GMAT Math : GMAT Quantitative Reasoning

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Example Questions

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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Example Question #31 : Solving Inequalities

Solve for :

|12 - 5x | < 160

Possible Answers:

$\left ( - \infty , 12 \frac{11}{12}\right ) \cup \left ( 13\frac{3}{4} , \infty \right )$

$\varnothing$

$\left (12 \frac{11}{12}, 13\frac{3}{4} \right )$

$\left ( - \infty , -29 \frac{3}{5} \right ) \cup \left ( 34 \frac{ 2}{5}, \infty \right )$

$\left ( -29 \frac{3}{5}, 34 \frac{ 2}{5} \right )$

Correct answer:

$\left ( -29 \frac{3}{5}, 34 \frac{ 2}{5} \right )$

Explanation:

|12 - 5x | < 160

can be rewritten as the inequality

-160<12 - 5x < 160

-160- 12 <12 - 5x - 12 < 160 - 12

-172 < - 5x < 148

$\frac{-172}{-5} > \frac{- 5x }{-5} > \frac{148}{-5}$ (note the change in direction of the inequality symbols)

$34 \frac{ 2}{5} > x > -29 \frac{3}{5}$

This is the set $\left ( -29 \frac{3}{5}, 34 \frac{ 2}{5} \right )$ .

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Example Question #31 : Solving Inequalities

Solve the following inequality:

$-3+9x+14\leq 6x+1-2x$

Possible Answers:

$x\leq-2$

$x\geq2$

$x\geq \frac{4}{3}$

$x\leq-\frac{4}{3}$

$x\leq1$

Correct answer:

$x\leq-2$

Explanation:

Like any other equation, we solve the inequality by first grouping like terms. Grouping the terms on the left side of the equation and the constants on the right side of the equation, we have:

$-3+9x+14\leq 6x+1-2x$

$9x-6x+2x\leq 1+3-14$

$5x\leq -10$

$x=\frac{-10}{5}$

$x\leq-2$

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Example Question #141 : Algebra

There is water tank already $\frac{4}{7}$ full. If Jose adds 5 gallons of water to the water tank, the tank will be $\frac{13}{14}$ full. How many gallons of water would the water tank hold if it were full?

Possible Answers:

$25$

$14$

$15$

$5$

$20$

Correct answer:

$14$

Explanation:

In this case, we need to solve for the volume of the water tank, so we set the full volume of the water tank as $x$ . According to the question, $\frac{4}{7}$ -full can be replaced as $\frac{4}{7}x$ . $\frac{13}{14}$ -full would be $\frac{13}{14}x$ . Therefore, we can write out the equation as:

$\frac{4}{7}x+5=\frac{13}{14}x$ .

Then we can solve the equation and find the answer is 14 gallons.

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Example Question #142 : Algebra

There exists a set = {1, 2, 3, 4}. Which of the following defines a function of ?

Possible Answers:

two are functions

$f=\left \{ (2,3),(1,4),(2,1),(3,2),(4,4) \right \}$

$h=\left \{ (2,1),(3,4),(1,4),(2,1),(4,4) \right \}$

$g=\left \{ (3,1),(4,2),(1,1) \right \}$

none are functions

Correct answer:

$h=\left \{ (2,1),(3,4),(1,4),(2,1),(4,4) \right \}$

Explanation:

Let's look at $f,g,and\ h$ and see if any of them are functions.

1. = {(2, 3), (1, 4), (2, 1), (3, 2), (4, 4)}: This cannot be a function of because two of the ordered pairs, (2, 3) and (2, 1) have the same number (2) as the first coordinate.

2. = {(3, 1), (4, 2), (1, 1)}: This cannot be a function of because it contains no ordered pair with first coordinate 2. Because the set = {1, 2, 3, 4}, we need an ordered pair of the form (2, ) .

3. = {(2, 1), (3, 4), (1, 4), (2, 1), (4, 4)}: This is a function. Even though two of the ordered pairs have the same number (2) as the first coordinate, is still a function of because (2, 1) is simply repeated twice, so the two ordered pairs with first coordinate 2 are equal.

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Example Question #1 : Understanding Functions

Let be a function that assigns $x^{2}$ to each real number . Which of the following is NOT an appropriate way to define ?

Possible Answers:

all are appropriate ways to define

$x\rightarrow x^{2}$

$f(x)=x^{2}$

$y=x^{2}$

$f(y)=x^{2}$

Correct answer:

$f(y)=x^{2}$

Explanation:

This is a definition question. The only choice that does not equal the others is $f(y)=x^{2}$ . This describes a function that assigns $x^{2}$ to some number , instead of assigning $x^{2}$ to its own square root, .

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Example Question #1 : Understanding Functions

If $f(x)=x^{2}$ , find $\frac{f(x+h)-f(x)}{h}$ .

Possible Answers:

$x^{2}$

2x+h

$x^{2}+2xh+h^{2}$

$x^{2}+4x+4$

Correct answer:

2x+h

Explanation:

We are given f(x) and h, so the only missing piece is f(x + h).

$f(x+h)=(x+h)^{2}=x^{2}+2xh+h^{2}$

Then $\frac{f(x+h)-f(x)}{h}= \frac{x^{2}+2xh+h^{2}-x^{2}}{h} = \frac{2xh+h^{2}}{h}=2x+h$

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Example Question #1 : Understanding Functions

Give the range of the function:

$f (x) = \left\{\begin{matrix} 2x - 1 & & x < -2\\ -3x& & -2\leq x\leq 2\ \\5 & & \; x > 2 \end{matrix}\right.$

Possible Answers:

$(-\infty , -6] \cup [-5, 6]$

$(-\infty , 5]$

$(-\infty , 6]$

$[-6, \infty )$

$[-5, \infty)$

Correct answer:

$(-\infty , 6]$

Explanation:

We look at the range of the function on each of the three parts of the domain. The overall range is the union of these three intervals.

On $(-\infty , -2)$ , f (x) takes the values:

x < -2

2x - 1 < 2 (-2) - 1

2x - 1 < -5

$f\left (x \right )< -5$

or $(-\infty , -5)$

On [-2, 2] , f (x) takes the values:

$-2 \leq x\leq 2$

$-3(-2) \geq -3x\geq -3 (2)$

$6 \geq -3x\geq -6$

$-6 \leq f(x)\leq 6$ ,

or [-6, 6]

On $(2, \infty)$ , f (x) takes only value 5.

The range of f (x) is therefore $(-\infty , -5) \cup [-6, 6] \cup \left \{ 5\right \}$ , which simplifies to $(-\infty , 6]$ .

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GMAT Math : GMAT Quantitative Reasoning

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #31 : Solving Inequalities

Example Question #32 : Solving Inequalities

Example Question #33 : Solving Inequalities

Example Question #31 : Solving Inequalities

Example Question #31 : Solving Inequalities

Example Question #141 : Algebra

Example Question #142 : Algebra

Example Question #1 : Understanding Functions

Example Question #1 : Understanding Functions

Example Question #1 : Understanding Functions

All GMAT Math Resources

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