Translating and Solving Algebraic Word Problems Practice Questions — ACT

Question 1Question →

A charity walkathon organizer pledges to donate a base amount of $\$150$ plus $\$2.50$ for every kilometer completed by each participant. On the day of the event, a participant completes a distance that is $3$ kilometers less than twice their training average distance. If the organizer's donation for this participant is $\$212.50$ , what is the participant's training average distance, in kilometers?

Show answer & explanation

Answer: 14

Answer

The participant's training average distance is

14

kilometers.

The correct answer is

14

. To find this value, we write the donation relationship as

150 + 2.50d = 212.50

, where

d

is the distance completed. Solving for

d

gives

d = 25

kilometers. Next, we translate the phrase '3 kilometers less than twice their training average distance' into the expression

2A - 3

, where

A

is the training average. Equating this to the completed distance gives

2A - 3 = 25

. Solving for

A

yields

A = 14

.

Step-by-Step Solution

1

Set up an equation for the total donation as a function of the distance completed,

d

, in kilometers.

150 + 2.50d = 212.50

The total donation is the sum of the flat base donation of

\$150

and the rate of

\$2.50

per kilometer completed.

2

Solve the equation for the completed distance,

d

.

d = 25

Subtracting

150

from both sides of the equation yields

2.50d = 62.50

. Dividing both sides by

2.50

yields

d = 25

.

3

Translate the relationship between the completed distance,

d

, and the training average distance,

A

, into an equation.

2A - 3 = 25

The phrase '3 kilometers less than twice their training average distance' translates mathematically to

2A - 3

. Since the completed distance is

25

kilometers, we set the expression equal to

25

.

4

Solve the equation for the training average distance,

A

.

A = 14

Adding

3

to both sides of the equation yields

2A = 28

. Dividing both sides by

2

yields

A = 14

.

Key Concept

Translating word problems into multi-step linear equations and solving for the unknown variable.

Estimated Time:1m 30s

Question 2Question →

A specialty coffee shop blends three types of coffee beans: Colombian ( $8.00$ dollars per pound), Ethiopian ( $11.00$ dollars per pound), and Sumatran ( $14.00$ dollars per pound). The shop manager wants to create a $50$ -pound blend that costs exactly $11.60$ dollars per pound. Due to supply constraints, the weight of the Ethiopian beans in the blend must be exactly $2$ pounds more than $\frac{1}{5}$ of the combined weight of the Colombian and Sumatran beans. What is the value of $2s - c$ , where $s$ is the amount of Sumatran beans, in pounds, and $c$ is the amount of Colombian beans, in pounds, in the blend?

Show answer & explanation

Answer: 35

Answer

The correct answer is

35

.

The correct answer of

35

is found by translating the given constraints into three linear equations, solving for

e = 10

using substitution of the sum

c+s = 50 - e

, and then solving the resulting two-variable system to find

s = 25

and

c = 15

. Evaluating

2s - c

yields

2(25) - 15 = 35

.

Step-by-Step Solution

1

Define variables and write the system of equations representing the constraints.

Let

c

,

e

, and

s

represent the weight in pounds of Colombian, Ethiopian, and Sumatran beans, respectively. The system is:
1)

c + e + s = 50

(total weight)
2)

8c + 11e + 14s = 50 \times 11.60 = 580

(total cost)
3)

e = \frac{1}{5}(c + s) + 2

(Ethiopian bean weight constraint)

This translates the word problem's conditions into algebraic expressions.

2

Solve for the variable

e

using substitution.

From equation (1), we have

c + s = 50 - e

. Substituting this expression into equation (3) yields:

e = \frac{1}{5}(50 - e) + 2

Multiply both sides by

5

:

5e = 50 - e + 10

6e = 60 \implies e = 10

Grouping

c + s

allows us to solve for

e

directly without dealing with three separate variable eliminations.

3

Substitute

e = 10

back into the first two equations to simplify the system to two variables.

Equation (1) becomes:

c + s = 40 \implies c = 40 - s

Equation (2) becomes:

8c + 11(10) + 14s = 580 \implies 8c + 14s = 470

This reduces the remaining problem to a standard system of two linear equations.

4

Solve for

c

and

s

by substitution.

Substitute

c = 40 - s

into the simplified cost equation:

8(40 - s) + 14s = 470

320 - 8s + 14s = 470

320 + 6s = 470 \implies 6s = 150 \implies s = 25

Then, find

c

:

c = 40 - 25 = 15

This gives the exact weight of both the Sumatran and Colombian beans.

5

Evaluate the expression

2s - c

requested by the prompt.

2s - c = 2(25) - 15 = 50 - 15 = 35

This satisfies the specific algebraic quantity requested in the question.

Key Concept

Translating word problems into systems of linear equations and solving them using algebraic substitution.

Question 3Question →

A rental car company charges a flat fee of $30$ dollars plus $0.20$ dollars per mile driven. If a customer's total rental bill is $56$ dollars, how many miles did they drive?

Show answer & explanation

Answer: 130

Answer

130 miles

Subtracting the flat fee of

30

dollars from the total bill of

56

dollars leaves

26

dollars representing the cost of the miles driven. Dividing this remaining cost of

26

dollars by the rate of

0.20

dollars per mile yields a total of

130

miles driven.

Step-by-Step Solution

1

Set up a linear equation for the total cost

30 + 0.20m = 56

The total cost of

56

dollars is the sum of the flat fee of

30

dollars and the variable cost of

0.20

dollars per mile multiplied by the number of miles

m

.

2

Subtract the flat fee from both sides of the equation

0.20m = 26

Isolating the variable term shows that the total amount spent on the mileage portion of the trip is

26

dollars.

3

Divide by the rate per mile to find the total miles

m = 130

Dividing the mileage portion of the cost by the cost per mile gives the total distance driven.

Key Concept

Translating and Solving Linear Word Problems

Question 4Question →

A local community theater group sells tickets for their upcoming play. A student ticket costs $8$ dollars, which is $4$ dollars less than half the price of an adult ticket. If $a$ represents the cost of an adult ticket in dollars, what is the value of $a$ ?

6

8

12

20

24

Show answer & explanation

Answer: $24$

Answer

The cost of the adult ticket is

24

dollars.

The correct answer is

24

. According to the problem, the student ticket price of

8

dollars is equal to

4

dollars less than half the price of the adult ticket (

a

). This relationship can be expressed algebraically as

8 = \frac{1}{2}a - 4

. Adding

4

to both sides of the equation results in

12 = \frac{1}{2}a

. Multiplying both sides by

2

yields

a = 24

.

Step-by-Step Solution

1

Translate the word problem into a linear equation.

8 = \frac{1}{2}a - 4

The student ticket price (

8

dollars) is equal to

4

dollars subtracted from half the price of the adult ticket (

a

).

2

Isolate the variable term by adding

4

to both sides of the equation.

12 = \frac{1}{2}a

To solve for

a

, we first eliminate the constant subtraction on the variable's side.

3

Solve for

a

by multiplying both sides of the equation by

2

.

a = 24

Multiplying by the reciprocal of

\frac{1}{2}

isolates the variable

a

.

Key Concept

Translating verbal statements into two-step linear equations and solving for the unknown variable.

Question 5Question →

A charity concert sells floor tickets and balcony tickets. The price of a floor ticket is $5$ dollars more than twice the price of a balcony ticket. If the concert organizers sell $80$ balcony tickets and $50$ floor tickets, their total revenue is $R$ dollars. Under a new promotional structure, the price of a balcony ticket is discounted by $25\%$ , the price of a floor ticket is increased by $10\%$ , and the organizers sell $120$ balcony tickets. In terms of the original price of a balcony ticket, $b$ , which of the following expressions represents the number of floor tickets they must sell under the new promotional structure to achieve the same total revenue, $R$ ?

\frac{1500b + 2500}{22b + 55}

\frac{900b + 2500}{22b - 55}

\frac{900b + 2500}{22b + 55}

\frac{2500 - 1100b}{55 - 22b}

\frac{900b + 5000}{22b + 110}

Show answer & explanation

Answer: $\frac{900b + 2500}{22b + 55}$

Answer

\frac{900b + 2500}{22b + 55}

The correct expression is derived by first writing the original revenue in terms of the balcony ticket price

b

, which yields

R = 180b + 250

. Under the new pricing, the balcony ticket costs

0.75b

and the floor ticket costs

1.1(2b + 5) = 2.2b + 5.5

. Selling

120

balcony tickets generates

90b

in revenue. Setting the new total revenue equal to

R

gives

90b + F(2.2b + 5.5) = 180b + 250

. Solving for the number of floor tickets,

F

, results in

F = \frac{90b + 250}{2.2b + 5.5}

. Multiplying the numerator and denominator by

10

to clear the decimals yields the correct expression.

Step-by-Step Solution

1

Express the original ticket prices and revenue in terms of

b

.

Balcony ticket price =

b

. Floor ticket price =

2b + 5

. Original revenue

R = 80b + 50(2b + 5) = 180b + 250

.

To set up the baseline total revenue equation using the algebraic descriptions.

2

Determine the new pricing for both balcony and floor tickets.

New balcony ticket price =

0.75b

. New floor ticket price =

1.1(2b + 5) = 2.2b + 5.5

.

To apply the

25\%

discount and

10\%

increase to the original ticket prices.

3

Set up the equation equating the new promotional revenue to the original revenue

R

.

New Revenue =

120(0.75b) + F(2.2b + 5.5) = 180b + 250

, where

F

is the number of floor tickets.

To represent the condition that the total revenue remains the same under the new structure.

4

Simplify the equation and isolate the variable

F

.

90b + F(2.2b + 5.5) = 180b + 250 \implies F(2.2b + 5.5) = 90b + 250 \implies F = \frac{90b + 250}{2.2b + 5.5}

.

To solve for the number of floor tickets algebraically.

5

Clear decimals from the rational expression.

F = \frac{10(90b + 250)}{10(2.2b + 5.5)} = \frac{900b + 2500}{22b + 55}

.

To match the standard fraction format of the options by multiplying the numerator and denominator by

10

.

Key Concept

Translating verbal descriptions into multi-step algebraic systems and isolating a target variable from rational equations.

Estimated Time:3m 0s

Question 6Question →

A chemist is preparing $120\text{ milliliters}$ of a chemical mixture with an overall acid concentration of $27.5\%$ by volume. To do this, she mixes Solution $X$ ( $10\%$ acid by volume), Solution $Y$ ( $25\%$ acid by volume), and Solution $Z$ ( $40\%$ acid by volume). The chemist decides that the volume of Solution $Y$ must be exactly $20\text{ milliliters}$ less than twice the volume of Solution $X$ used in the mixture. What is the positive difference, in milliliters, between the volume of Solution $Z$ and the volume of Solution $X$ in the final mixture?

Show answer & explanation

Answer: 20

Answer

The positive difference between the volume of Solution Z and Solution X is 20 milliliters.

The correct answer is 20, because solving the system of equations yields a volume of 30 milliliters for Solution X and 50 milliliters for Solution Z. The positive difference between these two volumes is 20 milliliters.

Step-by-Step Solution

1

Define variables for the volume of each solution used in the mixture.

Let

x

be the volume of Solution

X

,

y

be the volume of Solution

Y

, and

z

be the volume of Solution

Z

(all in milliliters).

Defining variables allows for translating the word problem's conditions into algebraic equations.

2

Translate the given information into a system of three linear equations.

Equation 1 (Total Volume):

x + y + z = 120

. Equation 2 (Total Acid):

0.10x + 0.25y + 0.40z = 33

(since

27.5\%

of

120\text{ ml}

is

33\text{ ml}

). Equation 3 (Volume Relation):

y = 2x - 20

.

These equations model the constraints and quantities described in the problem.

3

Express

z

in terms of

x

by substituting the expression for

y

into the total volume equation.

x + (2x - 20) + z = 120 \implies 3x - 20 + z = 120 \implies z = 140 - 3x

.

This substitution reduces the system to two variables (

x

and

z

), making it easier to solve.

4

Substitute the expressions for

y

and

z

in terms of

x

into the acid equation and solve for

x

.

0.10x + 0.25(2x - 20) + 0.40(140 - 3x) = 33 \implies 0.10x + 0.50x - 5 + 56 - 1.20x = 33 \implies -0.60x + 51 = 33 \implies -0.60x = -18 \implies x = 30

.

This isolates the single variable

x

so that its value can be calculated.

5

Determine the volume of Solution

Z

and calculate the final positive difference.

z = 140 - 3(30) = 50

. The positive difference between the volume of Solution

Z

and Solution

X

is

|50 - 30| = 20

.

This directly answers the question's requirement for the difference between the two solution volumes.

Key Concept

Translating word problems into a system of three linear equations and solving them using substitution.

Question 7Question →

A manufacturer of custom planners determines that the setup cost for a production run is $100$ dollars, and each planner costs $6$ dollars to produce. The planners sell for $10$ dollars each, except for the first $10$ planners sold, which are discounted by $2$ dollars each. If the manufacturer wants to achieve a net profit of exactly $300$ dollars for a single production run, how many planners must they produce and sell?

Show answer & explanation

Answer: 105

Answer

The manufacturer must produce and sell

105

planners to achieve a net profit of

300

dollars.

The correct answer is found by setting up the profit equation:

\text{Profit} = \text{Total Revenue} - \text{Total Cost}

. The cost is

100 + 6x

and the revenue is

10(8) + 10(x-10) = 10x - 20

. Equating their difference to

300

yields

(10x - 20) - (100 + 6x) = 300

, which simplifies to

4x - 120 = 300

. Solving for

x

results in

105

.

Step-by-Step Solution

1

Define the variable for the number of planners.

Let

x

be the number of planners produced and sold, where

x \geq 10

.

Establishing the variable is necessary to set up algebraic expressions for cost and revenue.

2

Write the total cost expression.

\text{Total Cost} = 100 + 6x

The cost combines the fixed setup fee of

100

dollars and the variable cost of

6

dollars per planner.

3

Write the total revenue expression.

\text{Total Revenue} = 10(8) + 10(x - 10) = 10x - 20

The first

10

planners sell for

8

dollars each, and the remaining

x - 10

planners sell for the regular price of

10

dollars each.

4

Set up the profit equation and solve for

x

.

(10x - 20) - (100 + 6x) = 300 \implies 4x - 120 = 300 \implies 4x = 420 \implies x = 105

Profit is the difference between total revenue and total cost, which must equal the target profit of

300

dollars.

Key Concept

Translating real-world pricing and cost constraints into a single-variable linear equation.

Question 8Question →

A gardener starts with $15$ flowers already planted in a garden. She plans to plant additional flowers at a constant rate of $8$ flowers per hour. How many hours will it take the gardener to have a total of $79$ flowers planted?

Show answer & explanation

Answer: 8

Answer

It will take the gardener

8

hours to have a total of

79

flowers planted.

The total number of flowers planted can be modeled by the linear equation

15 + 8h = 79

, where

h

is the number of hours. Subtracting

15

from both sides of the equation gives

8h = 64

. Dividing both sides by

8

reveals that

h = 8

hours.

Step-by-Step Solution

1

Set up the linear equation representing the total flowers planted over time.

15 + 8h = 79

, where

h

is the number of hours.

The gardener begins with

15

flowers and adds

8

flowers for each hour

h

, with the final goal of

79

total flowers.

2

Isolate the variable term by subtracting the initial number of flowers from the total.

8h = 64

Subtracting

15

from both sides of the equation isolates the term containing the variable

h

.

3

Divide by the rate to solve for the number of hours.

h = 8

Dividing the remaining flowers to be planted (

64

) by the planting rate (

8

flowers per hour) yields the total number of hours required.

Key Concept

Translating a real-world scenario into a linear equation and solving for the unknown variable.

Question 9Question →

A company produces solar-powered chargers. The daily production cost, in dollars, is a linear function of the number of chargers produced. The setup cost is the cost when $0$ chargers are produced. If the company produces $15$ chargers, the total daily cost is $400$ dollars. If the company produces $25$ chargers, the total daily cost is $580$ dollars. The company updates its production process, which reduces the cost per charger by $20\%$ but increases the setup cost by $50$ dollars. Under the updated process, what is the total daily cost, in dollars, to produce $30$ chargers?

562

dollars

586

dollars

612

dollars

714

dollars

828

dollars

Show answer & explanation

Answer: $612$ dollars

Answer

The correct daily cost under the updated process is

612

dollars.

The correct answer is

612

dollars. First, the relationship between unit cost and setup cost is modeled as a linear equation. Subtracting the cost of producing

15

chargers (

400

dollars) from the cost of producing

25

chargers (

580

dollars) gives the cost of producing the

10

additional units, which is

180

dollars. This determines the unit cost is

18

dollars per charger. Substituting this value back shows the original setup cost is

130

dollars. The updated rate decreases the unit cost to

14.40

dollars (

80\%

of

18

) and increases the setup cost to

180

dollars (

130 + 50

). For

30

chargers, the total daily cost is

14.40(30) + 180 = 612

dollars.

Step-by-Step Solution

1

Set up a system of linear equations using the cost function

C(c) = mc + S

, where

m

is the cost per charger,

S

is the setup cost, and

c

is the number of chargers.

Equation 1:

15m + S = 400

Equation 2:

25m + S = 580

This establishes the relationship between production volume and total cost under the initial process.

2

Solve the system of equations for the unit cost

m

and the setup cost

S

.

Subtracting Equation 1 from Equation 2 yields

10m = 180 \implies m = 18

.
Substituting

m = 18

into Equation 1 yields

15(18) + S = 400 \implies 270 + S = 400 \implies S = 130

.

This determines the original pricing parameters.

3

Apply the updates to the cost parameters.

New unit cost:

18 \times (1 - 0.20) = 14.40

dollars.
New setup cost:

130 + 50 = 180

dollars.

This accounts for the

20\%

decrease in the variable cost and the

50

dollar increase in the fixed setup cost.

4

Evaluate the new linear cost function for

30

chargers.

C_{\text{new}}(30) = 14.40(30) + 180 = 432 + 180 = 612

dollars.

This calculates the total daily cost under the updated process.

Key Concept

Translating and Solving Algebraic Word Problems

Alternative Method

Instead of solving for the setup cost first, note that under the original process, the cost of

30

chargers would be the cost of

15

chargers plus the cost of

15

more units:

400 + 15(18) = 670

dollars. The updated process reduces the rate of each of the

30

units by

20\%

of

18

dollars (saving

3.60 \times 30 = 108

dollars) and increases the setup cost by

50

dollars. Thus, the new cost is

670 - 108 + 50 = 612

dollars.

Estimated Time:2m 0s

Question 10Question →

A dog-walking service charges a flat fee of $10$ dollars per visit plus $20$ dollars for each dog walked. If a client was charged a total of $90$ dollars for a single visit, how many dogs were walked during that visit?

9

5

4

7

60

Show answer & explanation

Answer: 4

Answer

4

The correct answer is

4

. The relation between the total cost and the number of dogs walked is

10 + 20d = 90

. Subtracting

10

from both sides yields

20d = 80

, and dividing by

20

yields

d = 4

.

Step-by-Step Solution

1

Set up the equation based on the scenario details.

Let

d

be the number of dogs walked. The total cost is represented by the equation

10 + 20d = 90

.

The total cost of

90

dollars consists of a flat fee of

10

dollars plus

20

dollars per dog walked.

2

Subtract the flat fee from both sides of the equation.

20d = 80

Subtracting

10

from both sides isolates the cost of walking the dogs.

3

Divide both sides by the per-dog rate to solve for

d

.

d = 4

Dividing by

20

yields the number of dogs walked.

Key Concept

Translating verbal descriptions of multi-step scenarios into linear equations and solving them.

Question 11Question →

Jordan starts a walk with $2{}500$ steps already recorded on a fitness tracker. Jordan then walks at a constant rate of $120$ steps per minute. If the fitness tracker shows a total of $7{}900$ steps at the end of the walk, for how many minutes did Jordan walk?

Show answer & explanation

Answer: 45

Answer

Jordan walked for

45

minutes.

The correct answer is

45

. The scenario describes a linear relationship with a constant rate of

120

steps per minute and a starting baseline of

2{}500

steps. Let

m

represent the number of minutes Jordan walked. The equation is

2{}500 + 120m = 7{}900

. Subtracting

2{}500

from both sides gives

120m = 5{}400

, and dividing by

120

yields

m = 45

.

Step-by-Step Solution

1

Set up the algebraic equation based on the word problem details.

2{}500 + 120m = 7{}900

The total steps are the sum of the starting steps (

2{}500

) and the product of the rate (

120

steps/min) and time (

m

minutes).

2

Isolate the variable term by subtracting

2{}500

from both sides of the equation.

120m = 5{}400

Subtracting the initial steps gives the steps accumulated solely during the walk.

3

Solve for

m

by dividing both sides of the equation by

120

.

m = 45

Dividing the total steps walked by the rate per minute yields the duration of the walk in minutes.

Key Concept

Translating verbal descriptions of constant rates and starting values into linear equations

Alternative Method

Solve arithmetically by subtracting the baseline steps from the final count (

7{}900 - 2{}500 = 5{}400

steps) and dividing the remaining steps by the walking rate (

5{}400 \div 120 = 45

minutes).

Estimated Time:45s

Question 12Question →

A digital marketing firm runs advertisements on two platforms: SocialMedia and SearchEngine. The cost to run an advertisement on SocialMedia is $15$ dollars per day, and the cost to run an advertisement on SearchEngine is $25$ dollars per day. Last month, the firm ran advertisements on both platforms for a combined total of $60$ days. The total amount spent on SocialMedia advertisements was $350$ dollars more than half the total amount spent on SearchEngine advertisements. For how many days last month did the firm run advertisements on SocialMedia?

Show answer & explanation

Answer: 40

Answer

The firm ran advertisements on SocialMedia for 40 days.

Setting up the system of equations based on the problem description gives

S + E = 60

and

15S = 0.5(25E) + 350

. Substituting the first equation into the second yields

15S = 12.5(60 - S) + 350

. Solving this linear equation results in

S = 40

days.

Step-by-Step Solution

1

Define variables for the unknown quantities.

Let

S

be the number of days the firm ran advertisements on SocialMedia, and

E

be the number of days they ran advertisements on SearchEngine.

Establishing variables is the first step in translating a word problem into algebraic equations.

2

Express the relationship for the total number of days.

S + E = 60

, which simplifies to

E = 60 - S

.

This allows us to express one variable in terms of the other, making it easier to solve the system by substitution.

3

Translate the cost relationship statement into an algebraic equation.

15S = 12.5E + 350

The cost of running advertisements on SocialMedia is

15S

. The cost on SearchEngine is

25E

. Half of the SearchEngine cost is

12.5E

. Adding

350

to half of the SearchEngine cost gives the SocialMedia cost.

4

Substitute the expression for

E

into the cost equation and solve for

S

.

15S = 12.5(60 - S) + 350 \Rightarrow 15S = 750 - 12.5S + 350 \Rightarrow 27.5S = 1100 \Rightarrow S = 40

.

Solving the resulting linear equation yields the number of days spent on SocialMedia ads.

Key Concept

Translating and Solving Algebraic Word Problems

Estimated Time:2m 0s

Question 13Question →

A commercial bakery has two bread-kneading machines, Machine X and Machine Y. Machine X can knead dough at a constant rate of $p$ pounds per hour. Machine Y's kneading rate is $14$ pounds per hour more than half the kneading rate of Machine X. On a busy morning, Machine X starts kneading and operates for exactly $5$ hours. Machine Y starts operating $1$ hour after Machine X starts and operates for the next $4$ hours. Together, the two machines knead a total of $336$ pounds of dough. What is the total number of pounds of dough kneaded by Machine Y?

34

116

136

148

200

Show answer & explanation

Answer: 136

Answer

The total number of pounds of dough kneaded by Machine Y is 136.

To find the total dough kneaded by Machine Y, we first define the rates of both machines. Let Machine X's rate be

p

pounds per hour. Machine Y's rate is

14

more than half of Machine X's rate, which is written as

0.5p + 14

pounds per hour. Machine X runs for

5

hours, producing

5p

pounds of dough. Machine Y runs for

4

hours, producing

4(0.5p + 14) = 2p + 56

pounds of dough. Setting their sum equal to the total of

336

gives the equation

7p + 56 = 336

, which simplifies to

7p = 280

and yields

p = 40

. Machine Y's total work is then

4(0.5(40) + 14) = 136

pounds.

Step-by-Step Solution

1

Define the variable for Machine X's rate and write the expression for Machine Y's rate based on the text.

Machine X's rate is

p

pounds per hour. Machine Y's rate is

0.5p + 14

pounds per hour.

We need to express both rates in terms of a single variable to set up the equation.

2

Determine the operating duration for each machine.

Machine X operates for

5

hours. Machine Y operates for

4

hours.

Machine X starts first and runs for

5

hours. Machine Y starts

1

hour later and operates for the remaining

4

hours.

3

Write the equation for the total pounds of dough kneaded by both machines combined.

5p + 4(0.5p + 14) = 336

Total work is the sum of the work done by Machine X (rate times time) and Machine Y (rate times time).

4

Solve the equation for the variable

p

.

5p + 2p + 56 = 336 \implies 7p + 56 = 336 \implies 7p = 280 \implies p = 40

.

Distribute the

4

and combine like terms to isolate

p

.

5

Calculate the total work done specifically by Machine Y.

Machine Y's total work =

4(0.5(40) + 14) = 4(20 + 14) = 4(34) = 136

pounds.

The question asks for the total pounds of dough kneaded by Machine Y, which is its rate multiplied by its operating time.

Key Concept

Translating verbal descriptions of rates and times into algebraic equations and solving them.

Alternative Method

Instead of solving for

p

first, we can write the equation directly in terms of Machine Y's work. Let

y

be the work done by Machine Y. Since Machine Y worked for

4

hours, its rate is

\frac{y}{4}

. This rate is

14

more than half of Machine X's rate, so

\frac{y}{4} = 0.5R_X + 14 \implies R_X = 2(\frac{y}{4} - 14) = \frac{y}{2} - 28

. Machine X's work is

5 R_X = 5(\frac{y}{2} - 28) = 2.5y - 140

. Since the total work is

336

, we have

(2.5y - 140) + y = 336 \implies 3.5y = 476 \implies y = 136

.

Estimated Time:3m 0s

Question 14Question →

Two courier drones, Drone A and Drone B, fly in opposite directions along a straight path from the same distribution center. Drone A departs at 10:00 AM and flies at a constant speed of $40\text{ miles per hour}$ . Drone B departs from the same location at 10:30 AM and flies in the opposite direction at a constant speed of $60\text{ miles per hour}$ . At what time will the two drones be exactly $220\text{ miles}$ apart?

12:00 PM

12:12 PM

12:30 PM

12:42 PM

12:54 PM

Show answer & explanation

Answer: 12:30 PM

Answer

12:30 PM

The correct answer is 12:30 PM. To find this, define

t

as the travel time of Drone B in hours. Since Drone A departs 30 minutes (

0.5

hours) earlier, its travel time is

t + 0.5

hours. Using the relationship that the sum of the distances traveled in opposite directions equals the total distance, we set up the equation

40(t + 0.5) + 60t = 220

. Solving this yields

100t + 20 = 220

, which simplifies to

100t = 200

and

t = 2\text{ hours}

. Adding 2 hours to Drone B's departure time of 10:30 AM results in 12:30 PM.

Step-by-Step Solution

1

Define variables for the travel times of both drones relative to their departure times.

Let

t

be the number of hours Drone B travels after departing at 10:30 AM. Since Drone A departs 30 minutes (

0.5

hours) earlier at 10:00 AM, Drone A's travel time is

t + 0.5

hours.

Establishing a single variable for time allows us to write a single-variable linear equation for the total distance covered.

2

Express the distance traveled by each drone using the formula

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

and sum them to equal the total separation distance.

The distance traveled by Drone A is

40(t + 0.5)

miles, and the distance traveled by Drone B is

60t

miles. The equation is:

40(t + 0.5) + 60t = 220

.

Since the drones fly in opposite directions from the same starting point, the total distance between them is the sum of their individual distances.

3

Solve the equation for

t

.

40t + 20 + 60t = 220 \implies 100t + 20 = 220 \implies 100t = 200 \implies t = 2\text{ hours}

.

Solving for

t

gives the number of hours Drone B travels before they are

220\text{ miles}

apart.

4

Determine the final time by adding the calculated travel time to the corresponding departure time.

Drone B departs at 10:30 AM and travels for

2\text{ hours}

.

10:30\text{ AM} + 2\text{ hours} = 12:30\text{ PM}

. (Alternatively, Drone A departs at 10:00 AM and travels for

2 + 0.5 = 2.5\text{ hours}

, which also yields 12:30 PM).

Adding the elapsed time to the initial departure time gives the clock time when the condition is met.

Key Concept

Translating relative motion and time-offset word problems into linear equations

Question 15Question →

A food truck charges a flat fee of $6 for a custom lunch box container, plus$ 8 for each scoop of specialty salad added to the box. If a customer paid a total of $38 for one custom lunch box filled with specialty salad, how many scoops of specialty salad did they receive?

14

32

4

5

8

Show answer & explanation

Answer: 4

Answer

4 scoops of specialty salad

The correct answer is 4. The total cost is represented by the equation

6 + 8s = 38

, where

s

represents the number of scoops. Subtracting the flat container fee of

6 from the total of

38 leaves

32. Dividing

32 by the per-scoop cost of $8 yields 4 scoops.

Step-by-Step Solution

1

Set up the linear equation representing the total cost.

6 + 8s = 38

, where

s

is the number of scoops of specialty salad.

The total cost consists of a one-time flat fee of

6 plus

8 per scoop of salad.

2

Subtract the flat fee from both sides of the equation.

8s = 32

To isolate the variable term, we subtract the constant cost of the container from the total cost.

3

Divide both sides of the equation by the cost per scoop.

s = 4

Dividing the remaining cost by the price per scoop yields the number of scoops purchased.

Key Concept

Translating a verbal description of a linear cost scenario into a one-variable linear equation and solving it.

Question 16Question →

A chemist wants to create $100$ milliliters of a $42.5\%$ acid solution by mixing three different acid solutions: a $10\%$ acid solution, a $20\%$ acid solution, and an $80\%$ acid solution. She decides that the volume of the $20\%$ acid solution used must be exactly $3$ times the volume of the $10\%$ acid solution used. What is the difference, in milliliters, between the volume of the $80\%$ acid solution and the volume of the $10\%$ acid solution used in the final mixture?

Show answer & explanation

Answer: 25

Answer

The difference between the volume of the

80\%

acid solution and the volume of the

10\%

acid solution is

25

milliliters.

Solving the system of equations yields that

15

milliliters of the

10\%

solution,

45

milliliters of the

20\%

solution, and

40

milliliters of the

80\%

solution are needed. The difference between the volume of the

80\%

solution and the

10\%

solution is

40 - 15 = 25

milliliters.

Step-by-Step Solution

1

Define variables for the volume of each acid solution.

Let

x

be the volume of the

10\%

solution,

y

be the volume of the

20\%

solution, and

z

be the volume of the

80\%

solution.

This establishes algebraic representations for the unknowns.

2

Set up a system of linear equations based on the relationships given in the problem statement.

x + y + z = 100

(total volume)

y = 3x

(relationship between the

20\%

and

10\%

solutions)

0.10x + 0.20y + 0.80z = 42.5

(total acid content)

Translating word problems to mathematical equations allows us to solve for the variables systematically.

3

Reduce the system to a single equation in terms of

x

by substituting

y = 3x

and expressing

z

in terms of

x

.

4x + z = 100 \implies z = 100 - 4x

Substitute both into the acid equation:

0.10x + 0.20(3x) + 0.80(100 - 4x) = 42.5

Substitution simplifies the system of equations to a single linear equation with one variable.

4

Solve the simplified linear equation for

x

.

0.70x + 80 - 3.20x = 42.5 \implies -2.50x = -37.5 \implies x = 15

This determines the volume of the

10\%

acid solution.

5

Calculate the volume of the

80\%

solution,

z

.

z = 100 - 4(15) = 40

This determines the volume of the

80\%

acid solution.

6

Find the difference between

z

and

x

.

z - x = 40 - 15 = 25

The question asks for the difference between these two volumes.

Key Concept

Translating and solving systems of linear equations from verbal descriptions (mixture problems).

Estimated Time:2m 30s

Question 17Question →

A craft shop sells handmade candles. The price of a large candle is $3$ dollars more than twice the price of a small candle. If a large candle costs $15$ dollars, what is the price, in dollars, of a small candle?

Show answer & explanation

Answer: 6

Answer

The price of a small candle is

6

dollars.

By letting

s

represent the price of a small candle, the price of a large candle is

2s + 3

. Since the large candle costs

15

dollars, we write the equation

2s + 3 = 15

. Subtracting

3

from both sides gives

2s = 12

, and dividing by

2

yields

s = 6

. Therefore, the price of a small candle is

6

dollars.

Step-by-Step Solution

1

Define the variable and translate the verbal description into an algebraic expression.

Let

s

be the price of a small candle. The price of a large candle is expressed as

2s + 3

.

Translating 'twice the price of a small candle' to

2s

and '3 more than' to

+ 3

allows us to represent the large candle's cost algebraically.

2

Formulate an equation by setting the expression equal to the known cost of the large candle.

2s + 3 = 15

The problem states that the large candle costs

15

dollars.

3

Solve the equation for the variable

s

.

s = 6

Subtracting

3

from both sides gives

2s = 12

. Dividing both sides by

2

isolates

s

, resulting in

6

.

Key Concept

Translating verbal statements into linear equations and solving for a single variable.

Alternative Method

We can solve the problem by working backward from the price of the large candle. Since the large candle (

15

dollars) is

3

dollars more than twice the small candle's price, we subtract

3

dollars to find twice the price of the small candle:

15 - 3 = 12

dollars. Then, since

12

dollars is twice the price of the small candle, we divide by

2

to find the price of a single small candle:

12 \div 2 = 6

dollars.

Estimated Time:45s

Question 18Question →

At a local sports club, the initiation fee is $15$ dollars less than three times the monthly membership fee, $m$ . If the initiation fee is $75$ dollars, what is the monthly membership fee, in dollars?

20

25

30

40

90

Show answer & explanation

Answer: 30

Answer

30

The statement '15 less than three times the monthly membership fee,

m

' translates to the algebraic expression

3m - 15

. Since the initiation fee is

75

dollars, we can set up the equation

3m - 15 = 75

. Adding

15

to both sides yields

3m = 90

. Dividing both sides by

3

gives

m = 30

. Therefore, the monthly membership fee is

30

dollars.

Step-by-Step Solution

1

Translate the verbal description into an algebraic equation.

The expression '15 less than three times the monthly membership fee,

m

' is written as

3m - 15

. Setting this equal to the initiation fee of

75

gives the equation

3m - 15 = 75

.

To solve a word problem, you must first convert the written relationship into a mathematical statement.

2

Isolate the variable term by adding

15

to both sides of the equation.

3m = 90

Performing the inverse operation of subtraction (addition) simplifies the equation to isolate the term with the variable.

3

Solve for

m

by dividing both sides of the equation by

3

.

m = 30

Performing the inverse operation of multiplication (division) isolates the variable

m

completely.

Key Concept

Translating and Solving Algebraic Word Problems

Estimated Time:45s

Question 19Question →

An online retailer sells two types of subscription boxes: a Basic Box and a Premium Box. Last month, the retailer sold a total of $250$ boxes. The number of Basic Boxes sold was $10$ more than $3$ times the number of Premium Boxes sold. If the retailer made a total profit of $4,350$ dollars, and the profit from each Premium Box is $5$ dollars less than twice the profit from each Basic Box, what is the profit, in dollars, for a single Premium Box?

15

35

31

25

28

Show answer & explanation

Answer: 25

Answer

The profit for a single Premium Box is

25

dollars.

Solving the system of equations for the box quantities gives

190

Basic Boxes and

60

Premium Boxes. Setting up the profit relation

p = 2b - 5

and the total profit equation

190b + 60p = 4350

, we substitute

b = \frac{p+5}{2}

to get

155p = 3875

, which simplifies to

p = 25

dollars.

Step-by-Step Solution

1

Define variables for the quantities of boxes sold and set up a system of equations.

Let

B

be the number of Basic Boxes sold and

P

be the number of Premium Boxes sold. The given relationships are:

B + P = 250

and

B = 3P + 10

.

This translates the word problem statements about the number of boxes into solvable linear equations.

2

Solve for the quantities of each box sold.

Substitute

B = 3P + 10

into the first equation:

(3P + 10) + P = 250 \implies 4P + 10 = 250 \implies 4P = 240 \implies P = 60

. Then,

B = 3(60) + 10 = 190

.

Finding the exact number of each type of box sold is required to formulate the profit equation.

3

Define variables for the profit of each box and set up the profit equations.

Let

b

be the profit of a Basic Box and

p

be the profit of a Premium Box. The total profit equation is

190b + 60p = 4350

. The relationship between the profits is

p = 2b - 5

, which can be rearranged to

b = \frac{p+5}{2}

.

This sets up the system of equations representing the profit values.

4

Substitute and solve for the profit of a Premium Box (

p

).

Substitute

b = \frac{p+5}{2}

into the profit equation:

190\left(\frac{p+5}{2}\right) + 60p = 4350 \implies 95(p+5) + 60p = 4350 \implies 95p + 475 + 60p = 4350 \implies 155p = 3875 \implies p = 25

.

Solving this single-variable equation gives the final required value for the profit of a single Premium Box.

Key Concept

Translating and Solving Multi-Step Algebraic Word Problems

Question 20Question →

An online store sells digital songs for $\$1.20$ each. A customer uses a discount code to get $\$3.00$ off the total purchase. If the customer's total cost after using the discount code is $\$15.00$ , how many digital songs did the customer purchase?

Show answer & explanation

Answer: 15

Answer

The customer purchased 15 songs.

The word problem translates directly to the linear equation

1.20s - 3.00 = 15.00

, where

s

is the number of songs. Adding

3.00

to both sides of the equation yields

1.20s = 18.00

. Dividing both sides by

1.20

yields

s = 15

.

Step-by-Step Solution

1

Define the variable and set up the equation.

Let

s

be the number of digital songs purchased. The cost of

s

songs is

1.20s

. Subtracting the discount of

\$3.00

gives the equation:

1.20s - 3.00 = 15.00

.

We must represent the cost of the songs and the discount algebraically to equal the final payment.

2

Isolate the variable term.

1.20s = 18.00

Adding

3.00

to both sides of the equation simplifies the equation by canceling the subtraction of

3.00

.

3

Solve for the variable.

s = 15

Dividing both sides of the equation by

1.20

isolates

s

to find the total number of songs.

Key Concept

Translating real-world scenarios with linear relationships into algebraic equations and solving them.

Translating and Solving Algebraic Word Problems

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

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Common Mistakes

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Common Mistakes

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