frac{1}{3}\\)and x > \\frac{2}{5}\\)

@ INR

We can rewrite the above inequality as 15x - \\frac{2}{x}\\) - 1 > 0

i.e., \\frac{15x^2 - 2 - x}{x}\\) > 0

Split the middle term of the numerator as a precursor to factorizing it: \\frac{15x^2 - 6x + 5x - 2}{x}\\) > 0

Factorize the quadratic expression: \\frac{(5x - 2)(3x + 1)}{x}\\) > 0

The above inequality will hold good if the numerator and denominator are both positive or are both negative.

**Possibility 1:** When (5x - 2)(3x + 1) > 0 and x > 0

__Rule:__ (x - a)(x - b) > 0 when x does not lie between "a" and "b".

Applying the rule, the values of 'x' that will satisfy (5x - 2)(3x + 1) > 0 will not lie between -\\frac{1}{3}\\) and \\frac{2}{5}\\).

Combining the above result with the second condition that x > 0, we get x > \\frac{2}{5}\\)

**Possibility 2:** When (5x - 2)(3x + 1) < 0 and x < 0.

__Rule:__ (x - a)(x - b) < 0 when x lies between "a" and "b".

Applying the rule, the following values of 'x' will satisfy (5x - 2)(3x + 1) < 0: -\\frac{1}{3}\\) < x < \\frac{2}{5}\\).

Combining the above range of values with the second condition that x < 0, we get -\\frac{1}{3}\\) < x < 0.

Therefore, the range of values of x that will satisfy the inequality is: -\\frac{1}{3}\\) < x < 0 \\cup\\) x > \\frac{2}{5}\\).

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