Basic Algebraic Expressions and One-Step Equations Practice Questions — ACT

Question 1Question →

A bookstore has a sale where the price of any paperback book is reduced by $4$ dollars. If the sale price of a paperback book is $12$ dollars, and its original price is $p$ dollars, which of the following equations can be used to find $p$ ?

p - 4 = 12

p + 4 = 12

4 - p = 12

4p = 12

\frac{p}{4} = 12

Show answer & explanation

Answer: $p - 4 = 12$

Answer

The equation

p - 4 = 12

The original price of the book is

p

dollars. Since the price is reduced by

4

dollars, we subtract

4

from

p

, which is written as

p - 4

. Since this resulting sale price is equal to

12

dollars, the correct equation representing the relationship is

p - 4 = 12

.

Step-by-Step Solution

1

Identify the variable and the mathematical values given in the scenario.

The original price is represented by the variable

p

. The price reduction is

4

dollars, and the final sale price is

12

dollars.

This sets up the translation of the word problem into algebraic terms.

2

Translate the phrase 'reduced by

4

dollars' into an algebraic expression.

A reduction of

4

dollars from the original price

p

is written as

p - 4

.

A reduction or discount represents subtraction from the starting amount.

3

Equate the expression representing the sale price to the given sale price value.

The equation is

p - 4 = 12

.

Since the sale price is given as

12

dollars, the expression for the reduced price must equal

12

.

Key Concept

Translating real-world discount scenarios into one-step linear equations.

Estimated Time:45s

Question 2Question →

A runner completed a race in $45$ minutes. If this time is $12$ minutes faster than their previous race time, $t$ , which of the following equations can be used to find the previous race time?

t - 12 = 45

t + 12 = 45

12 - t = 45

12t = 45

\frac{t}{12} = 45

Show answer & explanation

Answer: $t - 12 = 45$

Answer

The equation

t - 12 = 45

correctly models the relationship between the previous and current race times.

The equation

t - 12 = 45

is correct because 'faster than' in running means a shorter duration of time. Thus, the previous time

t

must be

12

minutes greater than

45

minutes, which is algebraically expressed as

t - 12 = 45

(or

t = 45 + 12

).

Step-by-Step Solution

1

Identify the relationship between the current time and the previous time

t

.

The current time is

45

minutes, which is

12

minutes faster than

t

.

This establishes that the previous time

t

is larger than the current time of

45

minutes.

2

Set up the algebraic equation based on the time difference.

t - 12 = 45

Subtracting the difference of

12

minutes from the slower previous time

t

yields the faster current time of

45

minutes.

Key Concept

Translating verbal descriptions of differences into one-step subtraction equations

Estimated Time:45s

Question 3Question →

An art teacher divides a container of clay equally among $6$ students. If each student receives $15$ ounces of clay, and $c$ represents the total number of ounces of clay in the container, which of the following equations can be used to find $c$ ?

6c = 15

c - 6 = 15

\frac{c}{6} = 15

c + 6 = 15

\frac{6}{c} = 15

Show answer & explanation

Answer: $\frac{c}{6} = 15$

Answer

\frac{c}{6} = 15

The equation representing the total amount of clay divided by the number of students equals the amount of clay per student is correct. Since the teacher divides the clay

c

equally among

6

students, the expression

\frac{c}{6}

represents the amount each student gets, which is given as

15

ounces. Therefore, the correct equation is

\frac{c}{6} = 15

.

Step-by-Step Solution

1

Identify the relationship between the total amount of clay, the number of students, and the amount per student.

The total amount of clay,

c

, divided by the number of students,

6

, must equal the amount of clay each student receives.

Dividing a total quantity equally among a set number of groups is represented by division.

2

Write the relationship as an algebraic equation.

\frac{c}{6} = 15

Substitute the given values and variables into the division relationship.

Key Concept

Translating word problems into one-step algebraic division equations

Estimated Time:45s

Question 4Question →

A car travels at a constant speed of $r$ miles per hour. If the car travels $240$ miles in $4$ hours, which of the following equations can be used to determine the value of $r$ ?

4r = 240

r + 4 = 240

\frac{r}{4} = 240

r - 4 = 240

240r = 4

Show answer & explanation

Answer: $4r = 240$

Answer

The equation

4r = 240

represents the relationship because distance equals rate multiplied by time.

The relationship between distance, rate, and time is given by the formula

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

. Substituting the distance of

240

miles and the time of

4

hours into this formula yields

240 = r \times 4

, which is equivalent to

4r = 240

.

Step-by-Step Solution

1

Identify the formula relating distance, rate, and time.

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

(or

d = r \cdot t

)

This is the standard physical relationship governing constant speed motion.

2

Substitute the given values into the formula.

240 = r \cdot 4

The total distance

d

is given as

240

miles, the time

t

is

4

hours, and the speed is the variable

r

.

3

Rewrite the equation in standard algebraic format.

4r = 240

Multiplying the variable

r

by the coefficient

4

is conventionally written with the number first.

Key Concept

Translating a constant rate word problem into a one-step linear equation.

Estimated Time:45s

Question 5Question →

A store owner marks up the wholesale price of a jacket by $\$18$ to determine its retail price. If the retail price of the jacket is $\$75$ , and $w$ represents the wholesale price in dollars, which of the following equations can be used to find $w$ ?

w - 18 = 75

18w = 75

w + 18 = 75

\frac{w}{18} = 75

w + 75 = 18

Show answer & explanation

Answer: $w + 18 = 75$

Answer

w + 18 = 75

A markup of

\$18

means the retail price is

\$18

more than the wholesale price, which is represented by the variable

w

. Therefore, the retail price is

w + 18

. Since the retail price is given as

\$75

, the equation that represents this relationship is

w + 18 = 75

.

Step-by-Step Solution

1

Identify the relationship between wholesale price, markup, and retail price.

The retail price is equal to the wholesale price plus the markup.

A markup increases the cost of the item from its original wholesale value.

2

Substitute the given values and variables into the relationship.

Wholesale price is

w

, markup is

18

, and retail price is

75

, giving the equation

w + 18 = 75

.

This translates the verbal description directly into an algebraic equation.

Key Concept

Translating a real-world scenario into a one-step linear equation using addition.

Estimated Time:45s

Question 6Question →

A researcher is cataloging a collection of $T$ historic documents. She catalogs the documents at a constant rate of $r$ documents per day. After $d$ days of cataloging, the remaining documents represent a fraction, $f$ , of the entire collection. Which of the following equations correctly expresses the total number of documents in the collection, $T$ , in terms of $r$ , $d$ , and $f$ ?

T = \frac{rd}{f}

T = rd(1-f)

T = \frac{rd}{1-f}

T = \frac{r(1-f)}{d}

T = rdf

Show answer & explanation

Answer: $T = \frac{rd}{1-f}$

Answer

The correct equation is

T = \frac{rd}{1-f}

.

The correct equation is found by identifying that the total number of documents cataloged is the rate

r

multiplied by the number of days

d

, which equals

rd

. Because the remaining fraction of the collection is

f

, the completed fraction of the collection is

1-f

. The number of completed documents,

rd

, must equal the completed fraction of the total collection,

(1-f)T

. Setting these equal gives

(1-f)T = rd

. Dividing both sides by the group

(1-f)

isolates the total collection, resulting in

T = \frac{rd}{1-f}

.

Step-by-Step Solution

1

Calculate the total number of documents cataloged.

The number of cataloged documents is

rd

.

The total quantity completed is the constant rate

r

multiplied by the number of days

d

.

2

Determine the fraction of the collection that has been cataloged.

The cataloged fraction of the collection is

1-f

.

Since the remaining fraction is

f

, the completed fraction must be

1-f

because the sum of the completed and remaining fractions must equal 1.

3

Set up an equation relating the cataloged documents to the total collection.

(1-f)T = rd

The fraction of the total collection that is cataloged,

(1-f)T

, must equal the actual number of cataloged documents,

rd

.

4

Solve for the total number of documents,

T

.

T = \frac{rd}{1-f}

Dividing both sides of the equation by the coefficient

(1-f)

isolates

T

.

Key Concept

Translating a real-world scenario with rates and fractional parts into a solvable one-step equation.

Question 7Question →

A manufacturing assembly line produces a batch of $B$ electronic components in $h$ hours. To increase efficiency, a new assembly line is installed that can produce the same batch of components in $\frac{1}{3}$ of the time. Which of the following expressions represents the rate, in components per hour, at which the new assembly line produces these components?

\frac{3B}{h}

\frac{B}{3h}

\frac{3h}{B}

3Bh

\frac{h}{3B}

Show answer & explanation

Answer: $\frac{3B}{h}$

Answer

The expression

\frac{3B}{h}

represents the rate of production of the new assembly line.

The correct answer shows that the rate of production is the number of components divided by the new time, which is one-third of the original time. Since the original time is

h

hours, the new time is

\frac{h}{3}

hours. Dividing the number of components

B

by

\frac{h}{3}

yields

B \div \frac{h}{3} = B \times \frac{3}{h} = \frac{3B}{h}

components per hour.

Step-by-Step Solution

1

Determine the time it takes for the new assembly line to produce the batch of components.

The new assembly line takes

\frac{h}{3}

hours.

The problem states the new line produces the batch in

\frac{1}{3}

of the original time,

h

.

2

Set up the algebraic expression for the rate of production in components per hour.

\text{Rate} = \frac{\text{Total Components}}{\text{Total Time}} = \frac{B}{\frac{h}{3}}

Rate is defined as quantity produced divided by time elapsed.

3

Simplify the fraction to find the final expression.

\text{Rate} = B \times \frac{3}{h} = \frac{3B}{h}

Dividing by a fraction is equivalent to multiplying by its reciprocal.

Key Concept

Formulating algebraic rate expressions from word problems

Question 8Question →

A school club is raising money by selling rolls of wrapping paper. The club keeps $40\%$ of the total sales revenue as profit. If the club made a profit of $\$220$ from selling $r$ rolls of wrapping paper that cost $\$10$ each, which of the following equations can be used to find $r$ ?

0.40r = 220

10r = 220

4r = 220

10r + 0.40 = 220

25r = 220

Show answer & explanation

Answer: $4r = 220$

Answer

The equation expressing the profit is the profit per roll multiplied by the number of rolls, which yields

4r = 220

.

To find the correct equation, we calculate the profit per roll of wrapping paper. Since each roll is sold for

\$10

and the club keeps

40\%

of the revenue as profit, the profit per roll is

0.40 \times 10 = 4

dollars. For

r

rolls sold, the total profit is

4r

dollars. Setting this equal to the given profit of

\$220

yields the equation

4r = 220

.

Step-by-Step Solution

1

Determine the expression for the total sales revenue.

The total revenue is

10r

dollars.

Each of the

r

rolls of wrapping paper is sold for

\$10

.

2

Express the profit as a fraction of the total sales revenue.

The profit is

0.40 \times 10r = 4r

dollars.

The club keeps

40\%

(

0.40

) of the total revenue as profit.

3

Set the profit expression equal to the actual profit made to form the equation.

4r = 220

The total profit made by the club is given as

\$220

.

Key Concept

Translating a multi-step real-world scenario into a one-step linear equation

Alternative Method

Find the profit per roll first:

40\%

of

\$10

is

\$4

. Thus, the profit for

r

rolls is

4r

. Since the total profit is

\$220

, the equation is

4r = 220

.

Estimated Time:1m 30s

Question 9Question →

A tablet battery charges at a constant rate of $r$ percentage points per hour. If the battery charge increases by $48$ percentage points in $3$ hours, which of the following equations can be used to find $r$ ?

r + 3 = 48

3r = 48

\frac{r}{3} = 48

r - 3 = 48

\frac{3}{r} = 48

Show answer & explanation

Answer: $3r = 48$

Answer

The equation

3r = 48

can be used to find the charging rate.

The battery charges at a constant rate of

r

percentage points per hour. Over

3

hours, the total increase is the rate multiplied by the time, which is

3r

. Since the total increase is given as

48

percentage points, the correct relationship is expressed by the equation

3r = 48

.

Step-by-Step Solution

1

Identify the relationship between the charging rate, time, and total charge increase.

The total charge increase is equal to the charging rate multiplied by the number of hours.

Since the battery charges at a constant rate of

r

percentage points per hour, in

3

hours it will charge a total of

3 \times r

or

3r

percentage points.

2

Set the expression for the total charge increase equal to the given total increase value.

3r = 48

The total charge increase is given as

48

percentage points, so we set the expression

3r

equal to

48

.

Key Concept

Translating a real-world scenario involving a constant rate into a one-step linear equation.

Alternative Method

You can also solve for the rate first by dividing the total increase of

48

by

3

hours to get

r = 16

percentage points per hour. Substituting

r = 16

into the equations shows that only

3(16) = 48

is a true statement.

Estimated Time:45s

Question 10Question →

A commuter train travels at a constant speed to complete a trip of $D$ miles in $t$ hours. Due to track maintenance, the train's speed is decreased by 15 miles per hour for a 20-minute portion of the trip. Which of the following expressions represents the distance, in miles, the train traveled during this 20-minute portion of the trip?

\frac{D}{t} - 5

\frac{D}{3t} - 15

\frac{D}{3t} - 5

\frac{D}{3t} - 45

\frac{3D}{t} - 5

Show answer & explanation

Answer: $\frac{D}{3t} - 5$

Answer

The expression

\frac{D}{3t} - 5

represents the distance traveled during the 20-minute portion of the trip.

The train's original speed is

\frac{D}{t}

miles per hour. During the maintenance segment, its speed is reduced by 15 miles per hour, giving a speed of

\frac{D}{t} - 15

miles per hour. To find the distance traveled over a 20-minute interval (which represents

\frac{20}{60} = \frac{1}{3}

of an hour), the reduced speed is multiplied by the elapsed time. Applying the distributive property to

\left(\frac{D}{t} - 15\right) \times \frac{1}{3}

results in

\frac{D}{3t} - 5

.

Step-by-Step Solution

1

Determine the train's original speed.

Original speed is

\frac{D}{t}

miles per hour.

Speed is defined as distance divided by time.

2

Express the reduced speed during the track maintenance portion.

Reduced speed is

\frac{D}{t} - 15

miles per hour.

The speed was decreased by 15 miles per hour from the original speed.

3

Convert the duration of the maintenance portion from minutes to hours.

20 minutes is equal to

\frac{20}{60} = \frac{1}{3}

hour.

Since the speed is in miles per hour, the time must be converted to hours to ensure unit consistency.

4

Calculate the distance traveled during the 20-minute portion.

Distance is

\left(\frac{D}{t} - 15\right) \times \frac{1}{3} = \frac{D}{3t} - 5

miles.

Distance is equal to the product of speed and time. The distributive property must be applied to both terms in the speed expression.

Key Concept

Translating a verbal rate and time relationship into a simplified algebraic expression using the distributive property and unit conversion.

Alternative Method

Alternatively, you can assign plug-in values to make the problem concrete. Let the total distance

D = 120

miles and the total time

t = 2

hours. This gives an original speed of

60

miles per hour. The reduced speed is

60 - 15 = 45

miles per hour. Traveling at

45

miles per hour for

20

minutes (

\frac{1}{3}

hour) results in a distance of

45 \times \frac{1}{3} = 15

miles. Substituting

D = 120

and

t = 2

into the correct expression gives

\frac{120}{3(2)} - 5 = \frac{120}{6} - 5 = 20 - 5 = 15

miles, which matches our result.

Estimated Time:1m 30s

Question 11Question →

A group of friends agrees to share the total cost, $d$ dollars, of renting a cabin equally. There are 6 friends in the group, and each friend's share of the cost is $145. Which of the following equations can be solved to find$ d$?

d + 6 = 145

d - 6 = 145

6d = 145

\frac{d}{6} = 145

\frac{6}{d} = 145

Show answer & explanation

Answer: $\frac{d}{6} = 145$

Answer

The equation

\frac{d}{6} = 145

can be solved to find the total cost.

Because the total cost of renting the cabin,

d

dollars, is split equally among 6 friends, each friend's individual share is represented by the total cost divided by 6, which is written as

\frac{d}{6}

. Since each friend's share is given as

145, this expression is set equal to 145, resulting in the equation

\frac{d}{6} = 145$.

Step-by-Step Solution

1

Determine the relationship between the total cost, the number of friends sharing the cost, and each friend's individual share.

Since the cost is shared equally, the individual share is the total cost,

d

, divided by the number of friends, 6.

This translates the sharing scenario into the algebraic expression

\frac{d}{6}

.

2

Set the algebraic expression equal to the known individual share value.

\frac{d}{6} = 145

The problem states that each friend's share is $145, so the expression representing each share must equal 145.

Key Concept

Translating real-world sharing scenarios into one-step equations involving division.

Question 12Question →

To calibrate a temperature sensor, a technician uses the formula $T_c = T_r - e$ , where $T_c$ is the calibrated temperature, $T_r$ is the raw sensor reading, and $e$ is the calibration error of the sensor. For a particular sensor, the calibration error is $-2.4^\circ\text{C}$ . If the calibrated temperature is measured to be $18.6^\circ\text{C}$ , which of the following equations can be solved to find the raw sensor reading, $T_r$ , in degrees Celsius?

18.6 = T_r - 2.4

18.6 = T_r + 2.4

18.6 = 2.4 - T_r

18.6 = -2.4 - T_r

18.6 = \\frac{T_r}{-2.4}

Show answer & explanation

Answer: $18.6 = T_r + 2.4$

Answer

The equation

18.6 = T_r + 2.4

To find the correct equation, substitute the given values into the formula

T_c = T_r - e

. Substituting

T_c = 18.6

and

e = -2.4

yields

18.6 = T_r - (-2.4)

. Subtracting a negative number is equivalent to addition, which simplifies the equation to

18.6 = T_r + 2.4

.

Step-by-Step Solution

1

Identify the given variables and their values from the problem statement.

The calibrated temperature

T_c = 18.6

and the calibration error

e = -2.4

.

Knowing the values allows for direct substitution into the formula.

2

Substitute the identified values into the calibration formula

T_c = T_r - e

.

The equation becomes

18.6 = T_r - (-2.4)

.

This establishes the relationship between the raw reading

T_r

and the known values.

3

Simplify the subtraction of the negative number on the right side of the equation.

The equation simplifies to

18.6 = T_r + 2.4

.

Subtracting a negative number is mathematically equivalent to adding its positive counterpart.

Key Concept

Substituting signed values into algebraic expressions and simplifying one-step linear equations.

Question 13Question →

A logistics company distributes $T$ tons of cargo among a fleet of trucks. A standard truck has a maximum capacity of $c$ tons, and a heavy-duty truck has a maximum capacity of twice that of a standard truck. The fleet consists of $x$ standard trucks and $y$ heavy-duty trucks, all loaded to their maximum capacities. If the ratio of standard trucks to heavy-duty trucks in the fleet is exactly $3:1$ , which of the following equations correctly expresses the capacity of a standard truck, $c$ , in terms of $T$ and $y$ ?

c = \frac{T}{4y}

c = \frac{3T}{7y}

c = \frac{T}{5y}

c = \frac{T}{2y + 3}

c = T - 5y

Show answer & explanation

Answer: $c = \frac{T}{5y}$

Answer

The equation

c = \frac{T}{5y}

correctly expresses the capacity of a standard truck.

The total cargo

T

is the sum of the capacities of all trucks. Since there are

x

standard trucks carrying

c

tons each and

y

heavy-duty trucks carrying

2c

tons each, the equation is

T = cx + 2cy

. The

3:1

ratio of standard trucks to heavy-duty trucks means

x = 3y

. Substituting this relationship into the cargo equation gives

T = c(3y) + 2cy = 5cy

. To isolate the capacity of a standard truck,

c

, divide both sides by

5y

to get

c = \frac{T}{5y}

.

Step-by-Step Solution

1

Write the equation for the total cargo weight by summing the capacities of all trucks in the fleet.

T = cx + 2cy

The total cargo

T

is distributed among

x

standard trucks, each carrying

c

tons, and

y

heavy-duty trucks, each carrying

2c

tons.

2

Translate the ratio of standard trucks to heavy-duty trucks into an algebraic relationship.

x = 3y

The ratio of standard trucks (

x

) to heavy-duty trucks (

y

) is

3:1

, meaning there are 3 times as many standard trucks as heavy-duty trucks.

3

Substitute the ratio relationship

x = 3y

into the total cargo equation.

T = c(3y) + 2cy

Replacing

x

with

3y

eliminates the variable

x

and expresses

T

in terms of

c

and

y

.

4

Combine like terms to simplify the expression.

T = 5cy

Adding

3cy

and

2cy

yields

5cy

.

5

Isolate the variable

c

by dividing both sides of the equation by

5y

.

c = \frac{T}{5y}

Division is the inverse operation of multiplication, allowing us to solve the one-step equation

T = (5y)c

for

c

.

Key Concept

Translating verbal descriptions into algebraic equations and solving one-step literal equations

Estimated Time:2m 0s

Question 14Question →

An online retailer determines the sale price of a promotional item by reducing the original price, $P$ dollars, by a discount of $D$ dollars. The discount is defined as $D = \frac{1}{6} P$ . If the sale price of the item is $125$ dollars, which of the following equations can be solved to find the original price of the item, in dollars?

\frac{1}{6} P = 125

P - \frac{1}{6} = 125

\frac{5}{6} P = 125

P + \frac{1}{6} P = 125

\frac{6}{5} P = 125

Show answer & explanation

Answer: $\frac{5}{6} P = 125$

Answer

The equation stating that five-sixths of the original price equals one hundred twenty-five is correct.

The correct equation is found by expressing the sale price as the original price minus the discount. The discount is one-sixth of the original price, which is represented by the expression containing five-sixths of the original price. Setting this expression equal to the given sale price of one hundred twenty-five dollars results in the correct equation.

Step-by-Step Solution

1

Write the relationship for the sale price based on the scenario.

S = P - D

The sale price (

S

) is the original price (

P

) minus the discount amount (

D

).

2

Substitute the given expression for the discount (

D = \frac{1}{6} P

) into the sale price equation.

S = P - \frac{1}{6} P

This expresses the sale price in terms of the single variable

P

representing the original price.

3

Simplify the algebraic expression by combining the like terms.

S = \frac{5}{6} P

Combining

1P

and

-\frac{1}{6} P

gives

\frac{5}{6} P

because

1 - \frac{1}{6} = \frac{5}{6}

.

4

Substitute the known sale price of

125

dollars into the simplified equation.

\frac{5}{6} P = 125

This sets up the final one-step equation needed to find the original price.

Key Concept

Translating verbal descriptions of discounts into simplified one-step linear equations.

Estimated Time:1m 30s

Question 15Question →

A digital commercial printing press operates at a constant rate of $r$ pages per minute. During a scheduled production session lasting $t$ minutes, the machine was paused for $12$ minutes due to paper replenishment. If the printing press successfully printed $p$ pages during the operational portion of the session, which of the following equations correctly expresses the relationship between $r$ , $p$ , and $t$ ?

r = \frac{p}{t} - 12

r = \frac{p}{t - 12}

r = \frac{p}{t + 12}

r = \frac{p - 12}{t}

r = \frac{12p}{t}

Show answer & explanation

Answer: $r = \frac{p}{t - 12}$

Answer

The correct equation is

r = \frac{p}{t - 12}

.

Since the press was paused for 12 minutes during a session of

t

minutes, the total time spent actually printing was

(t - 12)

minutes. The printing rate

r

is defined as pages printed divided by active printing time, giving

r = \frac{p}{t - 12}

.

Step-by-Step Solution

1

Identify the active printing time in terms of

t

.

Active operating time =

(t - 12)

minutes.

The machine was paused for 12 minutes out of the total session duration of

t

minutes.

2

Use the basic rate relationship Rate = Quantity / Time.

r = \frac{p}{t - 12}

.

Dividing the total pages printed

p

by the active printing duration

(t - 12)

yields the rate

r

in pages per minute.

Key Concept

Formulating one-step rate and time algebraic relationships with adjusted variables.

Question 16Question →

An agricultural research facility divides a field with a total area of $A$ acres into $n$ individual test plots of equal size after reserving an area of 12 acres for administrative buildings. Which of the following equations correctly expresses the area, $p$ , in acres, of each individual test plot in terms of $A$ and $n$ ?

p = \frac{A - 12}{n}

p = \frac{A}{n} - 12

p = \frac{A + 12}{n}

p = n(A - 12)

p = \frac{n}{A - 12}

Show answer & explanation

Answer: $p = \frac{A - 12}{n}$

Answer

The equation representing the area of each test plot is

p = \frac{A - 12}{n}

.

Reserving 12 acres leaves

A - 12

acres available for testing. Dividing this remaining quantity equally into

n

test plots results in an area per plot of

p = \frac{A - 12}{n}

.

Step-by-Step Solution

1

Determine the remaining area allocated for test plots.

Available area =

A - 12

acres.

The 12 acres used for administrative buildings must be subtracted from the total field area

A

.

2

Divide the remaining area by the number of equal test plots

n

.

p = \frac{A - 12}{n}

.

Dividing the remaining acreage equally among

n

plots gives the area

p

of a single plot.

Key Concept

Translating contextual verbal scenarios into basic algebraic expressions using grouping and division.

Question 17Question →

A municipal water authority pumps water into a storage reservoir. During a drought conservation phase, the authority reduces its standard daily pumping volume by $35\%$ . If the resulting reduced daily pumping volume is $W$ acre-feet, which of the following equations correctly expresses the standard daily pumping volume, $S$ , in acre-feet, in terms of $W$ ?

S = \frac{W}{0.65}

S = 0.65W

S = \frac{W}{0.35}

S = 0.35W

S = W + 0.35

Show answer & explanation

Answer: $S = \frac{W}{0.65}$

Answer

The standard daily pumping volume is given by

S = \frac{W}{0.65}

.

Reducing the standard volume

S

by

35\%

leaves

65\%

of

S

, which can be written algebraically as

0.65S

. Since this reduced amount equals

W

, the relationship is

0.65S = W

. Dividing both sides by

0.65

isolates

S

to give the one-step equation solution

S = \frac{W}{0.65}

.

Step-by-Step Solution

1

Express the reduced volume in terms of the standard volume

S

Reduced volume

= S - 0.35S = 0.65S

A

35\%

reduction leaves

100\% - 35\% = 65\%

of the original standard volume

S

.

2

Set up the one-step algebraic equation matching the given variable

W

0.65S = W

The reduced volume is defined as

W

acre-feet.

3

Solve the one-step equation for

S

S = \frac{W}{0.65}

Divide both sides of the equation by

0.65

to isolate

S

.

Key Concept

Formulating and solving a one-step algebraic equation involving a percentage decrease.

Alternative Method

Convert

65\%

to the fraction

\frac{13}{20}

. Then

\frac{13}{20}S = W

, which yields

S = \frac{20}{13}W = \frac{W}{0.65}

.

Estimated Time:1m 30s

Question 18Question →

A commercial bakery uses an automated mixing system to prepare large batches of dough. The system processes flour at a constant rate such that $F$ kilograms of flour are mixed over $m$ minutes. If the mixer consumes flour at a rate of $\frac{3}{8}$ kilogram per minute, which of the following equations correctly expresses $m$ , the total mixing time in minutes, in terms of $F$ ?

m = \frac{3F}{8}

m = \frac{8F}{3}

m = F - \frac{3}{8}

m = \frac{3}{8F}

m = \frac{8}{3F}

Show answer & explanation

Answer: $m = \frac{8F}{3}$

Answer

The correct equation expressing

m

in terms of

F

is

m = \frac{8F}{3}

.

The total amount of flour used is determined by multiplying the rate per minute by the number of minutes, giving the equation

F = \frac{3}{8}m

. To express

m

in terms of

F

, solve this one-step equation for

m

by dividing both sides by

\frac{3}{8}

. Dividing by

\frac{3}{8}

is mathematically equivalent to multiplying by its reciprocal,

\frac{8}{3}

. Therefore,

m = \frac{8F}{3}

.

Step-by-Step Solution

1

Set up the one-step relationship between rate, total flour, and time.

F = \frac{3}{8}m

Total flour

F

equals the rate per minute multiplied by total minutes

m

.

2

Isolate the variable

m

by dividing both sides of the equation by

\frac{3}{8}

.

m = \frac{F}{\frac{3}{8}}

To solve a one-step multiplication equation, apply the inverse operation (division).

3

Simplify the fraction division by multiplying by the reciprocal of

\frac{3}{8}

.

m = F \cdot \frac{8}{3} = \frac{8F}{3}

Dividing by a fraction is equivalent to multiplying by its reciprocal.

Key Concept

Solving One-Step Equations with Fractional Coefficients

Estimated Time:1m 0s

Question 19Question →

A museum archivist digitizes a collection of historical glass-plate photographs at a constant rate of $p$ photographs per hour. If the archivist digitizes a total of $336$ photographs over a continuous $14$ -hour shift, which of the following equations can be solved to find $p$ ?

p + 14 = 336

\frac{p}{14} = 336

14p = 336

p - 14 = 336

14 + 336 = p

Show answer & explanation

Answer: $14p = 336$

Answer

The correct equation is

14p = 336

.

The total number of digitized photographs equals the rate per hour (

p

) multiplied by the number of hours (

14

). Setting this product equal to the total of

336

photographs yields

14p = 336

.

Step-by-Step Solution

1

Identify the relationship between rate, time, and total quantity.

\text{Total Quantity} = \text{Rate} \times \text{Time}

The total work done at a constant rate is found by multiplying the rate per unit of time by total time.

2

Substitute the given values into the relationship.

336 = p \times 14

, which simplifies to

14p = 336

.

The rate is

p

photographs per hour, time is

14

hours, and total quantity is

336

photographs.

Key Concept

Writing one-step multiplication equations from rate and time word problems

Estimated Time:1m 0s

Question 20Question →

A commercial bakery uses an industrial mixer that processes specialty flour at a constant rate. The mixer consumes $\frac{4}{7}$ of a full bag of flour in $12$ minutes. The equation $\frac{4}{7}b = 12$ represents this situation, where $b$ is the time, in minutes, it takes to process one full bag of flour. What is the value of $b$ ?

\frac{48}{7}

21

28

48

\frac{3}{7}

Show answer & explanation

Answer: $21$

Answer

The total time required to process one full bag of flour is 21 minutes.

To solve the one-step equation

\frac{4}{7}b = 12

, isolate

b

by multiplying both sides of the equation by the reciprocal of

\frac{4}{7}

, which is

\frac{7}{4}

. Computing

12 \times \frac{7}{4}

yields

\frac{84}{4} = 21

. Therefore, 21 minutes is the correct answer.

Step-by-Step Solution

1

Identify the given one-step equation

\frac{4}{7}b = 12

The equation models the relationship between the fraction of the bag used and the time taken.

2

Isolate the variable

b

by multiplying both sides by the reciprocal of

\frac{4}{7}

b = 12 \times \frac{7}{4}

Multiplying by the reciprocal undoes multiplication by a fraction.

3

Simplify the numerical calculation

b = \frac{12 \times 7}{4} = 3 \times 7 = 21

Dividing 12 by 4 yields 3, and multiplying 3 by 7 gives 21.

Key Concept

Solving One-Step Linear Equations with Fractional Coefficients

Estimated Time:1m 0s

Basic Algebraic Expressions and One-Step Equations

Answer

Step-by-Step Solution

Key Concept

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