GMAT Math : Algebra

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Example Questions

Example Question #21 : Solving Inequalities

Give the solution set of the inequality

$x^{2}+ 3x- 7 < 9x - 16$

Possible Answers:

$(-\infty, 3) \cup (3, \infty)$

$(-\infty, -3) \cup (-3, \infty)$

$(-\infty, -3) \cup (3, \infty)$

$\varnothing$

( -3,3 )

Correct answer:

$\varnothing$

Explanation:

To solve a quadratic inequality, move all expressions to the left first:

$x^{2}+ 3x- 7 < 9x - 16$

$x^{2}+ 3x- 7- 9x + 16 < 9x - 16 - 9x + 16$

$x^{2} - 6x + 9< 0$

$(x-3)^{2} < 0$

Since the square of any real number is nonnegative, there is no value of for which this is true. The solution set is the empty set.

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Example Question #1211 : Gmat Quantitative Reasoning

Give the solution set of the inequality

$x^{2}- x + 1> 3x- 3$

Possible Answers:

$(-\infty, \infty)$

(-2, 2)

$\varnothing$

$(-\infty, -2) \cup (-2, \infty)$

$(-\infty, 2) \cup (2, \infty)$

Correct answer:

$(-\infty, 2) \cup (2, \infty)$

Explanation:

To solve a quadratic inequality, move all expressions to the left first:

$x^{2}- x + 1> 3x- 3$

$x^{2}- x + 1- 3x + 3> 3x- 3 - 3x + 3$

$x^{2}- 4x + 4> 0$

$(x-2)^{2} > 0$

The only boundary point for the intervals in the solution set is the point at which x-2 = 0 - that is, x= 2 . This will be excluded from the solution set, as the symbol does not allow equality. But for any other value of , the square of the number must be positive. Therefore, the solution set is

$(-\infty, 2) \cup (2, \infty)$ .

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Example Question #21 : Solving Inequalities

Give the solution set of the inequality

$x^{2} + 6x + 2 > 6x - 7$

Possible Answers:

(-3, 3)

$(-\infty, -3) \cup ( -3 , \infty)$

$(-\infty, -3) \cup ( 3 , \infty)$

$(-\infty, 3) \cup ( 3 , \infty)$

$(-\infty, \infty)$

Correct answer:

$(-\infty, \infty)$

Explanation:

To solve a quadratic inequality, move all expressions to the left first

$x^{2} + 6x + 2 > 6x - 7$

$x^{2} + 6x + 2 - 6x + 7 > 6x - 7 - 6x + 7$

$x^{2} + 9 >0$

$x^{2} > -9$

Since the square of any number must be nonnegative, it follows that for any ,

$x^{2} \ge 0 > -9$

and the solution set is the set of all real numbers, $(-\infty, \infty)$ .

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Example Question #24 : Solving Inequalities

True or false: $N^{2} < N + 6$

Statement 1: N > 0

Statement 2: N < 3

Possible Answers:

STATEMENT 2 ALONE provides sufficient information to answer the question, but STATEMENT 1 ALONE does NOT provide sufficient information to answer the question.

EITHER STATEMENT ALONE provides sufficient information to answer the question.

BOTH STATEMENTS TOGETHER do NOT provide sufficient information to answer the question.

BOTH STATEMENTS TOGETHER provide sufficient information to answer the question, but NEITHER STATEMENT ALONE provides sufficient information to answer the question.

STATEMENT 1 ALONE provides sufficient information to answer the question, but STATEMENT 2 ALONE does NOT provide sufficient information to answer the question.

Correct answer:

BOTH STATEMENTS TOGETHER provide sufficient information to answer the question, but NEITHER STATEMENT ALONE provides sufficient information to answer the question.

Explanation:

First, solve the inequality:

$N^{2} < N + 6$

$N^{2} - N - 6 < N + 6 - N - 6$

$N^{2} - N - 6 < 0$

(N - 3 )(N+2)< 0

The boundary points of the intervals to be tested are the points at which:

N+2 = 0 ; that is, N = -2 , or

N - 3 = 0 ; that is, N = 3 .

Therefore, the intervals $(-\infty, -2), (-2, 3), (3, \infty)$ should be tested; do so by testing one value in each interval in the inequality and testing its truth:

$(-\infty, -2)$ - test N = -3

$N^{2} < N + 6$

$(-3)^{2} < (-3 )+ 6$

9 < 3

False - reject this interval.

(-2, 3) - test N = 0

$N^{2} < N + 6$

$0^{2} < 0 + 6$

0 < 6

True - accept this interval.

$(3, \infty)$ - test N = 4

$N^{2} < N + 6$

$4^{2} < 4 + 6$

16 < 10

False - reject this interval.

The solution set is the interval (-2, 3) ; therefore, $N^{2} < N + 6$ if and only if -2 < N < 3 .

From Statement 1 alone, we know that N > 0 ; this provides insufficient information, since, for example, N = 1 and N = 4 both fall in this range, but only the former is a solution of $N^{2} < N + 6$ . For similar reasons, Statement 2 alone provides infufficient information.

Assume both statements are true. Together, the two statements are euivalent to saying that 0 < N < 3 . Therefore, it holds that -2 < 0 < N < 3 , or -2 < N < 3 , and it can be established that $N^{2} < N + 6$ .

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Example Question #131 : Algebra

Find the solution set of the inequality

$x^{2}+ 8x \le 4x - 4$

Possible Answers:

$( -\infty, -2 ] \cup [2, \infty )$

$\left \{ -2 \right \}$

[-2, 2]

$\left \{ 2 \right \}$

$( -\infty, \infty )$

Correct answer:

$\left \{ -2 \right \}$

Explanation:

To solve a quadratic inequality, move all expressions to the left first:

$x^{2}+ 8x \le 4x - 4$

$x^{2}+ 8x - 4x + 4 \le 4x - 4 - 4x + 4$

$x^{2}+ 4x + 4 \le 0$

$(x+2) ^{2} \le 0$

The square of a real number cannot be less than 0, so

$(x+2) ^{2} = 0$

$\sqrt{(x+2) ^{2} }= \sqrt{0}$

x+2 = 0

x = -2 , the only solution.

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Example Question #1213 : Problem Solving Questions

Find the solution set of the inequality

$x^{2} + 6x \ge 14x - 16$

Possible Answers:

$\left \{ 4\right \}$

$( -\infty, -4 ] \cup [4, \infty )$

$\left \{ -4 \right \}$

$( -\infty, \infty )$

[-4, 4]

Correct answer:

$( -\infty, \infty )$

Explanation:

To solve a quadratic inequality, move all expressions to the left first:

$x^{2} + 6x \ge 14x - 16$

$x^{2} + 6x - 14x+16 \ge 14x - 16 - 14x+16$

$x^{2} -8x+16 \ge 0$

$(x-4)^{2} \ge 0$

The square of a real number must be nonnegative, so this is a true statement regardless of the value of . The solution set is the set of all real numbers $(-\infty, \infty)$

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Example Question #21 : Solving Inequalities

Solve for :

$| 47 - 3x | \ge 28$

Possible Answers:

$\left ( - \infty, 6 \frac{1}{3} \right ] \cup \left [25, \infty \right )$

$\left [- 25, -6 \frac{1}{3} \right ]$

$\left ( - \infty, -25 \right ] \cup \left [-6 \frac{1}{3} , \infty \right )$

$\left ( - \infty, \infty \right )$

$\left [6 \frac{1}{3} , 25\right ]$

Correct answer:

$\left ( - \infty, 6 \frac{1}{3} \right ] \cup \left [25, \infty \right )$

Explanation:

If $| 47 - 3x | \ge 28$ , then one of two things is true.

Either

$47 - 3x \ge 28$

$47 - 3x - 47 \ge 28 - 47$

$- 3x \ge -19$

$\frac{- 3x}{-3} \le \frac{-19}{-3}$ (note the change in direction of the inequality symbols)

$x \le \frac{19}{3}$

$x \le 6 \frac{1}{3}$

$47 - 3x \le -28$

$47 - 3x - 47 \le -28 - 47$

$- 3x \le -75$

$\frac{- 3x}{-3} \ge \frac{-75}{-3}$ (note the change in direction of the inequality symbols)

$x \ge 25$

The set is $\left ( - \infty, 6 \frac{1}{3} \right ] \cup \left [25, \infty \right )$ .

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Example Question #31 : Solving Inequalities

Give the solution set of the inequality

$x^{2} + 4x > 5x + 6$

Possible Answers:

$(-\infty, -2) \cup (3, \infty)$

$(-\infty, -3) \cup (2, \infty)$

$(-\infty, -1) \cup (6, \infty)$

$(-\infty, \infty)$

$(-\infty, -6) \cup (1, \infty)$

Correct answer:

$(-\infty, -2) \cup (3, \infty)$

Explanation:

To solve a quadratic inequality, move all expressions to the left first:

$x^{2} + 4x > 5x + 6$

$x^{2} + 4x- 5x - 6 > 5x + 6 - 5x - 6$

$x^{2} -x - 6 > 0$

(x-3)(x+2)> 0

The boundary points of the solution set will be the points at which:

x - 3 =0 : that is, x = 3 , or

x+ 2= 0 : that is, x = -2 .

None of these values will be included in the solution set, since equality is not allowed by the inequality symbol.

Test the intervals

$(-\infty, -2) , (-2, 3), (3, \infty)$

by choosing a value in each interval and testing the truth of the inequality.

$(-\infty, -2)$ : Test x = -3

$x^{2} + 4x > 5x + 6$

$(-3)^{2} + 4 (-3) > 5(-3) + 6$

9-12> -15+ 6

-3 > -9

True; include the interval $(-\infty, -2)$ .

(-2, 3) : Test x = 0

$x^{2} + 4x > 5x + 6$

$(0)^{2} + 4 (0) > 5(0) + 6$

0> 6

False; exclude the interval (-2, 3) .

$(3, \infty)$ : Test x = 4

$x^{2} + 4x > 5x + 6$

$(4)^{2} + 4 (4) > 5(4) + 6$

16+16> 20+6

32> 26

True; include the interval $(3, \infty)$ .

The solution set is $(-\infty, -2) \cup (3, \infty)$ .

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Example Question #32 : Solving Inequalities

Give the solution set of the inequality

$\frac{2}{x}< \frac{3}{x-3}$

Possible Answers:

$(-6, 0) \cup (3, \infty)$

$(- \infty, -6 ) \cup (0, 3)$

$(- \infty, 0) \cup ( 3, \infty)$

$(- \infty, \infty)$

(0, 3)

Correct answer:

$(-6, 0) \cup (3, \infty)$

Explanation:

To solve a rational inequality, move all expressions to the left first:

$\frac{2}{x}< \frac{3}{x-3}$

$\frac{2}{x}- \frac{3}{x-3} < \frac{3}{x-3}- \frac{3}{x-3}$

$\frac{2}{x}- \frac{3}{x-3} <0$

$\frac{2 (x-3)}{x(x-3)}- \frac{3 x}{x(x-3)} <0$

$\frac{2 x-6 }{x(x-3)}- \frac{3 x}{x(x-3)} <0$

$\frac{2 x-6 -3x }{x(x-3)} <0$

$\frac{-x-6 }{x(x-3)} <0$

$\frac{x+6 }{x(x-3)} >0$

The boundary points of the solution set will be the points at which:

x+6 = 0 - that is, x = -6 ;

x= 0 ; and

x-3 = 0 - that is, x = 3 .

None of these values will be included in the solution set, since equality is not allowed by the inequality symbol.

Test the intervals

$(- \infty, -6), (-6, 0), (0, 3),(3, \infty)$

by choosing a value in each interval and testing the truth of the inequality.

$(- \infty, -6)$ - test x= -7

$\frac{2}{x}< \frac{3}{x-3}$

$\frac{2}{-7}< \frac{3}{-7-3}$

$\frac{2}{-7}< \frac{3}{-10}$

$-\frac{2}{7}< -\frac{3}{10}$

False - exclude $(- \infty, -6)$

(-6, 0) - test x = -1

$\frac{2}{x}< \frac{3}{x-3}$

$\frac{2}{-1}< \frac{3}{-1-3}$

$\frac{2}{-1}< \frac{3}{-4}$

$-2<- \frac{3}{4}$

True - include (-6, 0)

(0, 3) - test x = 1

$\frac{2}{x}< \frac{3}{x-3}$

$\frac{2}{1}< \frac{3}{1-3}$

$\frac{2}{1}< \frac{3}{-2}$

$2<- \frac{3}{2}$

False - exclude (0, 3)

$(3, \infty)$ - test x = 4

$\frac{2}{x}< \frac{3}{x-3}$

$\frac{2}{4}< \frac{3}{4-3}$

$\frac{2}{4}< \frac{3}{1}$

$\frac{1}{2} < 3$

True - include $(3, \infty)$

The solution set is $(-6, 0) \cup (3, \infty)$ .

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Example Question #33 : Solving Inequalities

Give the solution set of the inequality

$x^{2} +10x + 7 \le 10x + 32$

Possible Answers:

$\varnothing$

[-5, 5]

$\left \{ -5\right \}$

$(-\infty, -5] \cup [5 , \infty)$

$\left \{ 5\right \}$

Correct answer:

[-5, 5]

Explanation:

The square of a real number must be nonnegative, so this is a true statement regardless of the value of . The solution set is the set of all real numbers $(-\infty, \infty)$

To solve a quadratic inequality, move all expressions to the left first

$x^{2} +10x + 7 \le 10x + 32$

$x^{2} +10x + 7 - 10x - 32 \le 10x + 32 - 10x - 32$

$x^{2} -25 \le 0$

$(x+5)(x-5) \le 0$

The boundary points of the solution set will be the points at which:

x+5 = 0 ; that is, x = -5 ; or,

x-5 = 0 ; that is, x = 5.

These values will be included in the solution set, since equality is allowed by the inequality symbol.

Test the intervals

$(-\infty, -5), (-5, 5), (5, \infty)$ by choosing a value in each interval and testing the truth of the inequality.

$(-\infty, -5)$ : test x = -6

$x^{2} +10x + 7 \le 10x + 32$

$(-6) ^{2} +10 \cdot (-6) + 7 \le 10 \cdot(-6) + 32$

$36 -60 + 7 \le -60 + 32$

$-17 \le -28$

False - exclude this interval

(-5, 5) : test x= 0

$x^{2} +10x + 7 \le 10x + 32$

$0^{2} +10 \cdot 0 + 7 \le 10 \cdot 0 + 32$

$7 \le 32$

True - include this interval

$(5, \infty)$ : test x = 6

$x^{2} +10x + 7 \le 10x + 32$

$6^{2} +10 \cdot 6 + 7 \le 10 \cdot 6 + 32$

$36 +60 + 7 \le 60 + 32$

$103 \le 92$

False - exclude this interval

[-5, 5] is the solution set.

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GMAT Math : Algebra

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #21 : Solving Inequalities

Example Question #1211 : Gmat Quantitative Reasoning

Example Question #21 : Solving Inequalities

Example Question #24 : Solving Inequalities

Example Question #131 : Algebra

Example Question #1213 : Problem Solving Questions

Example Question #21 : Solving Inequalities

Example Question #31 : Solving Inequalities

Example Question #32 : Solving Inequalities

Example Question #33 : Solving Inequalities

All GMAT Math Resources

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