NYS COMMON CORE MATHEMATICS CURRICULUM M1 Lesson 27 ALGEBRA II
Lesson 27: Word Problems Leading to Rational Equations
301
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Lesson 27: Word Problems Leading to Rational Equations
Student Outcomes
Students solve word problems using models that involve rational expressions.
Lesson Notes
In the preceding lessons, students learned to add, subtract, multiply, and divide rational expressions and solve rational
equations in order to develop the tools needed for solving application problems involving rational equations in this
lesson (A-REI.A.2). Students develop their problem-solving and modeling abilities by carefully reading the problem
description and converting information into equations (MP.1), thus creating a mathematical model of the problem
(MP.4).
Classwork
Exercise 1 (13 minutes)
This lesson turns to some applied problems that can be modeled with rational equations, strengthening studentsโ
problem-solving and modeling experience in alignment with standards MP.1 and MP.4. These equations can be solved
using the skills developed in previous lessons. Have students work in small groups to answer this set of four questions.
At the end of the work time, ask different groups to present their solutions to the class. Suggest to students that they:
(a) read the problem aloud, (b) paraphrase and summarize the problem in their own words, (c) find an equation that
models the situation, and (d) say how it represents the quantities involved. Check to make sure that students
understand the problem before they begin trying to solve it.
In Exercise 1, consider encouraging students to assign the variable ๐ to the unknown quantity, and ask if they can arrive
at an equation that relates ๐ to the known quantities.
NYS COMMON CORE MATHEMATICS CURRICULUM M1 Lesson 27 ALGEBRA II
Lesson 27: Word Problems Leading to Rational Equations
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Exercise 1
1. Anne and Maria play tennis almost every weekend. So far, Anne has won ๐๐ out of ๐๐
matches.
a. How many matches will Anne have to win in a row to improve her winning
percentage to ๐๐%?
Suppose that Anne has already won ๐๐ of ๐๐ matches, and let ๐ represent the
number of additional matches she must win to raise her winning percentage to ๐๐%.
After playing and winning all of those additional ๐ matches, she has won ๐๐ + ๐
matches out of a total of ๐๐ + ๐ matches played. Her winning percentage is then ๐๐+๐
๐๐+๐, and we want to find the value of ๐ that solves the equation
๐๐ + ๐
๐๐ + ๐= ๐. ๐๐.
Multiply both sides by ๐๐ + ๐.
๐๐ + ๐ = ๐. ๐๐(๐๐ + ๐)
๐๐ + ๐ = ๐๐ + ๐. ๐๐๐
Solve for ๐:
๐. ๐๐๐ = ๐
๐ = ๐๐
So, Anne would need to win ๐๐ matches in a row in order to improve her winning
percentage to ๐๐%.
b. How many matches will Anne have to win in a row to improve her winning percentage to ๐๐%?
This situation is similar to that for part (a), except that we want a winning percentage of ๐. ๐๐, instead of
๐. ๐๐. Again, we let ๐ represent the number of matches Anne must win consecutively to bring her winning
percentage up to ๐๐%.
๐๐ + ๐
๐๐ + ๐= ๐. ๐๐
Solve for ๐:
๐๐ + ๐ = ๐. ๐๐(๐๐ + ๐)
๐๐ + ๐ = ๐๐ + ๐. ๐๐๐
๐. ๐๐๐ = ๐
๐ = ๐๐
In order for Anne to bring her winning percentage up to ๐๐%, she would need to win the next ๐๐ consecutive
matches.
c. Can Anne reach a winning percentage of ๐๐๐%?
Allow students to come to the conclusion that Anne will never reach a winning percentage of ๐๐๐% because
she has already lost ๐ matches.
MP.2
Scaffolding:
Students may benefit from
having the problem read aloud
and summarized. They should
be encouraged to restate the
problem in their own words to
a partner.
If students are struggling,
present the equation 12+๐
20+๐= 0.75, and ask
students how this models the
situation.
Students who may be working
above grade level could be
challenged to write their own
word problems that result in
rational equations.
MP.1 &
MP.4
NYS COMMON CORE MATHEMATICS CURRICULUM M1 Lesson 27 ALGEBRA II
Lesson 27: Word Problems Leading to Rational Equations
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d. After Anne has reached a winning percentage of ๐๐% by winning consecutive matches as in part (b), how
many matches can she now lose in a row to have a winning percentage of ๐๐%?
Recall from part (b) that she had won ๐๐ matches out of ๐๐ to reach a winning percentage of ๐๐%. We will
now assume that she loses the next ๐ matches in a row. Then, she will have won ๐๐ matches out of ๐๐ + ๐
matches, and we want to know the value of ๐ that makes this a ๐๐% win rate.
๐๐
๐๐ + ๐= ๐. ๐๐
Solving the equation:
๐๐ = ๐. ๐๐(๐๐ + ๐)
๐๐ = ๐๐ + ๐. ๐๐๐
๐๐ = ๐. ๐๐๐
๐๐ = ๐
Thus, after reaching a ๐๐% winning percentage in ๐๐ matches, Anne can lose ๐๐ matches in a row to drop to
a ๐๐% winning percentage.
Example (5 minutes)
Work this problem at the front of the room, but allow the class to provide input and steer the discussion. Depending on
how students did with the first exercise, the teacher may lead a discussion of this problem as a class, ask students to
work in groups, or ask students to work independently while targeting instruction with a small group that struggled on
the first exercise.
Example
Working together, it takes Sam, Jenna, and Francisco two hours to paint one room. When Sam works alone, he can paint
one room in ๐ hours. When Jenna works alone, she can paint one room in ๐ hours. Determine how long it would take
Francisco to paint one room on his own.
Consider how much can be accomplished in one hour. Sam, Jenna, and Francisco together can paint half a room in one
hour. If Sam can paint one room in ๐ hours on his own, then in one hour he can paint ๐
๐ of the room. Similarly, Jenna can
paint ๐
๐ of the room in one hour. We do not yet know how much Francisco can paint in one hour, so we will say he can
paint ๐
๐ of the room. So, in one hour, Sam has painted
๐
๐ of the room, Jenna has painted
๐
๐ of the room, and all three
together can paint ๐
๐ the room, leading to the following equation for how much can be painted in one hour:
๐
๐+
๐
๐+
๐
๐=
๐
๐.
A common multiple of the denominators is ๐๐๐. Multiplying both sides by ๐๐๐ gives us:
๐๐๐
๐+
๐๐๐
๐+
๐๐๐
๐=
๐๐๐
๐
๐๐ + ๐๐ + ๐๐ = ๐๐,
which leads us to the value of ๐:
๐ = ๐๐.
So, Francisco can paint the room in ๐๐ hours on his own.
MP.4
NYS COMMON CORE MATHEMATICS CURRICULUM M1 Lesson 27 ALGEBRA II
Lesson 27: Word Problems Leading to Rational Equations
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Exercise 2 (5 minutes)
Remind students that distance equals rate times time (๐ = ๐ โ ๐ก) before having them work on this exercise in pairs or
small groups. Be sure to have groups share their results before continuing to the next exercise.
Exercises 2โ4
2. Melissa walks ๐ miles to the house of a friend and returns home on a bike. She averages ๐ miles per hour faster
when cycling than when walking, and the total time for both trips is two hours. Find her walking speed.
Using the relationship ๐ = ๐ โ ๐, we have ๐ =๐ ๐
. The time it takes for Melissa to walk to her friendโs house is ๐
๐, and
the time to cycle back is ๐
๐+๐. Thus, we can write an equation that describes the combined time for both trips:
๐
๐+
๐
๐ + ๐= ๐.
A common multiple of the denominators is ๐(๐ + ๐), so we multiply both sides of the equation by ๐(๐ + ๐).
๐(๐ + ๐) + ๐๐ = ๐๐(๐ + ๐)
๐๐ + ๐๐ + ๐๐ = ๐๐๐ + ๐๐
๐๐๐ + ๐๐ โ ๐๐ = ๐
๐(๐ โ ๐)(๐ + ๐) = ๐
Thus, ๐ = โ๐ or ๐ = ๐. Since ๐ represents Melissaโs speed, it does not make sense for ๐ to be negative. So, the only
solution is ๐, which means that Melissaโs walking speed is ๐ miles per hour.
Exercise 3 (10 minutes)
3. You have ๐๐ liters of a juice blend that is ๐๐% juice.
a. How many liters of pure juice need to be added in order to make a blend that is ๐๐% juice?
We start off with ๐๐ liters of a blend containing ๐๐% juice. Then, this blend contains ๐. ๐๐(๐๐) = ๐ liters of
juice in the ๐๐ liters of mixture. If we add ๐จ liters of pure juice, then the concentration of juice in the blend is ๐+๐จ
๐๐+๐จ. We want to know which value of ๐จ makes this blend ๐๐% juice.
๐ + ๐จ
๐๐ + ๐จ= ๐. ๐๐
๐ + ๐จ = ๐. ๐๐(๐๐ + ๐จ)
๐ + ๐จ = ๐. ๐ + ๐. ๐๐๐จ
๐. ๐๐๐จ = ๐. ๐
๐จ = ๐
Thus, if we add ๐ liters of pure juice, we have ๐๐ liters of a blend that contains ๐๐ liters of juice, meaning that
the concentration of juice in this blend is ๐๐%.
MP.1 &
MP.4
NYS COMMON CORE MATHEMATICS CURRICULUM M1 Lesson 27 ALGEBRA II
Lesson 27: Word Problems Leading to Rational Equations
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b. How many liters of pure juice need to be added in order to make a blend that is ๐๐% juice?
๐ + ๐จ
๐๐ + ๐จ= ๐. ๐๐
๐ + ๐จ = ๐. ๐(๐๐ + ๐จ)
๐ + ๐จ = ๐ + ๐. ๐๐จ
๐ = ๐. ๐๐จ
๐จ = ๐๐
Thus, if we add ๐๐ liters of pure juice, we will have ๐๐ liters of a blend that contains ๐๐ liters of pure juice,
meaning that the concentration of juice in this blend is ๐๐%.
c. Write a rational equation that relates the desired percentage ๐ to the amount ๐จ of pure juice that needs to
be added to make a blend that is ๐% juice, where ๐ < ๐ < ๐๐๐. What is a reasonable restriction on the set
of possible values of ๐? Explain your answer.
๐ + ๐จ
๐๐ + ๐จ=
๐
๐๐๐
We need to have ๐๐ < ๐ < ๐๐๐ for the problem to make sense. We already have ๐๐% juice; the percentage
cannot decrease by adding more juice, and we can never have a mixture that is more than ๐๐๐% juice.
d. Suppose that you have added ๐๐ liters of juice to the original ๐๐ liters. What is the percentage of juice in this
blend?
๐
๐๐๐=
๐ + ๐๐
๐๐ + ๐๐= ๐. ๐๐
So, the new blend contains ๐๐% pure juice.
e. Solve your equation in part (c) for the amount ๐จ. Are there any excluded values of the variable ๐? Does this
make sense in the context of the problem?
๐ + ๐จ
๐๐ + ๐จ=
๐
๐๐๐
๐๐๐(๐ + ๐จ) = ๐(๐๐ + ๐จ)
๐๐๐ + ๐๐๐๐จ = ๐๐๐ + ๐๐จ
๐๐๐๐จ โ ๐๐จ = ๐๐๐ โ ๐๐๐
๐จ(๐๐๐ โ ๐) = ๐๐๐ โ ๐๐๐
๐จ =๐๐๐ โ ๐๐๐
๐๐๐ โ ๐
We see from the equation for ๐จ that ๐ โ ๐๐๐. This makes sense because we can never make a ๐๐๐% juice
solution since we started with a diluted amount of juice.
Exercise 4 (5 minutes)
Allow students to work together in pairs or small groups for this exercise. This exercise is a bit different from the
previous example in that the amount of acid comes from a diluted solution and not a pure solution. Be sure that
students set up the numerator correctly. (If there is not enough time to do the entire problem, have students set up the
equations in class and finish solving them for homework.)
MP.4
MP.2
NYS COMMON CORE MATHEMATICS CURRICULUM M1 Lesson 27 ALGEBRA II
Lesson 27: Word Problems Leading to Rational Equations
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4. You have a solution containing ๐๐% acid and a solution containing ๐๐% acid.
a. How much of the ๐๐% solution must you add to ๐ liter of the ๐๐% solution to create a mixture that is ๐๐%
acid?
If we add ๐จ liters of the ๐๐% solution, then the new mixture is ๐ + ๐จ liters of solution that contains
๐. ๐ + ๐. ๐๐จ liters of acid. We want the final mixture to be ๐๐% acid, so we need to solve the equation:
๐. ๐ + ๐. ๐๐จ
๐ + ๐จ= ๐. ๐๐.
Solving this gives
๐. ๐ + ๐. ๐๐จ = ๐. ๐๐(๐ + ๐จ)
๐. ๐ + ๐. ๐๐จ = ๐. ๐๐ + ๐. ๐๐๐จ
๐. ๐๐๐จ = ๐. ๐๐
๐จ = ๐. ๐.
Thus, if we add ๐. ๐ liters of ๐๐% acid solution to ๐ liter of ๐๐% acid solution, the result is ๐. ๐ liters of ๐๐%
acid solution.
b. Write a rational equation that relates the desired percentage ๐ to the amount ๐จ of ๐๐% acid solution that
needs to be added to ๐ liter of ๐๐% acid solution to make a blend that is ๐% acid, where ๐ < ๐ < ๐๐๐.
What is a reasonable restriction on the set of possible values of ๐? Explain your answer.
๐. ๐ + ๐. ๐๐จ
๐ + ๐จ=
๐
๐๐๐
We must have ๐๐ < ๐ < ๐๐ because if we blend a ๐๐% acid solution and a ๐๐% acid solution, the blend will
contain an acid percentage between ๐๐% and ๐๐%.
c. Solve your equation in part (b) for ๐จ. Are there any excluded values of ๐? Does this make sense in the
context of the problem?
๐จ =๐๐ โ ๐
๐ โ ๐๐
We need to exclude ๐๐ from the possible range of values of ๐, which makes sense in context because we can
never reach a ๐๐% acid solution since we started with a solution that was ๐๐% acid.
d. If you have added some ๐๐% acid solution to ๐ liter of ๐๐% acid solution to make a ๐๐% acid solution, how
much of the stronger acid did you add?
The formula in part (c) gives ๐จ =๐๐โ๐๐๐๐โ๐๐
; therefore, ๐จ = ๐. We added ๐ liters of the ๐๐% acid solution to the
๐ liter of ๐๐% acid solution to make a ๐๐% acid mixture.
Closing (2 minutes)
Ask students to summarize the important parts of the lesson in writing, to a partner, or as a class. Use this as an
opportunity to informally assess understanding of the lesson.
Exit Ticket (5 minutes)
MP.4
MP.2
NYS COMMON CORE MATHEMATICS CURRICULUM M1 Lesson 27 ALGEBRA II
Lesson 27: Word Problems Leading to Rational Equations
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Name Date
Lesson 27: Word Problems Leading to Rational Equations
Exit Ticket
Bob can paint a fence in 5 hours, and working with Jen, the two of them painted a fence in 2 hours. How long would it
have taken Jen to paint the fence alone?
NYS COMMON CORE MATHEMATICS CURRICULUM M1 Lesson 27 ALGEBRA II
Lesson 27: Word Problems Leading to Rational Equations
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Exit Ticket Sample Solutions
Bob can paint a fence in ๐ hours, and working with Jen, the two of them painted a fence in ๐ hours. How long would it
have taken Jen to paint the fence alone?
Let ๐ represent the time it would take Jen to paint the fence alone. Then, Bob can paint the entire fence in ๐ hours;
therefore, in one hour he can paint ๐
๐ of the fence. Similarly, Jen can paint
๐
๐ of the fence in one hour. We know that it
took them two hours to complete the job, and together they can paint ๐
๐ of the fence in one hour. We then have to solve
the equation:
๐
๐+
๐
๐=
๐
๐
๐๐
๐๐๐+
๐๐
๐๐๐=
๐๐
๐๐๐
๐๐ + ๐๐ = ๐๐
๐ =๐๐
๐.
Thus, it would have taken Jen ๐ hours and ๐๐ minutes to paint the fence alone.
Problem Set Sample Solutions
1. If two inlet pipes can fill a pool in one hour and ๐๐ minutes, and one pipe can fill the pool in two hours and ๐๐
minutes on its own, how long would the other pipe take to fill the pool on its own?
๐
๐. ๐+
๐
๐=
๐
๐. ๐
We find that ๐ = ๐. ๐๐; therefore, it takes ๐ hours and ๐๐ minutes for the second pipe to fill the pool by itself.
2. If one inlet pipe can fill the pool in ๐ hours with the outlet drain closed, and the same inlet pipe can fill the pool in
๐. ๐ hours with the drain open, how long does it take the drain to empty the pool if there is no water entering the
pool?
๐
๐โ
๐
๐=
๐
๐. ๐
We find that ๐ = ๐๐; therefore, it takes ๐๐ hours for the drain to empty the pool by itself.
3. It takes ๐๐ minutes less time to travel ๐๐๐ miles by car at night than by day because the lack of traffic allows the
average speed at night to be ๐๐ miles per hour faster than in the daytime. Find the average speed in the daytime.
๐๐๐
๐ โ ๐๐=
๐๐๐
๐+
๐
๐
We find that ๐ = ๐๐๐. The time it takes to travel ๐๐๐ miles by car at night is ๐๐๐ minutes, which is ๐ hours. Since ๐๐๐
๐= ๐๐, the average speed in the daytime is ๐๐ miles per hour.
NYS COMMON CORE MATHEMATICS CURRICULUM M1 Lesson 27 ALGEBRA II
Lesson 27: Word Problems Leading to Rational Equations
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4. The difference in the average speed of two trains is ๐๐ miles per hour. The slower train takes ๐ hours longer to
travel ๐๐๐ miles than the faster train takes to travel ๐๐๐ miles. Find the speed of the slower train.
๐๐๐
๐โ
๐๐๐
๐ + ๐= ๐๐
We find that ๐ = ๐, so it takes ๐ hours for the faster train to travel ๐๐๐ miles, and it takes ๐ hours for the slower
train to travel ๐๐๐ miles. The average speed of the slower train is ๐๐ miles per hour.
5. A school library spends $๐๐ a month on magazines. The average price for magazines bought in January was ๐๐
cents more than the average price in December. Because of the price increase, the school library was forced to
subscribe to ๐ fewer magazines. How many magazines did the school library subscribe to in December?
๐๐
๐ + ๐. ๐๐=
๐๐
๐โ ๐
The solution to this equation is ๐. ๐๐, so the average price in December is $๐. ๐๐. Thus the school subscribed to ๐๐
magazines in December.
6. An investor bought a number of shares of stock for $๐, ๐๐๐. After the price dropped by $๐๐ per share, the investor
sold all but ๐ of her shares for $๐, ๐๐๐. How many shares did she originally buy?
๐๐๐๐
๐=
๐๐๐๐
๐ โ ๐+ ๐๐
This equation has two solutions: ๐๐ and ๐๐. Thus, the investor bought either ๐๐ or ๐๐ shares of stock.
7. Newtonโs law of universal gravitation, ๐ญ =๐ฎ๐๐๐๐
๐๐ , measures the force of gravity between two masses ๐๐ and ๐๐,
where ๐ is the distance between the centers of the masses, and ๐ฎ is the universal gravitational constant. Solve this
equation for ๐ฎ.
๐ฎ =๐ญ๐๐
๐๐๐๐
8. Suppose that =๐+๐
๐โ๐๐ .
a. Show that when ๐ =๐๐
and ๐ =๐๐โ๐๐+๐
, the value of ๐ does not depend on the value of ๐.
When simplified, we find that ๐ = ๐; therefore, the value of ๐ does not depend on the value of ๐.
b. For which values of ๐ do these relationships have no meaning?
If ๐ is ๐, then ๐ has no meaning. If ๐ = โ๐, then ๐ has no meaning.
9. Consider the rational equation ๐
๐น=
๐
๐+
๐
๐.
a. Find the value of ๐น when ๐ =๐๐
and ๐ =๐๐
.
๐
๐น=
๐
๐๐โ
+๐
๐๐โ
So ๐น =๐
๐๐.
NYS COMMON CORE MATHEMATICS CURRICULUM M1 Lesson 27 ALGEBRA II
Lesson 27: Word Problems Leading to Rational Equations
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b. Solve this equation for ๐น, and write ๐น as a single rational expression in lowest terms.
There are two approaches to solve this equation for ๐น.
The first way is to perform the addition on the
right:
The second way is to take reciprocals of both
sides and then simplify:
๐
๐น=
๐
๐+
๐
๐
=๐
๐๐+
๐
๐๐
=๐ + ๐
๐๐.
๐น =
๐
๐๐โ + ๐
๐โ
=๐
๐๐๐โ + ๐
๐๐โ
=๐
(๐ + ๐)๐๐โ
.
In either case, we find that ๐น =๐๐
๐+๐.
10. Consider an ecosystem of rabbits in a park that starts with ๐๐ rabbits and can sustain up to ๐๐ rabbits. An equation
that roughly models this scenario is
๐ท =๐๐
๐ +๐
๐ + ๐
,
where ๐ท represents the rabbit population in year ๐ of the study.
a. What is the rabbit population in year ๐๐? Round your answer to the nearest whole rabbit.
If ๐ = ๐๐, then ๐ท = ๐๐. ๐๐; therefore, there are ๐๐ rabbits in the park.
b. Solve this equation for ๐. Describe what this equation represents in the context of this problem.
๐ =๐๐ โ ๐๐ท
๐ท โ ๐๐
This equation represents the relationship between the number of rabbits, ๐ท, and the year, ๐. If we know how
many rabbits we have, ๐๐ < ๐ท < ๐๐, we will know how long it took for the rabbit population to grow that
much. If the population is ๐๐, then this equation says we are in year ๐ of the study, which fits with the given
scenario.
c. At what time does the population reach ๐๐ rabbits?
If ๐ท = ๐๐, then ๐ =๐๐โ๐๐๐๐๐โ๐๐
=โ๐๐๐โ๐๐
= ๐๐; therefore, the rabbit population is ๐๐ in year ๐๐ of the study.
NYS COMMON CORE MATHEMATICS CURRICULUM M1 Lesson 27 ALGEBRA II
Lesson 27: Word Problems Leading to Rational Equations
311
This work is derived from Eureka Math โข and licensed by Great Minds. ยฉ2015 Great Minds. eureka-math.org This file derived from ALG II-M1-TE-1.3.0-07.2015
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Extension:
11. Suppose that Huck Finn can paint a fence in ๐ hours. If Tom Sawyer helps him paint the fence, they can do it in
๐ hours. How long would it take for Tom to paint the fence by himself?
Huck paints the fence in ๐ hours, so his rate of fence painting is ๐
๐ fence per hour. Let ๐ป denote the percentage of the
fence Tom can paint in an hour. Then
๐ fence = ((๐
๐+ ๐ป) fence per hour) โ (๐ hours).
๐ =๐
๐๐
+ ๐ป=
๐
๐๐
+๐๐ป๐
=๐
๐ + ๐๐ป
๐(๐ + ๐๐ป) = ๐
๐๐๐ป = ๐
๐ป =๐
๐๐
So, Tom can paint ๐
๐๐ of the fence in an hour. Thus, Tom would take
๐๐
๐= ๐. ๐ hours to paint the fence by himself.
12. Huck Finn can paint a fence in ๐ hours. After some practice, Tom Sawyer can now paint the fence in ๐ hours.
a. How long would it take Huck and Tom to paint the fence together?
The amount of fence that Huck can paint per hour is ๐
๐, and the amount that Tom can paint per hour is
๐
๐. So,
together they can paint ๐
๐+
๐
๐ of the fence per hour. Suppose the entire job of painting the fence takes ๐
hours. Then, the amount of the fence that is painted is ๐ (๐๐
+๐๐
). Since the entire fence is painted, we need
to solve the equation ๐ (๐๐
+๐๐
) = ๐.
๐ (๐
๐+
๐
๐) = ๐
๐ =๐
๐๐
+๐๐
=๐๐
๐ + ๐=
๐๐
๐๐
So, together they can paint the fence in ๐๐
๐๐ hours, which is ๐ hours and ๐๐ minutes.
NYS COMMON CORE MATHEMATICS CURRICULUM M1 Lesson 27 ALGEBRA II
Lesson 27: Word Problems Leading to Rational Equations
312
This work is derived from Eureka Math โข and licensed by Great Minds. ยฉ2015 Great Minds. eureka-math.org This file derived from ALG II-M1-TE-1.3.0-07.2015
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
b. Tom demands a half-hour break while Huck continues to paint, and they finish the job together. How long
does it take them to paint the fence?
Suppose the entire job of painting the fence takes ๐ hours. Then, Huck paints at a rate of ๐
๐ of the fence per
hour for ๐ hours, so he paints ๐
๐ of the fence. Tom paints at a rate of
๐
๐ of the fence per hour for ๐ โ
๐๐
hour, so
he paints ๐
๐(๐ โ
๐
๐) of the fence. Together, they paint the whole fence; so, we need to solve the following
equation for ๐:
๐
๐๐ +
๐
๐(๐ โ
๐
๐) = ๐
๐
๐๐ +
๐
๐๐ โ
๐
๐๐= ๐
๐
๐๐ +
๐
๐๐ =
๐๐
๐๐
๐๐ (๐
๐๐ +
๐
๐๐) = ๐๐ โ
๐๐
๐๐
๐๐๐ + ๐๐๐ = ๐๐
๐ =๐๐
๐๐.
Thus, it takes ๐๐
๐๐ hours, which is ๐ hours ๐๐ minutes, to paint the fence with Tom taking a
๐
๐ hour break.
c. Suppose that they have to finish the fence in ๐๐๐
hours. Whatโs the longest break that Tom can take?
Suppose the entire job of painting the fence takes ๐
๐ hours, and Tom stops painting for ๐ hours for his break.
Then, Huck paints at a rate of ๐
๐ of the fence per hour for
๐
๐ hours, so he paints
๐
๐๐ of the fence. Tom paints at
a rate of ๐
๐ of the fence per hour for (
๐๐
โ ๐) hours, so he paints ๐
๐(
๐
๐โ ๐) of the fence. Together, they paint
the whole fence; so, we need to solve the following equation for ๐:
๐
๐๐+
๐
๐(
๐
๐โ ๐) = ๐
๐
๐๐+
๐
๐๐โ
๐
๐= ๐
๐๐ (๐
๐๐+
๐
๐๐โ
๐
๐) = ๐๐
๐๐ + ๐๐ โ ๐๐๐ = ๐๐
๐๐ + ๐๐ โ ๐๐ = ๐๐๐
๐ =๐๐
๐๐.
Thus, if Tom takes a break for ๐๐
๐๐ hours, which is ๐ hour and ๐๐ minutes, the fence will be painted in ๐
๐๐
hours.
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