## 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:

-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.

Answer:

b. Graph the data on the coordinate plane.

Answer:

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.