Eureka Math Grade 8 Module 4 Mid Module Assessment Answer Key

Engage NY Eureka Math 8th Grade Module 4 Mid Module Assessment Answer Key

Eureka Math Grade 8 Module 4 Mid Module Assessment Task Answer Key

Question 1.
Write and solve each of the following linear equations.
a. Ofelia has a certain amount of money. If she spends $12, then she has \(\frac{1}{5}\) of the original amount left. How much money did Ofelia have originally?
Answer:
Let x be the amount of money ofelia had
x – 12 = \(\frac{1}{5}\)
x – \(\frac{1}{5}\)x – 12 + 12 = \(\frac{1}{5}\) x – \(\frac{1}{5}\) x +12
\(\frac{4}{5}\) x = 12
x= 12 ∙ \(\frac{5}{4}\) = \(\frac{60}{4}\)
Ofelia had $15.00 originally.

b. Three consecutive integers have a sum of 234. What are the three integers?
Answer:
Let x be the first integer
x + x + 1 + x + 2 = 234
3x + 3 = 234
3x = 234 – 3
3x = 231
x = 77
The integers are 77, 78, and 79

c. Gil is reading a book that has 276 pages. He already read some of it last week. He plans to read 20 pages tomorrow. By then, he will be \(\frac{2}{3}\)of the way through the book. How many pages did Gil read last week?
Answer:
Let x be the number of pages gil read last week.
x + 20 = \(\frac{2}{3}\)(276)
x + 20 = 184
x + 20 – 20 = 184 – 20
x = 164
Gil read 164 pages last week

Question 2.
a. Without solving, identify whether each of the following equations has a unique solution, no solution,
or infinitely many solutions.
i. 3x + 5 = – 2
Answer:
Unique

ii. 6(x – 11) = 15 – 4x
Answer:
Unique

iii. 12x + 9 = 8x + 1 + 4x
Answer:
No solution

iv. 2(x – 3) = 10x – 6 – 8x
Answer:
Infinitely many solutions

v. 5x + 6 = 5x – 4
Answer:
No solution

b. Solve the following equation for a number x. Verify that your solution is correct.
– 15 = 8x + 1
Answer:
Engage NY Math 8th Grade Module 4 End of Module Assessment Answer Key 3
-2 = x

-15 = 8(-2) + 1
-15 = -16 + 1
-15 = -15

c. Solve the following equation for a number x. Verify that your solution is correct.
7(2x + 5) = 4x – 9 – x
Answer:
7(2x + 5) = 4x – 9 – x
14x + 35 = 4x – 9 – x
14x + 35 = 3x – 9
14x – 3x + 35 = 3x – 3x – 9
11x + 35 = -9
11x + 35 – 35 = -9 – 35
11x = -44
x = -4

7(2(-4) + 5) = 4(-4) – 9 – (-4)
7(-8 + 5) = -16 – 9 + 4
7(-3) = -25 + 4
-21 = -21

Question 3.
a. Parker paid $4.50 for three pounds of gummy candy. Assuming each pound of gummy candy costs the same amount, complete the table of values representing the cost of gummy candy in pounds.
Engage NY Math 8th Grade Module 4 End of Module Assessment Answer Key 1
Answer:
Engage NY Math 8th Grade Module 4 End of Module Assessment Answer Key 4

b. Graph the data on the coordinate plane.
Engage NY Math 8th Grade Module 4 End of Module Assessment Answer Key 2
Answer:
Engage NY Math 8th Grade Module 4 End of Module Assessment Answer Key 5

c. On the same day, Parker’s friend, Peggy, was charged $5 for 1 \(\frac{1}{2}\) lb. of gummy candy. Explain in terms of the graph why this must be a mistake.
Answer:
Even though 1\(\frac{1}{2}\) pounds of candy isn’t a point on the graph, it is reasonable to believe it will fall in line with the other points. The cost of 1\(\frac{1}{2}\) pounds of candy does not fit the pattern.

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