McGraw Hill Math

All the solutions provided in McGraw Hill Math Grade 5 Answer Key PDF Chapter 5 Lesson 4 Interpreting Numerical Expressions are as per the latest syllabus guidelines.

McGraw-Hill Math Grade 5 Answer Key Chapter 5 Lesson 4 Interpreting Numerical Expressions

Solve.

Write an expression for each group of steps. Remember the Order of Operations.

Question 1.
Multiply 3 times 2 squared and then subtract the product from 12.
12 – (3 × 2²)

Question 2.
Divide 18 by 2 and then subtract the quotient from 22.
Answer:
22 – (18÷2) = 13.

Explanation:
Here, we need to divide 18 by 2 and then subtract the quotient from 22. So the expression will be 22 – (18÷2)
= 22 – 9
= 13.

Question 3.
Multiply the difference between 5 and 2 by 3 squared.
Answer:
(5-2) × 3² = 27.

Explanation:
Here, we need to multiply the difference between 5 and 2 by 3 squared. So (5-2) × 3²
= 3 × 9
= 27.

Question 4.
Find the product of 6 and 9, then add 3.
Answer:
(6×9) + 3 = 57.

Explanation:
Here, we need to find the product of 6 and 9, then add 3. So (6×9) + 3 which is
= 54 + 3
= 57.

Write a description for each expression.

Question 5.
(4 + 2)² × (5 × 2)
Answer:
The addition of 4 and 2 which is squared and multiplied with the product of 5 and 2.

Explanation:
Given the expression is (4 + 2)² × (5 × 2) which is
= (6)² × (10)
= 36 × 10
= 360.
So it will be the addition of 4 and 2 which is squared and multiplied with the product of 5 and 2.

Question 6.
6 × (5 × 3) – 50
Answer:
6 × (5 × 3) – 50 = 40.

Explanation:
Given the expression is 6 × (5 × 3) – 50 which is
= 6 × (15) – 50
= 90 – 50
= 40.
So it will be the product of 5 and 3 multiplied and then subtract 50.

Question 7.
3² × 2² + (5 × 4)
Answer:
3² × 2² + (5 × 4) = 56.

Explanation:
Given the expression is 3² × 2² + (5 × 4) which is
= 9 × 4 + 20
= 36 + 20
= 56.
So it will be, multiply three squared and two squared and then add the product of 5 and 4.

Write < or >. Show how to simplify the expressions.

Question 8.
(45 ÷ 9) × 7 ____ 24 × 3 – 30
Answer:
(45 ÷ 9) × 7 < 24 × 3 – 30.

Explanation:
Given the expression is (45 ÷ 9) × 7 ____ 24 × 3 – 30 which is
= 5 × 7 ____  72 – 30
= 35 < 42.

All the solutions provided in McGraw Hill Math Grade 5 Answer Key PDF Chapter 5 Lesson 5 Problem Solving are as per the latest syllabus guidelines.

McGraw-Hill Math Grade 5 Answer Key Chapter 5 Lesson 5 Problem Solving

Solve.

Question 1.
Scott writes the following series of numbers: 0.007, 0.027, 0.04, 0.067, 0.087
What is the missing number? The difference between numbers is 0.020; 0.027 + 0.020 = 0.047

Question 2.
Kip is at an auction for antique motor bikes. The opening bid is $2,000. The next 3 bids are $2,500, $3,000, and $3,500. The bike sells on the eighth bid. If the pattern continues, what is the final, winning bid? Create a table.
Answer:
The final winning bid is $6,000.

Explanation:
Given that the next 3 bids are $2,500, $3,000, and $3,500. So for eighth bid it will be $2,500, $3,000, $3,500, $4,000, $4,500, $5,000, $5,500, $6,000. So eighth bid is $6,000.

DRAWING OF TABLE

Question 3.
Dylan has 8 U.S. coins. The coins have a total value of $0.56. What are the eight coins? _______________________
Answer:
The eight coins are 4 nickels, 1 dime, 1 quarter, and 1 penny.

Explanation:
Given that Dylan has 8 U.S. coins and the coins have a total value of $0.56. So the eight coins are 4 nickels, 1 dime, 1 quarter, and 1 penny which is 4 × 0.05 + 0.10 + 0.25 + 0.01 = $0.56.

All the solutions provided in McGraw Hill Math Grade 5 Answer Key PDF Chapter 5 Test are as per the latest syllabus guidelines.

McGraw-Hill Math Grade 5 Chapter 5 Test Answer Key

Follow the directions to simplify each expression.

Question 1.
(15 – 5)³ – (20 × 5) × (8 – 4) + 10² =
simplify inside parentheses: _____
____________________
Simplify exponents: ____________
Multiply: ___________
Add and subtract from left to right: ___________
Final answer: _____________
Answer:
simplify inside parentheses: (10)³ – (100) × (4) + 10²
Simplify exponents: 1000 – 100 × 4 + 100
Multiply: 1000 – 400 + 100
Add and subtract from left to right: 1000 – 500
Final answer: 500.

Explanation:
Given the expression is(15 – 5)³ – (20 × 5) × (8 – 4) + 10² . So first we will solve inside parentheses which is
(10)³ – (100) × (4) + 10² , now we will solve the exponents which is
= (10)³ – (100) × (4) + 10²
= 1000 – 100 × 4 + 100
= 1000 – 400 + 100
= 1000 – 500
= 500.

Question 2.
(30 × 5) × (6 – 2) + (10 – 5)² – 10¹ =
simplify inside parentheses: _____
____________________
Simplify exponents: ____________
Multiply: ___________
Add and subtract from left to right: ___________
Final answer: _____________
Answer:
simplify inside parentheses: (60) × (4) + (5)² – 10
Simplify exponents: 60 × 4 + 25 – 10
Multiply: 240 + 25 – 10
Add and subtract from left to right: 240 + 15
Final answer: 255.

Explanation:
Given the expression is (30 × 5) × (6 – 2) + (10 – 5)² – 10¹  . So first we will solve inside parentheses which is
(60) × (4) + (5)² – 10, now we will solve the exponents which is
= 60 × 4 + 25 – 10
= 240 + 15
= 255.

Use the order of operations to simplify and solve. Show your work.

Question 3.
(7 × 3) ÷ 3²
Answer:
(7 × 3) ÷ 3² = 2\(\frac{1}{3}\).

Explanation:
Given that the expression is
(7 × 3) ÷ 3²
= 21 ÷ 3²
= 21 ÷ 9
= 7 ÷ 3
= 2\(\frac{1}{3}\).

Question 4.
5 + 2² × (5 × 2) ÷ 5
Answer:
5 + 2² × (5 × 2) ÷ 5 = 13.

Explanation:
Given that the expression is 5 + 2² × (5 × 2) ÷ 5
= 5 + 4 × 10 ÷ 5
= 5 + 4 × 2
= 5 + 8
= 13.

Question 5.
(52 + 9) – 14
Answer:
(52 + 9) – 14 = 47.

Explanation:
Given that the expression is (52 + 9) – 14
= 61 – 14
= 47.

Question 6.
(60 ÷ 10) + (12 – 8)²
Answer:
(60 ÷ 10) + (12 – 8)² = 22.

Explanation:
Given that the expression is (60 ÷ 10) + (12 – 8)²
= 6 + (4)²
= 6 + 16
= 22.

Question 7.
12 + (4 × 5) × 8
Answer:
12 + (4 × 5) × 8 = 172.

Explanation:
Given that the expression is 12 + (4 × 5) × 8
= 12 + (20) × 8
= 12 + 160
= 172.

Question 8.
2² × 3² + 6 – (8 × 4)
Answer:
2² × 3² + 6 – (8 × 4) = 10.

Explanation:
Given that the expression is 2² × 3² + 6 – (8 × 4)
= 2² × 3² + 6 – 32
= 4 × 9 + 6 – 32
= 36 + 6 – 32
= 42 – 32
= 10.

Simplify each expression.

Question 9.
{[3(5 + 5) + 25] × 2}
Simplify inside the parentheses: ________________________
Simplify inside the brackets: _______________
Simplify inside the braces: _________
{[3(5 + 5) + 25] × 2) = _________
Answer:
Simplify inside the parentheses: [3(10) + 25] × 2
Simplify inside the brackets: [30 + 25] × 2
Simplify inside the braces: 55 × 2
{[3(5 + 5) + 25] × 2) = 110.

Explanation:
Given the expression is {[3(5 + 5) + 25] × 2} . So first we will solve inside parentheses which is
[3(10) + 25] × 2, now we will solve the exponents which is [30 + 25] × 2
= 55 × 2
= 110.

Question 2.
3{[2(4 + 5) + 8] – 20}
Simplify inside the parentheses: __________
Simplify inside the brackets: __________
Simplify inside the braces: _________
3{[2(4 + 5) + 8] – 20} = _________
Answer:
Simplify inside the parentheses: 3{[2(9) + 8] – 20}
Simplify inside the brackets: 3{[18 + 8] – 20}
Simplify inside the braces: 3{6}
3{[2(4 + 5) + 8] – 20} = 18.

Explanation:
Given the expression is 3{[2(4 + 5) + 8] – 20}. So first we will solve inside parentheses which is
3{[2(9) + 8] – 20}, now we will solve the exponents which is 3{[18 + 8] – 20}
= 3{[26] – 20}
= 3{26 – 20}
= 3{6}
= 18.

Question 11.
[(10 × 3 + 4) ÷ 17]²
Simplify inside the parentheses: __________
Simplify inside the brackets: __________
[(10 × 3 + 4) ÷ 17]² = _________
Answer:
Simplify inside the parentheses: [(30 + 4) ÷ 17]²
Simplify inside the brackets: [(34) ÷ 17]²
[(10 × 3 + 4) ÷ 17]² = 4

Explanation:
Given the expression is [(10 × 3 + 4) ÷ 17]². So first we will solve inside parentheses which is
[(30 + 4) ÷ 17]²,now we will solve the exponents which is [(34) ÷ 17]² =
= [2]²
= 4.

Write an expression for each description.

Question 12.
Multiply 4 times 2 squared, and then subtract the product from 15.
Answer:
15 – (4×4) = 1.

Explanation:
Here, we need to multiply 4 times 2 squared which is 4×4 = 16. And then we need to subtract the product from 15 which is 15-16 = -1.

Question 13.
Divide 36 by 6, and then subtract the quotient from 54.
Answer:
54 – (36÷6)  = 48.

Explanation:
Here, we need to divide 36 by 6 which is 36÷6 = 6. And then subtract the quotient from 54 which is 54-6 = 48.

Question 14.
Multiply 7 squared by the difference between 8 and 6.
Answer:
(8-6) × 7² = 98.

Explanation:
Here, we need to multiply 7 squared by the difference between 8 and 6 which is (8-6) × 7² =which is 2 × 49 = 98.

Question 15.
Find the product of 12 and 15, and then add 3
Answer:
(12×15) + 3 = 183.

Explanation:
The product of 12 and 15 is 12×15 = 180 and then add 3 which is 180 + 3 = 183.

Write a description for each expression.

Question 16.
(4 + 2)² – 16
Answer:
(4 + 2)² – 16 = 20.

Explanation:
Given the expression is (4 + 2)² – 16 which is (6)² – 16
= 36 – 16
= 20.

Question 17.
8 × (6 × 7) – 50
Answer:
8 × (6 × 7) – 50 = 286.

Explanation:
Given the expression is 8 × (6 × 7) – 50 which is
= 8 × (42) – 50
= 336 – 50
= 286.

Question 18.
48 – 3² × 2²
Answer:
48 – 3² × 2² = 12.

Explanation:
Given the expression is 48 – 3² × 2² which is
= 48 – 3² × 2²)
= 48 – 9 × 4
= 48 – 36
= 12.

Question 19.
7 × (9 × 2) – 2²
Answer:
7 × (9 × 2) – 2² = 122.

Explanation:
Given the expression is 7 × (9 × 2) – 2² which is
= 7 × (18) – 4
= 126 – 4
= 122.

Solve.

Question 20.
(150 – 25) – (25 – 5²)
Answer:
(150 – 25) – (25 – 5²) = 125.

Explanation:
Given that the expression is (150 – 25) – (25 – 5²) which is
= (125) – (25 – 25)
= 125 – 0
= 125.

Question 21.
{[4 × (17 × 8) + 27] × 4}
Answer:
{[4 × (17 × 8) + 27] × 4} = 2,284.

Explanation:
Given that the expression is {[4 × (17 × 8) + 27] × 4} which is
= {[4 × (136) + 27] × 4}
= {[544 + 27] × 4}
= 571 × 4
= 2,284.

Solve.

Question 22.
Elizbeth needs to find the missing number in the following series.
0.459, 0.859, ____, 1.659, 2.059
What is the missing number?
Answer:
0.459, 0.859, 1.259, 1.659, 2.059.

Explanation:
Given that the series is 0.459, 0.859, ____, 1.659, 2.059. As the series follows by adding 0.4 to the numbers. So the missing digit is 0.859+0.4 which is 1.259. So the series is 0.459, 0.859, 1.259, 1.659, 2.059.

Question 23.
Twins Don and Dan are working together to solve a math problem. Don finds the answer and tells
Dan: “Square the sum of 6 plus 2, and then subtract 30.” Write the numerical expression that matches Don’s directions.
Answer:
(6+2)² – 30 = 34.

Explanation:
Given that Dan Squares the sum of 6 plus 2 which is (6+2)²
= 8²
= 64, and then subtract 30 which is
= 64 – 30
= 34.

Question 24.
A gallon of fat-free milk costs $3.99. A year ago, the gallon of milk cost 3 percent (0.03) less. How much did a gallon of milk cost a year ago?
Answer:
The cost of a gallon of milk a year ago is $3.87.

Explanation:
Given that a gallon of fat-free milk costs $3.99. A year ago, the gallon of milk cost 3 percent (0.03) less. So the cost of a gallon of milk a year ago is 3.99×3 which is = 11.97 ÷ 100 = 0.12. So 3.99-0.12 = $3.87.

Question 25.
Devon works a total of 21 hours on Friday, Saturday, and Sunday. He works twice as many hours on Sunday as he does on Saturday. He works 6 hours on Friday. How many hours does he work on Saturday?
Answer:
The number of hours he works on Saturday is 5 hours.

Explanation:
Given that Devon works a total of 21 hours on Friday, Saturday, and Sunday, he works twice as many hours on Sunday as he does on Saturday. As he works 6 hours on Friday, so the number of hours does he work on Saturday is. Let the Saturday hours be X and the Sunday hours will be 2X. So 2X+X+6 = 21 which is
3X+6 = 21
3X = 21 – 6
3X = 15
X = 15÷3
= 5.
So the number of hours he works on Saturday is 5 hours.

All the solutions provided in McGraw Hill Math Grade 5 Answer Key PDF Chapter 6 Lesson 1 Finding Least Common Multiples are as per the latest syllabus guidelines.

McGraw-Hill Math Grade 5 Answer Key Chapter 6 Lesson 1 Finding Least Common Multiples

Solve

Write the multiples for each pair of numbers. Identify the LCM for each pair.

Question 1.
4, 5
Multiples: _______________
Multiples: _______________
LCM of 4, 5: _______________
Answer:
Multiples: 4, 8, 12, 16, 20, 24, 28
Multiples: 5, 10, 15, 20, 25, 30
LCM of 4, 5: 20

Question 2.
4, 6
Multiples: _______________
Multiples: _______________
LCM of 4, 6: _______________
Answer:
LCM of 4, 6: 12.

Explanation:
The multiples of 4 are 4, 8, 12, 16, 20
The multiplies of 6 are 6, 12, 18, 24, 30
LCM of 4, 6: 12.

Question 3.
8, 6
Multiples: _______________
Multiples: _______________
LCM of 8, 6: _______________
Answer:
LCM of 8, 6: 24.

Explanation:
The multiples of 8 are 8, 16, 24, 32
The multiplies of 6 are 6, 12, 18, 24
LCM of 8, 6: 24.

Question 4.
5, 6
Multiples: _______________
Multiples: _______________
LCM of 5, 6: _______________
Answer:
LCM of 5, 6: 30.

Explanation:
The multiples of 5 are 5, 10, 15, 20, 25, 30
The multiplies of 6 are 6, 12, 18, 24, 30
LCM of 5, 6: 30.

All the solutions provided in McGraw Hill Math Grade 5 Answer Key PDF Chapter 6 Lesson 10 Word Problems with Fractions are as per the latest syllabus guidelines.

McGraw-Hill Math Grade 5 Answer Key Chapter 6 Lesson 10 Word Problems with Fractions

Solve

First write an expression and then solve the problem. Simplify if needed.

Question 1.
Mr. Goldberg notices that \(\frac{1}{12}\) of the juice boxes hold apple juice and \(\frac{1}{16}\) are berry What fraction of the juice boxes are either apple or berry?
Expression: \(\frac{1}{12}\) + \(\frac{1}{16}\)
Solution: \(\frac{7}{48}\) are either apple or berry: \(\frac{1}{12}\) + \(\frac{1}{16}\) = \(\frac{4}{48}\) + \(\frac{3}{38}\) = \(\frac{7}{48}\)
Answer:
The fraction of the juice boxes are either apple or berry is \(\frac{7}{48}\).

Explanation:
Given that \(\frac{1}{12}\) of the juice boxes hold apple juice and \(\frac{1}{16}\) are berry. So the fraction of the juice boxes are either apple or berry is \(\frac{1}{12}\)+\(\frac{1}{16}\)
= \(\frac{4+3}{48}\)
= \(\frac{7}{48}\).

Question 2.
What fraction of juice boxes holds either orange juice or apple juice?
Expression: _______________
Solution: _____________________
Answer:
The fraction is \(\frac{7}{48}\).

Explanation:
The fraction of juice boxes holds either orange juice or apple juice is \(\frac{1}{12}\)+\(\frac{1}{16}\)
= \(\frac{4+3}{48}\)
= \(\frac{7}{48}\).

Question 3.
Explain how to use the number of juice boxes and the given fractions to find the actual number of orange, apple, and berry juice boxes.
Answer:
The actual number of orange, apple, and berry juice boxes is \(\frac{7}{48}\).

Explanation:
The number of juice boxes and the given fractions to find the actual number of orange, apple, and berry juice boxes is \(\frac{1}{12}\)+\(\frac{1}{16}\)
= \(\frac{4+3}{48}\)
= \(\frac{7}{48}\).

All the solutions provided in McGraw Hill Math Grade 5 Answer Key PDF Chapter 6 Lesson 11 Problem Solving are as per the latest syllabus guidelines.

McGraw-Hill Math Grade 5 Answer Key Chapter 6 Lesson 11 Problem Solving

First draw a diagram and then solve.

Question 1.
Ken and Pete built a bridge made of toothpicks. Ken worked 3\(\frac{1}{3}\) hours on the bridge, and Pete worked \(\frac{1}{2}\) hour more than Ken. How much total time did Pete spend working on the bridge?
Ken: 3\(\frac{1}{3}\) hours
Pete: Ken’s hours + \(\frac{1}{2}\) hours
Total Time = _____________________
Answer:
The total number of hours spent working on the bridge is 3\(\frac{5}{6}\) hours.

Explanation:
Given that Ken and Pete built a bridge made of toothpicks. Ken worked 3\(\frac{1}{3}\) hours on the bridge, and Pete worked \(\frac{1}{2}\) hour more than Ken. So the total time did Pete spend working on the bridge is 3\(\frac{1}{3}\)+\(\frac{1}{2}\) which is
= \(\frac{10}{3}\) + \(\frac{1}{2}\)
= \(\frac{20+3}{6}\)
= \(\frac{23}{6}\)
= 3\(\frac{5}{6}\) hours.

Question 2.
At Taylor School. 145 fifth-grade students display their projects in the science fair. 15 more girls than boys display their projects. How many girls and how many boys have entered the science fair?
Answer:
65 boys and 80 girls.

Explanation:
Given that 145 fifth-grade students display their projects in the science fair and 15 more girls than boys display their projects. So the number of girls. So the number of boys and girls who have entered the science fair is 65 boys and 65+15 = 80 girls.

All the solutions provided in McGraw Hill Math Grade 5 Answer Key PDF Chapter 6 Lesson 2 Finding Equivalent Fractions are as per the latest syllabus guidelines.

McGraw-Hill Math Grade 5 Answer Key Chapter 6 Lesson 2 Finding Equivalent Fractions

Solve

Find the equivalent fraction.

Question 1.
equivalent to \(\frac{1}{2}\), denominator 16 ______________
Answer:
\(\frac{8}{16}\)

Question 2.
equivalent to \(\frac{1}{3}\), denominator 12 ______________
Answer:
\(\frac{4}{12}\).

Explanation:
The equivalent fraction of \(\frac{1}{3}\) by denominator 12 is by multiplying numerator and denominator with 4. So \(\frac{1}{3}\) × \(\frac{4}{4}\) = \(\frac{4}{12}\).

Question 3.
equivalent to \(\frac{1}{5}\), denominator 20 ______________
Answer:
\(\frac{5}{20}\).

Explanation:
The equivalent fraction of \(\frac{1}{5}\) by denominator 20 is by multiplying numerator and denominator with 5. So \(\frac{1}{5}\) × \(\frac{5}{5}\) = \(\frac{5}{20}\).

Question 4.
equivalent to \(\frac{2}{3}\), denominator 15 ______________
Answer:
\(\frac{10}{15}\).

Explanation:
The equivalent fraction of \(\frac{2}{3}\) by denominator 15 is by multiplying numerator and denominator with 5. So \(\frac{2}{3}\) × \(\frac{5}{5}\) = \(\frac{10}{15}\).

Question 5.
equivalent to \(\frac{3}{4}\), denominator 12 ______________
Answer:
\(\frac{9}{12}\).

Explanation:
The equivalent fraction of \(\frac{3}{4}\) by denominator 12 is by multiplying numerator and denominator with 3. So \(\frac{3}{4}\) × \(\frac{3}{3}\) = \(\frac{9}{12}\).

Question 6.
equivalent to \(\frac{3}{4}\), denominator 16 ______________
Answer:
\(\frac{12}{16}\).

Explanation:
The equivalent fraction of \(\frac{3}{4}\) by denominator 16 is by multiplying numerator and denominator with 4. So \(\frac{3}{4}\) × \(\frac{4}{4}\) = \(\frac{12}{16}\).

Question 7.
equivalent to \(\frac{3}{5}\), denominator 20 ______________
Answer:
\(\frac{15}{20}\).

Explanation:
The equivalent fraction of \(\frac{3}{5}\) by denominator 20 is by multiplying numerator and denominator with 5. So \(\frac{3}{5}\) × \(\frac{5}{5}\) = \(\frac{15}{20}\).

Question 8.
equivalent to \(\frac{5}{12}\), denominator 24 ______________
Answer:
\(\frac{10}{24}\).

Explanation:
The equivalent fraction of \(\frac{1}{3}\) by denominator 24 is by multiplying numerator and denominator with 2. So \(\frac{5}{12}\) × \(\frac{2}{2}\) = \(\frac{10}{24}\).

Question 9.
equivalent to \(\frac{4}{5}\), denominator 30 ______________
Answer:
\(\frac{24}{30}\).

Explanation:
The equivalent fraction of \(\frac{4}{5}\) by denominator 30 is by multiplying numerator and denominator with 6. So \(\frac{4}{5}\) × \(\frac{6}{6}\) = \(\frac{24}{30}\).

Question 10.
equivalent to \(\frac{5}{7}\), denominator 49 ______________
Answer:
\(\frac{35}{49}\).

Explanation:
The equivalent fraction of \(\frac{5}{7}\) by denominator 49 is by multiplying numerator and denominator with 7. So \(\frac{5}{7}\) × \(\frac{7}{7}\) = \(\frac{35}{49}\).

All the solutions provided in McGraw Hill Math Grade 5 Answer Key PDF Chapter 6 Lesson 3 Adding Fractions with Like Denominators are as per the latest syllabus guidelines.

McGraw-Hill Math Grade 5 Answer Key Chapter 6 Lesson 3 Adding Fractions with Like Denominators

Solve

Add. Give the answer in simplest terms.

Question 1.
\(\frac{1}{4}\) + \(\frac{1}{4}\) ______________
Answer:
\(\frac{2}{4}\) = \(\frac{1}{2}\)

Question 2.
\(\frac{2}{9}\) + \(\frac{5}{9}\) ______________
Answer:
\(\frac{2}{9}\) + \(\frac{5}{9}\) = \(\frac{7}{9}\).

Explanation:
The simplest terms of \(\frac{2}{9}\) + \(\frac{5}{9}\) is \(\frac{2+5}{9}\)
= \(\frac{7}{9}\).

Question 3.
\(\frac{1}{7}\) + \(\frac{3}{7}\) _______________
Answer:
\(\frac{1}{7}\) + \(\frac{3}{7}\) = \(\frac{4}{7}\).

Explanation:
The simplest terms of \(\frac{1}{7}\) + \(\frac{3}{7}\) is \(\frac{1+3}{7}\)
= \(\frac{4}{7}\).

Question 4.

Answer:
\(\frac{3}{31}\) + \(\frac{4}{31}\) = \(\frac{7}{31}\).

Explanation:
The simplest terms of \(\frac{3}{31}\) + \(\frac{4}{31}\) is \(\frac{3+4}{31}\)
= \(\frac{7}{31}\).

Question 5.

Answer:
\(\frac{2}{9}\) + \(\frac{1}{9}\) = \(\frac{1}{3}\).

Explanation:
The simplest terms of \(\frac{2}{9}\) + \(\frac{1}{9}\) is \(\frac{2+1}{9}\)
= \(\frac{3}{9}\)
= \(\frac{1}{3}\).

Question 6.

Answer:
\(\frac{7}{39}\) + \(\frac{6}{39}\) = \(\frac{1}{3}\).

Explanation:
The simplest terms of \(\frac{7}{39}\) + \(\frac{6}{39}\) is \(\frac{7+6}{39}\)
= \(\frac{13}{39}\)
= \(\frac{1}{3}\).

Question 7.

Answer:
\(\frac{14}{23}\) + \(\frac{19}{23}\) = \(\frac{33}{23}\).

Explanation:
The simplest terms of \(\frac{14}{23}\) + \(\frac{19}{23}\) is \(\frac{14+19}{23}\)
= \(\frac{33}{23}\).

Question 8.

Answer:
\(\frac{5}{8}\) + \(\frac{5}{8}\) = \(\frac{5}{4}\).

Explanation:
The simplest terms of \(\frac{5}{8}\) + \(\frac{5}{8}\) is \(\frac{5+5}{8}\)
= \(\frac{10}{8}\)
= \(\frac{5}{4}\).

Question 9.

Answer:
\(\frac{2}{15}\) + \(\frac{8}{15}\) = \(\frac{2}{3}\).

Explanation:
The simplest terms of \(\frac{2}{15}\) + \(\frac{8}{15}\) is \(\frac{2+8}{15}\)
= \(\frac{10}{15}\)
= \(\frac{2}{3}\).

Question 10.

Answer:
\(\frac{4}{9}\) + \(\frac{2}{9}\) = \(\frac{2}{3}\).

Explanation:
The simplest terms of \(\frac{4}{9}\) + \(\frac{2}{9}\) is \(\frac{4+2}{9}\)
= \(\frac{6}{9}\)
= \(\frac{2}{3}\).

Question 11.

Answer:
\(\frac{2}{18}\) + \(\frac{1}{18}\) = \(\frac{1}{6}\).

Explanation:
The simplest terms of \(\frac{2}{18}\) + \(\frac{1}{18}\) is \(\frac{2+1}{18}\)
= \(\frac{3}{18}\)
= \(\frac{1}{6}\).

All the solutions provided in McGraw Hill Math Grade 5 Answer Key PDF Chapter 3 Lesson 3 Relating Multiplication and Division are as per the latest syllabus guidelines.

McGraw-Hill Math Grade 5 Answer Key Chapter 3 Lesson 3 Relating Multiplication and Division

Sketch each problem. List the related multiplication and division facts.

Question 1.
3 × 6 = 18
Answer:

Question 2.
5 × 4 = ____
Answer:
5 × 4 = 20.

Explanation:
By multiplying 5 and 4 we will get 20.

Question 3.
2 × 6 = _______
Answer:
2 × 6 = 12.

Explanation:
By multiplying 2 and 6 we will get 12.

Question 4.
2 × 1 = ___
Answer:
2 × 1 = 2.

Explanation:
By multiplying 2 and 1 we will get 2.

All the solutions provided in McGraw Hill Math Grade 5 Answer Key PDF Chapter 3 Lesson 4 Multiplying by 1 Digit Whole Numbers are as per the latest syllabus guidelines.

McGraw-Hill Math Grade 5 Answer Key Chapter 3 Lesson 4 Multiplying by 1 Digit Whole Numbers

Multiply and then describe the strategy you used to find the answer.

Question 1.
16 × 3 = ______
I changed 3 to 6 and 16 to 8. Then I multiplied 6 × 8 = 48.

Question 2.
21 × 5 = ___
Answer:
15 × 7 = 105.

Explanation:
Here, we have changed 5 to 15 and 21 to 7 and then multiplied 15 × 7 which is 105.

Question 3.
4 × 55 = ________
Answer:
20 × 11 = 220.

Explanation:
Here, we have changed 4 to 20 and 55 to 11 and then multiplied 20 × 11 which is 220.

Question 4.
225 × 3 ______
Answer:
225 × 3 = 675.

Explanation:
Here, we have changed 225 to 75 and 3 to 9 and then multiplied 75 × 9 which is 675.

Question 5.

Answer:
31 × 3 = 93.

Explanation:
Here, we have changed 31 to 93 and 3 to 1 and then multiplied 93 × 1 which is 93.

Question 6.

Answer:
42 .× 5 = 210

Explanation:
Here, we have changed 42 to 6 and 5 to 35 and then multiplied 6 × 35 which is 210.

Question 7.

Answer:
138 × 4 = 552.

Explanation:
Here, we have changed 138 to 276 and 4 to 2 and then multiplied 276 × 2 which is 552.

Question 8.

Answer:
21 × 8 = 168.

Explanation:
Here, we have changed 21 to 84 and 8 to 2 and then multiplied 84 × 2 which is 168.

Question 9.

Answer:
121 × 7 = 847.

Explanation:
Here, we have changed 121 to 11 and 7 to 77 and then multiplied 11 × 77 which is 847.

Question 10.

Answer:
125 × 4 = 500.

Explanation:
Here, we have changed 125 to 250 and 4 to 2 and then multiplied 250 × 2 which is 500.

Question 11.

Answer:
138 × 8 = 1,104.

Explanation:
Here, we have changed 138 to 276 and 8 to 4 and then multiplied 276 × 4 which is 1,104.

Question 12.

Answer:
303 × 6 = 1818.

Explanation:
Here, we have changed 303 to 101 and 6 to 18 and then multiplied 101 × 18 which is 1818.