## Engage NY Eureka Math 4th Grade Module 3 Lesson 11 Answer Key

### Eureka Math Grade 4 Module 3 Lesson 11 Problem Set Answer Key

Question 1.

Solve the following expressions using the standard algorithm, the partial products method, and the area model.

a. 4 2 5 Ã— 4

4 (400 + 20 + 5)

(4 Ã— __400___ ) + (4 Ã— _20____ ) + (4 Ã— __5___ )

Answer:

425 X 4 =

Standard Algorithm =

Â 1,2

425

X 4

1,700

Partial Products =

425

X 4

Â 20

80

+ 1600

1,700

Area Model =

425 X 4 = 1,700,

Explanation:

Solved the following expression 425 X 4 using the standard algorithmÂ =

we add same time of multiplying

1,2

425

X 4

1,700

(4 X 5 ones = 20 ones )+ (4 X 2 tens = 8 tens) +

(4 X 4 hundreds = 16 hundreds) = 20 + 80 + 1,600 = 1,700,

partial products =

425

X 4

Â 20— 4 X 5

80—-4 X 20

+ 1600–4 X 400

1,700

and the area model is as shown in the picture above,

So 425 X 4 = 1,700.

b. 5 3 4 Ã— 7

7 ( __500_ + _30___ + _4___ )

( _7_ Ã— __500___ ) + ( 7__ Ã— __30___ ) + ( 7__ Ã— _4___ )

Answer:

534Â X 7 =

Standard Algorithm =

Â 2,2

534

X 7

3,738

Partial Products =

534

X 7

Â 28

210

+3500

3,738

Area Model =

534Â X 7 = 3,738,

Explanation:

Solved the following expression 534 X 7 using the standard algorithmÂ =

we add same time of multiplying

2,2

534

X 7

3,738

(7 X 4 ones = 28 ones )+ (7 X 3 tens = 21 tens) +

(7 X 5 hundreds = 35 hundreds) = 28 + 210 + 3,500 = 3,738,

partial products =

534

X 7

Â 28— 7 X 4

210—-7 X 30

+3500–7 X 500

3,738

and the area model is as shown in the picture above

So 534 X 7 = 3,738.

c. 2 0 9 Ã— 8

8__ ( _200___ + _9___ )

( _8_ Ã— _200____ ) + ( _8_ Ã— __9___ )

Answer:

209Â X 8 =

Standard Algorithm =

Â 7,

209

X 8

1,672

Partial Products =

209

X 8

Â 72

0000

+1600

1,672

Area Model =

209Â X 8 = 1,672,

Explanation:

Solved the following expression 209 X 8 using the standard algorithmÂ =

we add same time of multiplying

7

209

X 8

1,672

(8 X 9 ones = 72 ones )+ (8 X 0 tens = 0 tens) +

(8 X 2 hundreds = 16 hundreds) = 72 + 0 + 1,600 = 1,672,

partial products =

209

X 8

Â 72— 8 X 9

000—-8 X 0

+1600–8 X 200

1,672

and the area model is as shown in the picture above

So 209 X 8 = 1,672.

Question 2.

Solve using the partial products method. Caylaâ€™s school has 258 students. Janetâ€™s school has 3 times as many students as Caylaâ€™s. How many students are in Janetâ€™s school?

Answer:

There are in Janet’s school are 774 students,

Explanation:

Given Caylaâ€™s school has 258 students. Janetâ€™s school has 3 times as many students as Caylaâ€™s.

So there are number of students in Janet’s school are

3 X 258 students =

Partial Products =

258

X 3

024

150

+600

774

So there are in Janet’s school are 774 students.

Question 3.

Model with a tape diagram and solve. 4 times as much as 467

Answer:

4 X 467 = 1,868,

Explanation:

Modeled with a tape diagram as shown above and 4 times as much as 467 is 4 X 467 =

2,2

467

X 4

1,868

So, 4 X 467 = 1,868.

Solve using the standard algorithm, the area model, the distributive property, or the partial products method.

Question 4.

5,131 Ã— 7

Answer:

Standard algorithm:

5,131

XÂ Â 7

35,917

Explanation:

Given expression as 5,131 X 7 solving using standard algorithm method as

2

5,131

XÂ Â 7

35,917

First we multiply (7 X 1 one = 7 ones) + (7 X 3 tens = 21 tens) +

(7 X 1 hundred = 7 hundreds) + (7 X 5 thousands = 35 thousands),

So 7 X 5,131 = 35,917.

Question 5.

3 times as many as 2,805,

Answer:

3 X 2,805 = 8,415,

Explanation:

Given to find 3 times as many as 2,805 using the distributive property as shown below

3 X 2,000 + 3 X 800 + 3 X 0 + 3 X 5 =

6,000 + 2,400 + 0 + 15 = 8,415,

So 3 times as many as 2,805 = 8,415.

Question 6.

A restaurant sells 1,725 pounds of spaghetti and 925 pounds of linguini every month. After 9 months, how many pounds of pasta does the restaurant sell?

Answer:

23,850 pounds of pasta the restaurant saled in 9 months,

Explanation:

Given a restaurant sells 1,725 pounds of spaghetti and 925 pounds of linguini every month. After 9 months, number of pounds of pasta does the restaurant sell is

9 X (1,725 +925) pounds = 9 X 2,650 =

Â 5,4

2,650

XÂ Â 9

23,850

Solved using the area model as shown above.

### Eureka Math Grade 4 Module 3 Lesson 11 Exit Ticket Answer Key

Question 1.

Solve using the standard algorithm, the area model, the distributive property, or the partial products method.

2,809 Ã— 4

Answer:

2,809

X 4

11,236

2,809 X 4 = 11,236,

Explanation:

Solved using the distributive property as shown below

(4 X 2,000) + (4 X 800) + (4 X 0) + (4 X 9) =

8,000 + 3,200 + 0 + 36 =11,236,

So, 2,809 X 4 = 11,236.

Question 2.

The monthly school newspaper is 9 pages long. Mrs. Smith needs to print 675 copies. What will be the total number of pages printed?

Answer:

Total number of pages printed are 6,075,

Explanation:

Given the monthly school newspaper is 9 pages long.

and Mrs. Smith needs to print 675 copies,

So the total number of pages printed are 9 X 6,075,

Solving using Partial Products as shown below

6,075

XÂ 9

45—9Â X 5

630–9 X 7

000– 9 X 0

5400–9 X 6

6,075

therefore, total number of pages printed are 6,075.

### Eureka Math Grade 4 Module 3 Lesson 11 Homework Answer Key

Question 1.

Solve the following expressions using the standard algorithm, the partial products method, and the area model.

a. 3 0 2 Ã— 8

8 (300 + 2)

(8 Ã— _300____ ) + (8 Ã— __2___ )

Answer:

302 X 8 =

Standard Algorithm =

Â 1

302

X 8

2,416

Partial Products =

302

X 8

Â 16

00

+2400

2,416

Area Model =

302 X 8 = 2,416,

Explanation:

Solved the following expression 302 X 8 using the standard algorithmÂ =

we add same time of multiplying

Â 1

302

X 8

2,416

(8 X 2 ones = 16 ones )+ (8 X 0 tens = 0 tens) +

(8 X 3 hundreds = 24 hundreds) = 16 + 0 + 2,400 = 2,416,

partial products =

302

X 8

16

00

+2400

2,416

and the area model is as shown in the picture above,

So 302 X 8 = 2,416.

b. 2 1 6 Ã— 5

5 ( _200___ + _10___ + _6___ )

( 5__ Ã— __200___ ) + ( _5_ Ã— __10___ ) + ( 5__ Ã— _6___ )

Answer:

216 X 5 =

Standard Algorithm =

Â 3

216

X 5

1,080

Partial Products =

216

X 5

Â 30

50

+1000

1,080

Area Model =

216 X 5 = 1,080,

Explanation:

Solved the following expression 302 X 8 using the standard algorithmÂ =

we add same time of multiplying

Â 3

216

X 5

1,080

(5 X 6 ones = 30 ones )+ (5 X 1 tens = 5 tens) +

(5 X 2 hundreds = 10 hundreds) = 30 + 50 + 1,000 = 1,080,

partial products =

216

X 5

Â 30

50

+1000

1,080

and the area model is as shown in the picture above,

So 216 X 5 = 1,080.

c. 5 9 3 Ã— 9

_9_ ( _500___ + _90___ + _3___ )

( _9_ Ã— __500___ ) + ( _9_ Ã— __90___ ) + ( _9_ Ã— __3__ )

Answer:

593 X 9 =

Standard Algorithm =

Â 8,2

593

X 9

5,337

Partial Products =

593

X 9

Â 27

810

+4500

5,337

Area Model =

593 X 9 = 5,337,

Explanation:

Solved the following expression 593 X 9 using

the standard algorithmÂ =

we add same time of multiplying

8,2

593

X 9

5,337

(9 X 3 ones = 27 ones )+ (9 X 9 tens = 81 tens) +

(9 X 5 hundreds = 45 hundreds)= 27 + 810 + 4,500 = 5,337,

partial products =

593

X 9

Â 27

810

+4500

5,337

and the area model is as shown in the picture above,

So 593 X 9 = 5,337.

Question 2.

Solve using the partial products method. On Monday, 475 people visited the museum. On Saturday, there were 4 times as many visitors as there were on Monday. How many people visited the museum on Saturday?

Answer:

Number of people visited the museum on Saturday are 1,900,

Partial Products =

475

X 4

Â 20

280

+1600

1,900

Explanation:

Given to solve using the partial products method. On Monday, 475 people visited the museum.

On Saturday, there were 4 times as many visitors as there were on Monday.

475

X 4

20

280

+1600

1,900

Number of people visited the museum on Saturday are 1,900.

Question 3.

Model with a tape diagram and solve. 6 times as much as 384

Answer:

6 X 384 = 2,304,

Explanation:

Modeled with a tape diagram as shown above and 6 times as much as 384 is 4 X 467 =

5,2

384

X 6

2,304

So, 4 X 467 = 2,304.

Solve using the standard algorithm, the area model, the distributive property, or the partial products method.

Question 4.

6,253 Ã— 3

Answer:

6,253 X 3 = 18,759,

Explanation:

Given expression 6,253 X 3 using the distributive property method

we solve

6000 X 3 + 200 X 3 + 50 X 3 + 3 X 3 =

18,000 + 600 + 150 + 9 = 18,759,

theerfore 6,253 X 3 = 18,759.

Question 5.

7 times as many as 3,073

Answer:

7 times as many as 3,073 is 21,511,

Explanation:

We solve 7 times as many as 3,073 the partial products method as

3,073

XÂ Â 7

Â 21—- 7 x 3

490—–7 X 70

0000—- 7 X 0

21000—-7 X 3,000

21,511

So, 7 times as many as 3,073 is 21,511.

Question 6.

A cafeteria makes 2,516 pounds of white rice and 608 pounds of brown rice every month. After 6 months, how many pounds of rice does the cafeteria make?

Answer:

After 6 months 18,744 pounds of rice the cafeteria make,

Explanation:

Given a cafeteria makes 2,516 pounds of white rice and 608 pounds of brown rice every month. Total number of pounds of rice every month is 2,516 + 608 = 3,124 pounds,

Now number of pounds of rice does the cafeteria make after 6 months is 6 X 3,124 using the standard algorithm we solve as

1,2

3,124

XÂ Â 6

18,744

First we multiply 6 X 4 ones = 24 ones, 6 X 2 tens = 12 tens,

6 X 1 hundred = 6 hundreds, 6 X 3 thousands = 18 thousands,

24 + 120 + 600 + 1,8000 = 18,744,

therefore, after 6 months 18,744 pounds of rice the cafeteria make.