Algebra Practice Questions — GRE General Test — Page 7

Question 121Question →

Let $P$ be the product of all positive real numbers $x$ that satisfy the exponential equation $x^{\sqrt{x}} = \left(x^2\sqrt{x}\right)^x$ . What is the value of $P$ ?

\frac{4}{25}

\frac{2}{5}

\frac{25}{4}

\frac{16}{625}

1

Show answer & explanation

Answer: $\frac{4}{25}$

Answer

\frac{4}{25}

Simplifying the right-hand side yields

x^{\sqrt{x}} = x^{\frac{5}{2}x}

. For

x = 1

, both sides equal 1, making

x = 1

a valid solution. For

x > 0

and

x \neq 1

, equating the exponents gives

\sqrt{x} = \frac{5}{2}x

, which reduces to

\sqrt{x} = \frac{2}{5}

, so

x = \frac{4}{25}

. Multiplying all valid solutions together yields

1 \cdot \frac{4}{25} = \frac{4}{25}

.

Step-by-Step Solution

1

Simplify the right-hand side of the equation using fractional exponent rules.

Since

x^2\sqrt{x} = x^2 \cdot x^{1/2} = x^{5/2}

, the equation becomes

x^{\sqrt{x}} = \left(x^{5/2}\right)^x = x^{\frac{5}{2}x}

.

Combining terms with the same base into a single exponent simplifies comparison between both sides.

2

Check for the base root

x = 1

.

Substituting

x = 1

gives

1^{\sqrt{1}} = 1^1 = 1

and

(1^2\sqrt{1})^1 = 1^1 = 1

. Thus,

x = 1

is a valid solution.

For any exponential equation of the form

x^{f(x)} = x^{g(x)}

,

x = 1

is always a candidate solution because

1^a = 1^b = 1

for all real exponents.

3

Equate the exponents for positive real solutions where

x \neq 1

.

Setting the exponents equal gives

\sqrt{x} = \frac{5}{2}x

.

When the base

x > 0

and

x \neq 1

,

x^{f(x)} = x^{g(x)}

implies

f(x) = g(x)

.

4

Solve the resulting radical equation for

x

.

Divide both sides by

\sqrt{x}

(since

x > 0

):

1 = \frac{5}{2}\sqrt{x} \implies \sqrt{x} = \frac{2}{5} \implies x = \left(\frac{2}{5}\right)^2 = \frac{4}{25}

.

Isolating

\sqrt{x}

and squaring both sides gives the non-trivial solution.

5

Compute the product

P

of all positive real solutions.

P = 1 \cdot \frac{4}{25} = \frac{4}{25}

.

The question requests the product of all positive real values of

x

satisfying the original equation.

Key Concept

Solving exponential equations with variable bases and radical powers

Question 122Question →

If $a$ is a real number greater than $1$ such that $a^x - a^{-x} = 2$ , what is the value of the expression $\frac{a^{3x} + a^{-3x}}{a^{2x} + a^{-2x}}$ ?

\frac{5\sqrt{2}}{3}

\frac{4}{3}

2

\frac{5\sqrt{6}}{6}

\frac{5}{3}

Show answer & explanation

Answer: $\frac{5\sqrt{2}}{3}$

Answer

\frac{5\sqrt{2}}{3}

Squaring the given relationship

a^x - a^{-x} = 2

gives

a^{2x} - 2 + a^{-2x} = 4

, which simplifies to

a^{2x} + a^{-2x} = 6

. Adding 4 to both sides gives

(a^x + a^{-x})^2 = 8

, so

a^x + a^{-x} = 2\sqrt{2}

. By the sum of cubes identity,

a^{3x} + a^{-3x} = (a^x + a^{-x})(a^{2x} + a^{-2x} - 1) = 2\sqrt{2}(6 - 1) = 10\sqrt{2}

. Dividing

10\sqrt{2}

by

6

gives the simplified result

\frac{5\sqrt{2}}{3}

.

Step-by-Step Solution

1

Square both sides of the given equation

a^x - a^{-x} = 2

.

(a^x - a^{-x})^2 = a^{2x} - 2(a^x)(a^{-x}) + a^{-2x} = 4 \implies a^{2x} + a^{-2x} = 6

.

Expanding the binomial square allows us to find the denominator

a^{2x} + a^{-2x}

directly.

2

Determine the value of

a^x + a^{-x}

.

(a^x + a^{-x})^2 = a^{2x} + 2 + a^{-2x} = 6 + 2 = 8 \implies a^x + a^{-x} = \sqrt{8} = 2\sqrt{2}

.

Since

a > 1

,

a^x > 0

and

a^{-x} > 0

, their sum must be positive.

3

Use the sum of cubes identity

u^3 + v^3 = (u + v)(u^2 - uv + v^2)

to evaluate

a^{3x} + a^{-3x}

.

a^{3x} + a^{-3x} = (a^x + a^{-x})(a^{2x} - 1 + a^{-2x}) = (2\sqrt{2})(6 - 1) = 10\sqrt{2}

.

Factoring the numerator breaks it into terms whose numerical values are known.

4

Compute the ratio of numerator to denominator.

\frac{a^{3x} + a^{-3x}}{a^{2x} + a^{-2x}} = \frac{10\sqrt{2}}{6} = \frac{5\sqrt{2}}{3}.

Simplify the fraction by dividing numerator and denominator by 2.

Key Concept

Algebraic transformations of exponential expressions using polynomial identities.

Estimated Time:2m 30s

Question 123Question →

A car rental agency charges a flat daily rate of $45 plus$ 0.20 per mile driven. If a customer rented a car for one day and the total rental cost before taxes was $75, how many miles did the customer drive?

150

225

375

600

60

Show answer & explanation

Answer: 150

Answer

150 miles

Subtracting the

45 fixed fee from the total charge of

75 leaves

30 attributable to mileage. Dividing

30 by the variable rate of $0.20 per mile gives 150 miles.

Step-by-Step Solution

1

Set up the linear equation representing total cost

45 + 0.20m = 75

, where

m

represents the number of miles driven.

The total cost consists of a fixed fee plus the variable per-mile charge.

2

Isolate the variable term by subtracting the fixed fee from both sides

0.20m = 75 - 45 \implies 0.20m = 30

This determines the portion of the total cost accrued strictly from mileage.

3

Solve for

m

by dividing by the per-mile rate

m = \frac{30}{0.20} = 150

Dividing total mileage cost by the cost per mile gives the total miles driven.

Key Concept

Linear Modeling and Single-Variable Equations

Estimated Time:45s

Question 124Question →

What is the minimum integer value of $x$ that satisfies the inequality $|2x - 7| \le 5$ ?

Show answer & explanation

Answer: 1

Answer

The minimum integer value of

x

that satisfies the inequality is 1.

To solve

|2x - 7| \le 5

, write it as the compound inequality

-5 \le 2x - 7 \le 5

. Adding 7 across all parts gives

2 \le 2x \le 12

. Dividing by 2 yields

1 \le x \le 6

. The integer solutions are 1, 2, 3, 4, 5, and 6. The minimum integer among these is 1.

Step-by-Step Solution

1

Convert the absolute value inequality into a compound inequality.

-5 \le 2x - 7 \le 5

An inequality of the form

|u| \le a

(where

a \ge 0

) is equivalent to

-a \le u \le a

.

2

Add 7 to all three parts of the inequality.

2 \le 2x \le 12

Adding a constant to an inequality preserves the direction of the inequality signs.

3

Divide all three parts by 2.

1 \le x \le 6

Dividing by a positive constant isolates

x

without reversing the inequality signs.

4

Determine the minimum integer within the solution set

[1, 6]

.

1

The solution set contains integers

\{1, 2, 3, 4, 5, 6\}

, making 1 the smallest integer value.

Key Concept

Solving absolute value inequalities using equivalent compound linear inequalities

Question 125Question →

Let $S$ be the set of all real numbers $x$ that satisfy the nested absolute value inequality $|3 - |2x - 1|| \le 4$ . Which of the following inequalities MUST be satisfied by every value of $x$ in $S$ ? Select all such inequalities.

Select all that apply

|x| \le 4

(x + 3)(x - 4) \le 0

|2x - 1| \le 4

-1 \le x \le 7

x \ge -2

Show answer & explanation

Answer: $|x| \le 4$ ; $(x + 3)(x - 4) \le 0$

Answer

The inequalities that must be satisfied by every value of

x

in

S

are

|x| \le 4

and

(x + 3)(x - 4) \le 0

.

The solution set to the nested inequality is

S = [-3, 4]

. For any value of

x

in

[-3, 4]

, the absolute value

|x|

ranges from

0

to

4

, so the inequality stating that the magnitude of

x

is at most 4 is satisfied. Additionally, the quadratic expression factored as

(x + 3)(x - 4)

has roots at

x = -3

and

x = 4

and opens upward, meaning it takes on values less than or equal to zero for all

x

between

-3

and

4

.

Step-by-Step Solution

1

Unfold the outer absolute value inequality

-4 \le 3 - |2x - 1| \le 4

By definition,

|u| \le k

(with

k \ge 0

) is equivalent to

-k \le u \le k

.

2

Isolate the inner absolute value expression by subtracting 3 and multiplying by -1

-1 \le |2x - 1| \le 7

Subtracting 3 yields

-7 \le -|2x - 1| \le 1

. Multiplying by

-1

flips the inequality signs, giving

-1 \le |2x - 1| \le 7

.

3

Simplify the compound absolute value bound and solve for

x

x \in [-3, 4]

Since an absolute value is non-negative,

|2x - 1| \ge -1

is satisfied for all real

x

. Thus, we only need

|2x - 1| \le 7

, which gives

-7 \le 2x - 1 \le 7 \implies -6 \le 2x \le 8 \implies -3 \le x \le 4

.

4

Test the solution set

S = [-3, 4]

against each given statement

Statements

|x| \le 4

and

(x + 3)(x - 4) \le 0

hold for all

x \in [-3, 4]

.

For

x \in [-3, 4]

, the extreme values of

x

yield

|x| \le 4

. Furthermore, a quadratic with roots at

-3

and

4

is non-positive on

[-3, 4]

.

Key Concept

Solving nested absolute value inequalities by systematic expansion and isolating valid intervals.

Question 126Question →

For all real numbers $x$ that satisfy the absolute value inequality $|3 - 2x| - 4 \le x$ , the rational expression $y = \frac{|x - 6|}{2 - x}$ is defined. Which of the following intervals represents the complete set of all possible real values of $y$ ?

(-\infty, 0] \cup [\frac{19}{7}, \infty)

(-\infty, -\frac{1}{5}] \cup [\frac{19}{7}, \infty)

(-\infty, 0) \cup (1, \infty)

[-\frac{1}{5}, \frac{19}{7}]

(-\infty, 1] \cup [\frac{12}{7}, \infty)

Show answer & explanation

Answer: $(-\infty, 0] \cup [\frac{19}{7}, \infty)$

Answer

The complete set of all possible real values of

y

is

(-\infty, 0] \cup [\frac{19}{7}, \infty)

.

The correct answer is

(-\infty, 0] \cup [\frac{19}{7}, \infty)

. Solving

|3 - 2x| - 4 \le x

gives

-\frac{1}{3} \le x \le 7

. Excluding

x = 2

where the denominator is zero, evaluating

y = \frac{|x - 6|}{2 - x}

on

[-\frac{1}{3}, 2)

yields

y \ge \frac{19}{7}

, on

(2, 6)

yields

y < 0

, and on

[6, 7]

yields

-\frac{1}{5} \le y \le 0

. Taking the union of these intervals gives

(-\infty, 0] \cup [\frac{19}{7}, \infty)

.

Step-by-Step Solution

1

Isolate the absolute value expression in the inequality.

|3 - 2x| \le x + 4

Adding

4

to both sides prepares the inequality for standard double-inequality solving.

2

Set up the compound inequality and solve for

x

.

-(x + 4) \le 3 - 2x \le x + 4

, yielding

x \le 7

from

-x - 4 \le 3 - 2x

, and

x \ge -\frac{1}{3}

from

3 - 2x \le x + 4

(remembering to flip the inequality sign when dividing by

-3

). Thus,

x \in [-\frac{1}{3}, 7]

.

An absolute value inequality

|A| \le B

(with

B \ge 0

) is equivalent to

-B \le A \le B

.

3

Identify domain restrictions for

y = \frac{|x - 6|}{2 - x}

.

The expression is undefined at

x = 2

. Therefore, the domain of

x

is divided into three sub-intervals:

[-\frac{1}{3}, 2)

,

(2, 6)

, and

[6, 7]

.

The denominator cannot be zero, and the absolute value

|x - 6|

changes definition at

x = 6

.

4

Analyze

y

on the first interval

[-\frac{1}{3}, 2)

.

For

x < 6

,

|x - 6| = 6 - x

. So

y = \frac{6 - x}{2 - x} = 1 + \frac{4}{2 - x}

. As

x

increases from

-\frac{1}{3}

towards

2

,

2 - x

decreases from

\frac{7}{3}

to

0^+

, so

y

increases from

1 + \frac{4}{7/3} = \frac{19}{7}

to

+\infty

. Hence

y \in [\frac{19}{7}, \infty)

.

As the positive denominator approaches zero from above, the positive fraction grows without bound towards

+\infty

.

5

Analyze

y

on the second interval

(2, 6)

.

Here

x < 6

, so

|x - 6| = 6 - x

and

y = 1 + \frac{4}{2 - x}

. As

x

increases from

2^+

to

6

,

2 - x

increases from

0^-

to

-4

. Thus

y

increases from

-\infty

up to

1 + \frac{4}{-4} = 0

. Hence

y \in (-\infty, 0)

.

As the negative denominator moves away from zero towards

-4

, the expression increases from

-\infty

to

0

.

6

Analyze

y

on the third interval

[6, 7]

.

For

x \ge 6

,

|x - 6| = x - 6

. So

y = \frac{x - 6}{2 - x} = -1 - \frac{4}{2 - x}

. At

x = 6

,

y = 0

. At

x = 7

,

y = \frac{1}{-5} = -\frac{1}{5}

. As

x

increases from

6

to

7

,

y

decreases continuously from

0

down to

-\frac{1}{5}

. Hence

y \in [-\frac{1}{5}, 0]

.

Combining

(-\infty, 0)

from the second interval and

[-\frac{1}{5}, 0]

from the third interval gives

(-\infty, 0]

.

7

Combine the ranges from all intervals.

y \in (-\infty, 0] \cup [\frac{19}{7}, \infty)

.

Taking the union of all output values across the valid domain yields the total range.

Key Concept

Solving absolute value inequalities and finding the range of rational expressions with absolute values over restricted domains.

Estimated Time:3m 0s

Question 127Question →

Which of the following inequalities represents all real values of $x$ that satisfy $-3x + 14 \le 2$ ?

x \ge 4

x \le 4

x \ge -4

x \le -4

x \ge \frac{16}{3}

Show answer & explanation

Answer: $x \ge 4$

Answer

x \ge 4

Subtracting 14 from both sides of

-3x + 14 \le 2

yields

-3x \le -12

. Dividing both sides by

-3

requires reversing the inequality sign from

\le

to

\ge

, giving

x \ge 4

. Thus, the inequality

x \ge 4

correctly represents all solution values.

Step-by-Step Solution

1

Isolate the variable term on the left side of the inequality.

Subtract 14 from both sides:

-3x \le 2 - 14

, which simplifies to

-3x \le -12

.

To solve for

x

, constant terms must first be removed from the variable side using inverse operations.

2

Divide both sides by the coefficient of

x

and apply the inequality rule for negative multipliers.

Divide by

-3

and reverse the inequality sign:

x \ge \frac{-12}{-3}

, which simplifies to

x \ge 4

.

Dividing or multiplying an inequality by a negative quantity changes the direction of the inequality sign.

Key Concept

Solving linear inequalities with negative coefficients

Estimated Time:45s

Question 128Question →

If $x$ is a real number such that $|2x - 7| \le 5$ , what is the minimum possible value of $|x - 8|$ ?

Show answer & explanation

Answer: 2

Answer

2

Solving the given inequality

|2x - 7| \le 5

yields the compound inequality

-5 \le 2x - 7 \le 5

. Adding 7 across the inequality gives

2 \le 2x \le 12

, which simplifies to

1 \le x \le 6

. Geometrically,

|x - 8|

represents the distance between

x

and 8 on the real number line. To minimize this distance for any

x

in the closed interval

[1, 6]

, we select the point in

[1, 6]

closest to 8, which is

x = 6

. Evaluating at

x = 6

produces

|6 - 8| = 2

.

Step-by-Step Solution

1

Unpack the absolute value inequality

1 \le x \le 6

The inequality

|2x - 7| \le 5

is equivalent to

-5 \le 2x - 7 \le 5

. Adding 7 gives

2 \le 2x \le 12

, and dividing by 2 yields

1 \le x \le 6

.

2

Determine the value in the domain

[1, 6]

that minimizes

|x - 8|

x = 6

The expression

|x - 8|

measures the distance from

x

to 8 on the number line. The value within

[1, 6]

nearest to 8 is

x = 6

.

3

Evaluate the expression at

x = 6

2

Substituting

x = 6

into

|x - 8|

gives

|6 - 8| = |-2| = 2

.

Key Concept

Properties of Linear Inequalities and Absolute Value as Distance

Estimated Time:1m 30s

Question 129Question →

If

x

satisfies the linear equation

\frac{3x - 1}{4} - \frac{2x + 3}{5} = \frac{x - 3}{4} + \frac{2}{5}

what is the value of

4x - 3

?

5

13

17

57

77

Show answer & explanation

Answer: 17

Answer

17

Solving the equation by grouping terms with like denominators gives

\frac{2x + 2}{4} = \frac{2x + 5}{5}

. Simplifying the left side to

\frac{x + 1}{2}

and cross-multiplying yields

5(x + 1) = 2(2x + 5)

, which simplifies to

5x + 5 = 4x + 10

, giving

x = 5

. Evaluating

4x - 3

at

x = 5

produces

4(5) - 3 = 17

, making this answer correct.

Step-by-Step Solution

1

Group like fractional terms with denominator 4 on one side and denominator 5 on the other side of the equation.

\frac{3x - 1}{4} - \frac{x - 3}{4} = \frac{2x + 3}{5} + \frac{2}{5}

Grouping terms with common denominators simplifies algebraic combination.

2

Combine the numerators over their common denominators, carefully distributing signs.

\frac{(3x - 1) - (x - 3)}{4} = \frac{(2x + 3) + 2}{5} \implies \frac{2x + 2}{4} = \frac{2x + 5}{5}

Subtracting

(x - 3)

requires distributing the negative sign to yield

-x + 3

.

3

Simplify the left side fraction and cross-multiply to eliminate denominators.

\frac{x + 1}{2} = \frac{2x + 5}{5} \implies 5(x + 1) = 2(2x + 5)

Simplifying

\frac{2x+2}{4}

to

\frac{x+1}{2}

reduces computation before cross-multiplication.

4

Expand both sides and solve for

x

.

5x + 5 = 4x + 10 \implies 5x - 4x = 10 - 5 \implies x = 5

Subtracting

4x

and

5

from both sides isolates

x

.

5

Substitute

x = 5

into the required expression

4x - 3

.

4(5) - 3 = 20 - 3 = 17

The question asks for the value of

4x - 3

, not

x

alone.

Key Concept

Solving linear equations in one variable involving fractional terms and evaluating targeted algebraic expressions.

Estimated Time:2m 0s

Question 130Question →

Working independently at their respective constant rates, Alex and Blair can complete a certain job together. Alex works alone for $4$ hours, after which Blair joins Alex, and together they work for an additional $6$ hours to finish the entire job. If Alex takes strictly less time to complete the job working alone than Blair takes working alone, which of the following statements must be true? Select all that apply.

Select all that apply

Alex working alone would take less than

16

hours to complete the job.

The time required for Alex and Blair to complete the job working together from start to finish is strictly between

8

hours and

10

hours.

Alex completes more than

60\%

of the entire job.

Blair working alone would take more than

24

hours to complete the job.

Blair completes more than

35\%

of the entire job.

Show answer & explanation

Answer: Alex working alone would take less than $16$ hours to complete the job.; The time required for Alex and Blair to complete the job working together from start to finish is strictly between $8$ hours and $10$ hours.; Alex completes more than $60\%$ of the entire job.

Answer

The true statements are that Alex working alone would take less than 16 hours, the combined time to finish the job working together is between 8 and 10 hours, and Alex completes more than 60 percent of the entire job.

By setting up the total work equation

10 r_A + 6 r_B = 1

and using the inequality

r_A > r_B > 0

, we find that

1/16 < r_A < 1/10

. This implies Alex's solo time is strictly less than 16 hours. The combined time

T_{together}

is constrained between 8 and 10 hours because

6/T_{together} = 1 - 4 r_A

. Additionally, Alex's total work contribution is

10 r_A > 10/16 = 62.5\%

, which is strictly greater than 60%.

Step-by-Step Solution

1

Set up the work equation using individual rates

r_A

and

r_B

.

4 r_A + 6(r_A + r_B) = 1 \implies 10 r_A + 6 r_B = 1

.

Alex works alone for 4 hours and then both Alex and Blair work together for 6 hours to complete 1 unit of work.

2

Apply the condition that Alex takes strictly less time alone than Blair (

r_A > r_B > 0

).

10 r_A + 6 r_A > 10 r_A + 6 r_B = 1 \implies 16 r_A > 1 \implies r_A > \frac{1}{16}

.

Since Alex's rate

r_A

is strictly greater than Blair's rate

r_B

, replacing

r_B

with

r_A

gives an upper bound on Alex's solo time

T_A = 1/r_A < 16

hours.

3

Determine the feasible range for Alex's rate

r_A

.

\frac{1}{16} < r_A < \frac{1}{10}

.

From

6 r_B = 1 - 10 r_A > 0

, we get

r_A < 1/10

. Combined with

r_A > 1/16

, we have

1/16 < r_A < 1/10

.

4

Calculate the combined time

T_{together} = \frac{1}{r_A + r_B}

.

8 < T_{together} < 10

.

Since

6(r_A + r_B) = 1 - 4 r_A

, substituting

1/16 < r_A < 1/10

gives

3/5 < 6(r_A + r_B) < 3/4

, which simplifies to

8 < T_{together} < 10

.

5

Calculate the fraction of total work performed by Alex.

Alex performs

10 r_A > 10 \times \frac{1}{16} = 0.625 = 62.5\%

of the total work.

Alex works for a total of 10 hours (

4 + 6

). Since

r_A > 1/16

, Alex completes over

62.5\%

of the job, which is greater than

60\%

.

Key Concept

Formulating algebraic inequalities for work rate problems when relative individual speeds are given.

Question 131Question →

For all real numbers $x$ satisfying the absolute value inequality $|4x - 12| \le 20$ , the maximum possible value of the expression $|2 - 3x|$ is $M$ . What is the value of $M$ ?

Show answer & explanation

Answer: 22

Answer

The maximum possible value

M

of the expression

|2 - 3x|

on the domain

-2 \le x \le 8

is 22.

Solving

|4x - 12| \le 20

yields

-20 \le 4x - 12 \le 20

, which simplifies to

-2 \le x \le 8

. Evaluating

|2 - 3x|

over this interval gives a minimum of

0

(at

x = 2/3

) and endpoint values of

|2 - 3(-2)| = 8

and

|2 - 3(8)| = |-22| = 22

. Thus, the maximum value

M

is 22.

Step-by-Step Solution

1

Unfold the given absolute value inequality into a compound linear inequality.

-20 \le 4x - 12 \le 20

The inequality

|u| \le k

for

k \ge 0

is equivalent to

-k \le u \le k

.

2

Isolate the variable

x

by adding 12 and dividing by 4.

-2 \le x \le 8

Adding 12 gives

-8 \le 4x \le 32

. Dividing by positive 4 preserves inequality signs, yielding

-2 \le x \le 8

.

3

Evaluate the target expression

|2 - 3x|

at the boundary points of the interval

[-2, 8]

.

For

x = -2

:

|2 - 3(-2)| = |8| = 8

. For

x = 8

:

|2 - 3(8)| = |-22| = 22

.

The expression

f(x) = |2 - 3x|

is convex and non-negative, reaching its local minimum of 0 at

x = \frac{2}{3}

. Its maximum over a closed interval must occur at one of the endpoints.

4

Compare the evaluated values to find the maximum

M

.

M = \max(8, 22) = 22

Comparing 8 and 22 shows that 22 is the absolute maximum value achievable within the domain.

Key Concept

Solving linear absolute value inequalities to determine variable bounds and evaluating extreme values of absolute value expressions.

Estimated Time:2m 0s

Question 132Question →

A corporate enterprise allocates a total annual budget of $\$52,000$ across three departments: Marketing, Operations, and Development. The amount allocated to Operations is $\$5,000$ less than twice the amount allocated to Marketing. The amount allocated to Development is $\$1,000$ more than half of the combined allocations of Marketing and Operations. What is the budget allocation for Development?

\$13,000

\$17,000

\$18,000

\$21,000

\$24,000

Show answer & explanation

Answer: $\$18,000$

Answer

The budget allocation for Development is

\$18,000

.

By defining the Marketing budget as

x

, Operations becomes

2x - 5,000

and Development becomes

\frac{1}{2}(3x - 5,000) + 1,000 = 1.5x - 1,500

. Adding these three expressions equals the total budget of

\$52,000

, giving

4.5x - 6,500 = 52,000

. Solving for

x

yields

x = 13,000

. Substituting

x = 13,000

into the expression for Development gives

1.5(13,000) - 1,500 = \$18,000

.

Step-by-Step Solution

1

Define the variable and express each department's budget in terms of that variable.

Let

x

be the Marketing budget in dollars. Then Operations

= 2x - 5,000

. The combined Marketing and Operations budget

= x + (2x - 5,000) = 3x - 5,000

. Therefore, Development

= \frac{1}{2}(3x - 5,000) + 1,000 = 1.5x - 1,500

.

Establishing a single linear variable allows all three department budgets to be combined into one linear equation.

2

Set up the total budget equation and solve for

x

.

x + (2x - 5,000) + (1.5x - 1,500) = 52,000 \implies 4.5x - 6,500 = 52,000 \implies 4.5x = 58,500 \implies x = 13,000

.

Summing the allocations for Marketing, Operations, and Development yields the total corporate budget of

\$52,000

.

3

Calculate the specific budget allocation requested for Development.

Development allocation

= 1.5(13,000) - 1,500 = 19,500 - 1,500 = 18,000

.

Substituting

x = 13,000

back into the algebraic expression for Development gives the final required dollar amount.

Key Concept

Formulating and solving multi-step linear equations in one variable from word problems.

Estimated Time:2m 30s

Question 133Question →

If

x

is a real number greater than

1

satisfying the exponential equation

\left(x^x\right)^{\sqrt{x}} = x^{\left(x^{\sqrt{x}}\right)}

what is the value of

x

?

\frac{3}{2}

\frac{4}{9}

\frac{9}{4}

\frac{27}{8}

\frac{81}{16}

Show answer & explanation

Answer: $\frac{9}{4}$

Answer

The value of

x

is

\frac{9}{4}

.

Applying the exponent rule

(a^b)^c = a^{bc}

simplifies the left side to

x^{x\sqrt{x}}

. Since the base

x > 1

is identical on both sides, we set the exponents equal:

x\sqrt{x} = x^{\sqrt{x}}

. Expressing

x\sqrt{x}

as

x^{3/2}

yields

x^{3/2} = x^{\sqrt{x}}

, which implies

\sqrt{x} = \frac{3}{2}

. Squaring both sides gives

x = \frac{9}{4}

.

Step-by-Step Solution

1

Apply the power rule

(a^b)^c = a^{bc}

to the left-hand side of the equation.

\left(x^x\right)^{\sqrt{x}} = x^{x \cdot \sqrt{x}} = x^{x^{3/2}}

Raising a power to another exponent requires multiplying the exponents:

x \cdot x^{1/2} = x^{1 + 1/2} = x^{3/2}

.

2

Equate the exponents of the expressions on both sides, as the bases are equal and

x > 1

.

x \sqrt{x} = x^{\sqrt{x}}

, which means

x^{3/2} = x^{\sqrt{x}}

If

x^A = x^B

and

x > 1

, then

A = B

.

3

Equate exponents once more for the base

x

.

\frac{3}{2} = \sqrt{x}

Since the bases are identical (

x > 1

), their exponents must be equal.

4

Square both sides to solve for

x

.

x = \left(\frac{3}{2}\right)^2 = \frac{9}{4}

Squaring

\sqrt{x}

isolates

x

.

Key Concept

Properties of exponents and nested power rules with radicals

Estimated Time:2m 0s

Question 134Question →

For the function $f(x) = x^2 - 5x$ , what is the value of $f(-3)$ ?

24

6

-6

-24

-16

Show answer & explanation

Answer: 24

Answer

24

Substituting

x = -3

into the function yields

(-3)^2 - 5(-3) = 9 + 15 = 24

, which correctly evaluates both the exponent and linear terms following standard sign rules.

Step-by-Step Solution

1

Substitute

x = -3

into the function expression

f(x) = x^2 - 5x

.

f(-3) = (-3)^2 - 5(-3)

Function notation

f(-3)

requires replacing every instance of variable

x

with

-3

.

2

Evaluate the exponent term

(-3)^2

.

9

Squaring any negative number produces a positive result:

(-3) \times (-3) = 9

.

3

Evaluate the product term

-5(-3)

.

15

Multiplying two negative numbers yields a positive product:

-5 \times (-3) = 15

.

4

Sum the evaluated components.

9 + 15 = 24

Combine the results to obtain the final value of

f(-3)

.

Key Concept

Evaluating functions with negative input values

Estimated Time:45s

Question 135Question →

How many integer values of $x$ satisfy the inequality $|3 - 2x| + |x + 4| \le 9$ ?

3

4

5

6

7

Show answer & explanation

Answer: 5

Answer

5 integer values (specifically -2, -1, 0, 1, and 2)

Solving the absolute value inequality using piecewise intervals yields the continuous solution set

[-2, \frac{8}{3}]

. The integers contained in this range are

-2, -1, 0, 1,

and

2

, giving a total of 5 integer values.

Step-by-Step Solution

1

Identify the critical points of the absolute value expressions.

The critical points are

x = \frac{3}{2}

and

x = -4

. These split the real number line into three intervals:

x < -4

,

-4 \le x \le \frac{3}{2}

, and

x > \frac{3}{2}

.

Absolute value expressions change definition at their zeroes.

2

Analyze Case 1:

x < -4

.

Here

|3 - 2x| = 3 - 2x

and

|x + 4| = -(x + 4)

. The inequality becomes

(3 - 2x) - (x + 4) \le 9 \implies -3x - 1 \le 9 \implies -3x \le 10 \implies x \ge -\frac{10}{3} \approx -3.33

. Since there is no overlap between

x < -4

and

x \ge -3.33

, no solutions exist in this interval.

Evaluating expressions according to the sign of terms when

x < -4

.

3

Analyze Case 2:

-4 \le x \le \frac{3}{2}

.

Here

|3 - 2x| = 3 - 2x

and

|x + 4| = x + 4

. The inequality becomes

(3 - 2x) + (x + 4) \le 9 \implies 7 - x \le 9 \implies -x \le 2 \implies x \ge -2

. Combining with the case interval gives

[-2, \frac{3}{2}]

.

Determining valid values of

x

within the middle interval.

4

Analyze Case 3:

x > \frac{3}{2}

.

Here

|3 - 2x| = 2x - 3

and

|x + 4| = x + 4

. The inequality becomes

(2x - 3) + (x + 4) \le 9 \implies 3x + 1 \le 9 \implies 3x \le 8 \implies x \le \frac{8}{3} \approx 2.67

. Combining with the case interval gives

(\frac{3}{2}, \frac{8}{3}]

.

Determining valid values of

x

within the upper interval.

5

Combine solution intervals and count integer solutions.

The total solution set is

[-2, \frac{8}{3}]

. The integer values within this interval are

-2, -1, 0, 1,

and

2

. Total count = 5.

Identifying all integer values within the bounded set

[-2, 2.67]

.

Key Concept

Solving absolute value inequalities with multiple absolute value terms using critical points and case analysis.

Estimated Time:2m 0s

Question 136Question →

What is the sum of all integer values of $x$ that satisfy both $|2x - 5| \le 9$ and $|x + 2| \ge 4$ ?

Show answer & explanation

Answer: 27

Answer

The sum of all integer values of

x

that satisfy both inequalities is 27.

First, solving

|2x - 5| \le 9

yields

-9 \le 2x - 5 \le 9 \implies -4 \le 2x \le 14 \implies -2 \le x \le 7

. Second, solving

|x + 2| \ge 4

yields

x + 2 \ge 4 \implies x \ge 2

or

x + 2 \le -4 \implies x \le -6

. Taking the intersection of

-2 \le x \le 7

and

(x \ge 2 \text{ or } x \le -6)

gives the solution set

2 \le x \le 7

. The integer values satisfying this condition are 2, 3, 4, 5, 6, and 7. Summing these integers gives

2 + 3 + 4 + 5 + 6 + 7 = 27

.

Step-by-Step Solution

1

Solve the inequality

|2x - 5| \le 9

-2 \le x \le 7

An inequality of the form

|u| \le a

(where

a \ge 0

) expands to

-a \le u \le a

. Adding 5 gives

-4 \le 2x \le 14

, and dividing by 2 yields

-2 \le x \le 7

.

2

Solve the inequality

|x + 2| \ge 4

x \ge 2 \text{ or } x \le -6

An inequality of the form

|u| \ge a

(where

a > 0

) expands to

u \ge a

or

u \le -a

. Subtracting 2 from both inequalities yields

x \ge 2

or

x \le -6

.

3

Determine the overlapping interval for both inequalities

2 \le x \le 7

Combining

-2 \le x \le 7

with

x \ge 2 \text{ or } x \le -6

eliminates

x \le -6

. The intersection of

[-2, 7]

and

[2, \infty)

is

[2, 7]

.

4

Identify the integer values in the solution interval and calculate their sum

27

The integers in the closed interval

[2, 7]

are 2, 3, 4, 5, 6, and 7. Adding them together gives

2 + 3 + 4 + 5 + 6 + 7 = 27

.

Key Concept

System of Linear Absolute Value Inequalities

Question 137Question →

For real constants

a

and

b

, consider the linear equation in one variable

x

:

\frac{a(x - 2)}{3} - \frac{2x + 1}{4} = \frac{(a - 3)x + b}{12}

If this equation has infinitely many solutions for

x

, which of the following statements must be true? Select all such statements.

Select all that apply

a + b = -10

ab < 0

2a - b = 13

a - b = -10

ab = 11

Show answer & explanation

Answer: $a + b = -10$ ; $ab < 0$ ; $2a - b = 13$

Answer

The statements

a + b = -10

,

ab < 0

, and

2a - b = 13

are all true.

Clearing denominators gives

4a(x - 2) - 3(2x + 1) = (a - 3)x + b

. Expanding both sides yields

(4a - 6)x - (8a + 3) = (a - 3)x + b

, which rearranges to

(3a - 3)x = 8a + b + 3

. For a linear equation in one variable to have infinitely many solutions, both the coefficient of

x

and the constant term must be zero (

0x = 0

). Setting

3a - 3 = 0

gives

a = 1

, and substituting

a = 1

into

8a + b + 3 = 0

yields

b = -11

. Evaluating the statements with

a = 1

and

b = -11

shows that

a + b = 1 + (-11) = -10

is true,

ab = -11 < 0

is true, and

2a - b = 2(1) - (-11) = 13

is true.

Step-by-Step Solution

1

Clear the denominators by multiplying both sides of the equation by 12.

4a(x - 2) - 3(2x + 1) = (a - 3)x + b

Eliminating fractions simplifies expanding terms and combining like variables.

2

Expand all terms and group terms containing

x

on the left-hand side.

4ax - 8a - 6x - 3 = (a - 3)x + b \implies (4a - 6)x - (8a + 3) = (a - 3)x + b

Preparing the linear equation to be expressed in standard form

Ax = B

.

3

Rearrange into standard form

(A)x = B

.

[(4a - 6) - (a - 3)]x = 8a + b + 3 \implies (3a - 3)x = 8a + b + 3

A linear equation has infinitely many solutions if and only if

A = 0

and

B = 0

simultaneously.

4

Set the coefficient of

x

and the constant term equal to zero to determine

a

and

b

.

3a - 3 = 0 \implies a = 1

; then

8(1) + b + 3 = 0 \implies b + 11 = 0 \implies b = -11

The equation reduces to

0 \cdot x = 0

, which is satisfied by every real number

x

.

5

Evaluate the given statements using

a = 1

and

b = -11

.

a + b = 1 + (-11) = -10

(True);

ab = (1)(-11) = -11 < 0

(True);

2a - b = 2(1) - (-11) = 13

(True).

Direct substitution confirms which statements hold true.

Key Concept

Conditions for a linear equation in one variable to have infinitely many solutions

Estimated Time:2m 30s

Question 138Question →

Train $X$ departs from a station traveling due east at a constant speed of $50$ miles per hour. Exactly $1$ hour later, Express Train $Y$ departs from the same station along the same track, traveling due east at a constant speed of $75$ miles per hour. How many hours after Express Train $Y$ departs will it catch up to Train $X$ ?

1

1.5

2

2.5

3

Show answer & explanation

Answer: $2$

Answer

Express Train

Y

will catch up to Train

X

exactly

2

hours after Express Train

Y

departs.

Let

t

represent the number of hours Express Train

Y

travels. Because Train

X

departed

1

hour earlier, it has been traveling for

t + 1

hours. For Express Train

Y

to catch up to Train

X

, both trains must cover the exact same distance from the starting station. Setting up the distance equation

75t = 50(t + 1)

gives

75t = 50t + 50

, which simplifies to

25t = 50

, yielding

t = 2

hours.

Step-by-Step Solution

1

Define variables for the time traveled by each train.

Let

t

be the time in hours that Express Train

Y

travels. Since Train

X

departed

1

hour earlier, Train

X

travels for

t + 1

hours.

Train

X

has a

1

-hour head start.

2

Express the distance traveled by each train using

\text{Distance} = \text{Rate} \times \text{Time}

.

Distance of Train

X = 50(t + 1)

miles; Distance of Express Train

Y = 75t

miles.

Both trains travel at constant rates along the same path.

3

Equate the two distance expressions to solve for

t

.

75t = 50(t + 1) \implies 75t = 50t + 50 \implies 25t = 50 \implies t = 2

.

Express Train

Y

catches Train

X

when both have covered the exact same distance.

Key Concept

Distance, Rate, and Time Modeling for Catch-up Scenarios

Question 139Question →

For how many integer values of $k$ does the inequality $|2x - k| + |x + 3| \le 7$ have at least one real solution $x$ such that $x \ge 1$ ?

8

9

10

11

13

Show answer & explanation

Answer: 10

Answer

10 integer values of

k

satisfy the given condition.

The correct answer is 10 because analyzing the condition

x \ge 1

simplifies the inequality to

|2x - k| \le 4 - x

, requiring

x \in [1, 4]

. The double inequality

k - 4 \le x \le \frac{k + 4}{3}

yields solutions overlapping with

[1, 4]

if and only if

-1 \le k \le 8

, which contains exactly 10 integers.

Step-by-Step Solution

1

Simplify the absolute value term

|x + 3|

using the given condition

x \ge 1

.

Since

x \ge 1

,

x + 3 > 0

, so

|x + 3| = x + 3

. The inequality becomes

|2x - k| + x + 3 \le 7

, which simplifies to

|2x - k| \le 4 - x

.

Establishing the sign of

x + 3

allows eliminating one set of absolute value bars.

2

Determine the valid range for

x

.

Since

|2x - k| \ge 0

, it must hold that

4 - x \ge 0

, which implies

x \le 4

. Combined with

x \ge 1

, any solution

x

must lie in the interval

[1, 4]

.

An absolute value quantity cannot be less than a negative number.

3

Unwrap the absolute value inequality

|2x - k| \le 4 - x

.

-(4 - x) \le 2x - k \le 4 - x

. Splitting into two linear inequalities:
1)

2x - k \le 4 - x \implies 3x \le k + 4 \implies x \le \frac{k + 4}{3}

.
2)

2x - k \ge x - 4 \implies x \ge k - 4

.
Thus,

k - 4 \le x \le \frac{k + 4}{3}

.

Rewriting absolute value inequalities as compound inequalities defines explicit bounds on

x

in terms of

k

.

4

Find the range of

k

for which

[k - 4, \frac{k + 4}{3}]

overlaps with

[1, 4]

.

For an overlapping solution to exist in

[1, 4]

:
1) The upper bound

\frac{k + 4}{3}

must be at least 1:

\frac{k + 4}{3} \ge 1 \implies k \ge -1

.
2) The lower bound

k - 4

must be at most 4:

k - 4 \le 4 \implies k \le 8

.
Combining these gives

-1 \le k \le 8

.

The solution interval for

x

must have a non-empty intersection with the allowed domain

[1, 4]

.

5

Count the total number of integer values of

k

in the interval

[-1, 8]

.

The integers are

-1, 0, 1, 2, 3, 4, 5, 6, 7, 8

, giving a total of

8 - (-1) + 1 = 10

integer values.

Counting inclusive integer endpoints gives the total count.

Key Concept

Solving absolute value inequalities involving parameters and restricted variable domains.

Estimated Time:2m 0s

Question 140Question →

An express train travels along a straight track between Station A and Station B. For the first $40\%$ of the total distance, the train travels at a constant speed of $v$ miles per hour. For the next $50\%$ of the remaining distance, due to track maintenance, it travels at a constant speed that is $25\%$ slower than $v$ . For the final leg of the journey, the train increases its speed to a constant rate that is $20\%$ faster than $v$ . If the average speed for the entire journey from Station A to Station B is $60$ miles per hour, what is the value of $v$ , in miles per hour?

57.5

60.0

63.0

68.0

69.0

Show answer & explanation

Answer: $63.0$

Answer

The base speed

v

is

63.0

miles per hour.

The correct answer is derived by setting up a model for distance, speed, and time across all three segments. The first leg covers

0.4D

at speed

v

(time

= 0.4D/v

). The second leg covers

50\%

of the remaining

0.6D

, which is

0.3D

, at speed

0.75v

(time

= 0.3D/(0.75v) = 0.4D/v

). The third leg covers the remaining

0.3D

at speed

1.2v

(time

= 0.3D/(1.2v) = 0.25D/v

). Summing these gives total time

T = 1.05D/v

. The average speed is

D / (1.05D/v) = v / 1.05 = 20v/21

. Setting

20v/21 = 60

yields

v = 63

.

Step-by-Step Solution

1

Define distances for each leg of the trip in terms of total distance

D

.

Leg 1 distance

d_1 = 0.40D

. Remaining distance is

D - 0.40D = 0.60D

. Leg 2 distance

d_2 = 0.50 \times 0.60D = 0.30D

. Leg 3 distance

d_3 = 0.60D - 0.30D = 0.30D

.

The problem specifies percentages of remaining distance, requiring step-by-step subtraction of completed distance.

2

Determine the speed for each leg in terms of

v

.

Leg 1 speed

v_1 = v

. Leg 2 speed

v_2 = v(1 - 0.25) = 0.75v

. Leg 3 speed

v_3 = v(1 + 0.20) = 1.20v

.

Percentage increases and decreases are applied to the base rate

v

.

3

Calculate time spent on each leg (

t = \frac{d}{v}

) and sum for total time

T

.

t_1 = \frac{0.40D}{v}

,

t_2 = \frac{0.30D}{0.75v} = \frac{0.40D}{v}

,

t_3 = \frac{0.30D}{1.20v} = \frac{0.25D}{v}

. Total time

T = \frac{0.40D + 0.40D + 0.25D}{v} = \frac{1.05D}{v} = \frac{21D}{20v}

.

Average speed requires total distance divided by total time.

4

Set up the average speed equation and solve for

v

.

Average speed

= \frac{D}{T} = \frac{D}{\frac{21D}{20v}} = \frac{20v}{21} = 60 \implies 20v = 1260 \implies v = 63

.

Equating the algebraic average speed expression to the given value of 60 mph yields the value of

v

.

Key Concept

Weighted Average Speed in Multi-Phase Motion Problems

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