# GMAT Algebra Practice - Inequalities

Concept: Finding solution to an algebraic inequality.

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

#### Question: What range of values of 'x' will satisfy the inequality 15x - $$frac{2}{x}\\$ > 1? 1. x > 0.4 2. x < $\frac{1}{3}\\$ 3. -$\frac{1}{3}\\$ < x < 0.4, x > $\frac{15}{2}\\$ 4. -$\frac{1}{3}\\$ < x < 0, x > $\frac{2}{5}\\$ 5. x < -$\frac{1}{3}\\$and x > $\frac{2}{5}\\$ #### Explanatory Answer Video explanation will be added soon 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 > $\frac{2}{5}\\$.

Choice D is the correct answer.

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