problem solving multiplication lesson 4 10 answer key

Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers

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Big Ideas 4th Grade Chapter 4 Multiply by Two-Digit Numbers Solution Key

Students can easily understand the concepts in-depth with the help of Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers. We have provided all the questions, answers, along with explanations. Also, most of the explanations are given with images for the best learning for students. In order to become a math expert, you must refer to the Big Ideas Multiply by Two-Digit Numbers chapter. Verify the step-by-step procedure to solve a problem to make your preparation easy.

Lesson 1: Multiply by Tens

Lesson 4.1 Multiply by Tens

Multiply by tens homework & practice 4.1.

Lesson 2: Estimate Products

Lesson 4.2 Estimate Products

Estimate products homework & practice 4.2.

Lesson 3: Use Area Models to Multiply Two-Digit Numbers

Lesson 4.3 Use Area Models to Multiply Two-Digit Numbers

Use area models to multiply two-digit numbers homework & practice 4.3.

Lesson 4: Use the Distributive Property to Multiply Two-Digit Numbers

Lesson 4.4 Use the Distributive Property to Multiply Two-Digit Numbers

Use the distributive property to multiply two-digit numbers homework & practice 4.4.

Lesson 5: Use Partial Products to Multiply Two-Digit Numbers

Lesson 4.5 Use Partial Products to Multiply Two-Digit Numbers

Use partial products to multiply two-digit numbers homework & practice 4.5.

Lesson 6: Multiply Two-Digit Numbers

Lesson 4.6 Multiply Two-Digit Numbers

Multiply two-digit numbers homework & practice 4.6.

Lesson 7: Practice Multiplication Strategies

Lesson 4.7 Practice Multiplication Strategies

Practice multiplication strategies homework & practice 4.7.

Lesson 8: Problem Solving: Multiplication with Two-Digit Numbers

Lesson 4.8 Problem Solving: Multiplication with Two-Digit Numbers

Problem solving: multiplication with two-digit numbers homework & practice 4.8.

Performance Task

Multiply by Two-Digit Numbers Performance Task

Multiply by two-digit numbers activity, multiply by two-digit numbers chapter practice, 4.3 use area models to multiply two-digit numbers, 4.4 use distributive property to multiply two-digit numbers, 4.5 use partial products to multiply two-digit numbers, 4.6 multiply two-digit numbers, 4.7 practice multiplication strategies.

Answer: The Place- value method. From the above pattern, we can conclude that the result has different place-values of 6

Explanation: The position of 3 is different in the given 4 multiplications. So, 2 × 3 = 6 2 × 30 = 60 2 × 300 = 600 2 × 3000 = 6,000 From the above pattern, we can conclude that the result has different place-values of 6 Repeated Reasoning How can the pattern above help you find 20 × 30?

Answer: 20 × 30 = 600

Explanation: You can think of 20 as two tens and 30 as Three tens. So, 20 × 30 = 2 × 1 ten × 3 × 1 ten = 2 tens × 2 × 3 = 6 × 2 tens = 600 Think and Grow: Multiply by Multiples of Tens You can use place value and properties to multiply two-digit numbers by multiples of ten. Example Find 40 × 20. One Way: Use place value. 40 × 20 = 40 × ____ tens = ____ tens = _____ So, 40 × 20 = _____.

Answer: 800

Explanation: By using the Place-value method, 40 × 20 = 40 × 2 tens = 80 tens = 800 So, 40 × 20 = 800 Another Way: Use the Associative Property of Multiplication. 40 × 20 = 40 × (2 × 10) Rewrite 20 as 2 × 10. = (40 × 2) × 10 Associative Property of Multiplication = ____ × 10 = ____ So, 40 × 20 = _____.

Answer: 360

Explanation: Using the Place-value method, 12 × 30 = 12 × 3 tens = 36 tens = 360 So, 12 × 30 = 360 Another Way: Use the Associative Property of Multiplication 12 × 30 = 12 × (3 × 10) Rewrite 30 as 3 × 10. = (12 × 3) × 10 Associative Property of Multiplication = ____ × 10 = _____ So, 12 × 30 = ____.

Explanation: By using the Associative Property of Multiplication, 12 × 30 = 12 × (3 × 10) = (12 × 3) × 10 = 36 × 10 = 360 Note : Associative Property of Multiplication Take 3 numbers a, b, c. By using Associative Property, we can write a × (b × c) = (a × b) × c Show and Grow Find the product.

Question 1. 70 × 40 = _____

Answer: 2800

Explanation: By using the Associative Property of Multiplication, 70 × 40 = 70 × (4 × 10) = (70 × 4) × 10 = 280 × 10 = 2800 Note :  Associative Property of Multiplication Take 3 numbers a, b, c. By using Associative Property, we can write a × (b × c) = (a × b) × c

Question 2. 50 × 80 = ____

Answer: 4,000

Explanation: By using the Associative Property of Multiplication, 50 × 80 = 50 × (8 × 10) = (50 × 8) × 10 = 400 × 10 = 4000 Note :  Associative Property of Multiplication Take 3 numbers a, b, c. By using Associative Property, we can write a × (b × c) = (a × b) × c

Question 3. 24 × 90 = _____

Answer: 2160

Explanation: By using the Associative Property of Multiplication, 24 × 90 = 24 × (9 × 10) = (24 × 9) × 10 = (8 × 3 × 9) × 10 = 216 × 10 = 2160 Note :  Associative Property of Multiplication Take 3 numbers a, b, c. By using Associative Property, we can write a × (b × c) = (a × b) × c

Question 4. 45 × 60 = _____

Answer: 2700

Explanation: By using the Associative Property of Multiplication, 45 × 60 = 45 × (6 × 10) = (45 × 6) × 10 = (5 × 9 × 6) × 10 = 270 × 10 = 2700 Note :  Associative Property of Multiplication Take 3 numbers a, b, c. By using Associative Property, we can write a × (b × c) = (a × b) × c Apply and Grow: Practice Find the product.

Question 5. 90 × 10 = _____

Answer: 900

Explanation: By using the place-value method, 90 × 10 = 10 × 9 tens = 90 tens = 900 So, 90 × 10 = 900

Question 6. 40 × 60 = ____

Answer: 2400 Explaination: By using the place-value method, 40 × 60 = 40 × 6 tens = 4 tens × 6 tens = 24 × tens × tens = 2400 So, 40 × 60 = 2400

Question 7. 20 × 70 = _____

Answer: 1400

Explanation: By using the place-value method, 70 × 20 = 70 × 2 tens = 7 tens × 2 tens = 14 × tens × tens = 1400 So, 70 × 20 = 1400

Question 8. 11 × 30 = ____

Answer: 330

Explanation: By using the place-value method, 11 × 30 = 11 × 3 tens = 33 tens = 330 So, 11 × 30 = 330

Question 9. 12 × 40 = ____

Answer: 480

Explanation: By using the place-value method, 12 × 40 = 12 × 4 tens = 48 tens = 480 So, 12 × 40 = 480

Question 10. 15 × 50 = _____

Answer: 750

Explanation: By using the place-value method, 15 ×50 = 15 × 5 tens = 75 tens = 750 So, 15 × 50 = 750

Question 11. 30 × 13 = _____

Answer: 390

Explanation: By using the place-value method, 13 × 30 = 13 × 3 tens = 39 tens = 390 So, 13 × 30 = 390

Question 12. 10 × 76 = _____

Answer: 760

Explanation: By using the place-value method, 10 × 76 = 76 × 1 ten = 76 tens = 760 So, 76 × 10 = 760

Question 13. 40 × 25 = ____

Answer: 1,000

Explanation: By using the place-value method, 25 × 40 = 25 × 4 tens = 5 × 5× 4 tens = 100 tens = 1,000 So, 25 × 40 = 1,000 Find the missing factor.

Question 14. 50 × ____ = 1,500

Answer: 30 Explanation; Let the missing number be X So, 50 × X = 1,500 X = 1,500 / 50 = 30 Hence, the value of X is: 30

Question 15. 20 × ____ = 1,800

Answer: 90 Explanation; Let the missing number be X So, 20 × X = 1,800 X = 1,800 /20 = 90 Hence, the value of X is: 90

Question 16. 60 × ___ = 4,200

Answer: 70 Explanation; Let the missing number be X So, 60 × X = 4,200 X = 4,200 / 60 = 70 Hence, the value of X is: 70

Question 17. ____ × 80 = 6,400

Answer: 80 Explanation; Let the missing number be X So, X × 80 = 6,400 X = 6,400 / 80 = 80 Hence, the value of X is: 80

Question 18. ____ × 90 = 3,600

Answer: 40 Explanation; Let the missing number be X So, X × 90 = 3,600 X = 3,600 / 90 = 40 Hence, the value of X is: 40

Question 19. ____ × 70 = 3,500

Answer: 50 Explanation; Let the missing number be X So, X × 70 = 3,500 X = 3,500 / 70 = 50 Hence, the value of X is: 50 Compare.

Answer: 60 × 30 is equal to 1,800

Explanation: 60 × 30 = 1,800 Given numbers are: 1,800 and 1,800 By comparing 2 values, we can conclude that 1,800 is equal to 1,800

Answer: 480 is greater than 460

Explanation: 40 × 12 = 480 Given numbers are: 480 and 460 By comparing 2 values, we can conclude that480 is greater than 460

Answer: 2,250 is less than 2,340

Explanation: 25 × 90 = 2,250 Given numbers are 2,250 and 2,340 By comparing 2 values, we can conclude that 2,250 is less than 2,340

Answer: 200 days

Explanation: Given that it takes 10 days to film 1 episode of a Television show. So, The number of days it will take to film a 20-episode season is: 20 × 10 = 200 days

Question 25. YOU BE THE TEACHER Newton says that the product of two multiples of ten will always have exactly two zeros. Is he correct? Explain.

Answer: He is correct

Explanation: Let us suppose 2 numbers 10 and 20 which are the two multiples of 10. Now, 10 × 20 = 200 According to Newton, the product of two multiples of ten will always have exactly two zeroes. So, from the above multiplication, we can say that Newton is correct. Think and Grow: Modeling Real Life

Question 26 A library has 2,124 new books. The library has 40 empty shelves. Each shelf holds 35 books. How many not books will fit on the empty shelves?

Answer: 724

Explanation: Multiply to find how many books will hold on the shelves. 40 × 35 = 35 × (4 × 10) Rewrite 40 as 5 × 10. = (35 × 4) × 10 Associative Property of Multiplication = 140 × 10 = 1,400 So, 1,400 cans fit in the boxes. Subtract the number of books that will hold on the shelves from the total number of books collected. The number of books that will not hold on the shelves = 2,124 – 1,400 = 724 So, 724 books will not hold on the shelves

Question 27. An apartment building has 15 floors. Each floor is 10 feet tall. An office building has 30 floors. Each floor is 13 feet tall. How much taller is the office building than the apartment building?

Answer: 240 feet

Explanation: Given that an apartment has 15 floors and in that, each floor is 10 feet tall. So, the height of the apartment = 15 × 10 = 150 feet Given that an office building has 30 floors and in that, each floor is 13 feet tall. So, the height of the office building = 30 × 13 = 390 feet Now, to calculate how much taller an office building than the apartment, we have to subtract both the heights of the apartment and the office building. So, The difference in height between the office building and the apartment = 390 feet – 150 feet = 240 feet. From the above, we can conclude that the office building is 240 feet taller than the apartment.

Question 28. You burn 35 calories each hour you spend reading and 50 calories each hour you spend playing board games. In 2 weeks, you spend14 hours reading and 28 hours playing board games. How many calories do you burn reading and playing board games?

Answer: The calories burned during reading in 2 weeks = 490 calories The calories burned during playing board games = 1400 calories

Explanation: Given that, The calories burned during reading is: 35 calories each hour The calories burned during playing board games is: 50 calories each hour It is also given that In 2 weeks, the time spend on reading and Playing Board Games are 14 hours and 28 hours So, to calculate the calorie consumption in these 2 weeks, we have to multiply the number of hours and the number of calories. So, The calories burned during reading in 2 weeks = 490 calories The calories burned during playing board games in 2 weeks = 1400 calories

Find the product.

Question 1. 30 × 10 = _____

Answer: 300

Explanation: The product of multiply by tens can be done in 2 ways. They are: A) The place-value method  B) The Associative Property of Multiplication A) By using the place-value method: 30 × 10 = 10 × 3 tens = 1 ten × 3 tens = 3 × 1 ten × 1 ten = 3 × 100 = 300 B) The Associative Property of Multiplication: 30 × 10 = 3 × (10 × 10) = (3 × 10) × 10 = 30 × 10 = 300 Note :  Associative Property of Multiplication Take 3 numbers a, b, c. By using Associative Property, we can write a × (b × c) = (a × b) × c

Question 2. 20 × 90 = _____

Answer: 1800

Explanation: The product of multiply by tens can be done in 2 ways. They are: A) The place-value method  B) The Associative Property of Multiplication A) By using the place-value method: 20 × 90 = 20 × 9 tens = 2 tens × 9 tens = 18 × 1 ten × 1 ten = 18 × 100 = 1800 B) The Associative Property of Multiplication: 20 × 90 = 20 × (9 × 10) = (20 × 9) × 10 =(5 × 4 × 9) × 10 = 180 × 10 = 1800 Note :  Associative Property of Multiplication Take 3 numbers a, b, c. By using Associative Property, we can write a × (b × c) = (a × b) × c

Question 3. 50 × 70 = _____

Answer: 3500

Explanation: The product of multiply by tens can be done in 2 ways. They are: A) The place-value method  B) The Associative Property of Multiplication A) By using the place-value method: 50 × 70 = 50 × 7 tens = 5 tens × 7 tens = 35 × 1 ten × 1 ten = 35 × 100 = 3500 B) The Associative Property of Multiplication: 50 × 70 = 50 × (7 × 10) = (50 × 7) × 10 =(5 × 10 × 7) × 10 = 350 × 10 = 3500 Note :  Associative Property of Multiplication Take 3 numbers a, b, c. By using Associative Property, we can write a × (b × c) = (a × b) × c

Question 4. 40 × 13 = ______

Answer: 520

Explanation: The product of multiply by tens can be done in 2 ways. They are: A) The place-value method  B) The Associative Property of Multiplication A) By using the place-value method: 40 × 13 = 13 × 4 tens = 13 × 4 tens = 52 × 1 ten = 52 × 10 = 520 B) The Associative Property of Multiplication: 40 × 13 = 13 × (4 × 10) = (13 × 4) × 10 =(13 ×2× 2) × 10 = 54 × 10 = 540 Note :  Associative Property of Multiplication Take 3 numbers a, b, c. By using Associative Property, we can write a × (b × c) = (a × b) × c

Question 5. 27 × 60 = _____

Answer: 1620

Explanation: The product of multiply by tens can be done in 2 ways. They are: A) The place-value method  B) The Associative Property of Multiplication A) By using the place-value method: 27 × 60 = 27 × 6 tens = 3 × 9 × 6 tens = 162 × 1 ten = 162 × 10 = 1620 B) The Associative Property of Multiplication: 27 × 60 = 27 × (6 × 10) = (27 × 6) × 10 =(6 × 3 × 9) × 10 = 162 × 10 = 1620 Note :  Associative Property of Multiplication Take 3 numbers a, b, c. By using Associative Property, we can write a × (b × c) = (a × b) × c

Question 6. 80 × 56 = _____

Answer: 4480

Explanation: The product of multiply by tens can be done in 2 ways. They are: A) The place-value method  B) The Associative Property of Multiplication A) By using the place-value method: 80 × 56 = 56 × 8 tens = 7 × 8 × 8 tens = 448 × 1 ten = 448 × 10 = 4480 B) The Associative Property of Multiplication: 56 × 80 = 56 × (8 × 10) = (56 × 8) × 10 =(8 × 7 × 8) × 10 = 448 × 10 = 4480 Note :  Associative Property of Multiplication Take 3 numbers a, b, c. By using Associative Property, we can write a × (b × c) = (a × b) × c Find the missing factor.

Question 7. 70 × ____ = 2,100

Explanation: Let the missing number be X So, 70 × X = 2,100 X = 2,100 / 70 = 30 Hence, the value of X is: 30

Question 8. ____ × 10 = 900

Answer: 90 Let the missing number be X So, X × 10 = 900 X = 900 / 10 =90 Hence, the value of X is: 90

Question 9. 40 × ____ = 1,600

Answer: 40 Let the missing number be X So, 40 × X = 1,600 X = 1,600 / 40 = 40 Hence, the value of X is: 40

Question 10. ____ × 20 = 1,600

Answer: 80 Let the missing number be X So, X× 20 = 1,600 X = 1,600 / 20 = 80 Hence, the value of X is: 80

Question 11. 30 × ____ = 1,800

Answer: 60 Let the missing number be X So, 30 × X = 1,800 X = 1,800 / 30 = 60 Hence, the value of X is: 60

Question 12. ____ × 50 = 3,000

Answer: 60 Let the missing number be X So, X × 50 = 3,000 X = 3,000 / 50 = 60 Hence, the value of X is: 60 Compare.

Answer: 7,200 is less than 8,100

Explanation: 90 × 80 = 7,200 Given numbers are: 7,200 and 8,100 By comparing 2 values, we can conclude that 7,200 is less than 8,100

Answer: 1,200 is greater  than 1,020

Explanation: 60 ×17 = 1,020 Given numbers are: 1,200 and 1,020 By comparing 2 values, we can conclude that 1,200 is greater than 1,020

Answer: 2,380 is equal to 2,380

Explanation: 34 × 70 = 2,380 Given numbers are:2,380 and 2,380 By comparing the 2 values, we can conclude that 2,380 is equal to2,380

Answer: 15 kilometers

Explanation: Given that a shallow moonquake occurred 20 kilometers below the moon’s surface and a deep moonquake occurs 35 meters deeper than a shallow moonquake. Now, to calculate how much distance the deep moonquake occurred from the surface, we have to subtract the distance that a shallow moonquake occurred from the deep moonquake occurred. Hence, The distance below the surface the deep moonquake occurred = 30 – 25 = 15 kilometers

Question 18. Writing Explain how you can use 20 × 10 = 200 to find 20 × 12.

Answer: By using the Associative Property of Multiplication, 20 × 10 = 10 × (2 × 10) = (10 × 2) × 10 = 20 × 10 = 200 By using the same method, we can also find the value of 20 × 12. Now, By using the Associative Property of Multiplication, 20 × 12 = 12 × (2 × 10) = (12 × 2) × 10 = 24 × 10 = 240 Note :  Associative Property of Multiplication Take 3 numbers a, b, c. By using Associative Property, we can write a × (b × c) = (a × b) × c

Question 19. DIG DEEPER! The product of a number and twice that number is 800. What are the numbers?

Answer: 20, 40

Explanation: Let the number be X Given that, X × 2X = 800 Take X as a common multiple. Hence, X × X ( 1 × 2) = 800 X × X × 2 = 800 X × X = 800/2 = 400 X × X = 20 × 20 From the above, we can conclude that the value of X is 20 Hence, the value of twice the X is  20 × 2 = 40 So, The numbers that can give the product 800 are 20, 40

Question 20. Modeling Real Life There are 506 new plants in a greenhouse. A worker programs a robot to arrange the plants into14 rows with 30 plants in each row. How many plants will fit in the rows?

Answer: 420

Explanation: Given that there are 506 new plants in a greenhouse and a robot can arrange the plants into 14 rows with 30 plants each. So, to find how many plants will fit in the row, we have to multiply 14 and 30( Since the robot arranges the plants in rows) Now, By using the Associative Property of Multiplication, 14 × 30 =  14 × (3 × 10) = (14 × 3) × 10 = (2 × 7 × 3) × 10 = 42 × 10 = 420 From the above, we can conclude that 420 plants will fit in the row.

Question 21. Modeling Real Life The world’s largest pool is 13 meters longer than the total length of 20 Olympic pools. An Olympic pool is 50 meters long. How long is the world’s largest pool?

Answer: 1013 meters

Explanation: Given that an Olympic pool has a length of 50 meters. But, there are 20 Olympic pools. So, to find the total length of the Olympic Pool, we have to multiply the number of pools and the length of each pool. By using the Associative Property of Multiplication, 50 × 20 = 50 × (2 × 10) = (50 × 2) × 10 = ( 5 × 10 × 2) × 10 = 100 × 10 = 1,000 meters So, the total length of the 20 Olympic pools = 1,000 meters The Question also mentions that the world’s largest pool is 13 meters longer than the total length of 20 Olympic pools. Hence, The length of the World’s largest pool = 1,000 + 13 = 1,013 meters. So, the length of the World’s largest pool is 1,013 meters Review & Refresh Find the value of the underlined digit.

Question 22. 52,61 8

Answer: The place-value of 8 in the given number is: 8

Explanation: We can find the position of any given number by using the place- value method. From this, we can conclude that the place-value of 8 is: 8

Question 23. 3 7 9,021

Answer: The place-value of 7 in the given number is: 70,000

Explanation: We can find the position of any given number by using the place- value method. From this, we can conclude that the place-value of 7 in the given number is: 70,000

Explanation:

Question 24. 2 03,557

Answer: The place-value of 2 in the given number is: 200,000

Explanation: We can find the position of any given number by using the place- value method. From this, we can conclude that the place-value of 2 in the given number is: 200,000

Question 25. 497, 3 84

Answer: The place-value of 3 in the given number is: 300

Explanation: We can find the position of any given number by using the place- value method. From this, we can conclude that the place-value of 3 in the given number is: 300

Answer: Let your Expression be 20 ×25. The given Partner Expressions are: A) 21 × 24  B) 26 × 38  C) 23 × 17  D) 42 × 23 By Comparing your Expression and your Partner Expression, A) 500 is less than 504.

Answer: 1,000 will be closer to 1,073

Question 1. 27 × 50

Answer: 1500

Explanation: Let 27 be Rounded to 30. Now, we have to find the result of 30 × 50. We can find the product of multiples of ten by using two methods. They are: A) The place-value method  B) Associative Property of Multiplication A) By using the Associative Property of Multiplication, 30 × 50 = 30 × (5 × 10) = (30 × 5) × 10 = ( 3 × 10 × 5) × 10 = 150 × 10 = 1,500 B) By using the place-value method, 30 × 50 = 30 × 5 tens = 3 tens × 5 tens = 15 × 1 ten × 1ten = 15 × 10 × 10 = 1500 So, 27 × 50 can be rounded to 1,500

Question 2. 42 × 14

Answer: 600

Explanation: Let 42 be Rounded to 40 Let 14 be Rounded to 15 Now, we have to find the result of 40 × 15. We can find the product of multiples of ten by using two methods. They are: A) The place-value method  B) Associative Property of Multiplication A) By using the Associative Property of Multiplication, 40 × 15 = 15 × (4 × 10) = (15 × 4) × 10 = ( 3 × 4 × 5) × 10 = 60 × 10 = 600 B) By using the place-value method, 40 × 15 = 15 × 4 tens = 60 tens = 60 × 10 = 600 So, 42 × 16 can be rounded to 600

Question 3. 61 × 73

Answer: 4,200

Explanation: Let 61 be Rounded to 60 Let 73 be Rounded to 70 Now, we have to find the result of 60 × 70. We can find the product of multiples of ten by using two methods. They are: A) The place-value method  B) Associative Property of Multiplication A) By using the Associative Property of Multiplication, 60 × 70 = 60 × (7 × 10) = (60 × 7) × 10 = ( 6 × 10 × 7) × 10 = 420 × 10 = 4,200 B) By using the place-value method, 60 × 70 = 60 × 7 tens = 6 tens ×7 tens = 42 × 1 ten × 1ten = 42 × 10 × 10 = 4,200 So, 61 ×73 can be rounded to 4,200 Use compatible numbers to estimate the product.

Question 4. 19 × 26

Answer: 500

Explanation: Let 19 be Rounded to 20 Let 26 be Rounded to 25 Now, we have to find the result of 20 × 25. We can find the product of multiples of ten by using two methods. They are: A) The place-value method  B) Associative Property of Multiplication A) By using the Associative Property of Multiplication, 20 × 25 = 25 × (2 × 10) = (25 × 2) × 10 = ( 5 × 5 ×2) × 10 = 50 × 10 = 500 B) By using the place-value method, 20 × 25 = 25 × 2 tens = 50 tens = 50 × 1 ten = 50 × 10 = 500 So, 19 ×26 can be rounded to 500

Question 5. 23 × 78

Answer: 2,000

Explanation: Let 23 be Rounded to 25 Let 78 be Rounded to 80 Now, we have to find the result of 25 × 80. We can find the product of multiples of ten by using two methods. They are: A) The place-value method  B) Associative Property of Multiplication A) By using the Associative Property of Multiplication, 25 × 80 = 25 × (8 × 10) = (25 × 8) × 10 = ( 5× 5 × 8) × 10 = 200 × 10 = 2,000 B) By using the place-value method, 25 × 80 = 25 × 8 tens = 200 tens = 200 × 10 = 2,000 So, 23 ×78 can be rounded to 2,000

Question 6. 74 × 20

Answer: 1,500

Explanation: Let 74 be Rounded to 75 Now, we have to find the result of 75 × 20. We can find the product of multiples of ten by using two methods. They are: A) The place-value method  B) Associative Property of Multiplication A) By using the Associative Property of Multiplication, 75 × 20 = 75 × (2 × 10) = (75 × 2) × 10 = ( 5 × 5 × 3 × 2) × 10 = 150 × 10 = 1,500 B) By using the place-value method, 75× 20 = 75 ×2 tens = 150 tens = 150× 10 = 1,500 So, 74 ×20 can be rounded to 1,500 Apply and Grow: Practice Estimate the product.

Question 7. 41 × 73

Answer:  2,800

Explanation: Let 41 be Rounded to 40 Let 73 be Rounded to 70 Now, we have to find the result of 40 × 70. We can find the product of multiples of ten by using two methods. They are: A) The place-value method  B) Associative Property of Multiplication A) By using the Associative Property of Multiplication, 40 × 70 = 40 × (7 × 10) = (40 × 7) × 10 = ( 4 × 10 × 7) × 10 = 280 × 10 = 2,800 B) By using the place-value method, 40 × 70 = 40 × 7 tens = 4 tens ×7 tens = 28 × 1 ten × 1ten = 28 × 10 × 10 = 2,800 So, 41 ×73 can be rounded to 2,800

Question 8. 52 × 84

Answer: 4,250

Explanation: Let 52 be Rounded to 50 Let 84 be Rounded to 85 Now, we have to find the result of 50 × 85. We can find the product of multiples of ten by using two methods. They are: A) The place-value method  B) Associative Property of Multiplication A) By using the Associative Property of Multiplication, 50 × 85 = 85 × (5× 10) = (85 × 5) × 10 = ( 17× 5 × 5) × 10 = 425 × 10 = 4,250 B) By using the place-value method, 50 × 85 = 85 × 5 tens = 425 tens = 425 × 10 = 4,250 So, 52 ×84 can be rounded to 4,250

Question 9. 26 × 68

Answer: 1,750

Explanation: Let 26 be Rounded to 25 Let 68 be Rounded to 70 Now, we have to find the result of 25 × 70. We can find the product of multiples of ten by using two methods. They are: A) The place-value method  B) Associative Property of Multiplication A) By using the Associative Property of Multiplication, 25 × 70 = 25 × (7 × 10) = (25 × 7) × 10 = ( 5 × 5 × 7) × 10 = 175 × 10 = 1,750 B) By using the place-value method, 25 × 70 = 25 × 7 tens = 175  tens = 175 × 10 = 1,750 So, 26 ×68 can be rounded to 1,750

Question 10. 38 × 17

Explanation: Let 38 be Rounded to 40 Let 17 be Rounded to 15 Now, we have to find the result of 40 × 15. We can find the product of multiples of ten by using two methods. They are: A) The place-value method  B) Associative Property of Multiplication A) By using the Associative Property of Multiplication, 15 × 40 = 15 × (4× 10) = (15 × 4) × 10 = ( 5 × 3 × 4) × 10 = 60 × 10 = 600 B) By using the place-value method, 15 × 40 = 15 × 4 tens = 60  tens = 60 × 10 = 600 So, 38 ×17 can be rounded to 600

Question 11. 75 × 24

Answer: 1,875

Explanation: Let 24 be Rounded to 25 Now, we have to find the result of 25 × 75. We can find the Product by using the simplification method. 25 × 75 = 5 × 5 × 25 × 3 = 5 × 5× 5 × 5 × 3 = 25 × 25 × 3 = 625 × 3 = 1,875 So, 38 × 17 can be Rounded to 1,875

Question 12. 93 × 53

Answer: 4,500

Explanation: Let 93 be Rounded to 90 Let 53 be Rounded to 50 Now, we have to find the result of 90 × 50. We can find the product of multiples of ten by using two methods. They are: A) The place-value method  B) Associative Property of Multiplication A) By using the Associative Property of Multiplication, 90 × 50 = 90 × (5 × 10) = (90 × 5) × 10 = ( 5 × 2 × 9 × 5) × 10 = 450 × 10 = 4,500 B) By using the place-value method, 90 × 50 = 90 × 5 tens = 9 tens × 5 tens = 45 × 1 ten × 1 ten = 45 × 10 × 10 = 4,500 So, 93 ×53 can be rounded to 4,500

Question 13. 44 × 78

Answer: 3,600

Explanation: Let 44 be Rounded to 45 Let 78 be Rounded to 80 Now, we have to find the result of 45 × 80. We can find the product of multiples of ten by using two methods. They are: A) The place-value method  B) Associative Property of Multiplication A) By using the Associative Property of Multiplication, 45 × 80 = 45 × (8 × 10) = (45 × 8) × 10 = ( 5 × 2 × 9 × 4) × 10 = 360 × 10 = 3,600 B) By using the place-value method, 45 × 80 = 45 × 8 tens = 5 × 9 × 8 tens = 45 × 8 × 10 = 360 × 10 = 3,600 So, 44 ×78 can be rounded to 4,500

Question 14. 21 × 33

Explanation: Let 21 be Rounded to 20 Let 33 be Rounded to 30 Now, we have to find the result of 20 × 30. We can find the product of multiples of ten by using two methods. They are: A) The place-value method  B) Associative Property of Multiplication A) By using the Associative Property of Multiplication, 20 × 30 = 20 × (3× 10) = (20 × 3) × 10 = ( 5× 4 × 3) × 10 = 60 × 10 = 600 B) By using the place-value method, 20 × 30 = 30 × 2 tens = 3 tens × 2 tens = 6 × 10 × 10 = 600 So, 21 ×33 can be rounded to 600

Question 15. 45 × 45

Answer: 2,500

Explanation: Let 45 be Rounded to 50 Now, we have to find the result of 50 × 50. We can find the product of multiples of ten by using two methods. They are: A) The place-value method  B) Associative Property of Multiplication A) By using the Associative Property of Multiplication, 50 × 50 = 50 × (5× 10) = (50 × 5) × 10 = ( 10× 5 × 5) × 10 = 250 × 10 = 2,500 B) By using the place-value method, 50 × 50 = 50 × 5 tens = 250 tens = 25 × 10 × 10 = 2,500 So, 45 ×45 can be rounded to 2,500 Open-Ended Write two possible factors that can be estimated as shown.

Explanation: The given number is: 2,400 The Products of 24 are: 4 × 6 = 24 6 × 4 =24 From the above two products, we can conclude that the two possible numbers that can give the product 2,400 are: 40, 60

Explanation: The given number is: 1,200 The Products of 12 are: 3 × 4 =12 4 × 3 =12 6 ×2 =12 2 ×6 =12 From the above two products, we can conclude that the two possible numbers that can give the product 2,400 are: 40, 30, and 20, 60

Question 18. DIG DEEPER! You use 50 × 30 to estimate 46 × 29. Will your estimate be greater than or less than the actual product? Explain.

Answer: We will Estimate the Product greater than the actual Product

Explanation: Given Product is: 46 × 29

Explanation: Let 46 be Rounded to 50 Let 29 be Rounded to 30 Now, we have to find the result of 50 × 30. We can find the product of multiples of ten by using two methods. They are: A) The place-value method  B) Associative Property of Multiplication A) By using the Associative Property of Multiplication, 50 × 30 = 50 × (3 × 10) = (50 × 3) × 10 = ( 5 × 10 × 3) × 10 = 150 × 10 = 1,500 B) By using the place-value method, 50 × 30 = 30 × 5 tens = 150  tens = 150× 10 = 1,500 So, 46 ×29 can be rounded to 1,500

Question 19. YOU BE THE TEACHER Your friend uses rounding to estimate 15 × 72. She gets a product of 700. Is your friend’s estimate correct? Explain.

Answer: Internet Service cost: $540 Cable television service cost: $1068

Explanation: In the given table, The Internet cost and cable television service costs are given per month. For 1 year, there are 12 months. So, The cost of Internet service for a year is: 12 × $45 = $ 540 The cost of cable television service for a year is: 12 × $89 = 1068

Answer: 476 Pounds

Explanation: Given that a giant panda eats 28 pounds of food each day and an Orca eats 17 times as much food as the panda eats each day. So, The amount of food eaten by an Orca = 17 × The amount of food eaten by panda = 17 × 28 = 476 pounds

Use rounding to estimate the product.

Question 1. 42 × 13

Answer:  600

Explanation: Let 42 be Rounded to 40 Let 13 be Rounded to 15 Now, we have to find the result of 40 × 15. We can find the product of multiples of ten by using two methods. They are: A) The place-value method  B) Associative Property of Multiplication A) By using the Associative Property of Multiplication, 40 × 15 = 15 × (4 × 10) = (15 × 4) × 10 = ( 3 × 4 × 5) × 10 = 60 × 10 = 600 B) By using the place-value method, 40 × 15 = 15 × 4 tens = 60 tens = 60 × 10 = 600 So, 42 × 13 can be rounded to 600

Question 2. 56 × 59

Answer: 3,300

Explanation: Let 56 be Rounded to 55 Let 59 be Rounded to 60 Now, we have to find the result of 55 × 60. We can find the product of multiples of ten by using two methods. They are: A) The place-value method  B) Associative Property of Multiplication A) By using the Associative Property of Multiplication, 55 × 60 = 55 × (6 × 10) = (55 × 6) × 10 = ( 5 × 11 × 6) × 10 = 330 × 10 = 3,300 B) By using the place-value method, 55 × 60 = 55 × 6 tens = 330 tens = 330 × 10 = 3,300 So, 42 × 16 can be rounded to 600

Question 3. 19 × 91

Answer: 1,800

Explanation: Let 19 be Rounded to 20 Let91 be Rounded to 90 Now, we have to find the result of 20 × 90. We can find the product of multiples of ten by using two methods. They are: A) The place-value method  B) Associative Property of Multiplication A) By using the Associative Property of Multiplication, 20 × 90 = 20 × (9× 10) = (20 × 9) × 10 = ( 5 × 4 × 9) × 10 = 180 × 10 = 1,800 B) By using the place-value method, 20 × 90 = 20 × 9 tens = 2 tens × 9 tens = 18 × 1 ten × 1 ten =18 × 10 × 10 = 18 × 100 = 1,800 So, 19 × 91 can be rounded to 1,800 Use compatible numbers to estimate the product.

Question 4. 23 × 78

Explanation: Let 23 be Rounded to 25 Let 78 be Rounded to 80 Now, we have to find the result of 25 × 80. We can find the product of multiples of ten by using two methods. They are: A) The place-value method  B) Associative Property of Multiplication A) By using the Associative Property of Multiplication, 25 × 80 = 25 × (8 × 10) = (25 × 8) × 10 = ( 5 × 5 × 8) × 10 = 200 × 10 = 2,000 B) By using the place-value method, 25 × 80 = 25 × 8 tens = 200 tens = 200 × 10 = 2,000 So, 23 × 78 can be rounded to 2,000

Question 5. 67 × 45

Answer:3,150

Explanation: Let 67 be Rounded to 70 Now, we have to find the result of 70 × 45. We can find the product of multiples of ten by using two methods. They are: A) The place-value method  B) Associative Property of Multiplication A) By using the Associative Property of Multiplication, 45 × 70 = 45 × (7 × 10) = (45 × 7) × 10 = ( 5 × 9 × 7) × 10 = 315 × 10 = 3,150 B) By using the place-value method, 45 × 70 = 45 × 7 tens = 315 tens = 315 × 10 = 3,150 So, 67 × 45 can be rounded to 3,150

Question 6. 19 × 24

Answer: 500 Explanation; Let 19 be Rounded to 20 Let 24 be Rounded to 25 Now, we have to find the result of 25 × 20. We can find the product of multiples of ten by using two methods. They are: A) The place-value method  B) Associative Property of Multiplication A) By using the Associative Property of Multiplication, 20 × 25 =25 × (2 × 10) = (25 × 2) × 10 = ( 5 × 5 × 2) × 10 = 50 × 10 = 500 B) By using the place-value method, 25 × 20 = 25 × 2 tens = 50 tens = 50 × 10 = 500 So, 19 × 24 can be rounded to 500 Estimate the product.

Question 7. 84 × 78

Answer: 6,800

Explanation: Let 84 be Rounded to 85 Let 78 be Rounded to 80 Now, we have to find the result of 85 × 80. We can find the product of multiples of ten by using two methods. They are: A) The place-value method  B) Associative Property of Multiplication A) By using the Associative Property of Multiplication, 85 × 80 = 85 × (8 × 10) = (85 × 8) × 10 = ( 5 × 17 × 8) × 10 = 680 × 10 = 6,800 B) By using the place-value method, 85 × 80 = 85 × 8 tens = 680 tens = 680 × 10 = 6,800 So, 84 × 78 can be rounded to 600

Question 8. 92 × 34

Answer: 3,150

Explanation: Let 92 be Rounded to 90 Let 34 be Rounded to 35 Now, we have to find the result of 90 × 35. We can find the product of multiples of ten by using two methods. They are: A) The place-value method  B) Associative Property of Multiplication A) By using the Associative Property of Multiplication, 90 × 35 = 35 × (9 × 10) = (35 × 9) × 10 = ( 5 × 7 × 9) × 10 = 315 × 10 = 3,150 B) By using the place-value method, 90 × 35 = 35 × 9 tens = 315 tens = 315 × 10 = 3,150 So, 92 × 34 can be rounded to 3,150

Question 9. 57 × 81

Answer: 4,800

Explanation: Let 57 be Rounded to 60 Let 81 be Rounded to 80 Now, we have to find the result of 80 × 60. We can find the product of multiples of ten by using two methods. They are: A) The place-value method  B) Associative Property of Multiplication A) By using the Associative Property of Multiplication, 80 × 60 = 80 × (6 × 10) = (80 × 6) × 10 = ( 8 × 10 × 6) × 10 = 480 × 10 = 4,800 B) By using the place-value method, 80 × 60 = 80 × 6 tens = 480 tens = 480 × 10 = 4,800 So, 57 × 81 can be rounded to 600 Open-Ended Write two possible factors that could be estimated as shown.

Explanation: The Products of 64 are: 8 × 8 = 64 16 × 4 =64 From the above two products, we can conclude that the two possible numbers that can give the product 6,400 are: 80,80 and. 160,40

Explanation: The Products of 16 are: 4 × 4 = 16 8 × 2 =16 From the above two products, we can conclude that the two possible numbers that can give the product 1,600 are: 40, 40 and, 80,20

Answer: Both Newton’s and Descartes’s estimates are reasonable

Explanation: According to Newton, The estimated values of 27 and 68 are 30 and 70 According to Descartes, The estimated values of 27 and 68 are 25 and 70 According to Newton: 27 × 68 = 2,100 We can find the product of multiples of ten by using two methods. They are: A) The place-value method  B) Associative Property of Multiplication A) By using the Associative Property of Multiplication, 30 × 70 = 30 × (7 × 10) = (30 × 7) × 10 = ( 3 × 10 × 7) × 10 = 210 × 10 = 2,100 B) By using the place-value method, 30× 70 = 30 ×7 tens = 210 tens = 210 × 10 = 2,100 So, 27 × 68 can be rounded to 2,100 ( According to Newton) According to Descartes: 27 × 68 = 1,750 We can find the product of multiples of ten by using two methods. They are: A) The place-value method  B) Associative Property of Multiplication A) By using the Associative Property of Multiplication, 25 × 70 = 25 × (7 × 10) = (25 × 7) × 10 = ( 5 × 5 × 7) × 10 = 175 × 10 = 1,750 B) By using the place-value method, 25× 70 = 25 ×7 tens = 175 tens = 175 × 10 = 1,750 So, 27 × 68 can be rounded to 1,750 ( According to Descartes)

Question 13. DIG DEEPER! You use 90 × 30 to estimate 92 × 34. Will your estimate be greater than or less than the actual product? Explain.

Answer: No Your friend is going to estimate 92 × 34 and she gets a product  2,700. So, your Friend’s estimate is not correct. Now, Let 34 be Rounded to 35 Let 92 be Rounded to 90 Now, we have to find the result of 35 × 90. We can find the product of multiples of ten by using two methods. They are: A) The place-value method  B) Associative Property of Multiplication A) By using the Associative Property of Multiplication, 35 × 90 = 35 × (9 × 10) = (35 × 9) × 10 = (9× 5 × 7) × 10 = 315 × 10 = 3,150 B) By using the place-value method, 35 × 90 = 35 × 9 tens = 315  tens = 315 × 10 = 3,150 So, 92 ×34 can be rounded to 3,150

Answer: 744 hours

Explanation: From the above table, The days of darkness in Barrow, Alaska are 31 days. We know that there are 24 hours in a day. So, The number of hours of darkness does Barrow, Alaska have in December = 31 × 24 = 744 hours Review & Refresh

Question 15. Round 253,490 to the nearest ten thousand.

Answer: 250,000 Explanation; The position of a given number is dependent on the place-value of that number. So, When 253,490 rounded off to the nearest ten thousand, the result ts 250,000

Question 16. Round 628,496 to the nearest hundred thousand.

Answer: 630,000 Explanation; The position of a given number is dependent on the place-value of that number. So, When 628,496 rounded off to the nearest ten thousand, the result ts 630,000

Answer: The Total Area of your Model = 400

Answer: Yes

Answer: 168

Answer: 255

Question 2. 34 × 22 = _____

Answer: 748

Answer: 247

Answer: 975

Answer:  143

Question 6. 23 × 26 = ______

Answer: 598

Question 7. 27 × 45 = ______

Answer: 1,215

Answer: 885 kilometers

Explanation: Given that the Perseid meteors travel 59 kilometers each second. So, The distance traveled by the Perseid meteors in 15 seconds = 59 × 15 = 885 kilometers From the above, We can conclude that the Perseid meteors travel 885 kilometers in 15 seconds.

So, the wind farm has 176  turbines. Show and Grow

Question 10. You can type 19 words per minute. Your cousin can type 33 words per minute. How many more words can your cousin type in 15 minutes than you?

Answer: 210 words

Explanation: Given that you can type 19 words per minute and your cousin can type 33 words per minute. So, The number of words you can type in 15 minutes = 19 × 15 = 285 The number of words your cousin can type in 15 minutes = 495 Hence, The number of words you have to type more than your cousin in 15 minutes = 495 – 285 =  310 words

Answer: 270 glasses

Explanation: Given that a store Owner buys 24 packs of solar eclipse glasses and each pack has 12 glasses. So, the total number of glasses that a store owner buy = 24 × 12 = 288 glasses But, It is also given that the store owner did not sell 18 glasses. Hence, the total number of glasses the store owner sold = 288 – 18 = 270 glasses

Use the area model to find the product.

Answer: 156

Answer: 912

Answer: 342

Answer: 575

Question 5. 26 × 31 = _____

Answer: 806

Question 6. 22 × 47 = ______

Answer: 1,034

Answer: Yes, your friend is correct.

Question 8. Writing Explain how to use an area model and partial products to multiply two-digit numbers.

Answer:  The arcade has 336 games

Answer: 84,016

Answer: 71,585

Answer: 569,821

Answer: 608

Explanation: Using the Distributive Property, we can find the product of 32 × 19 32 × 19 = 32 × ( 10 + 9) = ( 32 × 10 ) + ( 32 × 9) = ( 30 + 2 ) × 10 + ( 30 + 2 ) × 9 = ( 30 × 10 ) + ( 2 × 10 ) + ( 30 × 9 ) + ( 2 × 9) = 300 + 20 + 270 + 18 = 608 So, 32 × 19 = 608 Apply and Grow: Practice

Answer: 884

Explanation: Using the Distributive Property, we can find the product of 34 × 26 34 × 26 = 34 × ( 20 + 6) = ( 34 × 20 ) + ( 34 × 6) = ( 30 + 4 ) × 20 + ( 30 + 4 ) × 6 = ( 30 × 20 ) + ( 4 × 20 ) + ( 30 × 6 ) + ( 4 × 6) = 600 + 80 + 180 + 24 = 884 So, 34 × 26 = 884 Use the Distributive Property to find the product.

Question 3. 28 × 47 = 28 × (40 + 7) = (28 × 40) + (28 × 7) = (20 + 8) × 40  + (20 + 8) × 7 = (20 × 40) + (8 × 40) + (20 × 7) + (8 × 7) = 800 + 320 + 140 + 56 =1,316 So, 28 × 47 = 1,316

Answer: 28 × 47 = 1,316

Question 4. 39 × 41 = _____

Answer: 39 × 41 = 39 × (40 + 1) = (39 × 40) + (39 × 1) = (30 + 9) × 40  + (30 + 9) × 1 = (30 × 40) + (9 × 40) + (30 × 1) + (9 ×1) =1,200 + 360 + 30 + 9 =1,599 So, 39 × 41 = 1,599

Question 5. 74 × 12 = ______

Answer: 74 × 12 = 74 × (10 + 2) = (74 × 10) + (74 × 2) = (70 + 4) × 10  + (70 + 4) × 2 = (70 × 10) + (4 × 10) + (70 × 2) + (4 ×2) =700 + 40 + 140 + 8 =888 So, 74 × 12 = 888

Question 6. 83 × 65 = _____

Answer: 83 × 65 = 83 × (60 + 5) = (83 × 60) + (83 × 5) = (80 + 3) × 60  + (80 + 3) ×5 = (80 × 60) + (3 × 60) + (80 × 5) + (3 ×5) =4,800 + 180 + 400 + 15 =5,395 So, 83 × 65 = 5,395

Answer: Let the given Expressions be ordered as A), B), C) and D) From the Order, we can say that Expression C) does not belong to the other three.

Answer: No, each fan will not get a T-shirt.

Explanation: Given that an event coordinator orders 35 boxes of T-shirts to give away at a baseball game and there are 48 T-shirts in each box. So, we will get the total number of T-shirts due to the Product of 35 × 48. We will get the product by using the Distributive Property of Multiplication. 35 × 48 = 35 × (40 + 8) = (35 × 40) + (35 × 8) = (30 + 5) × 40  + (30 + 5) × 8 = (30 × 40) + (5 × 40) + (30 × 8) + (8 ×5) =1,200 + 20 + 240 + 40 =1,680 So, 35× 48 = 1,680 Note: The Distributive property is given as: a × (b  + c) = ( a ×  b) + ( a ×  c)

Answer: Yes, the pasture is large enough for 2 horses.

Explanation: Given that a horse owner must provide 4,046 square meters for each horse. But, it is also given that the area of pasture is 86 × 96 square meters. We have to find the Product of 86 × 96 by using the Distributive Property of Multiplication.. 86 × 96 = 86 × (90 + 6) = (86 × 90) + (86 × 6) = (80 + 6) × 90  + (80 + 6) × 6 = (80 × 90) + (6 × 90) + (80 × 6) + (6 ×6) =7,200 + 540 + 480 + 36 =8,256 So, 86× 96 = 8,256 By comparing the area of the pasture of each horse and the Product, we can conclude that the pasture is large enough for the 2 horses. Note: The Distributive property is given as: a × (b  + c) = ( a ×  b) + ( a ×  c)

Answer: 945

Explanation: By using the Distributive Property of Multiplication, 45 × 21 = 45 × (20 + 1) = (45 × 20) + (45 × 1) = (40 + 5) × 20  + (40 + 5) ×1 = (20 × 40) + (5 × 20) + (40 × 1) + (1 ×5) =800 + 100 + 40 + 5 =945 So,  45× 21 = 945 Note: The Distributive property is given as: a × (b  + c) = ( a ×  b) + ( a ×  c)

Question 2. Use the Distributive Property to find the product. 34 × 49 = 34 × (40 + 9) = (34 × 40) + (34 × 9) = (30 + 4) × 40 + (30 + 4) × 9 = (30 × 40) + (4 × 40) + (30 × 9) + (4 × 9) = 1,200 +160 + 270 + 36 = 1,666 So, 34 × 49 =1,666

Question 3. 14 × 27 = ______

Answer: 378

Explanation: Using the Distributive Property of Multiplication, 14 × 27 = 14 × (20 + 7) = (14 × 20) + (14 × 7) = (10 + 4) × 20  + (10 + 4) × 7 = (10 × 20) + (4 × 20) + (10 × 7) + (4 ×7) =200 + 80 + 70 + 28 =378 So,  14× 27 = 378 Note: The Distributive property is given as: a × (b  + c) = ( a ×  b) + ( a ×  c)

Question 4. 38 × 31 = ______

Answer: 1,178

Explanation: Using the Distributive Property of Multiplication, 38 × 31 = 38 × (30 + 1) = (38 × 30) + (38 × 1) = (30 + 8) × 30  + (30 + 8) ×1 = (30 × 30) + (8× 30) + (30 × 1) + (8 ×1) =900 + 240 + 30 + 8 =1,178 So,  38× 31 = 1,178 Note: The Distributive property is given as: a × (b  + c) = ( a ×  b) + ( a ×  c)

Question 5. 58 × 26 = ______

Answer: 1,508

Explanation: Using the Distributive Property of Multiplication, 58 × 26 = 58 × (20 + 6) = (58 × 20) + (58 × 6) = (50 + 8) × 20  + (50 + 8) × 6 = (50 × 20) + (8 × 20) + (50 × 6) + (8 ×6) =1,000 + 160 + 300 + 48 =1,508 So,  58× 26 = 1,508 Note: The Distributive property is given as: a × (b  + c) = ( a ×  b) + ( a ×  c)

Question 6. 56 × 32 = ______

Answer: 1,792

Explanation: Using the Distributive Property of Multiplication, 56 × 32 = 56 × (30 + 2) = (56 × 30) + (56× 2) = (50 + 6) × 30  + (50 + 6) × 2 = (30 × 50) + (6 × 30) + (50 × 2) + (6 ×2) =1,500 + 180 + 100 + 12 =1,792 So,  56× 32 = 1,792 Note: The Distributive property is given as: a × (b  + c) = ( a ×  b) + ( a ×  c)

Question 7. 87 × 23 = ______

Answer: 2,001

Explanation: Using the Distributive Property of Multiplication, 87 × 23 = 87 × (20 + 3) = (87 × 20) + (87 × 3) = (80 + 7) × 20  + (80 + 7) × 3 = (80 × 20) + (7 × 20) + (3 × 80) + (7 ×3) =1,600 + 140 + 240 + 21 =2,001 So,  87× 23 = 2,001 Note: The Distributive property is given as: a × (b  + c) = ( a ×  b) + ( a ×  c)

Question 8. 95 × 81 = ______

Answer: 7,695

Explanation: Using the Distributive Property of Multiplication, 95 × 81 = 95 × (80 + 1) = (95 × 80) + (95 × 1) = (90 + 5) × 80  + (90 + 5) × 1 = (90 × 80) + (5 × 80) + (90 × 1) + (1 ×5) =7,200 + 400 + 90 + 5 =7,695 So,  95× 81 = 7,695 Note: The Distributive property is given as: a × (b  + c) = ( a ×  b) + ( a ×  c)

Question 9. DIG DEEPER! Find 42 × 78 by breaking apart 42 first.

Answer: 3,276

Explanation: Using the Distributive Property of Multiplication, 42 × 78 = 78 × (40 + 2) = (78 × 40) + (78 × 2) = (70 + 8) × 40  + (70 + 8) × 2 = (70 × 40) + (8 × 40) + (70 × 8) + (8 ×2) =2,800 + 320 + 560 + 16 =3,276 So,  42× 78 = 3,276 Note: The Distributive property is given as: a × (b  + c) = ( a ×  b) + ( a ×  c)

Answer: The 29 real elephants would reach the top of the building even when they stand on top of each other.

Explanation: Give that the Elephant Building is 335 feet high. It is also given that a real Asian Elephant is 12 feet tall and there are 29 real elephants. So, we have to find the height of 29 real elephants. We can find it using the product of 29 × 12. We find the product by using the Distributive Property of Multiplication. 29 × 12 = 12 × (20 + 9) = (12 × 20) + (12 × 9) = (10 + 2) × 20  + (10 + 2) × 9 = (10 × 20) + (2 × 20) + (10 × 9) + (2 ×9) =200 + 40 + 90 + 18 =348 So,  29× 12 = 348 We get the result of the Product as 348 feet but given that the Elephant Building is 335 feet high. From this, we can conclude that if 29 real elephants could stand on top of each other,  they would reach the top of the building Note: The Distributive property is given as: a × (b  + c) = ( a ×  b) + ( a ×  c) Review & Refresh Find the difference. Then check your answer.

Answer: 25,259

Answer: 53,162

Answer: 140,938

Explore and Grow

Answer: 1,272

Answer:  585

Explanation: Using the Partial Products method, 39 × 15 = ( 30 + 9) × ( 10 + 5) = 30 × 10 + 9 × 10 + 30 × 5 + 9 × 5 = 300 + 90 + 150 + 45 = 585 Estimate: Let 39 be Rounded to 40. So, 40 × 15 = 600 As the Estimate and the actual answer are near, the answer is reasonable.

Answer: 5,166

Explanation: Using the Partial Products method, 82 × 63 = ( 80 + 2) × ( 60 + 3) = 80 × 60 + 2 × 60 + 80 × 3 + 2 × 3 = 4,800 + 120 + 240 + 6 = 5,166 Estimate: Let 82 be Rounded to 80. Let 63 be Rounded to 65. So, 8 0 × 65 = 5,200 As the Estimate and the actual answer are near, the answer is reasonable.

Answer: 3,976

Explanation: Using the Partial Products method, 56 × 71 = ( 50 + 6) × ( 70 + 1) = 50 × 70 + 1 × 50 + 70 × 6 + 6 × 1 = 3,500 + 50 + 420 + 6 = 3,976 Estimate: Let 56 be Rounded to 55. Let 71 be Rounded to 70. So, 55 × 70 = 3,850 As the Estimate and the actual answer are not  near, the answer is not  reasonable. Apply and Grow: Practice Find the product. Check whether your answer is reasonable.

Answer: 364

Explanation: Using the Partial Products method, 14 × 26 = ( 10 + 4) × ( 20 + 6) = 10 × 20 + 10 × 6 + 20 × 4 + 4 × 6 = 200 + 60 + 80 + 24 = 364 Estimate: Let 26 be Rounded to 25. Let 14 be Rounded to 15. So, 25 × 15 = 375 As the Estimate and the actual answer are near, the answer is reasonable.

Answer: 1,767

Explanation: Using the Partial Products method, 57 × 31 = ( 50 + 7) × ( 30 + 1) = 50 × 30 + 1 × 50 + 30 × 7 + 7 ×1 = 1,500 + 50 + 210 + 7 = 1,767 Estimate: Let 57 be Rounded to 55. (or) Let 57 be Rounded to 60. Let 31 be Rounded to 30. So, 3 0 × 55 = 1,650 (or) 30 × 60 = 1,800 As the Estimate and the actual answer are near, the answer is reasonable. ( Depending on the Estimate value).

Answer: 2,116

Explanation: Using the Partial Products method, 23 × 92 = ( 20 +3) × ( 90 + 2) = 20 × 90 + 2 × 20 + 90 × 3 + 3 ×2 = 1,800 + 40 + 270 + 6 = 2,116 Estimate: Let 23 be Rounded to 25. Let 92 be Rounded to 90. So, 9 0 × 25 = 2,250  As the Estimate and the actual answer are near, the answer is reasonable.

Question 7. Estimate: ______ 13 × 98 = ______

Answer: 1,274

Explanation: Using the Partial Products method, 13 × 98 = ( 10 + 3) × ( 90 +8) = 10 × 90 + 3 × 90 + 10 × 8 + 3 ×8 = 900 + 270 + 80 + 24 = 1,274 Estimate: Let 13 be Rounded to 15. Let 98 be Rounded to 100. So, 15 × 100 = 1,500  As the Estimate and the actual answer are not near, the answer is not  reasonable.

Question 8. Estimate: ______ 65 × 22 = ______

Answer: 1,430

Explanation: Using the Partial Products method, 65 × 22 = ( 60 + 5) × ( 20 + 2) = 60 × 20 + 60 × 2 + 20 × 5 + 5 ×2 = 1,200 + 120 + 100 + 10 = 1,430 Estimate: Let 22 be Rounded to 20. So, 20 × 65 = 1,300 As the Estimate and the actual answer are not  near, the answer is not reasonable.

Question 9. Estimate: ______ 72 × 81 = _____

Answer: 5,832

Explanation: Using the Partial Products method, 72 × 81 = ( 70 + 2) × (80 + 1) = 70 × 80 + 1 × 70 + 80 × 2 + 2 ×1 = 5,600 + 70 + 160 + 2 = 5,832 Estimate: Let 72 be Rounded to 70. Let 81 be Rounded to 80. So, 7 0 × 80 = 5,600 As the Estimate and the actual answer are not near, the answer is not  reasonable.

Answer: 1,682 liters

Explanation: Given that a farmer has 58 cows and each cow gives 29 liters of milk. So, to find the total quantity of milk, we have to find the Product of 58 × 29 by using the Partial Products method. Now, Using the Partial Products method, 58 × 29 = ( 50 + 8) × ( 20 + 9) = 50 × 20 + 9 × 50 + 20 ×8 + 8 ×9 = 1,000 + 450 + 160 + 72 = 1,682 From the above value, we can conclude that the total quantity of milk produces is 1,682 liters.

Question 11. Number Sense How much greater is the product of 12 and 82 than the product of 11 and 82? Explain how you know without multiplying.

Answer: The product of 12 and 82 is greater than 82 than the product of 11 and 82.

Explanation: The given Products are A) product of 12 and 82  B) product of 11 and 82 From the above, we can see that both the products have “82” as a Common number. So, find the difference between the remaining 2 numbers and we find that difference as ‘1’. From this, we can conclude that the product of 12 and 82 is greater than 82 than the product of 11 and 82. To verify this, we can find the products using the Partial Products method. Now, 12 × 82 = ( 10 + 2) × ( 80 +2) = 10 × 80 + 10 × 2 + 80 × 2 + 2 ×2 = 800 + 20 + 160 + 4 = 984 11 × 82 = ( 10 + 1) × ( 80 + 2) = 10 × 80 + 10 × 2 + 80 ×1 + 2 × 1 = 800 + 20 + 80 + 2 = 902 Now, 984 – 902 = 82 By multiplication also, we can conclude that the product of 12 and 82 is greater than 82 than the product of 11 and 82.

Question 12. DIG DEEPER! Write the multiplication equation shown by the partial products. 200 + 60 + 50 + 15

Answer:  13 × 25

Answer: 378 hours

Explanation: From the above table, we can conclude that the python sleeps for 18 hours a day. We know that, 1 week = 7 days So, 3 weeks = 3 × 7 = 21 days Hence, to find how many hours a python sleep in 3 weeks, we have to find the product of 18 × 21. Using the Partial Products method, 18 × 21 = ( 10 + 8) × ( 20 +1) = 10 × 20 + 10 × 1 + 20 × 8 + 8 ×1 = 200 + 10 + 160 + 8 = 378 Hence, from the above, we can conclude that the python sleeps for 378 hours in 3 weeks.

Answer:  102 pea seeds are less than cucumber seeds

Explanation: Given that there are 12 packets of pea seeds and 23 packets of cucumber seeds. From the table, The number of seeds in each packet of pea seeds = 12 × 6 = 72 The number of seeds in each packet of cucumber = 12 × 3 = 36 + 6 = 42 So, The total number of seeds in pea = 72  × 12 The total number of seeds in cucumber = 42 × 23 By using the Partial Products method, 72 × 12 = ( 10 + 2) × ( 70 +2) = 10 × 70 + 10 × 2 + 2 × 70 + 2 ×2 = 700 + 20 + 140 + 4 = 864 42 × 23 = ( 40 + 2) × ( 20 +3) = 40 × 20 + 40 × 3 + 20 × 2 + 2 ×3 = 800 + 120 + 40 + 6 = 966 So, The number of pea seeds less than the cucumber seeds = 966 – 864 = 102 seeds

Find the product. Check whether your answer is reasonable.

Answer: 442

Explanation: Using the Partial Products method, 17 × 26 = ( 10 + 7) × ( 20 + 6) = 10 × 20 + 10 × 6 + 20 × 7 + 7 × 6 = 200 + 140 + 60 + 24 = 424 Estimate: Let 17 be Rounded to 15. Let 26 be Rounded to 25. So, 25 × 15 = 375 As the Estimate and the actual answer are not near, the answer is not reasonable.

Answer: 2,356

Explanation: Using the Partial Products method, 38 × 62 = ( 30 + 8) × ( 60 + 2) = 30 × 60 + 30 × 2 + 60 × 8 + 8 × 2 = 1,800 + 60 + 480 + 16 = 2,356 Estimate: Let 38 be Rounded to 40. Let 62 be Rounded to 60. So, 40 × 60 = 2,400 As the Estimate and the actual answer are near, the answer is reasonable.

Answer: 3,913

Explanation: Using the Partial Products method, 91 × 43 = ( 90 + 1) × ( 40 + 3) = 90 × 40 + 90 × 3 + 40 × 1 + 3 × 1 = 3,600 + 270 + 40 + 3 = 3,913 Estimate: Let 91 be Rounded to 90. Let 43 be Rounded to 45. So, 90 × 45 = 4,050 As the Estimate and the actual answer are not near, the answer is not reasonable. Find the product. Check whether your answer is reasonable.

Answer: 3,774

Explanation: Using the Partial Products method, 51 × 74 = ( 50 + 1) × ( 70 + 4) = 50 × 70 + 50 × 4 + 70 × 1 + 4 × 1 = 3,500 + 200 + 70 + 4 = 3,774 Estimate: Let 51 be Rounded to 50. Let 74 be Rounded to 75. So, 50 × 75 = 3,750 As the Estimate and the actual answer are near, the answer is reasonable.

Answer: 532

Explanation: Using the Partial Products method, 28 × 19 = ( 20 + 8) × ( 10 + 9) = 10 × 20 + 20 × 9 + 10 × 8 + 9 × 8 = 200 + 180 + 80 + 72 = 532 Estimate: Let 28 be Rounded to 30. Let 19 be Rounded to 20. So, 30 × 15 = 600 As the Estimate and the actual answer are near, the answer is reasonable.

Answer: 1,575

Explanation: Using the Partial Products method, 35 × 45 = ( 30 + 5) × ( 40 + 5) = 30 × 40 + 30 × 5 + 40 × 5 + 5 × 5 = 1,200 + 150 + 200 + 25 = 1,575 There is no need for Estimate because they are already rounded numbers. So, the answer is reasonable. Question 7. Estimate: ______ 82 × 63 = ______

Explanation: Using the Partial Products method, 82 × 63 = ( 80 + 2) × ( 60 + 3) = 80 × 60 + 80 × 3 + 60 × 2 + 2 × 3 = 4,800 + 240 + 120 + 6 = 5,166 Estimate: Let 82 be Rounded to 80. Let 63 be Rounded to 65. So, 80 × 65 = 5,200 As the Estimate and the actual answer are near, the answer is reasonable.

Question 8. Estimate: ______ 36 × 93 = ______

Answer: 3,348

Explanation: Using the Partial Products method, 36 × 93 = ( 30 + 6) × ( 90 + 3) = 30 × 90 + 30 × 3 + 90 × 6 + 3 × 6 = 2,700 + 90 + 540 + 18 = 3,348 Estimate: Let 36 be Rounded to 35. Let 93 be Rounded to 95. So, 35 × 95 = 3,325 As the Estimate and the actual answer are near, the answer is reasonable.

Question 9. Estimate: _______ 57 × 22 = ______

Answer: 1,254

Explanation: Using the Partial Products method, 57 × 22 = ( 50 + 7) × ( 20 + 2) = 50 × 20 + 50 × 2 + 20 × 7 + 7 × 2 = 1,000 + 100 + 140 + 14 = 1,254 Estimate: Let 57 be Rounded to 55. Let 22 be Rounded to 20. So, 55 × 20 = 1,100 As the Estimate and the actual answer are not near, the answer is not reasonable.

Answer: The missing digits are 1, 1

Explanation: Using the Partial Products method, 100 + 50 + 60 + 30 = 10 × 10 + 10 × 5 + 10 × 6 + 5 × 6 = 10( 10 + 5) + 6( 10 + 5) = ( 10 + 5) × ( 10 + 6) = 15 × 16 So, From the above, we can conclude that the missing digits are: 1, 1

Answer: 396 liters of water Explanation; Given, Each ♦ = 6 liters From this, Half ♦ = 3 liters From the above table, Monthly water Intake by Newton = 6 × 5 + 3 = 33 liters So, The yearly intake by Newton can find out by the product of 33 × 12. ( Since a year has 12 months) Using the Partial Products method, 33 × 12 = ( 30 + 3) × ( 10 + 2) = 30 × 10 + 30 × 2 + 10 × 3 + 3 × 2 = 300 + 60 + 30 + 6 = 396 liters From the above, we can conclude that the yearly intake by Newton is: 396 liters

Answer: 468 liters Explanation; From the above table, Given that, The monthly intake of water by you = 6 × 7 + 3 = 45 liters The monthly intake of water by Descartes = 6 liters We know that 1 year consists of 12 months. So, The yearly intake of water by Descartes = 12 × 6 = 72 liters The yearly intake of water by you = 45 × 12 We have to find 45 × 12 using the Partial Products method. 45 × 12 = ( 40 + 5) × ( 10 +2) = 10 × 40 + 40 × 2 + 5 × 10 + 5 ×2 = 400 + 80 + 50 + 10 = 540 liters Hence, The amount of water you drink more than Descartes = 540 – 72 = 468 liters Review & Refresh Add or subtract. Then check your answer.

Question 13. 512,006 + 318,071 = ______

Answer: 830,077

Explanation: To find the sum, add the digits starting from the Right-most position. If there is “Carry”, then add that carry to the result of the next Position value.

Question 14. 746,620 – 529,706 = ______

Answer: 216,914

Explanation: The difference between the multi-digit numbers can be found by the difference taken from the left-most digit. If the number we want to subtract is less than the number to be subtracted, then we will take the carry from the Previous digit and the Previous digit contains 1 less number.

Show and Grow Find the product. Check whether your answer is reasonable.

Answer: 1,312

Explanation: Using the Partial Products method, 41 × 32 = ( 40 + 1) × ( 30 +2) = 30 × 40 + 40 × 2 + 1 × 30 + 1 ×2 = 1,200 + 80 + 30 + 2 = 1,312 Estimate: Let 41 be Rounded to 40. Let 32 be Rounded to 30. So,  40 × 30 = 1,200 As the Estimate and the actual answer are not near, the answer is not reasonable.

Answer: 2,392

Explanation: Using the Partial Products method, 52 × 46 = ( 50 + 2) × ( 40 +6) = 50 × 40 + 50 × 6 +2 × 40 + 6 ×2 = 2,000 + 300 + 80 + 12 = 2,392 Estimate: Let 52 be Rounded to 50. Let 46 be Rounded to 45. So,  45 × 50 = 2,250 As the Estimate and the actual answer are not near, the answer is not reasonable.

Answer: 2,730

Explanation: Using the Partial Products method, 78 × 35 = ( 70 + 8) × ( 30 +5) = 30 × 70 + 70 × 5 + 8 × 30 + 8 ×5 = 2,100 + 350 + 240 + 40 = 2,730 Estimate: Let 78 be Rounded to 80. So,  35 × 80 = 2,800 As the Estimate and the actual answer are  near, the answer is reasonable. Apply and Grow: Practice Find the product. Check whether your answer is reasonable.

Answer: 516

Explanation: Using the Partial Products method, 12 × 43 = ( 10 + 2) × ( 40 +3) = 10 × 40 + 10 × 3 + 2 × 40 + 3 ×2 = 400 + 30 + 80 + 6 = 516 Estimate: Let 12 be Rounded to 10. Let 43 be Rounded to 45. So,  45 × 10 = 450 As the Estimate and the actual answer are near, the answer is  reasonable.

Answer: 1,992

Explanation: Using the Partial Products method, 83 × 24 = ( 80 + 3) × ( 20 +4) = 80 × 20 + 80 × 4 + 3 × 20 + 3 ×4 = 1,600 + 320 + 60 + 12 = 1,992 Estimate: Let 83 be Rounded to 85. Let 24 be Rounded to 25. So,  85 × 25 = 2,152 As the Estimate and the actual answer are not near, the answer is not reasonable.

Answer: 4,484

Explanation: Using the Partial Products method, 59 × 76 = ( 50 + 9) × ( 70 +6) = 50 × 70 + 50 × 6 + 9 × 70 + 9 ×6 = 3,500 + 300 + 630 + 54 = 4,484 EStimate: Let 59 be Rounded to 60. Let 76 be Rounded to 75. So,  60 × 75 = 4,500 As the Estimate and the actual answer are near, the answer is  reasonable.

Question 7. Estimate: ______ 22 × 41 = ______

Answer: 902

Explanation: Using the Partial Products method, 41 × 22 = ( 40 + 1) × ( 20 +2) = 20 × 40 + 40 × 2 + 1 × 20 + 1 ×2 = 800 + 80 + 20 + 2 = 902 Estimate: Let 41 be Rounded to 40. Let 22 be Rounded to 20. So,  40 × 20 = 800 As the Estimate and the actual answer are not near, the answer is not reasonable.

Question 8. Estimate: _____ 94 × 32 = ______

Answer: 3,008

Explanation: Using the Partial Products method, 94 × 32 = ( 90 + 4) × ( 30 +2) = 30 × 90 + 90 × 2 + 4 × 30 + 4 ×2 = 2,700 + 180 + 120 + 8 = 3,008 Estimate: Let 94 be Rounded to 95. Let 32 be Rounded to 30. So,  95 × 30 = 2,850 As the Estimate and the actual answer are not near, the answer is not reasonable.

Question 9. Estimate: _____ 63 × 54 = _____

Answer: 3,402

Explanation: Using the Partial Products method, 63 × 54 = ( 60 + 3) × ( 50 +4) = 60 × 50 + 60 × 4 + 3 × 50 + 3 ×4 = 3,000 + 240 + 150 + 12 = 3,402 Estimate: Let 63 be Rounded to 65. Let 54 be Rounded to 55. So,  55 × 65 = 3,575 As the Estimate and the actual answer are not near, the answer is not reasonable.

Answer: 462 treat calories

Explanation: Given that Newton eats 14 treats each week and each treat has 33 calories. So, to find the total number of treat calories, we have to find the product of 14 × 33 by using the Partial Products method. 14 × 33 = ( 10 + 4) × ( 30 +3) = 10 × 30 + 10 × 3 + 30 × 4 + 3 ×4 = 300 + 30 + 120 + 12 = 462 From the above, we can conclude that Newton eats 462 treat calories each week.

Answer: Your friend is not correct.

Explanation: First, we will find 43 × 26 by using the Partial Products method. 43 × 26 = ( 40 + 3) × ( 20 +6) = 40 × 20 + 40 × 6 + 20 × 3 + 3 ×6 = 800 + 240 + 60 + 18 = 860 + 258 = 1,118 Fro the above, we can conclude that 258 must be placed instead of 248

Answer: The whale-shark is longer when compared to the hammerhead shark.

Explanation: Given that a ninja lantern-shark is 18 inches long and a whale shark is 16 times longer than the ninja lantern-shark So, to find the total length of a whale shark, we have to find the product of 18 × 16 using the Partial Products method. 18 × 16 = ( 10 + 8) × ( 10 +6) = 10 × 10 + 10 × 6 + 10 × 8 + 8 ×6 = 100 + 60 + 80 + 48 = 288 inches It is also given that the hammerhead shark is 228 inches long. From the above, we can conclude that the whale shark is longer than the hammerhead shark.

Answer: $4,032

Explanation: Given that there are 24 science classrooms in a school district and each classroom receives 3 hot plates. So, the total number of hot plates received in a school district = 24 × 3 = 72 hot plates It is also given that each hot plate costs $56. So, to find the total cost of the hot plates we have to find the product of 56 × 72 using the Partial Products method. 56 × 72 = (50 + 6) × ( 70 +2) = 50 × 70 + 50 × 2 + 70 × 6 + 2 ×6 = 3,500 + 100 + 420 + 12 = $4,032 From the above, we can conclude that the cost of all the hotplates in a school district is: $4,032

Answer: $1,656

Explanation: From the above table, we can see that the admission price of an adult is: $26 the admission price of a student is: $19 It is also given that there are 14 adults and 68 students. So, the total admission cost of the adults is: 14 × 26 the total admission cost of the children is: 68 × 19 Using  the Partial Products method, 14 × 26 = (20 + 6) × ( 10 +4) = 20 × 10 + 20 × 4 + 10 × 6 + 4 ×6 = 200 + 80 + 60 + 24 = $364 Using  the Partial Products method, 68 × 19 = (60 + 8) × ( 10 +9) = 60 × 10 + 60 × 9 + 10 × 8 + 8 ×9 = 600 + 540 + 80 + 72 = $1,292 Hence, The total cost of admission (Both adults and children) = 364 + 1,292 = $1,656

Answer: 1,196

Explanation: Using the Partial Products method, 31 × 92 = ( 30 + 1) × ( 90 +2) = 30 × 90 + 30 × 2 + 1 × 90 + 1 ×2 = 2,700 + 60 + 90 + 2 = 2,852 Estimate: Let 31 be Rounded to 30. Let 92 be Rounded to 90. So,  30 × 90 = 2,700 As the Estimate and the actual answer are not near, the answer is not reasonable.

Answer: 3,456

Explanation: Using the Partial Products method, 48 × 72 = ( 40 + 8) × ( 70 +2) = 70 × 40 + 40 × 2 + 8 × 70 + 8 ×2 = 2,800 + 80 + 560 + 16 = 3,456 Estimate: Let 48 be Rounded to 50. Let 72 be Rounded to 70. So,  50 × 70 = 3,500 As the Estimate and the actual answer are near, the answer is  reasonable.

Answer: 1,290

Explanation: Using the Partial Products method, 15 × 86 = ( 10 + 5) × ( 80 +6) = 10 × 80 + 10 × 6 + 5 × 80 + 5 × 6 = 800 + 60 + 400 + 30 = 1,290 Estimate: Let 86 be Rounded to 85. So,  85 × 15 = 1,275 As the Estimate and the actual answer are near, the answer is reasonable.

Answer: 4,374

Explanation: Using the Partial Products method, 81 × 54 = ( 80 + 1) × ( 50 +4) = 80 × 50 + 80 × 4 + 1 × 50 + 1 × 4 = 4,000 + 320 + 50 + 4 = 4,374 Estimate: Let 81 be Rounded to 80. Let 54 be Rounded to 55. So,  55 × 80 = 4,400 As the Estimate and the actual answer are near, the answer is reasonable.

Answer: 1,426

Explanation: Using the Partial Products method, 23 × 62 = ( 20 + 3) × ( 60 +2) = 20 × 60 + 20 × 2 + 3 × 60 + 3 ×2 = 1,200 + 40 + 180 + 6 = 1,426 Estimate: Let 23 be Rounded to 25. Let 62 be Rounded to 60. So,  25 × 60 = 1,500 As the Estimate and the actual answer are near, the answer is  reasonable.

Answer: 5,335

Explanation: Using the Partial Products method, 97 × 55 = ( 90 + 7) × ( 50 +5) = 90 × 50 + 90 × 5 + 7 × 50 + 7 × 5 = 4,500 + 450 + 350 + 35 = 5,335 Estimate: Let 97 be Rounded to 95. So,  55 × 95 = 5,225 As the Estimate and the actual answer are near, the answer is  reasonable. Find the product. Check whether your answer is reasonable.

Question 7. Estimate: ______ 51 × 62 = ______

Answer: 3,162

Explanation: Using the Partial Products method, 51 × 62 = ( 50 +1) × ( 60 +2) = 60 × 50 + 50 × 2 + 1 × 60 + 1 × 2 = 3,000 + 100 + 60 + 2 = 3,162 Estimate: Let 51 be Rounded to 50. Let 62 be Rounded to 60. So,  50 × 60 = 3,000 As the Estimate and the actual answer are near, the answer is  reasonable.

Question 8. Estimate: ______ 37 × 13 = ______

Answer: 481

Explanation: Using the Partial Products method, 37 × 13 = ( 10 +3) × ( 30 +7) = 10 × 30 + 10 × 7 + 3 × 30 + 3 × 7 = 300 + 70 + 90 + 21 = 481 Estimate: Let 37 be Rounded to 40. Let 13 be Rounded to 15. So,  40 × 15= 600 As the Estimate and the actual answer are not near, the answer is not reasonable.

Question 9. Estimate: _______ 49 × 78 = ______

Answer: 3,822

Explanation: Using the Partial Products method, 49 × 78 = ( 40 +9) × ( 70 +8) = 40 × 70 + 40 × 8 + 9 × 70 + 9 × 8 = 2,800 + 320 + 630 + 72 = 3,822 Estimate: Let 49 be Rounded to 50. Let 78 be Rounded to 80. So,  50 × 80 = 4,000 As the Estimate and the actual answer are near, the answer is  reasonable.

Answer: He scored 252 points.

Explanation: Given that Newton plays 21 basketball games and he scores 12 points each game. So, to find the total number of points in all games, we have to find the product of 21 × 12 By using the Partial Products method, 21 × 12 = ( 20 +1) × ( 10 + 2) = 20 × 10 + 20 × 2 + 1 × 10 + 1 × 2 = 200 + 40 + 10 + 2 = 252 points Hence, From the above, we can conclude that Newton had scored 252 points in 21 basketball games.

Question 11. DIG DEEPER! When you use regrouping to multiply two-digit numbers, why does the second partial product always end in 0?

Answer: Because we divide the partial Products in terms of 10 only. Ex: 13 × 15 = ( 10 + 3) × (10 + 5)

Answer: The missing numbers are: 6, 7, 0 and 1

Answer:  Otter drives deeper than the Walrus.

Explanation: Given that a tiger dives 12 feet underwater and an otter dives 25 times deeper than the tiger. So, The total distance dived by otter = 25 × 12 feet By using the Partial Products method, 25 × 12 = ( 20 +5) × ( 10 + 2) = 20 × 10 + 20 × 2 + 5 × 10 + 2 × 5 = 200 + 40 + 50 + 10 = 300 It is also given that a walrus dives 262 feet underwater. From the above, we can conclude that the otter dives deeper than the walrus. Review & Refresh

Answer: A) 6,835 Word Form: Six thousand, eight hundred thirty-five Expanded Form: 6,000 + 800 + 30 + 5

Explanation: Any number can be written in 3 forms. They are: A) Standard Form B) Word Form C) Expanded Form So, we can write the given form in the remaining 2 forms. B) 70,000 + 4,000 + 100 + 2 Standard Form: 74,102 Word Form: Seventy-four thousand, One hundred two

Explanation: Any number can be written in 3 forms. They are: A) Standard Form B) Word Form C) Expanded Form So, we can write the given form in the remaining 2 forms. C) Five hundred one thousand, three hundred twenty-nine Standard Form: 501,329 Expanded Form: 500,000 +0 + 1,000 + 100 + 0 + 2

Explanation: Any number can be written in 3 forms. They are: A) Standard Form B) Word Form C) Expanded Form So, we can write the given form in the remaining 2 forms.

Answer: The strategy to find 60 × 80 is: Associative Property of Multiplication

Explanation: By using the Associative Property of Multiplication, 60 ×80 =60 × (8 × 10) = (60 × 8) × 10 = (6 × 10 × 8) × 10 = 480 × 10 = 4,800 Choose any strategy to find 72 × 13.

Answer: The strategy to find 72 × 13 is: Distributive Property

Explanation: 72 × 13 = 72 × ( 10 + 3) = ( 72 × 10 ) + ( 72 ×3) = ( 70 + 2 ) × 10 + ( 70 + 2 ) × 3 = ( 70 × 10 ) + ( 2 × 10 ) + ( 70 × 3 ) + ( 2 × 3) = 700 + 20 + 210 + 6 = 936 So,  72 × 13 = 936 Reasoning Explain why you chose your strategies. Compare your strategies to your partner’s strategies. How are they the same or different?

Question 1. 90 × 37 = _____

Answer: 3,330

Explanation: Using the Associative Property of Multiplication, 37 × 90 = 37 × (9 × 10) = (37 × 9) × 10 = (3 × 3 × 37) × 10 = 333 × 10 = 3,330

Question 2. 78 × 21 = ______

Answer: 1,638

Explanation: Using the Distributive Property, 78 × 21 = 78 × ( 20 + 1) = ( 78 × 20 ) + ( 78 × 1) = ( 70 + 8 ) × 20 + ( 70 + 8 ) × 1 = ( 70 × 20 ) + ( 8 × 20 ) + ( 70 × 1 ) + ( 8 × 1) = 1,400 + 160 + 70 + 8 = 1,638 So,  78 × 21 = 1,638

Question 3. 14 × 49 = _____

Answer: 686

Explanation: Using the Associative Property of Multiplication, 14 × 49 = 14 × ( 40 + 9) = ( 14 × 40 ) + ( 14 × 9) = ( 10 + 4 ) × 40 + ( 10 + 4 ) × 9 = ( 40 × 10 ) + ( 4 × 40 ) + ( 10 × 9 ) + ( 4 × 9) = 400 + 160 + 90 + 36 = 686 So,  14 × 49 = 686 Apply and Grow: Practice Find the product.

Question 4. 74 × 30 = _____

Answer: 2,220

Explanation: Using the Associative Property of Multiplication, 74 × 30 = 74 ( 3 × 10) = ( 74 × 3) × 10 = 222 × 10 = 2,220 So,  74 × 30 = 2,220

Question 5. 51 × 86 = _____

Answer: 4,386

Explanation: Using the Distributive Property , 51 × 86 = 51 × ( 80 + 6) = ( 51 × 80 ) + ( 51 × 6) = ( 50 + 1 ) × 80 + ( 50 + 1 ) × 6 = ( 50 × 80 ) + ( 1 × 80 ) + ( 50 × 6 ) + ( 1 × 6) = 4,000 + 80 + 300 + 6 = 4,386 So,  51 × 86 = 4,386

Question 6. 40 × 29 = ______

Answer: 1,160

Explanation: Using the Associative Property of Multiplication, 40 × 29 = 29 × (4 × 10) = (29 × 4) × 10 = (29× 2 × 2) × 10 = 116 × 10 = 1,160

Question 7. 92 × 80 = _____

Answer: 7,360

Explanation: Using the Associative Property of Multiplication, 92 × 80 = 92 × (8 × 10) = (92 × 8) × 10 = (92 × 2 × 4) × 10 = 736 × 10 = 7,360

Question 8. 41 × 17 = ______

Answer: 697

Explanation: Using the Distributive Property, 41 × 17 = 41 × ( 10 + 7) = ( 41 × 10 ) + ( 41 × 7) = ( 40 + 1 ) × 10 + ( 40 + 1 ) × 7 = ( 40 × 10 ) + ( 1 × 10 ) + ( 40 × 7 ) + ( 1 × 7) = 400 + 10 + 280 + 7 = 697 So,  41 × 17 = 697

Question 9. 60 × 53 = _____

Answer: 3,180

Explanation: By using the Associative Property of Multiplication, 60 × 53 = 53 × (6 × 10) = (53 × 6) × 10 = (53× 3 × 2) × 10 = 318 × 10 = 3,180 Logic Find the missing factor.

Answer: The missing number is: 41

Answer: The missing number is: 34

Answer: The missing number is: 87

Question 13. Writing Explain why you start multiplying with the one’s place when using regrouping to multiply.

Answer: When using ” Regrouping” to multiply, we start multiplying from the rightmost position i.e.., one’s position because the place value of that position is 1.

Answer: The missing digit so that both products are the same is: 3

Question 15. A teacher orders 25 rock classification kits. Each kit has 4 rows with 9 rocks in each row. How many rocks are there in all?

Answer: 900 rocks

Explanation: Given that a teacher orders 25 rock classification kits and each kit has 4 rows with 9 rocks in each row. So, The total number of rows present in 25 rock classification kits = 25 × 4 = 100 rows We have to find the number of rocks in 100 rows by multiplying 100 × 9. Now, 100 × 9 = 900 rocks. From the above, We can conclude that there are 900 total rocks.

Question 16. A hotel has 12 floors with 34 rooms on each floor. 239 rooms are in use. How many rooms are not in use?

Answer: The total number of rooms that are not in use: 169

Explanation: Given that a hotel has 12 floors with 34 rooms on each floor. So, the total number of rooms in a hotel = 34 × 12 By using the Distributive method, 34 × 12 = 34 ( 10 + 2 ) = ( 34 × 10 ) + ( 34 × 2 ) = ( 30 + 4 ) × 10 + ( 30 + 4 ) × 2 = ( 30 × 10 ) + ( 4 × 10 ) + ( 30 × 2 ) + ( 4 × 2) = 300 + 40 + 60 + 8 = 408 But, it is also given that 239 rooms are in use. So, the number of rooms that are not in use = 408 – 239 = 169 rooms

Answer: $637

Explanation: Given that a child ticket for a natural history museum costs $13 and an adult ticket costs twice as much as a child ticket. So, the cost of an adult ticket = 2 × $13 = $26 So, to find the total cost for 21 children and 14 adults in a museum, we have to find 21 × $13  and 14 × $26 Now, 21 × 13 = 21 ( 10 + 3 ) = ( 21 × 10 ) + ( 21 × 3 ) = ( 20 + 1 ) × 10 + ( 20 +1 ) × 3 = ( 20 × 10 ) + ( 1 × 10 ) + ( 20 × 3 ) + ( 1 × 3) = 200 + 10 + 60 + 3 = 273 14 × 26 = 14 ( 20 + 6 ) = ( 14 × 20 ) + ( 14 × 6 ) = ( 10 + 4 ) × 20 + ( 10 +4 ) × 6 = ( 10 × 20 ) + ( 4 × 20 ) + ( 10 × 6 ) + ( 4 × 6) = 200 + 80 + 60 + 24 = 364 Hence, The total cost for children and adults in a museum = 273 + 364 = $637

Question 1. 16 × 13 = _____

Answer: 208

Explanation: By using the Distributive method, 16 × 13 = 16 ( 10 + 3 ) = ( 16 × 10 ) + ( 16 × 3 ) = ( 10 + 6 ) × 10 + ( 10 + 6 ) × 3 = ( 10 × 10 ) + ( 6 × 10 ) + ( 10 × 3 ) + ( 6 × 3) = 100 + 30 + 60 + 18 = 208

Question 2. 29 × 50 = _____

Answer: 1,450

Explanation: By using the Associative Property of Multiplication, 29 × 50 = 29 × ( 5 × 10 ) = ( 29 × 5 ) × 10 = 145 × 10 = 1,450

Question 3. 78 × 45 = _____

Answer: 3,510

Explanation: By using the Distributive method, 78 × 45 = 78 ( 40 + 5 ) = ( 78 × 40 ) + ( 78 × 5 ) = ( 70 + 8 ) × 40 + ( 70 + 8 ) × 5 = ( 70 × 40 ) + ( 8 × 40 ) + ( 70 × 5 ) + ( 8 × 5) = 2,800 + 320 + 350 + 40 = 3,510

Question 4. 30 × 71 = ______

Answer: 2,130

Explanation: By using the Associative Property of Multiplication, 30 × 71 = 71 × ( 3 × 10 ) = ( 71 × 3 ) × 10 = 213 × 10 = 2,130

Question 5. 62 × 14 = _____

Answer: 868

Explanation: By using the Distributive method, 62 × 14 = 62 ( 10 + 4 ) = ( 62 × 10 ) + ( 62 × 4 ) = ( 60 + 2 ) × 10 + ( 60 + 2 ) × 4 = ( 60 × 10 ) + ( 2 × 10 ) + ( 60 × 4 ) + ( 2 × 4) = 600 + 20 + 240 + 8 = 868

Question 6. 80 × 90 = _____

Answer: 7,200

Explanation: By using the Associative Property of Multiplication, 80 × 90 = 80 × ( 9 × 10 ) = ( 80 × 9 ) × 10 = 720 × 10 = 7,200 Find the product.

Question 7. 70 × 18 = _____

Answer: 1,260

Explanation: By using the Associative Property of Multiplication, 70 × 18 = 18 × ( 7 × 10 ) = ( 18 × 7 ) × 10 = 126 × 10 = 1,260

Question 8. 32 × 59 = _____

Answer: 1,888

Explanation: By using the Distributive method, 32 × 59 = 59 ( 30 + 2 ) = ( 59 × 30 ) + ( 59 × 2 ) = (50 + 9 ) × 30 + ( 50 + 9 ) × 2 = ( 50 × 30 ) + ( 9 × 30 ) + ( 50 × 2 ) + ( 2 × 9) = 1,500 + 270 + 100 + 18 = 1,888

Question 9. 67 × 20 = ____

Answer: 1,340

Explanation: By using the Associative Property of Multiplication, 67 × 20 = 67 × ( 2 × 10 ) = ( 67 × 2 ) × 10 = 134 × 10 = 1,340

Question 10. 51 × 84 = _____

Answer: 4,284

Explanation: By using the Distributive method, 51 × 84 = 51 ( 80 + 4 ) = ( 51 × 80 ) + ( 51 × 4 ) = (50 + 1 ) × 80 + ( 50 + 1 ) × 4 = ( 50 × 80 ) + ( 1 × 80 ) + ( 50 × 4 ) + ( 1 × 4) = 4,000 + 80 + 200 + 4 = 4,284

Question 11. 40 × 40 = _____

Answer: 1,600

Explanation: By using the Associative Property of Multiplication, 40 × 40 = 40 × ( 4 × 10 ) = ( 40 × 4 ) × 10 = 160 × 10 = 1,600

Question 12. 23 × 97 = ______

Answer: 2,231

Explanation: By using the Distributive method, 23 × 97 = 97 ( 20 + 3 ) = ( 97 × 20 ) + ( 97 × 3 ) = (90 + 7 ) × 20 + ( 90 + 7 ) × 3 = ( 90 × 20 ) + ( 7 × 20 ) + ( 90 × 3 ) + ( 7 × 3) = 1,800 + 140 + 270 + 21 = 2,231

Question 13. Writing Which strategy do you prefer to use when multiplying two-digit numbers? Explain.

Answer: The strategy you have to prepare when multiplying 2-digit numbers must depend on the numbers. some strategies to multiply 2-digit numbers are: A) The place-value method  B) The Associative Property of Multiplication  C) Distributive Property D) Partial Products method E) Regrouping method

Answer: The number you can multiply the number of tires by to find the total weight is: 20

Answer: 585 cups of popcorn

Explanation: Given that each bag of popcorn makes 13 cups and the principal brings 15 boxes of popcorn. It is also given that each box has 3 bags of popcorn. So, to find the total number of cups of popcorn the principal bring = 13 × 15 × 3 Now, By using the Distributive method, 13 × 45 = 45 ( 10 + 3 ) = (45 × 10 ) + ( 45 × 3 ) = (40 +5 ) × 10 + ( 40 + 5 ) × 3 = ( 40 × 10 ) + ( 5 × 10 ) + ( 40 × 3 ) + ( 5 × 3) = 400 + 50 + 120 + 15 = 585 From the above, we can conclude that the number of cups the principal bring is: 585 cups of popcorn Review & Refresh Find the product.

Question 16. 8 × 200 = _____

Explanation: By using the Associative Property of Multiplication, 8 × 200 = 8 × (20 × 10) = (8 × 20) × 10 = (8× 5 × 4) × 10 = 160 × 10 = 1,600

Question 17. 7 × 300 = _____

Answer: 2,100

Explanation: By using the Associative Property of Multiplication, 7 × 300 = 7 × (30 × 10) = (7 × 30) × 10 = (7× 5 × 6) × 10 = 210 × 10 = 2,100

Question 18. 6,000 × 5 = _____

Answer: 30,000

Explanation: By using the Associative Property of Multiplication, 5 × 6,000 = 5 × (600 × 10) = (5 × 600) × 10 = (6× 5 × 100) × 10 = 300 × 10 = 3,000

Question 19. 9 × 90 = ____

Answer: 810

Explanation: By using the Associative Property of Multiplication, 9 × 90 = 9 × (9 × 10) = (9 × 9) × 10 = 81 × 10 = 810

Question 20. 3,000 × 6 = _____

Answer: 18,000

Explanation: By using the Associative Property of Multiplication, 6 × 3,000 = 6 × (300 × 10) = (6 × 300) × 10 = (6× 3 × 100) × 10 = 1800 × 10 = 18,000

Question 21. 5 × 500 = _____

Explanation: By using the Associative Property of Multiplication, 5 × 500 = 5 × (50 × 10) = (5 × 50) × 10 = (5× 5 × 10) × 10 = 250 × 10 = 2,500

Answer: 1,408 cars can ferry transport in 1 day

Explanation: Given that a ferry can transport 64 cars each time it leaves a port and the ferry leaves a port 22 times in 1 day. From this, the total number of cars that can ferry transport in 1 day is: 64 × 22 By using the Distributive method, 64 × 22 = 64 ( 20 + 2 ) = (64 × 20 ) + ( 64 × 2 ) = (60 + 4 ) × 20 + ( 60 + 4 ) × 2 = ( 60 × 20 ) + ( 4 × 20 ) + ( 60 × 2 ) + ( 4 × 2) = 1,200 + 80 + 120 + 8 = 1,408 From the above, we can conclude that the total number of cars that ferry transport in 1 day is 1,408 cars. Construct Arguments Make a plan to find how many cars the ferry can transport in 1 week.

Answer: 9,856 cars

Answer: 4,608 ounces

Explanation: Given that the store receives 8 boxes and Each box has 18 bags of dog treats. It is also given that each bag weighs 32 ounces. The number of ounces of dog treats the pet store receives in the shipment = 18 × 8 × 32 It is also given that each box is 2 feet high. But it is not necessary for the calculation of a total number of ounces. First, we will find the weight of 1 box = 18 × 32 By using the Distributive method, 18 × 32 = 18 ( 30 + 2 ) = (18 × 30 ) + ( 18 × 2 ) = (10 + 8 ) × 30 + ( 10 + 8 ) × 2 = ( 10 ×30 ) + ( 8 × 30 ) + ( 10 × 2 ) + ( 8 × 2) = 300 + 240 + 20 + 16 = 576 So, The total weight of dog treats the pet store receives in the shipment = 576 × 8 = 4,608 ounces Hence, The pet store receives 4,608 ounces of dog treats. Show and Grow

Question 1. Show how to solve the problem above using one equation.

Answer: 18 × 32 × 8 Apply and Grow: Practice Understand the problem. What do you know? What do you need to find? Explain.

Question 2. Thirteen students create a petition for longer recess. They need 5,000 signatures in all. So far, each student has 99 signatures. How many more signatures do they need?

Answer: 3,713 signatures

Explanation: Given that there are 13 students that create a petition for longer recess and they need 5,000 signatures in all. It is also given that each student has 99 signatures. So, the total number of signatures that 13 students have are: 13 × 99 By using the Distributive method, 13 × 99 = 13 ( 90 + 9 ) = (13 × 90 ) + ( 13 × 9 ) = (10 +3 ) × 90 + ( 10 + 3 ) × 9 = ( 90 × 10 ) + ( 3 × 90 ) + ( 10 × 9 ) + ( 9 × 3) = 900 + 270 + 90 + 27 = 1,287 But, the students required 5000 signatures in total. So, The remainin number of signatures required by students = 5,000 – 1,287 = 3,713 signatures

Question 3. An activity book has 35 pages and costs $7. Each page has 4 puzzles. You have completed all of the puzzles on 16 of the pages. How many puzzles do you have left to complete?

Answer: 76 puzzles

Explanation: Given that an activity book has 35 pages and each page has 4 puzzles. So, The total number of puzzles = 35 × 4 = 140 puzzles But, it is also given that the puzzles completed in all of the 16 pages and we know that each page has 4 puzzles. So, The number of puzzles in 16 pages = 16 × 4 = 64 puzzles Now, The remaining number of puzzles in the remaining pages = 140 – 64 = 76 puzzles Understand the problem. Then make a plan. How will you solve it? Explain.

Answer: 552 items

Explanation: Given that there are 12 classes that provide items for a time capsule and there are 23 students in each class and each student puts 2 small items in the time capsule. First, we have to find the total number of students by the product 23 × 12 By using the Distributive method, 12 × 23 = 23( 10 + 2 ) = (23 × 10 ) + ( 23 × 2 ) = (20 +3 ) × 10 + ( 20 + 3 ) × 3 = ( 20 × 10 ) + ( 3 × 10 ) + ( 20 × 3 ) + ( 3 × 3) = 200 + 30 + 60 + 9 = 296 students Now, The total number of small items put by the total number of students = 296 × 2 = 552 small items

Question 5. A craftsman cuts letters and numbers from license plates to make signs. He has 15 Florida plates and 25 Georgia plates. Each plate has a total of 7 letters and numbers. How many letters and numbers does the craftsman cut in all?

Answer: 280 letters and numbers Explanation; Given that a craftsman has 15 Florida plates and 25 Georgia plates and each plate has a total of 7 letters and numbers. Hence, The total number of plates that a craftsman have = 25 + 15 = 40 plates The total number of letters and numbers each plate has = 40 × 7 = 280 letters and numbers

Answer: 1,673 kilowatt-hours of electricity Explanation; Given that each solar panel produces 30 kilowatt-hours and a wind turbine produces 833 kilowatt-hours. Now, the electricity produced by 28 solar panels = 28 × 30 By using the Associative Property of Multiplication, 28 × 30 = 28 × (3 × 10) = (28 × 3) × 10 = (7× 4 × 3) × 10 = 84 × 10 = 840 kilowatt-hours Hence, The total electricity produced by 28 solar panels and 1 wind turbine = 840 + 833 = 1,673 kilowatt-hours

Question 7. A theater has 45 rows of 72 seats on the floor level and 22 rows of 36 seats in the balcony. How many seats are there in all? How many more seats are on the floor level than on the balcony?

Answer: 4,032 seats Explanation; Given that a theater has 45 rows of 72 seats on the floor level and 22 rows of 36 seats in the balcony. So, The total number of seats on the floor level = 45 × 72 The total number of seats on the balcony = 22 × 36 By using the Distributive method, 72 × 45 = 45 ( 70 + 2 ) = (45 × 70 ) + ( 45 × 2 ) = (40 +5 ) × 70 + ( 40 + 5 ) × 2 = ( 40 × 70 ) + ( 5 × 70 ) + ( 40 × 2 ) + ( 5 × 2) = 2,800 + 350 + 80 + 6 = 3,240 By using the Distributive method, 22 × 36 = 36 ( 20 + 2 ) = (36 × 20 ) + ( 36× 2 ) = (30 +6 ) × 20 + ( 30 + 6) × 2 = ( 30 × 20 ) + ( 6 × 20 ) + ( 30 × 2 ) + ( 6 × 2) = 600 + 120 + 60 + 12 = 792 So, The total number of seats ( Floor level and balcony) = 3,240 + 792 = 4,032 seats

Understand the problem. Then make a plan. How will you solve it? Explain.

Answer: 1,152 dogs

Explanation: Given that there are 72 mushers compete in a sled-dog race and each musher has 16 dogs. So, The total number of dogs that compete in the race = 16 × 72 By using the Distributive method, 16 × 72 = 72 ( 10 + 6 ) = (72 × 10 ) + ( 72× 6 ) = (70 +2 ) × 10 + ( 70 + 2) × 6 = ( 70 × 10 ) + ( 2 × 10 ) + ( 70 × 6 ) + ( 6 × 2) = 700 + 20 + 420 + 12 = 1,152 Hence, The total number of dogs that compete in the race are: 1,152 dogs

Question 2. A photographer buys 3 USB drives that each cost $5. She puts 16 folders on each drive. Each folder has 75 photographs. How many photographs does the photographer put on the USB drives in all?

Answer: 3,600 photographs Explanation; Given that a photographer has 3 USB drives and there are 16 folders and 75 photographers on each drive. So, The total number of photographs on all the USB drives = 16 × 3 × 75 = 48 × 75 By using the Distributive method, 48 × 75 = 75 (40 + 8 ) = (75 × 40 ) + ( 75 × 8 ) = (70 +5 ) × 40 + ( 70 + 5) × 8 = ( 70 × 40 ) + ( 5 × 40 ) + ( 70 × 8 ) + ( 5 × 8) = 2,800 + 200 + 560 + 40 = 3,600 photographs Hence, The total number of photographs on all the USB drives are: 3,600 photographs

Question 3. A teacher has 68 students take 25 Question test. The teacher checks the answers for 9 of the tests. How many answers does the teacher have left to check?

Answer: 1,475 answers

Explanation: Given that a teacher has 68 students take 25 Question test. So, The total number of answers the teacher have to check = 68 × 25 By using the Distributive method, 68 × 25 = 68 ( 20 + 5 ) = (68 × 20 ) + ( 68× 5 ) = (60 +8 ) × 20 + ( 60 + 8) × 5 = ( 60 × 20 ) + ( 8 × 20 ) + ( 60 × 5 ) + ( 8 × 5) = 1,200 + 160 + 300 + 40 = 1,700 It is also given that the teacher checked 9 papers only. So. The number of answers checked = 9 × 25 = 225 Hence, The number of answers remained unchecked = 1,700 – 225 = 1,475 answers

Question 4. Each day, a cyclist bikes uphill for 17 miles and downhill for 18 miles. She drinks 32 fluid ounces of water after each bike ride. How many miles does the cyclist bike in 2 weeks?

Answer: 490 miles Explanation; Given that a cyclist bikes uphill for 17 miles and downhill for 18 miles. SO, The total distance that cyclist bikes each day = 17 + 18 = 35 miles We know that, 1 week = 7 days S0, 2 weeks = 2 × 7  = 14 days. So, the distance traveled by a cyclist in 2 weeks = 35 × 14 By using the Distributive method, 35 × 14 = 35 ( 10 + 4 ) = (35 × 10 ) + ( 35× 4 ) = (30 +5 ) × 10 + ( 30 + 5) × 4 = ( 30 × 10 ) + ( 5 × 10 ) + ( 30 × 4 ) + ( 5 × 4) = 300 + 50 + 120 + 20 = 490 miles Hence, The distance traveled by a cyclist in 2 weeks is: 490 miles

Question 5. Precision Which expressions can be used to solve the problem? Twelve friends play a game that has 308 cards. Each player receives 16 cards. How many cards are left? (308 – 12) × 16 308 – (16 × 12) 308 – (12 × 16) (308 – 16) – 12

Answer: Let the Expressions be named as A),  B), C) and D) Either B) or C) can be used to solve the problem.

Explanation: The given Expressions are; A) (308 – 12) × 16 B) 308 – (16 × 12) C) 308 – (12 × 16) D) (308 – 16) – 12 It is given that there are 12 friends that have 16 cards each and there are 308 cards in total. So, The number of cards that have left = 308 – ( 16 × 12 ) (or) 308 – ( 12 × 16 ) Hence, From the above, we can conclude that either B) or C) can be used to solve the problem.

Answer: The total ticket cost is: $2,617

Explanation: The given cost of an Adult ticket = $57 It is also given that a child ticket costs $14 less than an adult ticket. So, The given cost of a child ticket = 57 – 14 = $43 Now, The cost for 18 adults = 18 × 57 The cost for 37 children = 37 × 43 By using the Distributive method, 18 × 57 = 57 ( 10 + 8 ) = (57 × 10 ) + ( 57× 8 ) = (50 + 7 ) × 10 + ( 50 + 7) × 8 = ( 50 × 10 ) + ( 7 × 10 ) + ( 50 × 8 ) + ( 7 × 8) = 500 + 70 + 400 + 56 = 1,026 By using the Distributive method, 37 × 43 = 37 ( 40 + 3 ) = (37 × 40 ) + ( 37× 3 ) = (30 +7 ) × 40 + ( 30 + 7) × 3 = ( 30 × 40 ) + ( 7 × 40 ) + ( 30 × 3 ) + ( 7 × 3) = 1,200 + 280 + 90 + 21 = 1,591 Hence, The total cost of tickets = 1,591 + 1,026 = $2,617

Answer: 1,680 cm

Explanation: Given that an artist creates a pattern by alternating square and rectangular tiles and the pattern 14 square tiles and 13 rectangular tiles. From the given fig., Area of Square = 8 × 8 = 64 square cm Area of Rectangle = 8 × 14 = 112 square cm So, Area of 14 squares = 14 × 64 Area of 13 rectangles = 112 × 13 By using the Distributive method, 14 × 64 = 64 ( 10 + 4 ) = (64 × 10 ) + ( 64× 4 ) = (60 +4 ) × 10 + ( 60 + 4) × 4 = ( 60 × 10 ) + ( 4 × 10 ) + ( 60 × 4 ) + ( 4 × 4) = 600 + 40 + 240 + 16 = 896 By using the Distributive method, 112 × 13 = 13 ( 100 + 12 ) = (13 × 100 ) + ( 13× 12 ) = (10 +3 ) × 100 + ( 10 + 3) × 12 = ( 10 × 100 ) + ( 3× 100 ) + ( 10 × 12 ) + ( 12 × 3) = 1,000 + 300 + 120 + 36 = 1,456 Hence, The total length of the Pattern = 896 + 1,456 = 2,352 square cm

Answer: 1,858 crates are still on the ship.

Explanation: Given that a cargo ship has a go ship has 34 rows of crates and each row has 16 stacks of crates. It is also given that there are 5 crates in each stack. So, The total number of crates = 34 × 16 × 5 = 34 × 80 By using the Associative Property of Multiplication, 34 × 80 = 34 × (8 × 10) = (34 × 8) × 10 = (17× 2 × 8) × 10 = 272 × 10 = 2,720 It is also given that the ship workers unload 862 crates. Hence, The number of crates that the ship had = 2,720 – 862 = 1,858 crates Review & Refresh Estimate the product.

Question 1. 4 × 85

Answer: 340

Explanation: By using the partial products method, 4 × 85 = ( 80 + 5 ) × ( 2 + 2 ) = ( 80 × 2 ) + (80 × 2 ) + ( 5 × 2 ) + ( 5 × 2 ) = 160 + 160 + 10 + 10 = 340

Question 2. 6 × 705

Answer: 4,230

Explanation: By using the partial products method, 6 × 705 = ( 700 + 5 ) × ( 2 + 4 ) = ( 700 × 2 ) + (700 × 4 ) + ( 5 × 2 ) + ( 5 × 4 ) = 1,400 + 2,800 + 10 + 20 = 4,230

Question 3. 8 × 7,923

Answer: 63,384

Question 1. A wind turbine rotates between 15 and 40 times in 1 minute. a.What is the least number of times the turbine rotates in 1 hour? b.What is the greatest number of times the turbine rotates in 1 hour?

Answer: a) 900 times b) 2,400 times

Explanation: Given that a wind turbine rotates between 15 and 40 times in 1 minute. We know that, 1 hour = 60 minutes The wind turbine rotates minimum 15 times and maximum 40 times in a minute. So, The minimum number of times a wind turbine rotate in 1 hour = 15 × 60 The maximum number of times a wind turbine rotate in 1 hour = 40 × 60 Using the Associative Property of Multiplication, 15 × 60 = 15 ( 6 × 10) = ( 15 × 6) × 10 = 90 × 10 = 900 Using the Associative Property of Multiplication, 40 × 60 = 40 ( 6 × 10) = ( 40 × 6) × 10 = 240 × 10 = 2,400 Hence, The minimum number of times a wind turbine rotate in 1 hour = 900 times The maximum number of times a wind turbine rotate in 1 hour = 2,400 times

Question 2. The tips of the turbine blades spin 5 times faster than the speed of the wind. The speed of the wind is 22 miles per hour. How fast do the blade tips spin?

Answer: 110 miles per hour Explanation; Given that the tips of the turbine blades spin 5 times faster than the speed of the wind. It is also given that the speed of the wind is 22 miles per hour. So, The speed of the blade tips = 22 × 5 = 110 miles per hour

Question 3. A turbine farm has 7 large wind turbines. Each wind turbine can generate enough energy to power 1,485 houses. How many houses can the turbine farm power in all?

Answer: 10,395 houses

Explanation: Given that a turbine farm has 7 large wind turbines and each wind turbine can generate enough energy to power 1,485 houses. So, The total number of houses that a turbine farm generates enough energy = 1,485 × 7 = 10,395 houses

Answer: a) 2 times b) 33,000 houses Explanation; a) The given original length of each blade is 15 meters and 30 meters. It is also given that some houses are powered. Now, It is also given that the length of each blade is doubled. So, 15 meters becomes 30 meters and 30 meters becomes 60 meters. From this, we have to observe that the change in length also changes the energy powered in a directly proportional way. From this, we can conclude that The number of times more houses are powered when the length of each blade is doubled is: 2 times b) Given that the length f each wind turbine is 60 meters long. So, the number of houses that can be powered up = 60 × 550 Using the Associative Property of Multiplication, 550 × 60 = 550 × ( 10 × 6) = ( 550 × 6 ) × 10 = 3300 × 10 = 33,000 Hence, The number of houses can the wind turbine power is: 33,000 houses

4.1 Multiply by Tens

Question 1. 50 × 20 = _____

Explanation: By using the Place-value method, 50 × 20 = 50 × 2 tens = 5 tens × 2 tens = 10 × 10 × 10 = 1,000 So, 50 × 20 = 1,000

Question 2. 30 × 60 = _____

Explanation: By using the Place-value method, 30 × 60 = 30 × 6 tens = 3 tens × 6 tens = 18 × 10 × 10 = 1,800 So, 30 × 60 = 1,800

Question 3. 80 × 10 = _____

Explanation: By using the Place-value method, 80 × 10 = 10 × 8 tens = 1 ten × 8 tens = 8 × 10 × 10 = 800 So, 80 × 10 = 800

Question 4. 40 × 70 = _____

Answer: 2,800

Explanation: By using the Place-value method, 70 × 40 = 70 × 4 tens = 7 tens × 4 tens = 28 × 10 × 10 = 2,800 So, 70 × 40 = 2,800

Question 5. 60 × 50 = _____

Answer: 3,000

Explanation: By using the Place-value method, 50 × 60 = 50 × 6 tens = 5 tens × 6 tens = 30 × 10 × 10 = 3,000 So, 50 × 60 = 3,000

Question 6. 90 × 90 = _____

Answer: 8,100

Explanation: By using the Place-value method, 90 × 90 = 90 × 9 tens = 9 tens × 9 tens = 81 × 10 × 10 = 8,100 So, 90 × 90 = 8,100

Question 7. 70 × 11 = _____

Answer: 770

Explanation: By using the Place-value method, 70 × 11 = 11 × 7 tens = 77 tens = 77 × 10 = 770 So, 70 × 11 = 770

Question 8. 18 × 30 = _____

Answer: 540

Explanation: By using the Place-value method, 30 × 18 = 18 × 3 tens = 54 tens = 54 × 10 = 540 So, 30 × 18 = 540

Question 9. 20 × 75 = _____

Explanation: By using the Place-value method, 20 × 75 = 75 × 2 tens = 150 tens = 150 × 10 = 1,500 So, 20 × 75 = 1,500 Find the missing factor.

Question 10. 40 × ____ = 3,200

Answer: The missing number is: 80 Explanation; Let the missing number be X So, 40 × X = 3,200 X = 3,200 /40 = 80 Hence, the value of X is: 80

Question 11. _____ × 20 = 1,200

Answer: The missing number is: 60 Explanation; Let the missing number be X So, 20 × X = 1,200 X = 1,200 /20 = 60 Hence, the value of X is: 60

Question 12. 30 × ____ = 2,100

Answer: The missing number is: 70 Explanation; Let the missing number be X So, 30 × X = 2,100 X = 2,100 /30 = 70 Hence, the value of X is: 70

4.2 Estimate Products

Estimate the product.

Question 13. 25 × 74

Answer: 1,850

Explanation: Let 74 be rounded to 75 By using the Partial Fraction method, 25 × 75 = ( 20 + 5) × ( 70 + 5) = ( 20 × 70) + ( 20 × 5) + ( 5 × 70) + ( 5 × 5) = 1,400 + 100 + 350 + 25 = 1,875 So, 25 × 74 = 1,875

Question 14. 16 × 28

Answer: 448

Explanation: Let 16 be rounded to 15 Let 28 be rounded to 30 By using the Partial Fraction method, 15 × 30 = ( 10 + 5) × ( 25 + 5) = ( 10 × 25) + ( 10 × 5) + ( 5 × 25) + ( 5 × 5) = 200 + 50 + 125 + 25 = 400 So, 16 × 28 = 400

Question 15. 42 × 81

Explanation: Let 42 be rounded to 40 Let 81 be rounded to 80 By using the tens method, 40 × 80 = 40 × 8 tens = 4 tens × 8 tens = 32 × 1 ten × 1 ten = 32 × 10 × 10 = 3,200 So, 42 × 81 = 3,200 Open-Ended Write two possible factors that can be estimated as shown.

Explanation: The Products of 81 are: 9 × 9 = 81 From the above two products, we can conclude that the two possible numbers that can give the product 6,400 are: 90,90

Explanation: The Products of 64 are: 2 × 2 = 4 4 × 1 =4 From the above two products, we can conclude that the two possible numbers that can give the product 400 are: 20,20 and. 10,40

Draw an area model to find the product.

Question 18. 13 × 19 = _____

Explanation: By using the Partial Products method, 13 × 19 = ( 10 + 3) × ( 10 + 9) 10 × 10 +    10× 9 + 10 × 3 + 3 × 9 = 100 + 90 + 30 + 27 = 247 So, 19 × 13 = 247

Question 19. 21 × 36 = _____

Answer: 756

Explanation: By using the Partial Products method, 21 × 36 = ( 20 + 1) × ( 30 + 6) = 20 × 30 +   20× 6 + 30 × 1 + 1 × 6 = 600 + 120 + 30 + 6 = 756 So, 21 × 36 = 756

Use Distributive Property to find a product.

Question 21. 27 × 34 = 27 × (30 + 4) = (27 × 30) + (27 × 4) = (20 + 7) × 30 + (20 + 7) × 4 = (20 ×30) + (7 × 30) + (20 ×4) + (7 × 4) = 600 + 210 + 80 + 28 = 918 So, 27 × 34 = 918.

Question 22. 43 × 18 = _____

Answer: 774

Explanation: Using the Partial Products method, 43 × 18 = 43× (10 + 8) = (43 × 10) + (43 × 8) = (40 + 3) × 10 + (40 + 3) × 8 = (40 ×10) + (3 × 10) + (40 ×8) + (3 × 8) = 400 + 30 + 320 + 24 = 774 So, 43 × 18 = 774

Question 23. 35 × 57 = _____

Answer: 1,995

Explanation: Using the Partial products method, 35 × 57 = 35 × (50 + 7) = (35 × 50) + (35 × 7) = (30 + 5) × 50 + (30 + 5) × 7 = (50 ×30) + (5 × 50) + (30 ×7) + (7 × 5) = 1,500 + 250 + 210 + 35 = 1,995 So, 35 × 57 = 1,995

Question 24. 81 × 76 = _____

Answer: 6,156 81 × 76 = 81 × (70 + 6) = (81 × 70) + (81 × 6) = (80 + 1) × 70 + (80 + 1) × 6 = (80 ×70) + (70 × 1) + (80 ×6) + (1 × 6) = 5,600 + 70 + 480 + 6 = 6,156 So, 81 × 76 = 6,156

Answer: 396

Explanation: Using the Partial Products method, 18 × 22 = 22 × (10 + 8) = (22 × 10) + (22 × 8) = (20 + 2) × 10 + (20 + 2) × 8 = (20 ×10) + (2 × 10) + (20 ×8) + (2 × 8) = 200 + 20 + 160 + 16 = 396 So, 18 × 22 = 396

Answer: 3,358

Explanation: Using the Partial Products method, 73 × 46 = 73 × (40 + 6) = (73 × 40) + (73 × 6) = (70 + 3) × 40 + (70 + 3) × 6 = (70 ×40) + (3 × 40) + (70 ×6) + (3 × 6) = 2,800 + 120 + 420 + 18 = 3,358 So, 73 × 46 = 3,358.

Explanation: By using the Partial Products method, 39 × 84 = 39 × (80 + 4) = (39 × 80) + (39 × 4) = (30 + 9) × 80 + (30 + 9) × 4 = (80 ×30) + (9 × 80) + (30 ×4) + (9 × 4) = 2,400 + 720 + 120 + 36 = 3,276 So, 39 × 84 = 3,276

Question 28. 57 × 19 = _____

Answer: 1,083

Explanation: By using the Partial Products method, 57 × 19 = 57 × (10 + 9) = (57 × 10) + (57 × 9) = (50 + 7) × 10 + (50 + 7) × 9 = (50 ×10) + (7 × 10) + (50 ×9) + (7 × 9) = 500 + 70 + 450 + 63 = 1,083 So, 57 × 19 = 1,083

Question 29. 38 × 65 = _____

Answer: 2,470

Explanation: By using the Partial Products method, 38 × 65 = 65 × (30 + 8) = (65 × 30) + (65 × 8) = (60 + 5) × 30 + (60 + 5) × 8 = (60 ×30) + (5 × 30) + (60 ×8) + (5 × 8) = 1,800 + 150 + 480 + 40 = 2,470 So, 38 × 65 = 2,470

Question 30. 94 × 26 = _____

Answer: 2,444

Explanation: By using the Partial Products method, 94 ×26 = 94 × (20 + 6) = (94 × 20) + (94 × 6) = (90 + 4) × 20 + (90 + 4) × 6 = (90 ×20) + (4 × 20) + (90 ×6) + (4 × 6) = 1,800 + 80 + 540 + 24 = 2,444 So, 94 × 26 = 2,444 Reasoning Find the missing digits. Then find the product.

Answer: 2,108

Explanation: Using the Partial Products method, 34 × 62 = ( 60 + 2) × ( 30 + 4) = 30 × 60 + 4 × 60 + 30 × 2 + 2 × 4 = 1,800 + 240 + 60 + 8 = 2,108 Estimate: Let 62 be Rounded to 60. Let 34 be Rounded to 35. So, 60 × 35 = 2,100 As the Estimate and the actual answer are near, the answer is reasonable.

Answer: 7,917

Explanation: Using the Partial Products method, 87 × 91 = ( 80 + 7) × ( 90 + 1) = 80 × 90 + 7 × 90 + 80 × 1 + 7 × 1 = 7,200 + 630 + 80 + 7 = 7,917 Estimate: Let 87 be Rounded to 85. Let 91 be Rounded to 90. So, 90 × 85 = 7,650 As the Estimate and the actual answer are not near, the answer is  not reasonable.

Answer: 3,285

Explanation: Using the Partial Products method, 73 × 45 = ( 70 + 3) × ( 40 + 5) = 70 × 40 + 3 × 40 + 70 × 5 + 5 × 3 = 2,800 + 120 + 350 + 15 = 3,285 Estimate: Let 73 be Rounded to 75. So, 45 × 75 = 3,375 As the Estimate and the actual answer are near, the answer is reasonable.

Question 36. Estimate: ____ 13 × 21 = ______

Answer: 273

Explanation: Using the Partial Products method, 13 × 21 = ( 20 + 1) × ( 10 + 3) = 20 × 10 + 20 × 3 + 10 × 1 + 1 × 3 = 200 + 60 + 10 + 3 = 273 Estimate: Let 13 be Rounded to 15. Let 21 be Rounded to 20. So, 20 × 15 = 300 As the Estimate and the actual answer are near, the answer is reasonable.

Question 37. Estimate: _____ 42 × 53 = _____

Answer: 2,226

Explanation: Using the Partial Products method, 42 × 53 = ( 40 + 2) × ( 50 + 3) = 40 × 50 + 3 × 40 + 50 × 2 + 2 × 3 = 2,000 + 120 + 100 + 6 = 2,226 Estimate: Let 42 be Rounded to 40. Let 53 be Rounded to 55. So, 40 × 55 = 2,200 As the Estimate and the actual answer are near, the answer is reasonable.

Question 38. Estimate: _____ 29 × 66 = _____

Answer: 1,914

Explanation: Using the Partial Products method, 29 × 66 = ( 20 + 9) × ( 60 + 6) = 20 × 60 + 9 × 60 + 20 × 6 + 9 × 6 = 1,200 + 540 + 120 + 54 = 1,914 Estimate: Let 29 be Rounded to 30. Let 66 be Rounded to 65. So, 30 × 65 = 1,950 As the Estimate and the actual answer are near, the answer is reasonable.

Question 39. 80 × 30 = _____

Answer: 2,400

Explanation: Using the Place-value method, 80 × 30 = 80 × 3 tens = 240 tens = 2,400 So, 80 × 30 = 2,400

Question 40. 26 × 51 = _____

Answer: 1,326

Explanation: Using the Partial Products method, 26 × 51 = ( 20 + 6) × ( 50 + 1) = ( 20 × 50) + ( 20 × 1) + ( 6 × 50) + (6 × 1) = 1,000 + 20 + 300 + 6 = 1,326

Question 41. 94 × 70 = _____

Answer: 6,580

Explanation: Using the Place-value method, 94 × 70 = 94 × 7 tens = 658 tens = 6,580 So, 94 × 70 = 6,580

Question 42. 15 × 67 = _____

Answer: 1,005

Explanation: Using the Partial Products method, 15 × 67 = ( 10 + 5) × ( 60 + 7) = ( 10 × 60) + ( 10 × 7) + ( 5 × 60) + ( 5 × 7) = 600 + 70 + 300 + 35 =1,005

Question 43. 40 × 38 = _____

Answer: 1,520

Explanation: Using the Place-value method, 38 × 40 = 38 × 4 tens = 152 tens = 1,520 So, 38 × 40 = 1,520

Question 44. 29 × 92 = _____

Answer: 2,668

Explanation: Using the partial products method, 29 × 92 = ( 20 + 9) × ( 90 + 2) = ( 20 × 90 ) + ( 20 × 2) + ( 9 × 90) + ( 9 × 2) = 1,800 + 40 + 810 + 18 = 2,668

Answer: 2,560 people will ride the Ferris wheel in 1 day

Explanation: Given that a  Ferris wheel runs 40 times each day and it has 16 cars with 4 seats in each car. So, the total number of people that can ride a Ferris wheel = 40 × 16 × 4 = 40 × 64 By using the place-value method, 64  × 40  = 64 × 4 tens = 256 tens = 2,560 From the above, We can conclude that there are 2,560 people who will ride the Ferris wheel in 1 day 4.8 Problem Solving: Multiplication with Two-Digit Numbers

Answer: 136 songs

Explanation: Given that a music fan has to memorize all of the songs from 13 albums and there are 15 songs on each album. So, the total number of songs the music fan has to memorize = 13 × 15 By using the Partial Products method, 13 × 15 = ( 10 + 3) × ( 10 + 5) = ( 10 × 10 ) + ( 10 × 5) + ( 3 × 10 ) + ( 3 × 5) = 100 + 50 + 30 + 15 = 195 It is also given that the music fan memorized 59 songs for the Concert. Hence, The number of songs that has to memorize by the music fan = 195 – 59 = 136 songs

Answer: 2,592 inches Explanation; The given Jamaican flag is in the shape of a rectangle. We know that, Area of the rectangle = length × breadth So, Area of the Jamaican flag = 36 × ( 36 + 36) = 36 × 72 Using the Partial Products method, 36 × 72 = ( 30 + 6) × ( 70 + 2) = ( 30 × 70 ) + ( 30 × 2 ) + ( 6 × 70) + ( 6 × 2) = 2,100 + 60 + 720 + 12 = 2,592 inches From this, We can conclude that the area of the Jamaican flag is: 2,592 inches

Final Words:

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Math  /  4th Grade  /  Unit 2: Multi-Digit Multiplication

Multi-Digit Multiplication

Students are introduced to multi-digit multiplication and its applications, including multiplicative comparison.

Unit Practice

Unit summary.

In Unit 2, 4th grade students multiply up to four-digit numbers by one-digit numbers, relying on their understanding of place value and properties of operations, as well as visual models like an area model, to solve. 

As a foundation for their multi-year work with multiplication and division, students in 2nd grade learned to partition a rectangle into rows and columns and write a repeated addition sentence to determine the total. They also skip-counted by 5s, 10s, and 100s. Then, in  3rd grade , students developed a conceptual understanding of multiplication and division in relation to equal groups, arrays, and area. They developed a variety of strategies to build toward fluency with multiplication and division within 100 and applied that knowledge to the context of one- and two-step problems using the four operations. 

To begin the unit, students extend their understanding of multiplication situations that they learned in 3rd grade to include multiplicative comparison using the words “times as many.” This helps acquaint students with these types of problems before solving them with larger quantities later in the unit. It also gives students time to refresh their multiplication and division facts from 3rd grade, since the problems in Topic A are limited to those that involve multiplication and division within 100. Then, students move into two-digit by one-digit, three-digit by one-digit, four-digit by one-digit, and two-digit by two-digit multiplication, using the area model, partial products, and finally the standard algorithm, making connections between all methods as they go. The use of the area model serves to help students conceptually understand multiplication and as a connection to their work with area and perimeter (4.MD.3), a supporting cluster standard. Finally, with a full understanding of all multiplication cases, they then apply their new multiplication skills to solve multi-step word problems using multiplication, addition, and subtraction, including cases involving multiplicative comparison (4.NBT.5, 4.OA.3, 4.MD.3), allowing for many opportunities to connect content across multiple domains. 

This unit affords lots of opportunities to deepen students’ mathematical practices. For example, “when students decompose numbers into sums of multiples of base-ten units to multiply them, they are seeing and making use of structure (MP.7). Further, “by reasoning repeatedly (MP.8) about the connection between math drawings and written numerical work, students can come to see multiplication and division algorithms as abbreviations or summaries of their reasoning about quantities” (NBT Progression, p. 14). Lastly, as students solve multi-step word problems involving addition, subtraction, and multiplication, they are modeling with mathematics (MP.4).

Students’ work in this unit will prepare them for fluency with the multiplication algorithm in 5th grade  (5.NBT.5). Students also learn about new applications of multiplication in future grades, including scaling quantities up and down in 5th grade (5.NF.5), all the way up to rates and slopes in the middle grades (6.RP, 7.RP). Every subsequent grade level depends on the understanding of multiplication and its algorithm, making this unit an important one for students in 4th grade.

Pacing: 21 instructional days (18 lessons, 2 flex days, 1 assessment day)

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The following assessments accompany Unit 2.

Have students complete the Pre-Unit Assessment and Pre-Unit Student Self-Assessment before starting the unit. Use the Pre-Unit Assessment Analysis Guide to identify gaps in foundational understanding and map out a plan for learning acceleration throughout the unit.

Pre-Unit Student Self-Assessment

Have students complete the Mid-Unit Assessment after lesson 10.

Use the resources below to assess student understanding of the unit content and action plan for future units.

Post-Unit Assessment

Post-Unit Assessment Answer Key

Post-Unit Student Self-Assessment

Use student data to drive instruction with an expanded suite of assessments. Unlock Pre-Unit and Mid-Unit Assessments, and detailed Assessment Analysis Guides to help assess foundational skills, progress with unit content, and help inform your planning.

Intellectual Prep

Suggestions for how to prepare to teach this unit

Before you teach this unit, unpack the standards, big ideas, and connections to prior and future content through our guided intellectual preparation process. Each Unit Launch includes a series of short videos, targeted readings, and opportunities for action planning to ensure you're prepared to support every student.

Intellectual Prep for All Units

• Read and annotate “Unit Summary” and “Essential Understandings” portion of the unit plan. 
• Do all the Target Tasks and annotate them with the “Unit Summary” and “Essential Understandings” in mind. 
• Take the Post-Unit Assessment. 

Unit-Specific Intellectual Prep

• Read pp. 14–15 in Progressions for the Common Core State Standards in Mathematics  Number and Operations in Base Ten, K-5  (starting at the section titled “Use place value understanding and properties of operations to perform multi-digit arithmetic”).
• Read the document “ Situation Types for Operations in Word Problems ” for multiplication and division. Identify the word problem types of any applicable assessment questions. 
• Read the following table that includes methods used throughout the unit.

Essential Understandings

The central mathematical concepts that students will come to understand in this unit

• In an additive comparison, the underlying question is what amount would be added to one quantity in order to result in the other. In a multiplicative comparison, the underlying question is what factor would multiply one quantity in order to result in the other.
• One component of understanding general methods for multiplication is understanding how to compute products of one-digit numbers and multiples of 10, 100, and 1,000. Another part of understanding general base-ten methods for multi-digit multiplication is understanding the role played by the distributive property.
• Rounding numbers can help one to determine whether an answer is reasonable based on whether the estimate is close to the computed answer or not.
• Making sense of problems and persevering in solving them is an important practice when solving word problems. Key words do not always indicate the correct operation.

Terms and notation that students learn or use in the unit

partial product

To see all the vocabulary for Unit 2, view our 4th Grade Vocabulary Glossary .

The materials, representations, and tools teachers and students will need for this unit

• Optional: Base ten blocks (Maximum of 18 thousands, 18 hundreds, 18 tens, 18 ones per student or small group) — Students might not need these depending on their reliance on concrete materials. See Lesson 4 Anchor Task 1 Notes for more information.

Help students strengthen their application and fluency skills with daily word problem practice and content-aligned fluency activities.

Topic A: Multiplicative Comparison

Solve multiplicative comparison problems with a larger unknown. Distinguish multiplicative comparison from additive comparison.

4.OA.A.1 4.OA.A.2

Solve multiplicative comparison problems with a smaller unknown.

Solve multiplicative comparison problems with an unknown multiplier. Interpret a multiplication equation as a comparison. 

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Topic B: Multiplication of up to Four-Digit Whole Numbers by One-Digit Whole Numbers

Multiply 10, 100, and 1,000 by one- and two-digit numbers.

Multiply multiples of 10, 100, and 1,000 by one-digit numbers. Estimate multi-digit products by rounding numbers to their largest place value.

Multiply two-, three-, and four-digit numbers by one-digit numbers using a variety of mental strategies.

Multiply two-digit numbers by one-digit numbers.

Multiply three-digit numbers by one-digit numbers.

Multiply four-digit numbers by one-digit numbers.

Multiply two-, three-, and four-digit numbers by one-digit numbers and assess the reasonableness of the product. 

Topic C: Multiplication of Two-Digit Whole Numbers by Two-Digit Whole Numbers

Multiply two-digit multiples of 10 by two-digit multiples of 10. Estimate multi-digit products by rounding numbers to their largest place value.

Multiply two-digit multiples of 10 by two-digit numbers.

Multiply two-digit numbers by two-digit numbers using a variety of mental strategies.

Multiply two-digit by two-digit numbers using four partial products.

Multiply two-digit by two-digit numbers using two partial products and assess the reasonableness of the product.

Topic D: Multi-Step Word Problems

Abstract the formulas for the area and perimeter of a rectangle and apply those formulas in real-world and mathematical problems involving multiplication, addition, and subtraction.

4.MD.A.3 4.OA.A.3

Solve two-step word problems involving multiplication, addition, and subtraction.

4.OA.A.2 4.OA.A.3

Solve multi-step word problems involving multiplication, addition, and subtraction.

Common Core Standards

Major Cluster

Supporting Cluster

Additional Cluster

Core Standards

The content standards covered in this unit

Measurement and Data

4.MD.A.3 — Apply the area and perimeter formulas for rectangles in real world and mathematical problems. For example, find the width of a rectangular room given the area of the flooring and the length, by viewing the area formula as a multiplication equation with an unknown factor.

Number and Operations in Base Ten

4.NBT.B.5 — Multiply a whole number of up to four digits by a one-digit whole number, and multiply two two-digit numbers, using strategies based on place value and the properties of operations. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.

Operations and Algebraic Thinking

4.OA.A.1 — Interpret a multiplication equation as a comparison, e.g., interpret 35 = 5 × 7 as a statement that 35 is 5 times as many as 7 and 7 times as many as 5. Represent verbal statements of multiplicative comparisons as multiplication equations.

4.OA.A.2 — Multiply or divide to solve word problems involving multiplicative comparison, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem, distinguishing multiplicative comparison from additive comparison.

4.OA.A.3 — Solve multistep word problems posed with whole numbers and having whole-number answers using the four operations, including problems in which remainders must be interpreted. Represent these problems using equations with a letter standing for the unknown quantity. Assess the reasonableness of answers using mental computation and estimation strategies including rounding.

Foundational Standards

Standards covered in previous units or grades that are important background for the current unit

3.MD.C.7.B — Multiply side lengths to find areas of rectangles with whole-number side lengths in the context of solving real world and mathematical problems, and represent whole-number products as rectangular areas in mathematical reasoning.

3.MD.C.7.C — Use tiling to show in a concrete case that the area of a rectangle with whole-number side lengths a and b + c is the sum of a × b and a × c. Use area models to represent the distributive property in mathematical reasoning.

3.MD.D.8 — Solve real world and mathematical problems involving perimeters of polygons, including finding the perimeter given the side lengths, finding an unknown side length, and exhibiting rectangles with the same perimeter and different areas or with the same area and different perimeters.

3.NBT.A.3 — Multiply one-digit whole numbers by multiples of 10 in the range 10—90 (e.g., 9 × 80, 5 × 60) using strategies based on place value and properties of operations.

4.NBT.A.1 — Recognize that in a multi-digit whole number, a digit in one place represents ten times what it represents in the place to its right. For example, recognize that 700 ÷ 70 = 10 by applying concepts of place value and division.

4.NBT.A.2 — Read and write multi-digit whole numbers using base-ten numerals, number names, and expanded form. Compare two multi-digit numbers based on meanings of the digits in each place, using >, =, and < symbols to record the results of comparisons.

4.NBT.A.3 — Use place value understanding to round multi-digit whole numbers to any place.

4.NBT.B.4 — Fluently add and subtract multi-digit whole numbers using the standard algorithm.

3.OA.A.1 — Interpret products of whole numbers, e.g., interpret 5 × 7 as the total number of objects in 5 groups of 7 objects each. For example, describe a context in which a total number of objects can be expressed as 5 × 7.

3.OA.A.2 — Interpret whole-number quotients of whole numbers, e.g., interpret 56 ÷ 8 as the number of objects in each share when 56 objects are partitioned equally into 8 shares, or as a number of shares when 56 objects are partitioned into equal shares of 8 objects each. For example, describe a context in which a number of shares or a number of groups can be expressed as 56 ÷ 8.

3.OA.A.3 — Use multiplication and division within 100 to solve word problems in situations involving equal groups, arrays, and measurement quantities, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem.

3.OA.A.4 — Determine the unknown whole number in a multiplication or division equation relating three whole numbers. Determine the unknown whole number in a multiplication or division equation relating three whole numbers. For example, determine the unknown number that makes the equation true in each of the equations 8 × ? = 48, 5 = _ ÷ 3, 6 × 6 = ?

3.OA.B.5 — Apply properties of operations as strategies to multiply and divide. Students need not use formal terms for these properties. Example: Knowing that 8 × 5 = 40 and 8 × 2 = 16, one can find 8 × 7 as 8 × (5 + 2) = (8 × 5) + (8 × 2) = 40 + 16 = 56. (Distributive property.) Example: If 6 × 4 = 24 is known, then 4 × 6 = 24 is also known (Commutative property of multiplication.) 3 × 5 × 2 can be found by 3 × 5 = 15, then 15 × 2 = 30, or by 5 × 2 = 10, then 3 × 10 = 30. (Associative property of multiplication.)

3.OA.B.6 — Understand division as an unknown-factor problem. For example, find 32 ÷ 8 by finding the number that makes 32 when multiplied by 8.

3.OA.C.7 — Fluently multiply and divide within 100, using strategies such as the relationship between multiplication and division (e.g., knowing that 8 × 5 = 40, one knows 40 ÷ 5 = 8) or properties of operations. By the end of Grade 3, know from memory all products of two one-digit numbers.

3.OA.D.8 — Solve two-step word problems using the four operations. Represent these problems using equations with a letter standing for the unknown quantity. Assess the reasonableness of answers using mental computation and estimation strategies including rounding. This standard is limited to problems posed with whole numbers and having whole-number answers; students should know how to perform operations in the conventional order when there are no parentheses to specify a particular order (Order of Operations).

Future Standards

Standards in future grades or units that connect to the content in this unit

4.NBT.B.6 — Find whole-number quotients and remainders with up to four-digit dividends and one-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.

5.NBT.B.5 — Fluently multiply multi-digit whole numbers using the standard algorithm.

Number and Operations—Fractions

4.NF.A.1 — Explain why a fraction a/b is equivalent to a fraction (n × a)/(n × b) by using visual fraction models, with attention to how the number and size of the parts differ even though the two fractions themselves are the same size. Use this principle to recognize and generate equivalent fractions.

4.NF.B.4 — Apply and extend previous understandings of multiplication to multiply a fraction by a whole number.

5.NF.B.3 — Interpret a fraction as division of the numerator by the denominator (a/b = a ÷ b). Solve word problems involving division of whole numbers leading to answers in the form of fractions or mixed numbers, e.g., by using visual fraction models or equations to represent the problem. For example, interpret 3/4 as the result of dividing 3 by 4, noting that 3/4 multiplied by 4 equals 3, and that when 3 wholes are shared equally among 4 people each person has a share of size 3/4. If 9 people want to share a 50-pound sack of rice equally by weight, how many pounds of rice should each person get? Between what two whole numbers does your answer lie?

5.NF.B.4 — Apply and extend previous understandings of multiplication to multiply a fraction or whole number by a fraction.

5.NF.B.5 — Interpret multiplication as scaling (resizing), by:

5.NF.B.5.A — Comparing the size of a product to the size of one factor on the basis of the size of the other factor, without performing the indicated multiplication.

5.NF.B.5.B — Explaining why multiplying a given number by a fraction greater than 1 results in a product greater than the given number (recognizing multiplication by whole numbers greater than 1 as a familiar case); explaining why multiplying a given number by a fraction less than 1 results in a product smaller than the given number; and relating the principle of fraction equivalence a/b = (n×a)/(n×b) to the effect of multiplying a/b by 1.

5.NF.B.6 — Solve real-world problems involving multiplication of fractions and mixed numbers, e.g., by using visual fraction models or equations to represent the problem.

Standards for Mathematical Practice

CCSS.MATH.PRACTICE.MP1 — Make sense of problems and persevere in solving them.

CCSS.MATH.PRACTICE.MP2 — Reason abstractly and quantitatively.

CCSS.MATH.PRACTICE.MP3 — Construct viable arguments and critique the reasoning of others.

CCSS.MATH.PRACTICE.MP4 — Model with mathematics.

CCSS.MATH.PRACTICE.MP5 — Use appropriate tools strategically.

CCSS.MATH.PRACTICE.MP6 — Attend to precision.

CCSS.MATH.PRACTICE.MP7 — Look for and make use of structure.

CCSS.MATH.PRACTICE.MP8 — Look for and express regularity in repeated reasoning.

Place Value, Rounding, Addition, and Subtraction

Multi-Digit Division

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Word Problems and Multiplication

In our Word Problems and Multiplication lesson plan, students learn strategies for solving word problems that contain multiplication. They will practice writing word problems that require multiplication and will solve problems using multiplication.

Included with this lesson are some adjustments or additions that you can make if you’d like, found in the “Options for Lesson” section of the Classroom Procedure page. One of the optional additions to this lesson is to have students solve problems as a class using each of the strategies

Description

Additional information, what our word problems and multiplication lesson plan includes.

Lesson Objectives and Overview: Word Problems and Multiplication lesson plan lists several strategies for solving word problems using multiplication. Some of the strategies include making a diagram, eliminating possibilities, and looking for patterns. Several clue words are listed to help students know when multiplication is necessary to solve a word problem. At the end of the lesson, students will be able to solve word problems using multiplication. This lesson is for students in 3rd grade and 4th grade.

Classroom Procedure

Every lesson plan provides you with a classroom procedure page that outlines a step-by-step guide to follow. You do not have to follow the guide exactly. The guide helps you organize the lesson and details when to hand out worksheets. It also lists information in the blue box that you might find useful. You will find the lesson objectives, state standards, and number of class sessions the lesson should take to complete in this area. In addition, it describes the supplies you will need as well as what and how you need to prepare beforehand.

Options for Lesson

Included with this lesson is an “Options for Lesson” section that lists a number of suggestions for activities to add to the lesson or substitutions for the ones already in the lesson. If you’d like to add to the activity worksheet, you can have students draw cards with pictures on them to determine what their multiplication problems should be about. For an additional activity, you can have students solve problems as a class using each of the strategies. Finally, you can also have students work together to create a poster that shows their strategy for solving a word problem.

Teacher Notes

The teacher notes page includes lines that you can use to add your own notes as you’re preparing for this lesson.

WORD PROBLEMS AND MULTIPLICATION LESSON PLAN CONTENT PAGES

The Word Problems and Multiplication lesson plan includes two content pages. The lesson begins by reminding students that they complete math problems every day without thinking about it. There are many real-world situations that require you to use different math operations, like multiplication. You can use different strategies to solve multiplication word problems.

The first thing you should do when solving a multiplication word problem is to explore! You should figure out what you already know, what information the problem gave to you, what you need to find out, and what question is being asked.

The second step is to plan. You need to decide what strategy you’re going to use to solve the problem. The lesson lists some problem solving strategies that you can think about using, including guess and check, look for a pattern, make an organized list, and draw a diagram.

The third step is to solve by using your chosen problem solving strategy. During this step, you’ll do all of the actual math. It’s important to have already identifies the key words in the word problem, because these will tell you which operation you will need to use (this lesson only uses multiplication, but other word problems will require the use of addition, subtraction, and division). Some words and phrases that indicate that you need to use multiplication are times, rate, each , and product of . The lesson also lists a few more.

The fourth step is to examine. You should re-read the question to make sure your answer makes sense, and check your work. If your answer doesn’t make sense for the problem, try solving using a different strategy.

The lesson then delves into two different example problems and shows how to solve them step-by-step using the four steps described above.

WORD PROBLEMS AND MULTIPLICATION LESSON PLAN WORKSHEETS

The Word Problems and Multiplication lesson plan includes three worksheets: an activity worksheet, a practice worksheet, and a homework assignment. You can refer to the guide on the classroom procedure page to determine when to hand out each worksheet.

CREATE YOUR OWN ACTIVITY WORKSHEET

For the activity worksheet, students will write their own multiplication problems using the pictures shown on the worksheet and numbers 100 or less.

SOLVING WORD PROBLEMS PRACTICE WORKSHEET

The practice worksheet asks students to solve four word problems.

WORD PROBLEMS AND MULTIPLICATION HOMEWORK ASSIGNMENT

For the homework assignment, students will write their own multiplication word problems (like the activity worksheet). They should be creative, use numbers 100 or less, and attach a picture if they want.

This lesson includes a quiz to test students’ understanding of the lesson material. For the quix, students will solve a multiplication word problem.

Worksheet Answer Keys

This lesson plan includes answer keys for the practice worksheet and the quiz. If you choose to administer the lesson pages to your students via PDF, you will need to save a new file that omits these pages. Otherwise, you can simply print out the applicable pages and keep these as reference for yourself when grading assignments.

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4.10 Problem Solving Multiplication

By alisa olsen on jan 02, 2014.

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Math Expressions Grade 4 Unit 4 Lesson 3 Answer Key Situation and Solution Equations for Multiplication and Division

Solve the questions in Math Expressions Grade 4 Homework and Remembering Answer Key Unit 4 Lesson 3 Answer Key Situation and Solution Equations for Multiplication and Division to attempt the exam with higher confidence. https://mathexpressionsanswerkey.com/math-expressions-grade-4-unit-4-lesson-3-answer-key/

Math Expressions Common Core Grade 4 Unit 4 Lesson 3 Answer Key Situation and Solution Equations for Multiplication and Division

Math Expressions Grade 4 Unit 4 Lesson 3 Homework

Division equation 90 ÷ 15 = 6 90 ÷  6 = 15 90 ÷ (5 x 3) = 6 90 ÷  (3 x 2) = 15

Solve each equation.

Explanation: r = 200 ÷ 5 r = 40

Situation And Solution Equations Grade 4 Unit 4 Lesson 3 Math Expressions Question 3. 12 × d = 84 d = _________ Answer: 12 x d = 84 d = 84 ÷ 12 d = 7 (12 x 7 = 84)

Math Expressions Grade 4 Answer Key Unit 4 Lesson 3 Math Expressions Question 4. 80 ÷ 10 = n n = __________ Answer: 80 ÷ 10 = n n = 80 ÷ 10 n = 8 (10 x 8 = 80)

Situation And Solution Equation For Multiplication And Division Grade 4 Unit 4 Question 5. 120 = 10 × m m = _________ Answer: 120 = 10 x m 10 x m = 120 m = 120 ÷ 10 m =12 ( 120 x 10 = 120)

Unit 4 Lesson 3 Math Expressions Question 6. 88 = 8 × c c = __________ Answer: 88 = 8 × c 8 x c = 88 c = 88 ÷  8 c = 11 ( 11 x 8 = 88)

Question 7. 100 ÷ q = 20 q = __________ Answer: 100 ÷ q = 20 100 ÷ 20 = q q = 5 ( 20 x 5 = 100)

Write an equation to solve the problem. Draw a model if you need to. Show your work.

Explanation: Lucy bought some shrubs to plant in her garden.Each shrub cost $9. If Lucy spent $216 in all. Divide 216 by 9 to find how many shurbs lucy bought 216 ÷ 9 = 24 Therefore, Lucy buys 24 shurbs.

Explanation: Jeremiah has 592 flyers in stacks of 8 flyers each Divide 892 by 8 to find the number of stacks 892 ÷ 8 = 74 Therefore, Jeremiah made 74 stacks of flyers.

Question 10. The apples from an average-sized tree will fill 20 baskets. If an orchard has 17 average-sized trees, how many baskets of apples can it produce? Answer: The orchard can produce 340 baskets of apples.

Math Expressions Grade 4 Unit 4 Lesson 3 Remembering

Use the Algebraic Notation Method to solve the problem. Complete the steps.

Explanation: 5 • 68 = ______ • (_____ + _____) 5 • 68 = 5 • (60 +8) = 5 • 60 + 5 • 8 = 300 + 40 = 340 5 • 68 = 5 • (60 +8).

Solve. Use the Place Value Sections and the Expanded Notation Methods for division.

Write = or ≠ to make each statement true.

Explanation: 9 x p = 450 p = 450 ÷ 9 p = 50 Therefore, if 9 x p = 450 then p = 45.

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Texas Go Math Grade 4 Lesson 10.4 Answer Key Place the First Digit

Refer to our Texas Go Math Grade 4 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 4 Lesson 10.4 Answer Key Place the First Digit.

Essential Question

How can you use place value to know where to place the first digit in the quotient? Answer: place value is the value of the digit in its place.

Unlock the Problem

Jaime took 144 photos on a digital camera. The photos are to be placed equally in 6 photo albums. asked to find. How many photos will be in each album?

• Underline what you are
• Circle what you need to use.

We need to find the number of photos will be there in each album.

Example Divide. 144 ÷ 6

The first digit of the quotient is in tens place.

So, 14 – 12 = 2

Math Idea: After you divide each place, the remainder should be less than the divisor.

So, there are 24 photos in each album.

Mathematical Processes Explain how the answer change if Jaime had 146 photos. Answer:

If Jamie had 146 photos, then

Quotient : 24

Remainder : 2

So, The answer in remainder changes if Jacob have 146 photos.

Share and Show

Quotient = 113

Remainder = 0

Quotient = 041

Remainder= 2

Quotient = 155

Remainder= 0

Mathematical Processes Explain how you placed the first digit of the quotient in Exercise 2. Answer:

We can place the first digit in the first place.

Problem Solving

Practice: Copy and Solve Divide.

Question 4. 516 ÷ 2 Answer:

Quotient = 258

Go Math Lesson 10.4 4th Grade Place The First Digit Question 5. 516 ÷ 3 Answer:

Quotient = 172

Question 6. 516 ÷ 4 Answer:

Quotient = 129

Question 7. 516 ÷ 5 Answer:

Quotient = 103

Remainder= 1

Question 8. H.O.T. Look hack at your answers to Exercises 4-7. What happens to the quotient when the divisor increases? Explain. Answer:

The value of remainder increases when the value of divisor increases.

Question 9. Multi-Step Reggie has 192 pictures of animals. He wants to keep half and then divide the rest equally among three friends. How many pictures will each friend get? (A) 96 (B) 32 (C) 48 (D) 14 Answer:

Total number of animal pictures Reggie have = 192

Given, he kept half of them

Which means,

Now, 96 animal pictures are shared equally among 3 friends.

Therefore, each friend gets 32 pictures.

Go Math Lesson 10.4 Homework Answers Grade 4 Question 10. H.O.T. Multi-Step There are 146 students, 5 teachers, and 8 chaperones going to the theater. To reserve their seats, they need to reserve entire rows. Each row has 8 seats. How many rows must they reserve? (A) 20 (B) 18 (C) 19 (D) 58 Answer: C

Number of students = 146

Number of teachers = 5

Number of chaperones = 8

Total number of people = 146+5+8

Also give, the number of people for each row = 8

Therefore, They must reserve 19 rows.

=  Unlock the Problem

Number of pictures Nan have = 234

Pictures per blue page = 4

Therefore, 58 full pages are there in her album.

a. What do you need to find? Answer:

We need to find the number of full pages in Nan album

b. Communicate How will you use division to find the number of full pages? Answer:

We can find the number of full pages by dividing number of pictures Nan have by the number of pictures for blue page.

c. Show the steps you will use to solve the problem. Answer:

d. Complete the following sentences. Nan has ________ pictures. She wants to put the pictures in an album with pages that each hold ________ pictures. She will have an album with ________ full pages and ________ pictures on another page. Answer:

Nan has 234 pictures. She wants to put the pictures in an album with pages that each hold 4 pictures. She will have an album with 58 full pages and 4 pictures on another page.

Question 12. Apply Juan wants to put his 672 pictures in an album with a green cover. how many full pages will he have in his album? Answer:

For green color, pictures per page = 6

Number of pictures Juan in an album = 672

Therefore, he have 112 full pages in his album.

Go Math Answer Key Grade 4 Lesson 10.4 Place The First Digit Question 13. H.O.T. Multi-Step Talla has 162 pictures to put in a photo album. If she wants only full pages, which color albums could she use? Answer:

Number of pictures Talla have = 162

For blue color, number of pictures per page = 4

Remainder = 0 Which means there are only full pages.

Therefore, She should use blue color.

Daily Assessment Task

Fill in the bubble completely to show your answer.

Question 14. Danny makes a website to showcase images of unusual manhole covers. He takes 264 pictures for the site. Danny places 8 pictures on each page of the site. How many pages of pictures does Danny’s site have? (A) 33 pages (B) 330 pages (C) 3 pages (D) 30 pages Answer:

Number of pictures Danny takes = 264

Number of pictures he kept on each page of the site = 8

Therefore, 33 pages of pictures does Danny’s site have

Question 15. Debi celebrates her birthday with friends at an arcade. Debi’s parents want to share 300 tokens equally among the 9 friends. Debi uses division to find how many tokens each person gets. In which place is the first digit of the quotient? (A) thousands (B) hundreds (C) tens (D) ones Answer:

Total number of token = 300

Number of friends = 9

Also given, Debi’s parents give each person the same number of tokens.

Quotient = 033

Remainder = 3

The 0 is in hundreds place, 3 is in tens place.

Question 16. Multi-Step Debi’s parents want to share 300 tokens equally among the 9 friends. After Debi’s parents give each person the same number of tokens, how many tokens are left over? (A) 0 tokens (B) 2 tokens (C) 1 token (D) 3 tokens Answer:

Therefore, the number of tokens left = 3

TEXAS Test Prep

Question 17. Kat wants to put 485 pictures in an album with a red cover. There are 9 pictures per page. She uses division to find out how many full pages she will have. In which place is the first digit of the quotient? (A) thousands (B) hundreds (C) tens (D) ones Answer:

The number of pictures Kat wants to keep in album = 485

Number of pictures per page = 9

The first digit in quotient is in tens place.

Texas Go Math Grade 4 Lesson 10.4 Homework and Practice Answer Key

Quotient = 090

Quotient = 043

Remainder = 1

Quotient = 073

Quotient = 046

Quotient = 064

Remainder = 4

Quotient = 038

Quotient = 019

Remainder = 7

Question 9. 428 ÷ 2 ____________ Answer:

Quotient = 214

Question 10. 428 ÷ 3 ____________ Answer:

Quotient = 142

Remainder = 2

Question 11. 428 ÷ 4 ____________ Answer:

Quotient = 107

Question 12. 428 ÷ 5 ____________ Answer:

Question 13. Camp Mesquite will provide 4 buses for 212 campers. If each bus carries the same number of campers, how many campers will ride in each bus? Answer:

Total number of campers = 212

Number of buses = 4

If each bus carries the same number of campers,

Therefore, 53 campers will ride in each bus.

Question 14. The garden center received a shipment of 132 daisies. If each table displays the same number of daisies, how many daisies will be placed on each of 3 tables? Answer:

Number of daisies shipped = 132

Number of tables = 3

If each table displays the same number of daisies,

Therefore, 44 daisies will be placed on each of 3 tables.

Lesson Check

Question 15. Lauren collected 532 pennies in a donation box. She wants to divide the pennies equally into 4 containers. Lauren uses division to find how many pennies to put into each container. In which place is the first digit of the quotient? (A) thousands (B) hundreds (C) tens (D) ones Answer:

The number of pennies Lauren collected = 532

Number of boxes = 5

The digit of the quotient is in Hundreds place.(1)

Question 16. Mark opens a box of 500 cups for the football concession stand. Mark uses division to divide the cups equally among windows. In which place is the first digit of the quotient? (A) ones (B) tens (C) hundreds (D) thousands Answer:

Question 17. Mrs. Samson bought 294 craft sticks for a class project. She divides her class into 7 groups and gives each group an equal number of sticks. If she uses all of the sticks, how many will each group receive? (A) 40 (B) 43 (C) 41 (D) 42 Answer:

Number of sticks Samson bought to the class = 294

Number of groups she divides her class = 7

Also given, She uses all the sticks in equal number.

Therefore, Each group receive 42 sticks.

Question 18. Cassie read 483 pages of her hook in one week. if she read the same number of pages each day, how many pages did she read in 1 day? (A) 78 (B) 79 (C) 69 (D) 68 Answer:

Total number of pages Cassie read  = 483

Number of days in a week = 7

If she read the same number of pages each day,

Therefore, Cassie read 69 pages in 1 day.

Question 19. Multi-Step Mr. Parsons bought 293 apples to make pies for his shop. Six apples are needed for each pie. If Mr. Parsons makes the most apple pies possible, how many apples will be left over? (A) 5 (B) 1 (C) 3 (D) 6 Answer:

Number of apples Parsons bought = 293

Number of apples required to make 1 apple pie = 6

So, 293/6 = 48

Now, 6 x 48 = 288

So, Total – Number of apples required = leftover

293-288 = 5

Therefore, the number of apples left = 5.

Question 20. Multi-Step At an art school, 112 students are in 4 equal-sized drawing classes. There are 120 students in 5 equal-sized painting classes. I low many more students are in one drawing class than are in one painting class? (A) 28 (B) 5 (C) 25 (D) 4 Answer:

Total number of students in 4 equal sized drawing class = 112

Total number of students in 5 equal sized drawing class = 120

Now, The number of more students = 28 – 24

Therefore,4 students are in one drawing class than are in one painting class.

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McGraw Hill My Math Grade 5 Chapter 10 Lesson 8 Answer Key Multiplication as Scaling

All the solutions provided in McGraw Hill Math Grade 5 Answer Key PDF Chapter 10 Lesson 8 Multiplication as Scaling  will give you a clear idea of the concepts.

McGraw-Hill My Math Grade 5 Answer Key Chapter 10 Lesson 8 Multiplication as Scaling

Scaling is the process of resizing a number when you multiply by a fraction that is greater than or less than 1.

Draw It Multiply the number 2 by three fractions greater than 1.

Multiply the number 2 by three fractions less than 1.

Talk About it

Question 1. Predict whether the product of 3 and \(\frac{4}{5}\) is greater than, less than, or equal to 3. Explain. Answer: The above-given equation: product of 3 and 4/5. This can be written as: 3 x 4/5 = 12/5 = 2 2/5. 12/5 = 2.4 which is less than 3 Therefore, the product is less than 3.

Question 2. Mathematical PRACTICE 3 Draw a Conclusion Predict whether the product of 2 and 2\(\frac{1}{5}\) is greater than, less than, or equal to 2. Explain. Answer: The above-given equation: product of 2 and 2 1/5 2 x 2 1/5 2 1/5 is a mixed fraction. Convert it into an improper fraction. 2 1/5 = 11/5 Now multiply: 2 x 11/5 = 22/5 = 4.4 Therefore, the product is greater than 2.

Practice It

Without multiplying, circle whether each product is greater than, less than, or equal to the whole number.

Algebra Without multiplying, circle whether the unknown in each equation is greater than, less than, or equal to the whole number.

For Exercises 13-15, analyze each product in the table.

Question 13. Why is the first product less than \(\frac{3}{4}\)? Answer: The above-given: The first product and its factors are given in the table. 1/2 x 3/4 = 3/8 3/8 is smaller than 3/4 3/8 = 0.375 3/4 = 0.75 3/8 is 50% smaller than 3/4. Question 14. Why is the product of 1 and \(\frac{3}{4}\) equal to \(\frac{3}{4}\)? Answer: The above-given: 1 x 3/4 = 3/4 if we multiply anything by 1 we get the same product So, they are equal.

Write About It

Question 18. How can I use scaling to help predict the product of a number and a fraction? Answer: I can predict the size of the product based on the size of the factors. For example fraction × fraction = smaller fraction, fraction × whole number = smaller number, whole number × mixed number = larger than the original whole number.

McGraw Hill My Math Grade 5 Chapter 10 Lesson 8 My Homework Answer Key

Problem Solving

Vocabulary Check

Question 7. Fill in the blank with the correct term or number to complete the sentence. ____ is the process of resizing a number when you multiply by a fraction that is greater than or less than 1. Answer: Scaling – Scaling is the factor which is used to represent the object size. The size of the object can be shown by increasing or decreasing its original size. In general, the represented size of the object increased for the small object whereas it decreased for the bigger object. Scaling is used for better viewing of an object. The ratio by which an object’s size increases or decreases is called a scale factor. – Scale factor is the ratio between the corresponding measurement of an object and the representation of that object. A scale factor is a whole number or greater than 1 to make a larger copy. The scale factor is a fraction or less than 1 to make a smaller copy. Therefore, the scale factor can be expressed by any number or fraction. Also, the scale factor can be expressed using the colon (:), such as Original sized object: Representation-sized object where, Representation sized object = Any number x Original sized object (in case of larger copy) or, Representation sized object = 1/ any number x Original sized object (in case of smaller copy)

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