## Engage NY Eureka Math 7th Grade Module 4 Lesson 17 Answer Key

### Eureka Math Grade 7 Module 4 Lesson 17 Example Answer Key

Example 1.
A 5 – gallon container of trail mix is 20% nuts. Another trail mix is added to it, resulting in a 12 – gallon container of trail mix that is 40% nuts.
a. Write an equation to describe the relationships in this situation.
Let j represent the percent of nuts in the second trail mix that is added to the first trail mix to create the resulting 12 – gallon container of trail mix.
0.4(12) = 0.2(5) + j(12 – 5)

b. Explain in words how each part of the equation relates to the situation.
Quantity = Percent×Whole
(Resulting gallons of trail mix)(Resulting % of nuts) = (1st trail mix in gallons)(% of nuts) + (2nd trail mix in gallons)(% of nuts)

c. What percent of the second trail mix is nuts?
4.8 = 1 + 7j
4.8 – 1 = 1 – 1 + 7j
3.8 = 7j
j ≈ 0.5429
About 54% of the second trail mix is nuts.

Example 2.
Soil that contains 30% clay is added to soil that contains 70% clay to create 10 gallons of soil containing 50% clay. How much of each of the soils was combined?
Let x be the amount of soil with 30% clay.
(1st soil amount)(% of clay) + (2nd soil amount)(% of clay) = (resulting amount)(resulting % of clay)
(0.3)(x) + (0.7)(10 – x) = (0.5)(10)
0.3x + 7 – 0.7x = 5
– 0.4x + 7 – 7 = 5 – 7
– 0.4x = – 2
x = 5
5 gallons of the 30% clay soil and 10 – 5 = 5, so 5 gallons of the 70% clay soil must be mixed to make 10 gallons of 50% clay soil.

### Eureka Math Grade 7 Module 4 Lesson 17 Exercise Answer Key

Opening Exercise
Imagine you have two equally – sized containers. One is pure water, and the other is 50% water and 50% juice. If you combined them, what percent of juice would be the result?  25% of the resulting mixture is juice because $$\frac{0.5}{2}$$ = $$\frac{1}{4}$$.

If a 2 – gallon container of pure juice is added to 3 gallons of water, what percent of the mixture is pure juice? Let x represent the percent of pure juice in the resulting juice mixture. If a 2 – gallon container of juice mixture that is 40% pure juice is added to 3 gallons of water, what percent of the mixture is pure juice?  → How many gallons of the juice mixture is pure juice?
(2 gallons)(0.40) = 0.8 gallons

→ What percent is pure juice out of the resulting mixture?
16%

→ Does this make sense relative to the prior problem?
Yes, because the mixture should have less juice than in the prior problem

If a 2 – gallon juice cocktail that is 40% pure juice is added to 3 gallons of pure juice, what percent of the resulting mixture is pure juice?  What is the difference between this problem and the previous one?
Instead of adding water to the two gallons of juice mixture, pure juice is added, so the resulting liquid contains 3.8 gallons of pure juice.
What percent is pure juice out of the resulting mixture?
Let x represent the percent of pure juice in the resulting mixture.
x(5) = 40%(2) + 100%(3)
5x = 0.8 + 3
5x = 3.8
x = 0.76
The mixture is 76% pure juice.

Exercise 1.
Represent each situation using an equation, and show all steps in the solution process.
a. A 6 – pint mixture that is 25% oil is added to a 3 – pint mixture that is 40% oil. What percent of the resulting mixture is oil?
Let x represent the percent of oil in the resulting mixture.
0.25(6) + 0.40(3) = x(9)
1.5 + 1.2 = 9x
2.7 = 9x
x = 0.3
The resulting 9 – pint mixture is 30% oil.

b. An 11 – ounce gold chain of 24% gold was made from a melted down 4 – ounce charm of 50% gold and a golden locket. What percent of the locket was pure gold?
Let x represent the percent of pure gold in the locket.
0.5(4) + (x)(7) = 0.24(11)
2 + 7x = 2.64
2 – 2 + 7x = 2.64 – 2
$$\frac{7x}{7}$$ = $$\frac{0.64}{7}$$
x≈0.0914
The locket was about 9% gold.

c. In a science lab, two containers are filled with mixtures. The first container is filled with a mixture that is 30% acid. The second container is filled with a mixture that is 50% acid, and the second container is 50% larger than the first. The first and second containers are then emptied into a third container. What percent of acid is in the third container?
Let m represent the total amount of mixture in the first container.
0.3m is the amount of acid in the first container.
0.5(m + 0.5m) is the amount of acid in the second container.
0.3m + 0.5(m + 0.5m) = 0.3m + 0.5(1.5m) = 1.05m is the amount of acid in the mixture in the third container.
m + 1.5m = 2.5m is the amount of mixture in the third container. So, $$\frac{1.05 m}{2.5 m}$$ = 0.42 = 42% is the percent of acid in the third container.

Exercise 2.
The equation (0.2)(x) + (0.8)(6 – x) = (0.4)(6) is used to model a mixture problem.
a. How many units are in the total mixture?
6 units

b. What percents relate to the two solutions that are combined to make the final mixture?
20% and 80%

c. The two solutions combine to make 6 units of what percent solution?
40%

d. When the amount of a resulting solution is given (for instance, 4 gallons) but the amounts of the mixing solutions are unknown, how are the amounts of the mixing solutions represented?
If the amount of gallons of the first mixing solution is represented by the variable x, then the amount of gallons of the second mixing solution is 4 – x.

### Eureka Math Grade 7 Module 4 Lesson 17 Problem Set Answer Key

Question 1.
A 5 – liter cleaning solution contains 30% bleach. A 3 – liter cleaning solution contains 50% bleach. What percent of bleach is obtained by putting the two mixtures together?
Let x represent the percent of bleach in the resulting mixture.
0.3(5) + 0.5(3) = x(8)
1.5 + 1.5 = 8x
3 ÷ 8 = 8x ÷ 8
x = 0.375
The percent of bleach in the resulting cleaning solution is 37.5%.

Question 2.
A container is ﬁlled with 100 grams of bird feed that is 80% seed. How many grams of bird feed containing 5% seed must be added to get bird feed that is 40% seed?
Let x represent the amount of bird feed, in grams, to be added.
0.8(100) + 0.05x = 0.4(100 + x)
80 + 0.05x = 40 + 0.4x
80 – 40 + 0.05x = 40 – 40 + 0.4x
40 + 0.05x = 0.4x
40 + 0.05x – 0.05x = 0.4x – 0.05x
40 ÷ 0.35 = 0.35x ÷ 0.35
x ≈ 114.3
About 114.3 grams of the bird seed containing 5% seed must be added.

Question 3.
A container is ﬁlled with 100 grams of bird feed that is 80% seed. Tom and Sally want to mix the 100 grams with bird feed that is 5% seed to get a mixture that is 40% seed. Tom wants to add 114 grams of the 5% seed, and Sally wants to add 115 grams of the 5% seed mix. What will be the percent of seed if Tom adds 114 grams? What will be the percent of seed if Sally adds 115 grams? How much do you think should be added to get 40% seed?
If Tom adds 114 grams, then let x be the percent of seed in his new mixture. 214x = 0.8(100) + 0.05(114). Solving, we get the following:
x = $$\frac{80 + 5.7}{214}$$ = $$\frac{85.7}{214}$$ ≈ 0.4005 = 40.05%.
If Sally adds 115 grams, then let y be the percent of seed in her new mixture. 215y = 0.8(100) + 0.05(115). Solving, we get the following:
y = $$\frac{80 + 5.75}{215}$$ = $$\frac{85.75}{215}$$ ≈ 0.3988 = 39.88%.
The amount to be added should be between 114 and 115 grams. It should probably be closer to 114 because 40.05% is closer to 40% than 39.88%.

Question 4.
Jeanie likes mixing leftover salad dressings together to make new dressings. She combined 0.55 L of a 90% vinegar salad dressing with 0.45 L of another dressing to make 1 L of salad dressing that is 60% vinegar. What percent of the second salad dressing was vinegar?
Let c represent the percent of vinegar in the second salad dressing.
0.55(0.9) + (0.45)(c) = 1(0.6)
0.495 + 0.45c = 0.6
0.495 – 0.495 + 0.45c = 0.6 – 0.495
0.45c = 0.105
0.45c ÷ 0.45 = 0.105 ÷ 0.45
c ≈ 0.233
The second salad dressing was around 23% vinegar.

Question 5.
Anna wants to make 30 mL of a 60% salt solution by mixing together a 72% salt solution and a 54% salt solution. How much of each solution must she use?
Let s represent the amount, in milliliters, of the first salt solution.
0.72(s) + 0.54(30 – s) = 0.60(30)
0.72s + 16.2 – 0.54s = 18
0.18s + 16.2 = 18
0.18s + 16.2 – 16.2 = 18 – 16.2
0.18s = 1.8
s = 10
Anna needs 10 mL of the 72% solution and 20 mL of the 54% solution.

Question 6.
A mixed bag of candy is 25% chocolate bars and 75% other filler candy. Of the chocolate bars, 50% of them contain caramel. Of the other filler candy, 10% of them contain caramel. What percent of candy contains caramel?
Let c represent the percent of candy containing caramel in the mixed bag of candy.
0.25(0.50) + (0.75)(0.10) = 1(c)
0.125 + 0.075 = c
0.2 = c
In the mixed bag of candy, 20% of the candy contains caramel.

Question 7.
A local fish market receives the daily catch of two local fishermen. The first fisherman’s catch was 84% fish while the rest was other non – fish items. The second fisherman’s catch was 76% fish while the rest was other non – fish items. If the fish market receives 75% of its catch from the first fisherman and 25% from the second, what was the percent of other non – fish items the local fish market bought from the fishermen altogether?
Let n represent the percent of non – fish items of the total market items.
0.75(0.16) + 0.25(0.24) = n
0.12 + 0.06 = n
0.18 = n
The percent of non – fish items in the local fish market is 18%.

### Eureka Math Grade 7 Module 4 Lesson 17 Exit Ticket Answer Key

A 25% vinegar solution is combined with triple the amount of a 45% vinegar solution and a 5% vinegar solution resulting in 20 milliliters of a 30% vinegar solution.
Question 1.
Determine an equation that models this situation, and explain what each part represents in the situation.
Let s represent the number of milliliters of the first vinegar solution.
(0.25)(s) + (0.45)(3s) + (0.05)(20 – 4s) = (0.3)(20)
(0.25)(s) represents the amount of the 25% vinegar solution.
(0.45)(3s) represents the amount of the 45% vinegar solution, which is triple the amount of the 25% vinegar solution.
(0.05)(20 – 4s) represents the amount of the 5% vinegar solution, which is the amount of the remainder of the solution.
(0.3)(20) represents the result of the mixture, which is 20 mL of a 30% vinegar solution.

Question 2.
Solve the equation and find the amount of each of the solutions that were combined.