Eureka Math Algebra 2 Module 1 Lesson 27 Answer Key

Engage NY Eureka Math Algebra 2 Module 1 Lesson 27 Answer Key

Eureka Math Algebra 2 Module 1 Lesson 27 Example Answer Key

Working together, it takes Sam, Jenna, and Francisco two hours to paint one room. When Sam works alone, he can paint one room in 6 hours. When Jenna works alone, she can paint one room in 4 hours. Determine how long It would take Francisco to paint one room on his own.
Answer:
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 6 hours on his own, then in one hour he can paint \(\frac{1}{6}\) of the room. Similarly, Jenna can paint \(\frac{1}{4}\) 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 \(\frac{1}{f}\) of the room. So, in one hour, Sam has painted \(\frac{1}{6}\) of the room, Jenna has painted \(\frac{1}{4}\) of the room, and all three together can paint \(\frac{1}{2}\) the room, leading to the following equation for how much can be painted in one hour:
\(\)
A common multiple of the denominators is 12f. Multiplying both sides by 12f gives us:
\(\frac{12 f}{6}+\frac{12 f}{4}+\frac{12 f}{f}=\frac{12 f}{2}\)
2f + 3f + 12 = 6f,
which leads us to the value off:
f = 12.
So, Francisco can paint the room in 12 hours on his own.

Eureka Math Algebra 2 Module 1 Lesson 27 Exercise Answer Key

Exercise 1
Anne and Maria play tennis almost every weekend. So far, Anne has won 12 out of 20 matches.
a. How many matches will Anne have to win in a row to improve her winning percentage to 75%?
Answer:
Suppose that Anne has already won 12 of 20 matches, and let m represent the number of additional matches she must win to raise her winning percentage to 75%. After playing and winning all of those additional m matches, she has won 12 + m matches out of a total of 20 + m matches played. Her winning percentage is then \(\frac{12+m}{20+m}\), and we want to find the value of m that solves the equation
\(\frac{12+m}{20+m}\) = 0.75.
Multiply both sides by 20 + m.
12 + m = 0.75(20 + m)
12 + m = 15 + 075m
Solve for m:
0.25m = 3
m = 12
So, Anne would need to win 12 matches in a row in order to improve her winning percentage to 75%.

b. How many matches will Anne have to win In a row to improve her winning percentage to 90%?
Answer:
This situation is similar to that for part (a), except that we want a winning percentage of 0.90, instead of 0.75. Again, we let m represent the number of matches Anne must win consecutively to bring her winning percentage up to 90%.
\(\frac{12+m}{20+m}\) = 0.90.
Solve for m:
12 + m = 0.90(20 +m)
12 + m = 18 + 0.90m
0. 10m = 6
m = 60
In order for Anne to bring her winning percentage up to 90%, she would need to win the next 60 consecutive matches.

c. Can Anne reach a winning percentage of 100%?
Answer:
Allow students to come to the conclusion that Anne will never reach a winning percentage of 100% because she has already lost 8 matches.

d. After Anne has reached a winning percentage of 90% by winning consecutive matches as in part (b), how many matches can she now lose In a row to have a winning percentage of 50%?
Answer:
Recall from part (b) that she had won 72 matches out of 80 to reach a winning percentage of 90%. We will now assume that she loses the next k matches in a row. Then, she will have won 72 matches out of 80 + k matches, and we want to know the value of k that makes this a 50% win rafe.
\(\frac{72}{80+k}\) = 0.50
Solving the equation:
72 = 0.50(80 + k)
72 = 40 + 0.50k
32 = 0.50k
64 = k
Thus, after reaching a 90% winning percentage in 80 matches, Anne can lose 64 matches in a row to drop to a 50% winning percentage.

Exercise 2.
Melissa walks 3 miles to the house of a friend and returns home on a bike. She averages 4 miles per hour faster when cycling than when walking, and the total time for both trips Is two hours. Find her walking speed.
Answer:
Using the relationship d = r ∙ t, we have t = \(\frac{d}{r}\). The time it takes for Melissa to walk to her friend’s house is \(\frac{3}{r}\), and the time to cycle back is \(\frac{3}{r+4}\). Thus, we can write an equation that describes the combined time for both trips:
\(\frac{3}{r}+\frac{3}{r+4}\) = 2
A common multiple of the denominators is r(r + 4), so we multiply both sides of the equation by r(r + 4).
3(r + 4) + 3r = 2r(r + 4)
3r + 12 + 3r = 2r2 + 8r
2r2 + 2r – 12 = 0
2(r – 2) (r + 3) = 0
Thus, r = – 3 or r = 2. Since r represents Melissa’s speed, It does not make sense for r to be negative. So, the only solution is 2, which means that Melissa’s walking speed is 2 miles per hour.

Exercise 3.
You have 10 liters of a juice blend that is 60% juice.
a. How many liters of pure juice need to be added in order to make a blend that Is 75% juice?
Answer:
We start off with 10 liters of a blend containing 60% juice. Then, this blend contains 0. 60(10) = 6 liters of juice in the 10 liters of mixture. If we add A liters of pure juice, then the concentration of juice in the blend is \(\frac{6+A}{10+A}\). We want to know which value of A makes this blend 75% juice.
\(\frac{6+A}{10+A}\) = 0.75
6 + A = 0.75(10 + A)
6 + A = 7.5 + 0.75A
0.25A = 1.5
A = 6
Thus, if we add 6 liters of pure juice, we have 16 liters of a blend that contains 12 liters of juice, meaning that the concentration of juice in this blend is 75%.

b. How many liters of pure juice need to be added in order to make a blend that is 90% juice?
Answer:
\(\frac{6+A}{10+A}\) = 0.90
6 + A = 0.9(10 + A)
6 + A = 9 + 0.9A
3 = 0.1A
A = 30
Thus, if we add 30 liters of pure juice, we will have 40 liters of a blend that contains 36 liters of pure juice, meaning that the concentration of juice in this blend is 90%.

c. Write a rational equation that relates the desired percentage p to the amount A of pure juice that needs to be added to make a blend that is p% juice, where 0 < p < 100. What is a reasonable restriction on the set of possible values of p? Explain your answer.
Answer:
\(\frac{6+A}{10+A}=\frac{p}{100}\)
We need to have 60 < p < 100 for the problem to make sense. We already have 60% juice; the percentage cannot decrease by adding more juice, and we can never have a mixture that is more than 100% juice.

d. Suppose that you have added 15 liters of juice to the original 10 liters. What is the percentage of juice in this blend?
Answer:
\(\frac{p}{100}=\frac{6+15}{10+15}\) = 0.84
So, the new blend contains 84% pure juice.

e. Solve your equation in part (c) for the amount A. Are there any excluded values of the variable p? Does this make sense in the context of the problem?
Answer:
\(\frac{6+A}{10+A}=\frac{p}{100}\)
100 (6 + A) = p(10 + A)
600 + 100A = 10p + pA
100A – pA = 10p – 600
A(100 – p) = 10p – 600
\(\frac{10 p-600}{100-p}\)
We see from the equation for A that p ≠ 100. This makes sense because we can never make a 100% juice solution since we started with a diluted amount of juice.

Exercise 4.
You have a solution containing 10% acid and a solution containing 30% acid.
a. How much of the 30% solution must you add to 1 liter of the 10% solution to create a mixture acid?
Answer:
If we add A liters of the 30% solution, then the new mixture is 1 + A liters of solution that contains 0.1 + 0. 3A liters of acid. We want the final mixture to be 22% acid, so we need to solve the equation:
\(\frac{0.1+0.3 A}{1+A}\) = 0.22
Solving this gives
0.1 + 0. 3A = 0.22(1 + A)
0.1 + 0. 3A = 0.22 + 0. 22A
0.08A = 0.12
A = 1.5.
Thus, if we add 1.5 liters of 30% acid solution to 1 liter of 10% acid solution, the result is 2.5 acid solution.

b. Write a rational equation that relates the desired percentage p to the amount A of 30% acid needs to be added to 1 liter of 10% acid solution to make a blend that is p% acid, where 0 < p < 100. What is a reasonable restriction on the set of possible values of p? Explain your answer.
Answer:
\(\frac{0.1+0.3 A}{1+A}=\frac{p}{100}\)
We must have 10 < p < 30 because if we blend a 10% acid solution and a 30% acid solution, the blend will contain an acid percentage between 10% and 30%.

c. Solve your equation in part (b) for A. Are there any excluded values of p? Does this make sense in the context of the problem?
Answer:
\(\frac{10-p}{p-30}\)
We need to exclude 30 from the possible range of values of p, which makes sense in context because we can never reach a 30% acid solution since we started with a solution that was 10% acid.

d. If you have added some 30% acid solution to 1 liter of 10% acid solution to make a 26% acid solution, how much of the stronger acid did you add?
Answer:
The formula in part (c) gives A = \(\frac{10-26}{26-30}\), therefore, A = 4. We added 4 liters of the 30% acid solution to the 1 liter of 10% acid solution to make a 26% acid mixture.

Eureka Math Algebra 2 Module 1 Lesson 27 Problem Set Answer Key

Question 1.
If two inlet pipes can fill a pool in one hour and 30 minutes, and one pipe can fill the pool In two hours and 30 minutes on its own, how long would the other pipe take to fill the pool on its own?
Answer:
\(\frac{1}{2.5}+\frac{1}{x}=\frac{1}{1.5}\)
We find that x = 3.75; therefore, it takes 3 hours and 45 minutes for the second pipe to fill the pool by itself.

Question 2.
If one inlet pipe can fill the pool in 2 hours with the outlet drain closed, and the same inlet pipe can fill the pool in 2. 5 hours with the drain open, how long does it take the drain to empty the pool if there is no water entering the pool?
Answer:
\(\frac{1}{2}-\frac{1}{x}=\frac{1}{2.5}\)
We find that x = 10; therefore, it takes 10 hours for the drain to empty the pool by itself.

Question 3.
It takes 36 minutes less time to travel 120 miles by car at night than by day because the lack of traffic allows the average speed at night to be 10 miles per hour faster than in the daytime. Find the average speed in the daytime.
Answer:
\(\frac{120}{t-36}=\frac{120}{t}+\frac{1}{6}\)
We find that t = 180. The time it takes to travel 120 miles by car at night is 180 minutes, which is 3 hours. Since \(\frac{120}{3}\) = 40, the average speed in the daytime is 40 miles per hour.

Question 4.
The difference in the average speed of two trains is 16 miles per hour. The slower train takes 2 hours longer to travel 170 miles than the faster train takes to travel 150 miles. Find the speed of the slower train.
Answer:
\(\frac{150}{t}-\frac{170}{t+2}\) = 16
We find that t = 3, s0 it takes 3 hours for the faster train to travel 150 miles, and it takes 5 hours for the slower train to travel 170 miles. The average speed of the slower train is 34 miles per hour.

Question 5.
A school library spends $80 a month on magazines. The average price for magazines bought in January was 70 cents more than the average price in December. Because of the price increase, the school library was forced to subscribe to 7 fewer magazines. How many magazines did the school library subscribe to in December?
Answer:
\(\frac{80}{x+0.70}=\frac{80}{x}-7\)
The solution to this equation is 2.50, so the average price in December is $2. 50. Thus the school subscribed to 32 magazines in December.

Question 6.
An investor bought a number of shares of stock for $1,600. After the price dropped by $10 per share, the investor sold all but 4 of her shares for $1, 120. How many shares did she originally buy?
Answer:
\(\frac{1600}{x}=\frac{1120}{x-4}\) + 10
This equation has two solutions: 32 and 20. Thus, the investor bought either 32 or 20 shares of stock.

Question 7.
Newton’s law of universal gravitation, F = \(\frac{G m_{1} m_{2}}{r^{2}}\) measures the force of gravity between two masses m1 and m2, where r is the distance between the centers of the masses, and G is the universal gravitational constant. Solve this equation for G.
Answer:
G = \(\frac{F r^{2}}{m_{1} m_{2}}\)

Question 8.
Suppose that = \(\frac{x+y}{1-x y}\).
a. Show that when x = \(\frac{1}{a}\) and y = \(\frac{2 a-1}{a+2}\) the value of t does not depend on the value of a.
Answer:
When simplified, we find that t = 2; therefore, the value of t does not depend on the value of a.

b. For which values of a do these relationships have no meaning?
Answer:
If a is 0, then x has no meaning. If a = – 2, then y has no meaning.

Question 9.
Consider the rational equation \(\frac{1}{R}=\frac{1}{x}+\frac{1}{y}\).
a. Find the value of R when x = \(\frac{2}{5}\) and y = \(\frac{3}{4}\)
Answer:
\(\frac{1}{R}=\frac{1}{2 / 5}+\frac{1}{3 / 4}\)
So R = \(\frac{6}{23}\).

b. Solve this equation for R, and write R as a single rational expression in lowest terms.
Answer:
There are two approaches to solve this equation for R.
The first way is to perform the addition on the right:
Eureka Math Algebra 2 Module 1 Lesson 27 Problem Set Answer Key 1

The second way is to take reciprocals of both sides and then simplify:
Eureka Math Algebra 2 Module 1 Lesson 27 Problem Set Answer Key 2
In either case, we find that R = \(\frac{x y}{x+y}\)

Question 10.
Consider an ecosystem of rabbits In a park that starts with 10 rabbits and can sustain up to 60 rabbits. An equation that roughly models this scenario is
P = \(\frac{60}{1+\frac{5}{t+1}}\)
where P represents the rabbit population in year t of the study.
a. What is the rabbit population in year 10? Round your answer to the nearest whole rabbit.
Answer:
If t = 10, then P = 41.25; therefore, there are 41 rabbits in the park.

b. Solve this equation for t. Describe what this equation represents In the context of this problem.
Answer:
t = \(\frac{60-6 P}{P-60}\)
This equation represents the relationship between the number of rabbits, P, and the year, t. If we know how many rabbits we have, 10 < P < 60, we will know how long it took for the rabbit population to grow that much. If the population is 10, then this equation says we are in year 0 of the study, which fits with the given scenario.

c. At what time does the population reach 50 rabbits?
Answer:
If P = 50, then t = \(\frac{60-300}{50-60}=\frac{-240}{-10}\) = 24; therefore, the rabbit population is 50 in year 24 of the study.

Eureka Math Algebra 2 Module 1 Lesson 27 Exit Ticket Answer Key

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?
Answer:
Let x represent the time it would take Jen to paint the fence alone. Then, Bob can paint the entire fence in 5 hours; therefore, in one hour he can paint \(\frac{1}{5}\) of the fence. Similarly, Jen can paint \(\frac{1}{x}\) of the fence in one hour. We know that it took them two hours to complete the job, and together they can paint \(\frac{1}{2}\) of the fence in one hour. We then have to solve the equation:
\(\frac{1}{5}+\frac{1}{x}=\frac{1}{2}\)
\(\frac{2 x}{10 x}+\frac{10}{10 x}=\frac{5 x}{10 x}\)
2x + 10 = 5x
x = \(\frac{10}{3}\).
Thus, it would have taken Jen 3 hours and 20 minutes to paint the fence alone.

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