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NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 11
Lesson 11: Constant Rate
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Lesson 11: Constant Rate
Student Outcomes
Students know the definition of constant rate in varied contexts as expressed using two variables where one is
π‘ representing a time interval.
Students graph points on a coordinate plane related to constant rate problems.
Classwork
Example 1 (6 minutes)
Give students the first question below, and allow them time to work. Ask them to share their solutions with the class,
and then proceed with the discussion, table, and graph to finish Example 1.
Example 1
Pauline mows a lawn at a constant rate. Suppose she mows a ππ-square-foot lawn in π. π minutes. What area, in square
feet, can she mow in ππ minutes? π minutes?
What is Paulineβs average rate in 2.5 minutes?
Paulineβs average rate in 2.5 minutes is 35
2.5 square feet per minute.
What is Paulineβs average rate in 10 minutes?
Let π΄ represent the square feet of the area mowed in 10 minutes. Paulineβs average rate in 10 minutes
is π΄
10 square feet per mintute.
Let πΆ be Paulineβs constant rate in square feet per minute; then, 35
2.5= πΆ, and
π΄
10= πΆ. Therefore,
ππ
π. π=
π¨
ππ
πππ = π. ππ¨
πππ
π. π=
π. π
π. ππ¨
πππ = π¨
Pauline mows πππ square feet of lawn in ππ minutes.
If we let π¦ represent the number of square feet Pauline can mow in π‘ minutes, then Paulineβs average rate in π‘
minutes is π¦
π‘ square feet per minute.
Write the two-variable equation that represents the area of lawn, π¦, Pauline can mow in π‘ minutes.
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 11
Lesson 11: Constant Rate
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ππ
π. π=
π
π
π. ππ = πππ
π. π
π. ππ =
ππ
π. ππ
π =ππ
π. ππ
What is the meaning of 35
2.5 in the equation π¦ =
352.5
π‘?
The number 35
2.5 represents the constant rate at which Pauline can mow a lawn.
We can organize the information in a table.
π (time in
minutes)
Linear Equation:
π =ππ
π. ππ
π (area in
square feet)
π π =ππ
π. π(π) π
π π =ππ
π. π(π)
ππ
π. π= ππ
π π =ππ
π. π(π)
ππ
π. π= ππ
π π =ππ
π. π(π)
πππ
π. π= ππ
π π =ππ
π. π(π)
πππ
π. π= ππ
On a coordinate plane, we will let the π₯-axis represent time π‘, in minutes, and the π¦-axis represent the area of
mowed lawn in square feet. Then we have the following graph.
MP.7
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Lesson 11: Constant Rate
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Because Pauline mows at a constant rate, we would expect the square feet of mowed lawn to continue to rise
as the time, in minutes, increases.
Concept Development (6 minutes)
In the last lesson, we learned about average speed and constant speed. Constant speed problems are just a
special case of a larger variety of problems known as constant rate problems. Some of these problems were
topics in Grade 7, such as water pouring out of a faucet into a tub, painting a house, and mowing a lawn.
First, we define the average rate:
Suppose π gallons of water flow from a faucet in a given time interval π‘ (minutes). Then, the average rate of
water flow in the given time interval is π
π‘ in gallons per minute.
Then, we define the constant rate:
Suppose the average rate of water flow is the same constant πΆ for any given time interval. Then we say that
the water is flowing at a constant rate, πΆ.
Similarly, suppose π΄ square feet of lawn are mowed in a given time interval π‘ (minutes). Then, the average
rate of lawn mowing in the given time interval is π΄
π‘ square feet per minute. If we assume that the average rate
of lawn mowing is the same constant, πΆ, for any given time interval, then we say that the lawn is mowed at a
constant rate, πΆ.
Describe the average rate of painting a house.
Suppose π΄ square feet of house are painted in a given time interval π‘ (minutes). Then the average rate
of house painting in the given time interval is π΄
π‘ square feet per minute.
Describe the constant rate of painting a house.
If we assume that the average rate of house painting is the same constant, πΆ, over any given time
interval, then we say that the wall is painted at a constant rate, πΆ.
What is the difference between average rate and constant rate?
Average rate is the rate in which something can be done over a specific time interval. Constant rate
assumes that the average rate is the same over any time interval.
As you can see, the way we define average rate and constant rate for a given situation is very similar. In each
case, a transcription of the given information leads to an expression in two variables.
Example 2 (8 minutes)
Example 2
Water flows at a constant rate out of a faucet. Suppose the volume of water that comes out in three minutes is ππ. π
gallons. How many gallons of water come out of the faucet in π minutes?
Write the linear equation that represents the volume of water, π, that comes out in π‘ minutes.
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 11
Lesson 11: Constant Rate
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Let πͺ represent the constant rate of water flow.
ππ.π
π= πͺ, and
π½
π= πͺ; then,
ππ.π
π=
π½
π.
ππ. π
π=
π½
π
ππ½ = ππ. ππ
π
ππ½ =
ππ. π
ππ
π½ =ππ. π
ππ
What is the meaning of the number 10.5
3 in the equation π =
10.53
π‘?
The number 10.5
3 represents the constant rate at which water flows from a faucet.
Using the linear equation π =10.5
3π‘, complete the table.
π (time in minutes) Linear Equation:
π½ =ππ. π
ππ
π½ (in gallons)
π π½ =ππ. π
π(π) π
π π½ =ππ. π
π(π)
ππ. π
π= π. π
π π½ =ππ. π
π(π)
ππ
π= π
π π½ =ππ. π
π(π)
ππ. π
π= ππ. π
π π½ =ππ. π
π(π)
ππ
π= ππ
On a coordinate plane, we will let the π₯-axis represent time π‘ in minutes and the π¦-axis represent the volume
of water. Graph the data from the table.
MP.7
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 11
Lesson 11: Constant Rate
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Using the graph, about how many gallons of water do you think would flow after 112
minutes? Explain.
After 112
minutes, between 312
and 7 gallons of water will flow. Since the water is flowing at a constant
rate, we can expect the volume of water to rise between 1 and 2 minutes. The number of gallons that
flow after 112
minutes then would have to be between the number of gallons that flow out after 1
minute and 2 minutes.
Using the graph, about how long would it take for 15 gallons of water to flow out of the faucet? Explain.
It would take between 4 and 5 minutes for 15 gallons of water to flow out of the faucet. It takes 4
minutes for 14 gallons to flow; therefore, it must take more than 4 minutes for 15 gallons to come out.
It must take less than 5 minutes because 312
gallons flow out every minute.
Graphing proportional relationships like these last two constant rate problems provides us more information
than simply solving an equation and calculating one value. The graph provides information that is not so
obvious in an equation.
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 11
Lesson 11: Constant Rate
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Exercises (15 minutes)
Students complete Exercises 1β3 independently.
Exercises
1. Juan types at a constant rate. He can type a full page of text in πππ
minutes. We want to know how many pages, π,
Juan can type after π minutes.
a. Write the linear equation in two variables that represents the number of pages Juan types in any given time
interval.
Let πͺ represent the constant rate that Juan types in pages per minute. Then,
π
π.π= πͺ, and
π
π= πͺ; therefore,
π
π.π=
π
π.
π
π. π=
π
π
π. ππ = π
π. π
π. ππ =
π
π. ππ
π =π
π. ππ
b. Complete the table below. Use a calculator, and round your answers to the tenths place.
π (time in minutes) Linear Equation:
π =π
π. ππ
π (pages typed)
π π =π
π. π(π) π
π π =π
π. π(π)
π
π. πβ π. π
ππ π =π
π. π(ππ)
ππ
π. πβ π. π
ππ π =π
π. π(ππ)
ππ
π. πβ π. π
ππ π =π
π. π(ππ)
ππ
π. πβ π. π
c. Graph the data on a coordinate plane.
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d. About how long would it take Juan to type a π-page paper? Explain.
It would take him between ππ and ππ minutes. After ππ minutes, he will have typed π. π pages. In ππ
minutes, he can type π. π pages. Since π pages is between π. π and π. π, then it will take him between ππ and
ππ minutes.
2. Emily paints at a constant rate. She can paint ππ square feet in π minutes. What area, π¨, in square feet, can she
paint in π minutes?
a. Write the linear equation in two variables that represents the number of square feet Emily can paint in any
given time interval.
Let πͺ be the constant rate that Emily paints in square feet per minute. Then,
ππ
π= πͺ, and
π¨
π= πͺ; therefore,
ππ
π=
π¨
π.
ππ
π=
π¨
π
ππ¨ = πππ
π
ππ¨ =
ππ
ππ
π¨ =ππ
ππ
b. Complete the table below. Use a calculator, and round answers to the tenths place.
π (time in minutes) Linear Equation:
π¨ =ππ
ππ
π¨ (area painted in
square feet)
π π¨ =ππ
π(π) π
π π¨ =ππ
π(π)
ππ
π= π. π
π π¨ =ππ
π(π)
ππ
π= ππ. π
π π¨ =ππ
π(π)
ππ
π= ππ. π
π π¨ =ππ
π(π)
πππ
π= ππ. π
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 11
Lesson 11: Constant Rate
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c. Graph the data on a coordinate plane.
d. About how many square feet can Emily paint in πππ
minutes? Explain.
Emily can paint between ππ. π and ππ. π square feet in πππ
minutes. After π minutes, she paints ππ. π square
feet, and after π minutes, she will have painted ππ. π square feet.
3. Joseph walks at a constant speed. He walked to a store that is one-half mile away in π minutes. How many miles,
π, can he walk in π minutes?
a. Write the linear equation in two variables that represents the number of miles Joseph can walk in any given
time interval, π.
Let πͺ be the constant rate that Joseph walks in miles per minute. Then,
π.π
π= πͺ, and
π
π= πͺ; therefore,
π.π
π=
π
π.
π. π
π=
π
π
ππ = π. ππ
π
ππ =
π. π
ππ
π =π. π
ππ
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 11
Lesson 11: Constant Rate
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b. Complete the table below. Use a calculator, and round answers to the tenths place.
π (time in minutes)
Linear Equation:
π =π. π
ππ
π (distance in miles)
π π =π. π
π(π) π
ππ π =π. π
π(ππ)
ππ
π= π. π
ππ π =π. π
π(ππ)
ππ
π= π
ππ π =π. π
π(ππ)
ππ
π= π. π
πππ π =π. π
π(πππ)
ππ
π= ππ
c. Graph the data on a coordinate plane.
d. Josephβs friend lives π miles away from him. About how long would it take Joseph to walk to his friendβs
house? Explain.
It will take Joseph a little less than an hour to walk to his friendβs house. Since it takes ππ minutes for him to
walk π. π miles and ππ minutes to walk π miles, and π is closer to π than π. π, it will take Joseph less than an
hour to walk the π miles.
Closing (5 minutes)
Summarize, or ask students to summarize, the main points from the lesson:
Constant rate problems appear in a variety of contexts like painting a house, typing, walking, or water flow.
We can express the constant rate as a two-variable equation representing proportional change.
We can graph the constant rate situation by completing a table to compute data points.
X`
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 11
Lesson 11: Constant Rate
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Exit Ticket (5 minutes)
Lesson Summary
When constant rate is stated for a given problem, then you can express the situation as a two-variable equation.
The equation can be used to complete a table of values that can then be graphed on a coordinate plane.
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Lesson 11: Constant Rate
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Name Date
Lesson 11: Constant Rate
Exit Ticket
Vicky reads at a constant rate. She can read 5 pages in 9 minutes. We want to know how many pages, π, Vicky can read
after π‘ minutes.
a. Write a linear equation in two variables that represents the number of pages Vicky reads in any given time
interval.
b. Complete the table below. Use a calculator, and round answers to the tenths place.
π (time in minutes)
Linear Equation:
π (pages read)
0
20
40
60
c. About how long would it take Vicky to read 25 pages? Explain.
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 11
Lesson 11: Constant Rate
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Exit Ticket Sample Solutions
Vicky reads at a constant rate. She can read π pages in π minutes. We want to know how many pages, π, Vicky can read
after π minutes.
a. Write a linear equation in two variables that represents the number of pages Vicky reads in any given time
interval.
Let πͺ represent the constant rate that Vicky reads in pages per minute. Then,
π
π= πͺ, and
π
π= πͺ; therefore,
π
π=
π
π.
π
π=
π
π
ππ = ππ
π
ππ =
π
ππ
π =π
ππ
b. Complete the table below. Use a calculator, and round answers to the tenths place.
π (time in minutes)
Linear Equation:
π =π
ππ
π (pages read)
π π =π
π(π) π
ππ π =π
π(ππ)
πππ
πβ ππ. π
ππ π =π
π(ππ)
πππ
πβ ππ. π
ππ π =π
π(ππ)
πππ
πβ ππ. π
c. About how long would it take Vicky to read ππ pages? Explain.
It would take her a little over ππ minutes. After ππ minutes, she can read about ππ. π pages, and after π
hour, she can read about ππ. π pages. Since ππ pages is between ππ. π and ππ. π, it will take her between ππ
and ππ minutes to read ππ pages.
Problem Set Sample Solutions
Students practice writing two-variable equations that represent a constant rate.
1. A train travels at a constant rate of ππ miles per hour.
a. What is the distance, π , in miles, that the train travels in π hours?
Let πͺ be the constant rate the train travels. Then, ππ
π= πͺ, and
π
π= πͺ; therefore,
ππ
π=
π
π.
ππ
π=
π
π
π = πππ
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 11
Lesson 11: Constant Rate
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b. How many miles will it travel in π. π hours?
π = ππ(π. π)
= πππ. π
The train will travel πππ. π miles in π. π hours.
2. Water is leaking from a faucet at a constant rate of π
π gallons per minute.
a. What is the amount of water, π, in gallons per minute, that is leaked from the faucet after π minutes?
Let πͺ be the constant rate the water leaks from the faucet in gallons per minute. Then,
ππ
π= πͺ, and
π
π= πͺ; therefore,
ππ
π=
π
π.
ππ
π=
π
π
π =π
ππ
b. How much water is leaked after an hour?
π =π
ππ
=π
π(ππ)
= ππ
The faucet will leak ππ gallons in one hour.
3. A car can be assembled on an assembly line in π hours. Assume that the cars are assembled at a constant rate.
a. How many cars, π, can be assembled in π hours?
Let πͺ be the constant rate the cars are assembled in cars per hour. Then,
π
π= πͺ, and
π
π= πͺ; therefore,
π
π=
π
π.
π
π=
π
π
ππ = π
π
ππ =
π
ππ
π =π
ππ
b. How many cars can be assembled in a week?
A week is ππ Γ π = πππ hours. So, π =ππ
(πππ) = ππ. Twenty-eight cars can be assembled in a week.
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 11
Lesson 11: Constant Rate
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4. A copy machine makes copies at a constant rate. The machine can make ππ copies in πππ
minutes.
a. Write an equation to represent the number of copies, π, that can be made over any time interval in minutes,
π.
Let πͺ be the constant rate that copies can be made in copies per minute. Then,
ππ
πππ
= πͺ, and π
π= πͺ; therefore,
ππ
πππ
=π
π.
ππ
πππ
=π
π
ππ
ππ = πππ
π
ππ = πππ
π
πβ
π
ππ =
π
πβ πππ
π = πππ
b. Complete the table below.
π (time in minutes) Linear Equation:
π = πππ π (number of copies)
π π = ππ(π) π
π. ππ π = ππ(π. ππ) π
π. π π = ππ(π. π) ππ
π. ππ π = ππ(π. ππ) ππ
π π = ππ(π) ππ
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 11
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c. Graph the data on a coordinate plane.
d. The copy machine runs for ππ seconds and then jams. About how many copies were made before the jam
occurred? Explain.
Since ππ seconds is approximately π. π of a minute, then the number of copies made will be between π and
ππ because π. π is between π. ππ and π. π.
5. Connor runs at a constant rate. It takes him ππ minutes to run π miles.
a. Write the linear equation in two variables that represents the number of miles Connor can run in any given
time interval in minutes, π.
Let πͺ be the constant rate that Connor runs in miles per minute, and let π represent the number of miles he
ran in π minutes. Then,
π
ππ= πͺ, and
π
π= πͺ; therefore,
π
ππ=
π
π.
π
ππ=
π
π
πππ = ππ
ππ
πππ =
π
πππ
π =π
πππ
π =π
πππ
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 11
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b. Complete the table below. Use a calculator, and round answers to the tenths place.
π (time in minutes) Linear Equation:
π =π
πππ
π (distance in miles)
π π =π
ππ(π) π
ππ π =π
ππ(ππ)
ππ
ππβ π. π
ππ π =π
ππ(ππ)
ππ
ππβ π. π
ππ π =π
ππ(ππ)
ππ
ππβ π. π
ππ π =π
ππ(ππ)
πππ
ππβ π. π
c. Graph the data on a coordinate plane.
d. Connor ran for ππ minutes before tripping and spraining his ankle. About how many miles did he run before
he had to stop? Explain.
Since Connor ran for ππ minutes, he ran more than π. π miles but less than π. π miles. Since ππ is between ππ
and ππ, then we can use those reference points to make an estimate of how many miles he ran in ππ minutes,
probably about π miles.