Practicing the **Bridges in Mathematics Grade 4 Student Book Answer Key Unit 7 Module 4** will help students analyze their level of preparation.

## Bridges in Mathematics Grade 4 Student Book Answer Key Unit 7 Module 4

**Bridges in Mathematics Grade 4 Student Book Unit 7 Module 4 Session 1 Answer Key**

**Two-Part Multiplication**

Question 1.

For the following problems:

- Label the frame to show the 2 numbers that are being multiplied.
- Sketch in the rectangle and divide it into 2 parts.
- Label each of the parts with a multiplication equation.
- Add the partial products to get the answer.

Answer:

Question 2.

For the following problems:

- Multiply the top number by the ones and then by the tens.
- Add the partial products to get the answer.

Example

Answer:

**Reasonable Estimates & Partial Products**

Question 1.

Fill in a bubble for each problem to show which of the estimates is best.

a. 21 × 19

400

600

4,000

6,000

Answer:

400.

Explanation:

The multiplication of 21 × 19 is 399. So the estimation will be 400.

b. 20 × 31

600

700

6,000

7,000

Answer:

600.

Explanation:

The multiplication of 20 × 31 is 620. So the estimation will be 600.

c. 33 × 39

600

800

1,000

2,000

Answer:

1,000.

Explanation:

The multiplication of 33 × 39 is 1,287. So the estimation will be 1,000.

d. 96 × 22

180

1,000

2,000

18,000

Answer:

2,000.

Explanation:

The multiplication of 96 × 22 is 2,112. So the estimation will be 2,000.

Question 2.

Use partial products to solve each problem below. Draw lines between the digits to show which numbers you multiplied. The first one has been done as an example.

Answer:

**Bridges in Mathematics Grade 4 Student Book Unit 7 Module 4 Session 2 Answer Key**

**Roll Your Own Double-Digit Multiplication Problems**

Directions:

- Choose a die numbered 1-6 or 4-9.
- Roll it as many times as you need to fill in each of the boxes below. You can write each number you roll in any box on the sheet, but once all the boxes are filled, you can’t change them.
- Use the method you just learned in class to solve your problems.
- When you’re finished, trade papers with a classmate and have him or her check your answers.

**CHALLENGE**

Answer:

**School Supplies**

Question 1.

Mr. Wu got 35 boxes of crayons for his fourth graders. Every box had 24 crayons in it. When he got home from the store, he decided to give each of his own 3 children one of the boxes of crayons. How many crayons did he take to his fourth graders?

a. Circle the equation below that best represents this problem. (The letter c stands for crayons.)

(35 × 24) – 3 = c

35 × (24 – 3) = c

(35 – 3) × 24 = c

(35 + 3) × 24 = c

Answer:

(35 – 3) × 24 = c.

Explanation:

Given that Mr. Wu got 35 boxes of crayons for his fourth graders and every box had 24 crayons in it. When he got home from the store, he decided to give each of his own 3 children one of the boxes of crayons. So the number of crayons did he take to his fourth graders is (35 – 3) × 24 = c. So the equation will be (35 – 3) × 24 = c.

b. Use the standard multiplication algorithm or partial products to solve the problem. Show all your work.

Answer:

768 crayons.

Explanation:

The standard multiplication of (35 – 3) × 24 = c. So the value of c is

32 × 24 = 768 crayons.

Question 2.

Ms. Penny got 18 packs of felt markers for her fifth graders. Each pack had 36 markers in it. When she got back to her classroom, she put half the packs away. She dumped out the rest of the markers and divided them into 6 equal sets, one for each table. How many markers did each table get?

a. Write an equation to represent this problem.

Answer:

The equation is (9×36)÷6 = 54.

Explanation:

Given that Ms. Penny got 18 packs of felt markers for her fifth graders and each pack had 36 markers in it. She put half the packs away which is 18÷2 = 9 and she dumped out the rest of the markers which is 9×36 =324 markers and divided them into 6 equal sets which is 324÷6 = 54 markers. So each table will get 54 markers. The equation will be (9×36)÷6.

b. Solve the problem. Show all your work.

Answer:

(9×36)÷6 = 54.

Explanation:

The equation is (9×36)÷6

= 324÷6

= 54.

Question 3.

**CHALLENGE** The office got 15 cartons of envelopes. Each carton had 12 boxes of envelopes in it. Each box had 54 envelopes in it. How many envelopes did they get in all?

a. Write an equation to represent this problem.

Answer:

The equation is (15×12)×54.

Explanation:

Given that the office got 15 cartons of envelopes and each carton had 12 boxes of envelopes in it. So the total boxes of envelopes are 15×12 which is 180 boxes. As each box had 54 envelopes in it. So the number of envelopes did they get in all is 54×180 which is 9,720 envelopes. The equation will be (15×12)×54.

b. Use the standard multiplication algorithm or partial products to solve the problem. Show all your work.

Answer:

(15×12)×54 = 9,720.

Explanation:

The standard multiplication of the equation is (15×12)×54

= 180×54

= 9,720.

**Bridges in Mathematics Grade 4 Student Book Unit 7 Module 4 Session 3 Answer Key**

**Reviewing Multiplication Methods**

Read and review these multiplication methods with your class. Then complete the example in each strategy’s box.

Method A

Use basic fact strategies.

ex: 4 × 124

Double it and then double it again.

Answer:

4 × 124 = 496.

Explanation:

Given the equation is 4 × 124. So by using double strategy it will be

4 × 124 =

Double 124 which is 124+124 = 248,

Double 248 which is 248+248 = 496.

Method B

Multiply to get 4 partial products and add them up.

ex: 27 × 34

Answer:

27 × 34 = 918.

Explanation:

Method C

Multiply by the tens and then by the ones. Add the partial products.

ex: 16 × 25

Answer:

16 × 25 = 400.

Explanation:

Method D

Use the over strategy.

ex: 3 × 299

Answer:

3 × 299 = 897.

Explanation:

Method E

Use the standard algorithm.

ex: 46 × 73

Answer:

46 × 73 = 3,358.

Explanation:

**Evaluating Multiplication Methods**

For each problem on this page and the next,

- Write the letter of the method you think will work best.
- Use the method to solve the problem. Show all your work.

Question 1.

People need to drink about 8 cups of water each day. Zoo elephants need to drink about 158 quarts of water each day. How many cups of water are there in 158 quarts of water? (Remember that there are 4 cups in a quart.)

I think method ______________ will work best for this problem.

Answer:

The number of cups will be 40 cups.

I think method division will work best for this problem.

Explanation:

Given that people need to drink about 8 cups of water each day and Zoo elephants need to drink about 158 quarts of water each day. The number of cups of of water are there in 158 quarts of water is 158÷4 which is 40 cups approx. As there are 4 cups in a quart. So the number of cups will be 40 cups.

Question 2.

So far, the elephant keeper has brought in 40 gallons of water for the elephants. How many cups of water are there in 40 gallons? (Remember that there are 16 cups in a gallon.)

I think method ______________ will work best for this problem.

Answer:

The number of cups of water is 640 cups.

I think method standard multiplication will work best for this problem.

Explanation:

Given that the elephant keeper has brought in 40 gallons of water for the elephants. So the number of cups of water are there in 40 gallons is 40×16 which is 640 cups.

Question 3.

Zoo elephants eat about 175 pounds of food a day. Most of their food is hay, but they also eat fruits and vegetables. How many pounds of food would it take to feed 26 elephants for one day?

I think method ______________ will work best for this problem.

Answer:

The number of pounds of food 4,550 pounds.

I think method standard multiplication will work best for this problem.

Explanation:

Given that Zoo elephants eat about 175 pounds of food a day and most of their food is hay, but they also eat fruits and vegetables. So the number of pounds of food would it take to feed 26 elephants for one day is 175×26 which is 4,550 pounds.

Question 4.

Each elephant at our zoo gets about 45 pounds of vegetables a day. How many pounds of vegetables does it take to feed one elephant for 49 days (7 weeks)?

I think method ______________ will work best for this problem.

Answer:

The number of pounds will be 2,205 pounds.

Explanation:

Given that each elephant at our zoo gets about 45 pounds of vegetables a day. So the number of pounds of vegetables does it take to feed one elephant for 49 days will be 45×49 which is 2,205 pounds.

Question 5.

**CHALLENGE** An elephant can spend up to 18 hours a day eating. How many hours would that total in one year? About how many months’ worth of time is that?

Answer:

The number of hours is 6,570 hours and in months it will be 9 months.

Explanation:

Given that an elephant can spend up to 18 hours a day eating and how many hours would that total in one year. So for one year their will be 365 days. So in one year elephant can eat 18×365 which is 6,570 hours. So to get the number of months we will divide hours with 730 which is 6,570÷730 = 9 months.

**Addition & Subtraction Review**

Question 1.

Use the standard addition algorithm to solve the problems below.

Answer:

Question 2.

Use the standard subtraction algorithm to solve the problems below.

Answer:

Question 3.

Fill in the missing numbers to make each equation true.

100 = _________ + (30 – 5)

100 × 2 × _____________ = 1,000

4 = _____________ ÷ (3 × 2)

___________ = 100 – (28 + 13)

100 = _________ + (30 – 5)

Answer:

100 = 75 + (30 – 5).

Explanation:

Let the missing number be X, so the equation will be 100 = X + (30 – 5),

100 = X + 25

X = 100-25

= 75.

100 × 2 × _____________ = 1,000

Answer:

100 × 2 × 5 = 1,000

Explanation:

Let the missing number be X.

Given 100 × 2 × _____________ = 1,000 which is

100 × 2 × X = 1,000

200 × X = 1,000

X = 1,000/200

= 5.

4 = _____________ ÷ (3 × 2)

Answer:

4 = 24 ÷ (3 × 2)

Explanation:

Let the missing number be X.

Given 4 = _____________ ÷ (3 × 2) which is

4 = X ÷ (3 × 2)

4 = X ÷ 6

X = 4×6

= 24.

100 – (28 + 13)

Answer:

100 – (28 + 13) = 59.

Explanation:

Let the missing number be X.

Given 100 – (28 + 13) = X which is

X = 100 – (41)

X = 59.

CHALLENGE

18 × 2 = ___________ × 4

90 ÷ ____________ = 5 × 9

18 × 2 = ___________ × 4

Answer:

18 × 2 = 9 × 4

Explanation:

Let the missing number be X.

Given 18 × 2 = ___________ × 4 which is

18 × 2 = X × 4

36 = X × 4

X = 36÷4

= 9.

90 ÷ ____________ = 5 × 9

Answer:

90 ÷ 2 = 5 × 9.

Explanation:

Let the missing number be X.

Given 90 ÷ X = 5 × 9 which is

90 ÷ X = 45

X = 90 ÷ 45

= 2.

Question 4.

**CHALLENGE** Fill in the missing digits.

Answer:

**Bridges in Mathematics Grade 4 Student Book Unit 7 Module 4 Session 4 Answer Key**

**Secret Paths & Multiplication Tables**

Question 1.

Use multiplication and division to find the secret path through each maze. The starting and ending points are marked for you. You can only move one space up, down, over, or diagonally each time. Write four equations to explain the path through the maze.

Answer:

Question 2.

Complete the multiplication charts below.

Answer: