Engage NY Eureka Math 4th Grade Module 4 Mid Module Assessment Answer Key
Eureka Math Grade 4 Module 3 Mid Module Assessment Answer Key
Question 1.
Draw an area model to solve the following. Find the value of the following expressions.
a. 30 × 60
Answer:
Explanation:
Drawn an area model to solve the following expression 30 X 60 = 1,800 as shown above.
b. 3 × 269
Answer:
Explanation:
Drawn an area model to solve the following expression 3 X 269 = 807 as shown above.
Question 2.
Use any place value strategy to multiply.
a. 3 × 68
Answer:
Explanation:
Explanation:
Used place value strategy to multiply 3 X 68 = 204 as shown above.
b. 4 × 371
Answer:
Explanation:
Used place value strategy to multiply 4 X 371 = 1,484 as shown above.
c. 7 × 1,305
Answer:
Explanation:
Used place value strategy to multiply 7 X 1,305 = 9,135 as shown above.
d. 6,034 × 5
Answer:
Explanation:
Used place value strategy to multiply 6,034 X 5 = 30,170 as shown above.
Solve using a model or equation. Show your work and write your answer as a statement.
Question 3.
A movie theater has two rooms. Room A has 9 rows of seats with 18 seats in each row. Room B has three times as many seats as Room A. How many seats are there in both rooms?
Answer:
There are 648 seats in both Room A & Room B,
Explanation:
Given a movie theater has two rooms. Room A has 9 rows of seats with 18 seats in each row. Room B has three times as many seats as Room A.
Number of seats are there in both rooms are
First Room A :
9 X 18 =
7
18
X 9
162
and Room B has three times as many seats as Room A means 3 X 162 =
1
162
X 3
486
Number of seats in Room B are 486 seats,
Now, total number of seats are there in both rooms are
162 + 486 =
162
+486
648 seats in both Room A & Room B.
Question 4.
The high school art teacher has 9 cases of crayons with 52 boxes in each case. The elementary school art teacher has 6 cases of crayons with 104 boxes in each case. How many total boxes of crayons do both teachers have? Is your answer reasonable? Explain.
Answer:
Total boxes of crayons do both teachers have are 1,092,
Yes, answer is reasonable,
Explanation:
Given the high school art teacher has 9 cases of crayons with 52 boxes in each case. The elementary school art teacher has 6 cases of crayons with 104 boxes in each case. Number of total boxes of crayons do both teachers have are High school art teacher have 9 X 52 =
1
52
X 9
468 boxes of caryons,
The elementary school art teacher has 6 X 104 =
2
104
X 6
624 boxes of crayons, therefore total boxes of crayons
do both teachers have are 468 + 624 =
468
+624
1,092, Yes, answer is reasonable,
Question 5.
Last year, Mr. Petersen’s rectangular garden had a width of 5 meters and an area of 20 square meters. This year, he wants to make the garden three times as long and two times as wide.
a. Solve for the length of last year’s garden using the area formula. Then, draw and label the measurements of this year’s garden.
Answer:
Length of last year’s garden is 4 m
Drawn and labeled the measurements of this year’s garden as shown
Explanation:
Given last year, Mr. Petersen’s rectangular garden had a width of 5 meters and an area of 20 square meters.
a. The length of last year’s garden using the area formula. is
5 meters X length = 20 square meters,
length = 20 ÷ 5 = 4 meters
Then, drawn and labeled the measurements of this year’s garden as shown above.
b. How much area for planting will Mr. Petersen have in the new garden?
Answer:
120 square meters area is needed for planting Mr. Petersen have in the new garden,
Explanation:
As the new garden has given this year,
he wants to make the garden three times as long and
two times as wide so length of new garden is 3 X 4 m = 12 meters,
and wide is 2 X 5 m = 10 meters, So area of new garden is
12 meters X 10 meters = 120 square meters.
c. Last year, Mr. Petersen had a fence all the way around his garden. He can reuse all of the fence he had around the garden last year, but he needs to buy more fencing to go around this year’s garden. How many more meters of fencing is needed for this year’s garden than last year’s?
Answer:
26 meters of fencing is needed for this year’s garden than last year’s,
Explanation:
Last year, Mr. Petersen had a fence all the way around his garden.
He can reuse all of the fence he had around the garden last year,
but he needs to buy more fencing to go around this year’s garden.
More meters of fencing is needed for this year’s garden than last year’s
is last year fence is 2( 5 m + 4 m) = 2 X 9 meters = 18 meters,
this year’s garden fencing is 2(10m + 12 m) = 2 X 22 meters =
44 meters, So number of meters of fencing is needed for this year’s
garden than last year’s is 44 meters – 18 meters = 22 meters.
d. Last year, Mr. Petersen was able to plant 4 rows of carrots with 13 plants in each row. This year, he plans to plant twice as many rows with twice as many carrot plants in each. How many carrot plants will he plant this year?
Write a multiplication equation to solve. Assess the reasonableness of your answer.
Answer:
208 carrot plants he will plant this year,
Multiplication equation is 8 X 26,
Assessed the reasonableness of my answer below,
Explanation:
Last year, Mr. Petersen was able to plant 4 rows of carrots with 13 plants in each row. This year, he plans to plant twice as many rows with twice as many carrot plants in each so last
Mr.Petersen planted 4 X 13 =
1
13
X 4
52, this year Mr.Petersen will plant 8 X 26 =
4
26
X 8
208 carrot plants he will plant this year,
So multiplication equation is 8 X 26,
Assessed the reasonableness of my answer above.