Algebra and Functions Practice Questions — GMAT — Page 10

Question 181Question →

An event catering service charges a fixed setup fee plus a constant hourly rate for event staff. For a 4-hour event, the total charge is $680. For a 7-hour event with the same staff requirements, the total charge is$ 1,070. What is the fixed setup fee, in dollars, charged by the catering service?

Show answer & explanation

Answer: 160

Answer

The fixed setup fee charged by the catering service is 160 dollars.

Modeling the total cost as

C = S + rt

, where

S

is the fixed setup fee and

r

is the hourly rate, yields the system of equations

S + 4r = 680

and

S + 7r = 1070

. Subtracting the first equation from the second yields

3r = 390

, so

r = 130

. Substituting

r = 130

back into the first equation yields

S + 520 = 680

, which gives

S = 160

.

Step-by-Step Solution

1

Define the algebraic model for total cost.

Let

S

be the fixed setup fee in dollars and

r

be the hourly rate in dollars per hour. The total cost

C

for

t

hours is modeled by

C = S + rt

.

The problem presents a linear relationship between time and total cost, consisting of a fixed component and a variable component.

2

Construct a system of linear equations.

S + 4r = 680

and

S + 7r = 1070

.

Substitute the two given combinations of time (

t = 4

and

t = 7

) and total cost (

C = 680

and

C = 1070

) into the algebraic model.

3

Solve for the variable hourly rate

r

.

(S + 7r) - (S + 4r) = 1070 - 680 \implies 3r = 390 \implies r = 130

.

Subtracting the first equation from the second eliminates the fixed setup fee variable

S

.

4

Calculate the fixed setup fee

S

.

S + 4(130) = 680 \implies S + 520 = 680 \implies S = 160

.

Substitute

r = 130

back into the first equation to find the value of

S

.

Key Concept

Linear Equation Modeling and Systems of Equations

Question 182Question →

If $x$ is a real number that satisfies the equation $|5 - 2x| = 3x - 10$ , what is the value of $|x - 4| + 2x$ ?

5

7

11

14

18

Show answer & explanation

Answer: 11

Answer

11

Solving the equation

|5 - 2x| = 3x - 10

yields two potential values:

x = 3

and

x = 5

. Substituting

x = 3

into the original equation yields

|-1| = -1

, which is impossible because an absolute value cannot be negative. Therefore,

x = 3

is an extraneous root. Substituting

x = 5

yields

|-5| = 5

, which is true. Evaluating the requested expression

|x - 4| + 2x

at

x = 5

gives

|5 - 4| + 2(5) = 1 + 10 = 11

.

Step-by-Step Solution

1

Set up the two algebraic cases for the absolute value equation

|5 - 2x| = 3x - 10

.

Case 1:

5 - 2x = 3x - 10

when

5 - 2x \ge 0

(

x \le 2.5

).
Case 2:

-(5 - 2x) = 3x - 10

when

5 - 2x < 0

(

x > 2.5

).

An absolute value expression

|u|

equals

u

when

u \ge 0

and

-u

when

u < 0

.

2

Solve Case 1 algebraically.

5 - 2x = 3x - 10 \implies 15 = 5x \implies x = 3

.

Isolate the variable

x

on one side of the equation.

3

Solve Case 2 algebraically.

2x - 5 = 3x - 10 \implies 5 = x \implies x = 5$.

Simplify and isolate

x

.

4

Check both candidate solutions in the original equation to filter out extraneous roots.

For

x = 3

:

|5 - 2(3)| = 3(3) - 10 \implies |-1| = -1 \implies 1 = -1

(False, extraneous).
For

x = 5

:

|5 - 2(5)| = 3(5) - 10 \implies |-5| = 5 \implies 5 = 5

(True, valid).

Since the right-hand side

3x - 10

must be non-negative for the absolute value to hold, candidate solutions must be tested in the original equation.

5

Substitute the valid solution

x = 5

into the targeted expression

|x - 4| + 2x

.

|5 - 4| + 2(5) = |1| + 10 = 1 + 10 = 11

.

Evaluate the target expression using the single real value

x = 5

that satisfies the given linear absolute value equation.

Key Concept

Solving absolute value linear equations and identifying extraneous solutions

Estimated Time:2m 0s

Question 183Question →

If $y$ is a real number satisfying the inequality $\frac{|3y - 6|}{y + 2} < 1$ , which of the following represents the complete set of all possible values of $y$ ?

y < -2

or

1 < y < 4

1 < y < 4

-2 < y < 4

y < 1

or

y > 4

y < -2

or

y > 1

Show answer & explanation

Answer: $y < -2$ or $1 < y < 4$

Answer

y < -2

or

1 < y < 4

To evaluate

\frac{|3y - 6|}{y + 2} < 1

, analyze the sign of the denominator

y + 2

. If

y < -2

, the denominator is negative, so multiplying both sides by

y + 2

reverses the inequality to

|3y - 6| > y + 2

. Because absolute values are always non-negative and

y + 2

is negative for

y < -2

, this inequality holds for all

y < -2

. If

y > -2

, the denominator is positive, yielding

|3y - 6| < y + 2

, which expands to

-y - 2 < 3y - 6 < y + 2

. Solving the left inequality gives

y > 1

, and solving the right gives

y < 4

, producing

1 < y < 4

. Combining both valid cases gives

y < -2

or

1 < y < 4

.

Step-by-Step Solution

1

Determine the domain restriction and split into cases based on the denominator's sign.

The expression is undefined when

y = -2

. We analyze Case 1 (

y > -2

) and Case 2 (

y < -2

).

Multiplying an inequality by an algebraic expression requires knowing its sign to preserve or reverse the inequality direction.

2

Solve Case 1 where

y > -2

(positive denominator).

Multiplying by

y + 2

gives

|3y - 6| < y + 2

, which expands to

-(y + 2) < 3y - 6 < y + 2

. Solving

-y - 2 < 3y - 6

gives

4 < 4y \implies y > 1

. Solving

3y - 6 < y + 2

gives

2y < 8 \implies y < 4

. Combining gives

1 < y < 4

.

Since

y + 2 > 0

, multiplying preserves the inequality sign.

3

Solve Case 2 where

y < -2

(negative denominator).

Multiplying by

y + 2

flips the inequality sign to

|3y - 6| > y + 2

. Since

|3y - 6| \ge 0

for all real

y

and

y + 2 < 0

when

y < -2

, a non-negative number is always strictly greater than a negative number. Thus, all

y < -2

are valid solutions.

Any non-negative real value is strictly greater than any negative value.

4

Combine the valid intervals from both cases.

y < -2

or

1 < y < 4

.

The complete solution set is the union of solutions from Case 1 and Case 2.

Key Concept

Solving Rational Absolute Value Inequalities via Denominator Sign Case Analysis

Estimated Time:2m 0s

Question 184Question →

An artisan workshop produces custom wooden chairs and tables. Each chair requires $3$ hours of carving and $2$ hours of finishing, while each table requires $5$ hours of carving and $4$ hours of finishing. If the workshop logged a total of $110$ hours of carving and $84$ hours of finishing last week, how many tables were produced?

10

14

16

18

22

Show answer & explanation

Answer: 16

Answer

The workshop produced 16 tables.

By defining

c

as the number of chairs and

t

as the number of tables, we can set up two linear equations representing total hours:

3c + 5t = 110

for carving and

2c + 4t = 84

for finishing. Multiplying the second equation by

1.5

yields

3c + 6t = 126

. Subtracting

3c + 5t = 110

from

3c + 6t = 126

leaves

t = 16

. Thus, the workshop produced 16 tables.

Step-by-Step Solution

1

Define variables and translate the word problem into a system of linear equations.

Let

c

be the number of chairs and

t

be the number of tables. Carving equation:

3c + 5t = 110

. Finishing equation:

2c + 4t = 84

.

The total hours for each activity equal the sum of hours spent on chairs and tables.

2

Eliminate variable

c

to solve for

t

.

Multiply the finishing equation by

1.5

to get

3c + 6t = 126

. Subtract the carving equation (

3c + 5t = 110

) from this equation:

(3c + 6t) - (3c + 5t) = 126 - 110 \implies t = 16

.

Aligning the coefficient of

c

in both equations allows direct elimination of

c

to isolate

t

.

3

Verify the solution by calculating

c

and checking both original equations.

Substitute

t = 16

into

2c + 4(16) = 84 \implies 2c + 64 = 84 \implies 2c = 20 \implies c = 10

. Check carving:

3(10) + 5(16) = 30 + 80 = 110

.

Ensures that

t = 16

and

c = 10

satisfy both resource constraints without calculation errors.

Key Concept

Setting up and solving a system of two linear equations in two variables

Hints

1

Set up two separate linear equations: one for total carving hours and one for total finishing hours.

2

Let

c

be the number of chairs and

t

be the number of tables. Your system is

3c + 5t = 110

and

2c + 4t = 84

.

3

Multiply

2c + 4t = 84

by

1.5

to get

3c + 6t = 126

, then subtract

3c + 5t = 110

to find

t

directly.

Practice More

Try solving a similar problem where the total revenue and total unit count are given to practice standard linear system modeling.

Alternative Method

Divide the finishing equation

2c + 4t = 84

by

2

to get

c + 2t = 42 \implies c = 42 - 2t

. Substitute this into the carving equation:

3(42 - 2t) + 5t = 110 \implies 126 - 6t + 5t = 110 \implies -t = -16 \implies t = 16

.

Estimated Time:1m 30s

Question 185Question →

If $x$ is a positive real number such that $x^{\frac{1}{2}} + x^{-\frac{1}{2}} = 4$ , what is the value of $x^{\frac{3}{2}} + x^{-\frac{3}{2}}$ ?

48

52

56

61

64

Show answer & explanation

Answer: 52

Answer

52

Cubing both sides of

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

yields

(x^{\frac{1}{2}} + x^{-\frac{1}{2}})^3 = 64

. By the identity

(a+b)^3 = a^3 + b^3 + 3ab(a+b)

, the left side expands to

(x^{\frac{3}{2}} + x^{-\frac{3}{2}}) + 3(x^{\frac{1}{2}} \cdot x^{-\frac{1}{2}})(x^{\frac{1}{2}} + x^{-\frac{1}{2}})

. Substituting

x^{\frac{1}{2}} \cdot x^{-\frac{1}{2}} = 1

and

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

gives

(x^{\frac{3}{2}} + x^{-\frac{3}{2}}) + 12 = 64

. Subtracting 12 yields 52.

Step-by-Step Solution

1

Set up the cubic identity for the sum of fractional exponents.

Let

a = x^{\frac{1}{2}}

and

b = x^{-\frac{1}{2}}

. Then

a + b = 4

and

ab = x^{\frac{1}{2}} \cdot x^{-\frac{1}{2}} = 1

.

Recognizing that

x^{\frac{3}{2}} = a^3

and

x^{-\frac{3}{2}} = b^3

allows the use of binomial expansion.

2

Cube both sides of the given equation

a + b = 4

.

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

.

Expanding the cube retains the desired expression

a^3 + b^3

alongside simpler terms.

3

Simplify and solve for

x^{\frac{3}{2}} + x^{-\frac{3}{2}}

.

64 = (x^{\frac{3}{2}} + x^{-\frac{3}{2}}) + 12 \implies x^{\frac{3}{2}} + x^{-\frac{3}{2}} = 64 - 12 = 52.

Subtracting 12 from 64 gives the exact value of the expression.

Key Concept

Algebraic identities with fractional exponents

Estimated Time:2m 0s

Question 186Question →

If $x$ is an integer such that $|x - 3| + |x + 5| \le 12$ , how many possible values of $x$ exist?

Show answer & explanation

Answer: 13

Answer

13

Evaluating the inequality

|x - 3| + |x + 5| \le 12

across the three intervals defined by critical points

x = -5

and

x = 3

yields the continuous real solution set

[-7, 5]

. Counting all integers in this inclusive range gives

5 - (-7) + 1 = 13

possible values.

Step-by-Step Solution

1

Identify the critical points of the absolute value expressions.

The critical points where the expressions inside the absolute values change sign are

x = 3

and

x = -5

.

Setting

x - 3 = 0

gives

x = 3

, and setting

x + 5 = 0

gives

x = -5

.

2

Solve the inequality for the region

x < -5

.

-(x - 3) - (x + 5) \le 12 \implies -2x - 2 \le 12 \implies -2x \le 14 \implies x \ge -7

. Thus,

-7 \le x < -5

.

When

x < -5

, both

x - 3 < 0

and

x + 5 < 0

, so

|x - 3| = -(x - 3)

and

|x + 5| = -(x + 5)

.

3

Solve the inequality for the region

-5 \le x \le 3

.

-(x - 3) + (x + 5) \le 12 \implies 8 \le 12

, which is universally true for all

x

in this interval.

When

-5 \le x \le 3

,

x - 3 \le 0

and

x + 5 \ge 0

, so

|x - 3| = -(x - 3)

and

|x + 5| = x + 5

.

4

Solve the inequality for the region

x > 3

.

(x - 3) + (x + 5) \le 12 \implies 2x + 2 \le 12 \implies 2x \le 10 \implies x \le 5

. Thus,

3 < x \le 5

.

When

x > 3

, both

x - 3 > 0

and

x + 5 > 0

.

5

Combine the valid intervals and count the integer solutions.

The full solution range is

[-7, 5]

. The number of integer values is

5 - (-7) + 1 = 13

.

The number of integers in an inclusive range

[a, b]

is given by

b - a + 1

.

Key Concept

Solving multi-term absolute value inequalities by dividing the domain at critical points into distinct cases.

Question 187Question →

For what value of the constant $k$ does the system of linear equations below have no solution?

\begin{aligned} (k - 2)x + 3y &= 6 \\ 4x + (k + 2)y &= 12 \end{aligned}

-2

4

-4

Both

-4

and

4

16

Show answer & explanation

Answer: $-4$

Answer

The constant value

k = -4

results in a system with no solution.

For a

2 \times 2

linear system

a_1 x + b_1 y = c_1

and

a_2 x + b_2 y = c_2

to have no solution, the equations must have proportional variable coefficients but non-proportional constant terms:

\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}

. Solving

\frac{k-2}{4} = \frac{3}{k+2}

gives

k^2 - 4 = 12

, so

k = \pm 4

. Substituting

k = -4

yields

-6x + 3y = 6

and

4x - 2y = 12

, which simplify to

-2x + y = 2

and

-2x + y = -6

. Since the constants

2

and

-6

differ, the lines are parallel and distinct, meaning the system has no solution.

Step-by-Step Solution

1

Set up the condition for parallel lines (equal slopes) by equating the ratio of the coefficients of

x

and

y

.

\frac{k - 2}{4} = \frac{3}{k + 2}

A system of two linear equations has either zero solutions (parallel, non-intersecting lines) or infinitely many solutions (coincident lines) when the slopes are equal.

2

Cross-multiply and solve the quadratic equation for

k

.

(k - 2)(k + 2) = 12 \implies k^2 - 4 = 12 \implies k^2 = 16 \implies k = 4 \text{ or } k = -4

Finding all values of

k

where the coefficient matrix determinant is zero.

3

Test

k = 4

in the original system.

(4 - 2)x + 3y = 6 \implies 2x + 3y = 6 \quad \text{and} \quad 4x + (4 + 2)y = 12 \implies 4x + 6y = 12

Dividing

4x + 6y = 12

by

2

yields

2x + 3y = 6

, which is identical to the first equation. Thus,

k = 4

yields infinitely many solutions.

4

Test

k = -4

in the original system.

(-4 - 2)x + 3y = 6 \implies -6x + 3y = 6 \quad \text{and} \quad 4x + (-4 + 2)y = 12 \implies 4x - 2y = 12

Simplifying both equations gives

-2x + y = 2

and

-2x + y = -6

. The lines have identical slopes but different constants, so they are parallel and distinct, producing no solution.

Key Concept

System Solvability and Linear Consistency

Estimated Time:2m 0s

Question 188Question →

A beverage producer creates two liquid mixtures, Mixture $A$ and Mixture $B$ . Mixture $A$ consists of $30\%$ fruit concentrate by volume, and Mixture $B$ consists of $70\%$ fruit concentrate by volume. A lab technician combines $x$ liters of Mixture $A$ with $y$ liters of Mixture $B$ to prepare an $80$ -liter batch that contains $45\%$ fruit concentrate by volume. What is the value of $x$ ?

Show answer & explanation

Answer: 50

Answer

The value of

x

is 50.

To find

x

, we construct two linear equations based on total liquid volume and total fruit concentrate. The total volume equation is

x + y = 80

, which gives

y = 80 - x

. The concentrate equation is

0.30x + 0.70y = 0.45(80) = 36

. Substituting

y = 80 - x

yields

0.30x + 0.70(80 - x) = 36

. Expanding gives

0.30x + 56 - 0.70x = 36

, so

-0.40x = -20

, which results in

x = 50

.

Step-by-Step Solution

1

Formulate a system of two linear equations representing total volume and total concentrate volume.

System equations:

x + y = 80

and

0.30x + 0.70y = 36

.

The sum of the component volumes equals the total mixture volume, and the sum of the pure concentrate from each component equals the total concentrate in the final mixture.

2

Substitute

y = 80 - x

into the concentrate equation to eliminate

y

.

0.30x + 0.70(80 - x) = 36

.

Substituting one variable reduces the system to a single linear equation in one variable.

3

Simplify the single-variable linear equation and solve for

x

.

0.30x + 56 - 0.70x = 36 \implies -0.40x = -20 \implies x = 50

.

Combining like terms isolates the variable

x

.

Key Concept

Solving systems of two linear equations formed by weighted mixture word problems.

Question 189Question →

For all positive real numbers $a$ and $b$ such that $b = 4a$ , if $a^b = b^a$ , then the value of $a$ is $\sqrt{2}$ .

True

False

Show answer & explanation

Answer: False

Answer

False. The statement is false because the exact value of

a

is

\sqrt[3]{4}

, which is not equal to

\sqrt{2}

.

The statement is false because simplifying

a^{4a} = (4a)^a

leads directly to

a^3 = 4

, giving

a = \sqrt[3]{4} = 2^{2/3}

. The claimed value

\sqrt{2} = 2^{1/2}

fails to satisfy the original equation, making the assertion mathematically false.

Step-by-Step Solution

1

Substitute

b = 4a

into the given exponent equation

a^b = b^a

.

a^{4a} = (4a)^a

Eliminate variable

b

to express the equation solely in terms of

a

.

2

Raise both sides of the equation to the power of

\frac{1}{a}

.

(a^{4a})^{1/a} = ((4a)^a)^{1/a} \implies a^4 = 4a

Apply power of a power exponent rule

(x^m)^n = x^{mn}

to simplify the exponents.

3

Divide both sides by

a

(since

a > 0

) and solve for

a

.

\frac{a^4}{a} = \frac{4a}{a} \implies a^3 = 4 \implies a = \sqrt[3]{4} = 2^{2/3}

Isolate

a

using standard division and radical extraction rules.

4

Compare the calculated value of

a

with

\sqrt{2}

.

a = 2^{2/3} \approx 1.587

, whereas

\sqrt{2} = 2^{1/2} \approx 1.414

. Thus

a \neq \sqrt{2}

.

Determine the truth value of the claimed conclusion.

Key Concept

Solving variable exponent equations of the form

a^b = b^a

using exponent power rules and root extraction.

Question 190Question →

What is the sum of all real solutions to the equation $|x^2 - 10| = 3x$ ?

Show answer & explanation

Answer: 7

Answer

The sum of all real solutions to the equation is 7.

To solve

|x^2 - 10| = 3x

, note that

3x \ge 0

(so

x \ge 0

). Splitting into two algebraic cases gives

x^2 - 3x - 10 = 0

(yielding

x = 5

and extraneous

x = -2

) and

x^2 + 3x - 10 = 0

(yielding

x = 2

and extraneous

x = -5

). The valid real solutions are

x = 5

and

x = 2

, whose sum is 7.

Step-by-Step Solution

1

Determine domain restriction based on the absolute value definition.

x \ge 0

Because an absolute value expression cannot be negative,

|x^2 - 10| = 3x

requires

3x \ge 0

, which means

x \ge 0

.

2

Solve Case 1:

x^2 - 10 = 3x

.

x = 5

Rearranging yields

x^2 - 3x - 10 = 0

, which factors as

(x - 5)(x + 2) = 0

. The roots are

x = 5

and

x = -2

. Reject

x = -2

because

x \ge 0

.

3

Solve Case 2:

-(x^2 - 10) = 3x

.

x = 2

Rearranging yields

x^2 + 3x - 10 = 0

, which factors as

(x + 5)(x - 2) = 0

. The roots are

x = 2

and

x = -5

. Reject

x = -5

because

x \ge 0

.

4

Sum all valid real solutions.

7

The valid real solutions are

x = 5

and

x = 2

. Their sum is

5 + 2 = 7

.

Key Concept

Solving absolute value equations with a variable expression on one side requires verifying non-negativity constraints to eliminate extraneous roots.

Estimated Time:1m 30s

Question 191Question →

A biotechnology laboratory formulates three custom reagent mixtures—Solution $X$ , Solution $Y$ , and Solution $Z$ —using three chemical compounds: Alpha, Beta, and Gamma.

- Solution $X$ contains $2\text{ mL}$ of Alpha, $3\text{ mL}$ of Beta, and $1\text{ mL}$ of Gamma, and costs $\$13.00$ .
- Solution $Y$ contains $1\text{ mL}$ of Alpha, $2\text{ mL}$ of Beta, and $4\text{ mL}$ of Gamma, and costs $\$11.00$ .
- Solution $Z$ contains $3\text{ mL}$ of Alpha, $1\text{ mL}$ of Beta, and $2\text{ mL}$ of Gamma, and costs $\$13.00$ .

What is the cost of a mixture containing $5\text{ mL}$ of Alpha, $4\text{ mL}$ of Beta, and $3\text{ mL}$ of Gamma?

\$22.00

\$24.00

\$25.00

\$26.00

\$30.00

Show answer & explanation

Answer: $\$26.00$

Answer

\$26.00

The target quantity consists of

5\text{ mL}

of Alpha,

4\text{ mL}

of Beta, and

3\text{ mL}

of Gamma. By inspecting the given system, adding Solution

X

(

2\text{ mL}

Alpha,

3\text{ mL}

Beta,

1\text{ mL}

Gamma) and Solution

Z

(

3\text{ mL}

Alpha,

1\text{ mL}

Beta,

2\text{ mL}

Gamma) directly yields

5\text{ mL}

Alpha,

4\text{ mL}

Beta, and

3\text{ mL}

Gamma. Therefore, the required cost is simply the sum of the costs of Solution

X

and Solution

Z

:

\$13.00 + \$13.00 = \$26.00

.

Step-by-Step Solution

1

Set up the linear system of equations representing the cost of each solution.

Let

a

,

b

, and

g

be the cost per mL of Alpha, Beta, and Gamma, respectively.
Equation 1:

2a + 3b + g = 13

Equation 2:

a + 2b + 4g = 11

Equation 3:

3a + b + 2g = 13

Translate the given word problem into algebraic equations representing system relationships.

2

Identify the requested quantity and evaluate whether it can be formed as a linear combination of the given equations.

Target expression:

5a + 4b + 3g

Recognizing linear combinations avoids solving for individual variable values when not required.

3

Add Equation 1 and Equation 3.

(2a + 3b + g) + (3a + b + 2g) = 13 + 13 \implies 5a + 4b + 3g = 26

The sum of coefficients for Alpha (

2+3=5

), Beta (

3+1=4

), and Gamma (

1+2=3

) exactly matches the target mixture.

Key Concept

Solving systems of linear equations using linear combinations without full variable elimination

Estimated Time:2m 0s

Question 192Question →

If $x$ is a real number satisfying the equation $\sqrt{15 - 2x} = 2x - 3$ , what is the sum of all real values of $x$ that satisfy this equation?

3

\frac{5}{2}

-\frac{1}{2}

\frac{7}{2}

1

Show answer & explanation

Answer: $3$

Answer

The sum of all real values of

x

satisfying the equation is

3

.

Squaring both sides of

\sqrt{15 - 2x} = 2x - 3

gives

15 - 2x = 4x^2 - 12x + 9

, which simplifies to

2x^2 - 5x - 3 = 0

. Factoring yields candidate roots

x = 3

and

x = -1/2

. Substituting

x = 3

into the original equation gives

\sqrt{9} = 3

, which is true. Substituting

x = -1/2

yields

\sqrt{16} = -4

, which is false because principal radicals cannot evaluate to negative values. Therefore,

x = 3

is the sole valid solution, making the sum equal to

3

.

Step-by-Step Solution

1

Isolate the radical and state domain constraints.

The principal square root

\sqrt{15 - 2x}

must be non-negative, requiring

15 - 2x \ge 0 \implies x \le 7.5

, and

2x - 3 \ge 0 \implies x \ge 1.5

.

Radical expressions produce non-negative principal square roots.

2

Square both sides of the equation.

15 - 2x = (2x - 3)^2 \implies 15 - 2x = 4x^2 - 12x + 9

.

Eliminate the radical to form a polynomial equation.

3

Rearrange into standard quadratic form and solve for

x

.

4x^2 - 10x - 6 = 0 \implies 2x^2 - 5x - 3 = 0 \implies (2x + 1)(x - 3) = 0

, giving candidate solutions

x = 3

and

x = -1/2

.

Solve the quadratic equation using factoring.

4

Test candidate solutions in the original equation to eliminate extraneous roots.

For

x = 3

:

\sqrt{15 - 2(3)} = \sqrt{9} = 3

and

2(3) - 3 = 3

(Valid). For

x = -1/2

:

\sqrt{15 - 2(-1/2)} = \sqrt{16} = 4

, but

2(-1/2) - 3 = -4

(Extraneous).

Squaring an equation can introduce extraneous roots.

Key Concept

Solving Radical Equations and Identifying Extraneous Solutions

Question 193Question →

An arithmetic sequence $a_1, a_2, a_3, \dots$ has a first term $a_1 = 4$ and a common difference $d = 3$ . A geometric sequence $b_1, b_2, b_3, \dots$ has a first term $b_1 = 2$ and a common ratio $r = 2$ . If the $k$ -th term of the arithmetic sequence and the $m$ -th term of the geometric sequence are both equal to $64$ , what is the value of $k + m$ ?

Show answer & explanation

Answer: 27

Answer

The value of

k + m

is 27.

For the arithmetic sequence, the

k

-th term is

a_k = a_1 + (k-1)d

. Setting

4 + 3(k-1) = 64

gives

3(k-1) = 60

, so

k-1 = 20

and

k = 21

. For the geometric sequence, the

m

-th term is

b_m = b_1 r^{m-1}

. Setting

2 \cdot 2^{m-1} = 64

gives

2^m = 64

, which implies

m = 6

. Adding the two values gives

k + m = 21 + 6 = 27

.

Step-by-Step Solution

1

Determine the term position

k

in the arithmetic sequence.

k = 21

Using

a_k = a_1 + (k-1)d

, set

4 + 3(k-1) = 64 \implies 3(k-1) = 60 \implies k - 1 = 20 \implies k = 21

.

2

Determine the term position

m

in the geometric sequence.

m = 6

Using

b_m = b_1 \cdot r^{m-1}

, set

2 \cdot 2^{m-1} = 64 \implies 2^m = 64 \implies m = 6

.

3

Compute the sum of the two position indices

k

and

m

.

27

k + m = 21 + 6 = 27

.

Key Concept

Calculating term indices in arithmetic and geometric sequences using general term formulas

Estimated Time:1m 30s

Question 194Question →

A web hosting company charges each enterprise client a one-time fixed setup fee plus a constant monthly maintenance fee per server. A client operating $4$ servers pays a total of $\$1,100$ for the setup fee and the first $6$ months of server maintenance. A client operating $9$ servers pays a total of $\$1,850$ for the setup fee and the first $6$ months of server maintenance. What is the one-time fixed setup fee, in dollars?

Show answer & explanation

Answer: 500

Answer

500

The fixed setup fee is

\$500

. Letting

S

represent the fixed setup fee and

M

represent the 6-month maintenance fee per server, the two given scenarios yield

S + 4M = 1100

and

S + 9M = 1850

. Subtracting the first equation from the second gives

5M = 750

, which simplifies to

M = 150

. Substituting

M = 150

into

S + 4(150) = 1100

gives

S + 600 = 1100

, so

S = 500

.

Step-by-Step Solution

1

Define variables for the unknown fixed cost and per-server cost, and construct the system of linear equations.

Let

S

be the fixed setup fee in dollars and

M

be the 6-month maintenance fee per server in dollars. The equations are

S + 4M = 1100

and

S + 9M = 1850

.

Modeling the situational relationships as a linear system allows isolated solution of each unknown.

2

Subtract the two linear equations to eliminate the fixed fee

S

and solve for

M

.

5M = 750 \implies M = 150

.

Since the coefficient of

S

is 1 in both equations, elimination by subtraction directly isolates

M

.

3

Substitute the value of

M

back into the first equation to solve for

S

.

S + 4(150) = 1100 \implies S + 600 = 1100 \implies S = 500

.

Replacing

M

with 150 yields a linear equation in one variable for the setup fee.

Key Concept

Linear Equations in One and Two Variables

Estimated Time:1m 30s

Question 195Question →

How many integer values of $x$ satisfy the inequality $|x^2 - 5x| \le 6$ ?

Show answer & explanation

Answer: 8

Answer

The correct answer is 8.

To solve

|x^2 - 5x| \le 6

, express it as

-6 \le x^2 - 5x \le 6

. Solving

x^2 - 5x - 6 \le 0

gives

[-1, 6]

, while solving

x^2 - 5x + 6 \ge 0

gives

(-\infty, 2] \cup [3, \infty)

. Taking their intersection yields the solution set

[-1, 2] \cup [3, 6]

. The integers contained in this set are

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

, which gives 8 distinct integer values.

Step-by-Step Solution

1

Rewrite the absolute value inequality

-6 \le x^2 - 5x \le 6

For any real expression

A

and constant

k \ge 0

,

|A| \le k

is equivalent to

-k \le A \le k

.

2

Solve the upper bound condition

x^2 - 5x \le 6

-1 \le x \le 6

Subtract 6 from both sides to obtain

x^2 - 5x - 6 \le 0

. Factoring gives

(x - 6)(x + 1) \le 0

.

3

Solve the lower bound condition

x^2 - 5x \ge -6

x \le 2

or

x \ge 3

Add 6 to both sides to obtain

x^2 - 5x + 6 \ge 0

. Factoring gives

(x - 2)(x - 3) \ge 0

.

4

Combine the solution sets

[-1, 2] \cup [3, 6]

The intersection of

[-1, 6]

with

(-\infty, 2] \cup [3, \infty)

is the set of intervals

[-1, 2]

and

[3, 6]

.

5

Count all integer solutions in the combined set

8 integer values

The integer values in

[-1, 2]

are

-1, 0, 1, 2

(4 integers), and in

[3, 6]

are

3, 4, 5, 6

(4 integers), totaling

4 + 4 = 8

integers.

Key Concept

Solving quadratic absolute value inequalities using compound inequality decomposition.

Question 196Question →

A specialized coffee roastery produces three custom blends—Roast Alpha, Roast Beta, and Roast Gamma—using three varieties of single-origin beans: Grade A, Grade B, and Grade C.

- One batch of Roast Alpha requires 3 kg of Grade A, 1 kg of Grade B, and 2 kg of Grade C beans, and has a total raw material cost of $64. - One batch of Roast Beta requires 1 kg of Grade A, 4 kg of Grade B, and 2 kg of Grade C beans, and has a total raw material cost of$ 64.
- One batch of Roast Gamma requires 2 kg of Grade A, 2 kg of Grade B, and 5 kg of Grade C beans, and has a total raw material cost of $90.

What is the cost, in dollars, of 1 kg of Grade A beans?

Show answer & explanation

Answer: 12

Answer

The cost of 1 kg of Grade A beans is 12 dollars.

Setting up the 3-variable linear system

3x + y + 2z = 64

,

x + 4y + 2z = 64

, and

2x + 2y + 5z = 90

allows us to eliminate

z

by subtracting the second equation from the first, yielding

2x - 3y = 0

, or

x = 1.5y

. Substituting this relationship back into the system leads to

y = 8

and

x = 12

. Thus, 1 kg of Grade A beans costs 12 dollars.

Step-by-Step Solution

1

Formulate linear equations representing the total cost of each coffee blend batch.

Let

x

be the cost per kg of Grade A beans,

y

be the cost per kg of Grade B beans, and

z

be the cost per kg of Grade C beans:

\begin{aligned} (1)\quad 3x + y + 2z &= 64 \\ (2)\quad x + 4y + 2z &= 64 \\ (3)\quad 2x + 2y + 5z &= 90 \end{aligned}

Translating the word problem into a system of 3 linear equations with 3 variables.

2

Eliminate variable

z

by subtracting Equation (2) from Equation (1).

(3x + y + 2z) - (x + 4y + 2z) = 64 - 64 \implies 2x - 3y = 0 \implies x = 1.5y

Since both equations (1) and (2) contain the term

+2z

, subtracting them removes

z

directly and provides a simple relation between

x

and

y

.

3

Substitute

x = 1.5y

into Equation (1) and Equation (3) to obtain a system in terms of

y

and

z

.

From Equation (1):

3(1.5y) + y + 2z = 64 \implies 5.5y + 2z = 64 \implies 11y + 4z = 128 \quad (4)

From Equation (3):

2(1.5y) + 2y + 5z = 90 \implies 5y + 5z = 90 \implies y + z = 18 \implies z = 18 - y

Reducing the system from 3 variables down to 2 variables.

4

Substitute

z = 18 - y

into Equation (4) to solve for

y

, and subsequently calculate

x

.

11y + 4(18 - y) = 128 \implies 7y + 72 = 128 \implies 7y = 56 \implies y = 8

Using

x = 1.5y

:

x = 1.5 \times 8 = 12

Solving the single-variable linear equation for

y

, then substituting back to find the required cost

x

for Grade A beans.

Key Concept

Solving a 3-Variable System of Linear Equations via Variable Elimination

Question 197Question →

In an arithmetic sequence, the sum of the first $4$ terms is $28$ and the sum of the first $8$ terms is $88$ . What is the $10\text{th}$ term of this sequence?

18

20

22

24

26

Show answer & explanation

Answer: $22$

Answer

The 10th term of the sequence is 22.

Using the arithmetic series sum formula

S_n = \frac{n}{2}(2a_1 + (n-1)d)

, the conditions yield two linear equations:

2a_1 + 3d = 14

and

2a_1 + 7d = 22

. Subtracting these equations gives

4d = 8

, so

d = 2

. Substituting

d = 2

back into

2a_1 + 3(2) = 14

yields

a_1 = 4

. The 10th term is then calculated as

a_1 + 9d = 4 + 9(2) = 22

.

Step-by-Step Solution

1

Express the given sums using the arithmetic series sum formula

S_n = \frac{n}{2}(2a_1 + (n-1)d)

.

For

S_4 = 28

:

\frac{4}{2}(2a_1 + 3d) = 28 \implies 2a_1 + 3d = 14

.
For

S_8 = 88

:

\frac{8}{2}(2a_1 + 7d) = 88 \implies 2a_1 + 7d = 22

.

Setting up linear equations in terms of the first term

a_1

and common difference

d

allows us to solve for both sequence parameters.

2

Subtract the first equation from the second equation to solve for

d

.

(2a_1 + 7d) - (2a_1 + 3d) = 22 - 14 \implies 4d = 8 \implies d = 2

.

Eliminating

a_1

isolates the common difference

d

.

3

Substitute

d = 2

back into the first equation to solve for

a_1

.

2a_1 + 3(2) = 14 \implies 2a_1 + 6 = 14 \implies 2a_1 = 8 \implies a_1 = 4

.

Finding

a_1

completes the essential parameters of the sequence.

4

Calculate the 10th term using the formula

a_n = a_1 + (n-1)d

.

a_{10} = 4 + (10 - 1)(2) = 4 + 18 = 22

.

Evaluating the formula at

n = 10

provides the target term.

Key Concept

Arithmetic sequence term and series sum formulas.

Estimated Time:2m 0s

Question 198Question →

An event design company offers three distinct decorative bundles—Bundle X, Bundle Y, and Bundle Z—for corporate gala setups.

• Bundle X contains 3 floral arrangements, 2 LED uplights, and 1 table runner.
• Bundle Y contains 1 floral arrangement, 3 LED uplights, and 2 table runners.
• Bundle Z contains 2 floral arrangements, 1 LED uplight, and 3 table runners.

To decorate a venue, a coordinator orders a combination of these bundles containing a total of 26 floral arrangements, 23 LED uplights, and 23 table runners. If every bundle ordered is used in its entirety, what is the total number of bundles ordered by the coordinator?

10

12

15

18

24

Show answer & explanation

Answer: 12

Answer

The total number of bundles ordered by the coordinator is 12.

Summing all three equations gives

6x + 6y + 6z = 72

. Factoring out

6

yields

6(x + y + z) = 72

, so dividing by

6

gives

x + y + z = 12

. Alternatively, solving the system explicitly gives

x = 5

,

y = 4

, and

z = 3

, whose sum is

5 + 4 + 3 = 12

.

Step-by-Step Solution

1

Define variables and set up the system of linear equations based on item counts.

Let

x

,

y

, and

z

represent the number of Bundle X, Bundle Y, and Bundle Z ordered, respectively.

\begin{aligned} 3x + y + 2z &= 26 \quad \text{(Floral arrangements)} \\ 2x + 3y + z &= 23 \quad \text{(LED uplights)} \\ x + 2y + 3z &= 23 \quad \text{(Table runners)} \end{aligned}

Translate the word problem into a standard system of 3 linear equations in 3 variables.

2

Sum the three equations to find a direct linear combination for

(x + y + z)

.

(3x + 2x + x) + (y + 3y + 2y) + (2z + z + 3z) = 26 + 23 + 23

6x + 6y + 6z = 72

Notice that the sum of coefficients for each variable across all three equations is identical (

3 + 2 + 1 = 6

).

3

Factor out 6 and solve for the total number of bundles

(x + y + z)

.

6(x + y + z) = 72 \implies x + y + z = 12

Dividing both sides by 6 directly gives the required total quantity without needing to solve for individual variables

x

,

y

, and

z

separately.

Key Concept

Linear combinations in systems of equations

Estimated Time:2m 0s

Question 199Question →

If $9^{x+1} - 9^x = 216$ , what is the value of $4^x$ ?

Show answer & explanation

Answer: 8

Answer

The value of

4^x

is 8.

Factoring

9^x

from

9^{x+1} - 9^x

gives

9^x(9 - 1) = 216

, or

8 \cdot 9^x = 216

. Dividing by 8 yields

9^x = 27

. Expressing both sides with base 3 gives

3^{2x} = 3^3

, so

2x = 3

and

x = \frac{3}{2}

. Substituting this value into

4^x

results in

4^{3/2} = (\sqrt{4})^3 = 2^3 = 8

.

Step-by-Step Solution

1

Factor out common exponent terms from the left side of the equation

9^x(9^1 - 1) = 216 \implies 8 \cdot 9^x = 216

Using the exponent property

a^{m+n} = a^m \cdot a^n

, rewrite

9^{x+1}

as

9^x \cdot 9^1

to factor out

9^x

.

2

Isolate the exponential term

9^x = 27

Dividing both sides of

8 \cdot 9^x = 216

by 8 yields

27

.

3

Convert both sides to a common prime base of 3

(3^2)^x = 3^3 \implies 3^{2x} = 3^3 \implies 2x = 3 \implies x = \frac{3}{2}

Since bases are equal, exponents must be equal.

4

Evaluate the target expression

4^x

4^{3/2} = (4^{1/2})^3 = 2^3 = 8

Substitute

x = \frac{3}{2}

into

4^x

and apply the rule

a^{m/n} = (\sqrt[n]{a})^m

.

Key Concept

Solving exponential equations by factoring and equating powers with a common base.

Question 200Question →

If

x

is a real number that satisfies the absolute value equation

|2x - 7| = 3x - 11

what is the value of

x^2 - 3x

?

2.16

4

16

28

270

Show answer & explanation

Answer: $4$

Answer

The value of

x^2 - 3x

is

4

.

Solving the absolute value equation

|2x - 7| = 3x - 11

yields two potential roots:

x = 4

and

x = 3.6

. Checking both in the original equation shows that

x = 3.6

makes the right-hand side negative (

-0.2

), making it an extraneous solution. The only valid solution is

x = 4

. Substituting

x = 4

into

x^2 - 3x

yields

4^2 - 3(4) = 16 - 12 = 4

.

Step-by-Step Solution

1

Set up equations based on the definition of absolute value

Case 1:

2x - 7 = 3x - 11

; Case 2:

2x - 7 = -(3x - 11)

An absolute value equation

|A| = B

splits into

A = B

or

A = -B

, with the requirement that

B \ge 0

.

2

Solve Case 1:

2x - 7 = 3x - 11

3x - 2x = 11 - 7 \implies x = 4

Isolate the variable

x

by algebraic rearrangement.

3

Solve Case 2:

2x - 7 = -3x + 11

5x = 18 \implies x = \frac{18}{5} = 3.6

Isolate

x

for the negative case.

4

Check both potential solutions in the original equation

|2x - 7| = 3x - 11

For

x = 4

:

|2(4) - 7| = |1| = 1

and

3(4) - 11 = 1

(Valid).
For

x = 3.6

:

|2(3.6) - 7| = |0.2| = 0.2

but

3(3.6) - 11 = -0.2

(Extraneous).

The right side

3x - 11

must be non-negative. Since

3(3.6) - 11 = -0.2 < 0

,

x = 3.6

is extraneous.

5

Evaluate the target expression

x^2 - 3x

using the valid root

x = 4

4^2 - 3(4) = 16 - 12 = 4

Substitute the single valid root into the requested expression.

Key Concept

Absolute Value Linear Equations and Extraneous Solution Checking

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