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McGraw Hill Math Grade 6 Lesson 8.1 Answer Key Dividing Fractions by Whole Numbers

Practice questions available in McGraw Hill Math Grade 6 Answer Key PDF Lesson 8.1 Dividing Fractions by Whole Numbers will engage students and is a great way of informal assessment.

McGraw-Hill Math Grade 6 Answer Key Lesson 8.1 Dividing Fractions by Whole Numbers

Exercises Divide

Question 1.
\(\frac{1}{2}\) ÷ 4
Answer:
Dividing \(\frac{1}{2}\) by 4 ,we get the quotient \(\frac{1}{8}\)

Explanation:
\(\frac{1}{2}\) ÷ 4
= \(\frac{1}{2}\) × \(\frac{1}{4}\)
= \(\frac{1}{8}\)

Question 2.
\(\frac{3}{5}\) ÷ 4
Answer:
Dividing \(\frac{3}{5}\) by 4,,we get the quotient \(\frac{3}{20}\)

Explanation:
\(\frac{3}{5}\) ÷ 4
= \(\frac{3}{5}\) × \(\frac{1}{4}\)
= \(\frac{3}{20}\)

Question 3.
\(\frac{6}{7}\) ÷ 3
Answer:
Dividing \(\frac{6}{7}\) by 3,,we get the quotient \(\frac{6}{21}\)

Explanation:
\(\frac{6}{7}\) ÷ 3
= \(\frac{6}{7}\) × \(\frac{1}{3}\)
= \(\frac{6}{21}\)

Question 4.
\(\frac{1}{2}\) ÷ 6
Answer:
Dividing \(\frac{1}{2}\) by 6,,we get the quotient \(\frac{1}{12}\)

Explanation:
\(\frac{1}{2}\) ÷ 6
= \(\frac{1}{2}\) × \(\frac{1}{6}\)
= \(\frac{1}{12}\)

Question 5.
\(\frac{5}{19}\) ÷ 2
Answer:
Dividing \(\frac{5}{19}\) by 2,,we get the quotient \(\frac{5}{38}\)

Explanation:
\(\frac{5}{19}\) ÷ 2
= \(\frac{5}{19}\) × \(\frac{1}{2}\)
= \(\frac{5}{38}\)

Question 6.
\(\frac{4}{5}\) ÷ 7
Answer:
Dividing \(\frac{4}{5}\) by 7,we get the quotient \(\frac{4}{35}\)

Explanation:
\(\frac{4}{5}\) ÷ 7
= \(\frac{4}{5}\) × \(\frac{1}{7}\)
= \(\frac{4}{35}\)

Question 7.
\(\frac{1}{9}\) ÷ 9
Answer:
Dividing \(\frac{1}{9}\) by 9,we get the quotient \(\frac{1}{81}\)

Explanation:
\(\frac{1}{9}\) ÷ 9
= \(\frac{1}{9}\) × \(\frac{1}{9}\)
= \(\frac{1}{81}\)

Question 8.
\(\frac{3}{11}\) ÷ 12
Answer:
Dividing \(\frac{3}{11}\) by 12,we get the quotient \(\frac{3}{132}\)

Explanation:
\(\frac{3}{11}\) ÷ 12
= \(\frac{3}{11}\) × \(\frac{1}{12}\)
= \(\frac{3}{132}\)

Question 9.
\(\frac{17}{18}\) ÷ 4
Answer:
Dividing \(\frac{17}{18}\) by 4,we get the quotient \(\frac{17}{72}\)

Explanation:
\(\frac{17}{18}\) ÷ 4
= \(\frac{17}{18}\) × \(\frac{1}{4}\)
= \(\frac{17}{72}\)

Question 10.
\(\frac{12}{13}\) ÷ 3
Answer:
Dividing \(\frac{12}{13}\) by 3,we get the quotient \(\frac{4}{13}\)

Explanation:
\(\frac{12}{13}\) ÷ 3
= \(\frac{12}{13}\) × \(\frac{1}{3}\)
= \(\frac{12}{39}\)
= \(\frac{4}{13}\)

Question 11.
\(\frac{2}{3}\) ÷ 6
Answer:
Dividing \(\frac{2}{3}\) by 6,we get the quotient \(\frac{1}{9}\)

Explanation:
\(\frac{2}{3}\) ÷ 6
= \(\frac{2}{3}\) × \(\frac{1}{6}\)
= \(\frac{2}{18}\)
= \(\frac{1}{9}\)

Question 12.
\(\frac{5}{11}\) ÷ 20
Answer:
Dividing \(\frac{5}{11}\) by 20,we get the quotient \(\frac{1}{44}\)

Explanation:
\(\frac{5}{11}\) ÷ 20
= \(\frac{5}{11}\) × \(\frac{1}{20}\)
= \(\frac{5}{220}\)
= \(\frac{1}{44}\)

Question 13.
\(\frac{3}{7}\) ÷ 11
Answer:
Dividing \(\frac{3}{7}\) by 11,we get the quotient \(\frac{3}{77}\)

Explanation:
\(\frac{3}{7}\) ÷ 11
= \(\frac{3}{7}\) × \(\frac{1}{11}\)
= \(\frac{3}{77}\)

Question 14.
\(\frac{1}{3}\) ÷ 9
Answer:
Dividing \(\frac{1}{3}\) by 9,we get the quotient \(\frac{1}{27}\)

Explanation:
\(\frac{1}{3}\) ÷ 9
= \(\frac{1}{3}\) × \(\frac{1}{9}\)
= \(\frac{1}{27}\)

Question 15.
\(\frac{10}{11}\) ÷ 5
Answer:
Dividing \(\frac{10}{11}\) by 5,we get the quotient \(\frac{10}{55}\)

Explanation:
\(\frac{10}{11}\) ÷ 5
= \(\frac{10}{11}\) × \(\frac{1}{5}\)
= \(\frac{10}{55}\)

Question 16.
\(\frac{10}{13}\) ÷ 3
Answer:
Dividing \(\frac{10}{13}\) by 3,we get the quotient \(\frac{10}{39}\)

Explanation:
\(\frac{10}{13}\) ÷ 3
= \(\frac{10}{13}\) × \(\frac{1}{3}\)
= \(\frac{10}{39}\)

Question 17.
\(\frac{4}{5}\) ÷ 4
Answer:
Dividing \(\frac{4}{5}\) by 4,we get the quotient \(\frac{1}{5}\)

Explanation:
\(\frac{4}{5}\) ÷ 4
= \(\frac{4}{5}\) × \(\frac{1}{4}\)
= \(\frac{1}{5}\) × \(\frac{1}{1}\)
= \(\frac{1}{5}\)

Question 18.
\(\frac{12}{13}\) ÷ 5
Answer:
Dividing \(\frac{12}{13}\) by 5,we get the quotient \(\frac{12}{65}\)

Explanation:
\(\frac{12}{13}\) ÷ 5
= \(\frac{12}{13}\) × \(\frac{1}{5}\)
= \(\frac{12}{65}\)

Question 19.
\(\frac{2}{11}\) ÷ 4
Answer:
Dividing \(\frac{2}{11}\) by 4,we get the quotient \(\frac{1}{22}\)

Explanation:
\(\frac{2}{11}\) ÷ 4
= \(\frac{2}{11}\) × \(\frac{1}{4}\)
= \(\frac{1}{11}\) × \(\frac{1}{2}\)
= \(\frac{1}{22}\)

Question 20.
\(\frac{3}{4}\) ÷ 7
Answer:
Dividing \(\frac{3}{4}\) by 7,we get the quotient \(\frac{3}{28}\)

Explanation:
\(\frac{3}{4}\) ÷ 7
= \(\frac{3}{4}\) × \(\frac{1}{7}\)
= \(\frac{3}{28}\)

Question 21.
Mrs. Spinosa brought \(\frac{15}{19}\) pounds of chocolate for her fourth grade class to enjoy. If all 15 students want an equal portion of chocolate, how much will each student receive?
Answer:
Number of pounds of of chocolate each student receives = \(\frac{1}{19}\)

Explanation:
Number of pounds of of chocolate Mrs. Spinosa brought = \(\frac{15}{19}\)
Number of students = 15.
Number of pounds of of chocolate each student receives = Number of pounds of of chocolate Mrs. Spinosa brought ÷ Number of students
= \(\frac{15}{19}\) ÷ 15
= \(\frac{15}{19}\) × \(\frac{1}{15}\)
= \(\frac{1}{19}\) × \(\frac{1}{1}\)
= \(\frac{1}{19}\)

Question 22.
Mr. Lewis bought \(\frac{10}{13}\) liters of apple cider to divide equally among his three children. How much apple cider will each child receive?
Answer:
Number of liters of apple cider each child receives = \(\frac{10}{39}\)

Explanation:
Number of liters of apple cider Mr. Lewis bought = \(\frac{10}{13}\)
Number of children = 3.
Number of liters of apple cider each child receives = Number of liters of apple cider Mr. Lewis bought ÷ Number of children
= \(\frac{10}{13}\)  ÷ 3
= \(\frac{10}{13}\) × \(\frac{1}{3}\)
= \(\frac{10}{39}\)

McGraw Hill Math Grade 6 Lesson 8.1 Answer Key Dividing Fractions by Whole Numbers Read More »

McGraw Hill Math Grade 6 Lesson 6.4 Answer Key Subtracting Fractions with Like Denominators

Practice questions available in McGraw Hill Math Grade 6 Answer Key PDF Lesson 6.4 Subtracting Fractions with Like Denominators will engage students and is a great way of informal assessment.

McGraw-Hill Math Grade 6 Answer Key Lesson 6.4 Subtracting Fractions with Like Denominators

Exercises Subtract Fractions

Question 1.
\(\frac{5}{7}\) – \(\frac{2}{7}\)
Answer:
Difference between \(\frac{5}{7}\) and \(\frac{2}{7}\), we get \(\frac{3}{7}\).

Explanation:
\(\frac{5}{7}\) – \(\frac{2}{7}\) = \(\frac{3}{7}\)
Exercises Subtract Fractions

Question 2.
\(\frac{11}{13}\) – \(\frac{7}{13}\)
Answer:
Difference between \(\frac{11}{13}\) and \(\frac{7}{13}\), we get \(\frac{4}{13}\)

Explanation:
\(\frac{11}{13}\) – \(\frac{7}{13}\) =
McGraw-Hill-Math-Grade-6-Answer-Key-Lesson-6.4-Subtracting-Fractions-with-Like-Denominators-Exercises-Subtract-Fractions-2

Question 3.
\(\frac{10}{11}\) – \(\frac{9}{11}\)
Answer:
Difference between \(\frac{10}{11}\) and \(\frac{9}{11}\), we get \(\frac{1}{11}\)

Explanation:
\(\frac{10}{11}\) – \(\frac{9}{11}\)
McGraw-Hill-Math-Grade-6-Answer-Key-Lesson-6.4-Subtracting-Fractions-with-Like-Denominators-Exercises-Subtract-Fractions-3

Question 4.
\(\frac{13}{21}\) – \(\frac{8}{21}\)
Answer:
Difference between \(\frac{13}{21}\) and \(\frac{8}{21}\), we get \(\frac{5}{21}\)

Explanation:
\(\frac{13}{21}\) – \(\frac{8}{21}\)
McGraw-Hill-Math-Grade-6-Answer-Key-Lesson-6.4-Subtracting-Fractions-with-Like-Denominators-Exercises-Subtract-Fractions-4

Question 5.
\(\frac{2}{3}\) – \(\frac{1}{3}\)
Answer:
Difference between \(\frac{2}{3}\) and \(\frac{1}{3}\), we get \(\frac{1}{3}\)

Explanation:
\(\frac{2}{3}\) – \(\frac{1}{3}\) =
McGraw-Hill-Math-Grade-6-Answer-Key-Lesson-6.4-Subtracting-Fractions-with-Like-Denominators-Exercises-Subtract-Fractions-5

Question 6.
\(\frac{10}{47}\) – \(\frac{10}{47}\)
Answer:
Difference between \(\frac{10}{47}\) and\(\frac{10}{47}\), we get 0.

Explanation:
\(\frac{10}{47}\) – \(\frac{10}{47}\)  =
McGraw-Hill-Math-Grade-6-Answer-Key-Lesson-6.4-Subtracting-Fractions-with-Like-Denominators-Exercises-Subtract-Fractions-6

Question 7.
\(\frac{43}{96}\) – \(\frac{18}{96}\)
Answer:
Difference between \(\frac{43}{96}\) and \(\frac{18}{96}\), we get \(\frac{25}{96}\)

Explanation:
\(\frac{43}{96}\) – \(\frac{18}{96}\) =
McGraw-Hill-Math-Grade-6-Answer-Key-Lesson-6.4-Subtracting-Fractions-with-Like-Denominators-Exercises-Subtract-Fractions-7

Question 8.
\(\frac{12}{13}\) – \(\frac{11}{13}\)
Answer:
Difference between \(\frac{12}{13}\) and \(\frac{11}{13}\), we get \(\frac{1}{13}\)

Explanation:
\(\frac{12}{13}\) – \(\frac{11}{13}\) =
McGraw-Hill-Math-Grade-6-Answer-Key-Lesson-6.4-Subtracting-Fractions-with-Like-Denominators-Exercises-Subtract-Fractions-8

Question 9.
McGraw Hill Math Grade 6 Chapter 6 Lesson 6.4 Answer Key Subtracting Fractions with Like Denominators 1
Answer:
Difference between \(\frac{47}{49}\) and \(\frac{32}{49}\), we get \(\frac{15}{49}\)

Explanation:
\(\frac{47}{49}\) – \(\frac{32}{49}\) =
McGraw-Hill-Math-Grade-6-Answer-Key-Lesson-6.4-Subtracting-Fractions-with-Like-Denominators-Exercises-Subtract-Fractions-9

Question 10.
McGraw Hill Math Grade 6 Chapter 6 Lesson 6.4 Answer Key Subtracting Fractions with Like Denominators 2
Answer:
Difference between \(\frac{43}{31}\) and \(\frac{23}{31}\), we get \(\frac{20}{31}\)

Explanation:
\(\frac{43}{31}\) – \(\frac{23}{31}\) =
McGraw-Hill-Math-Grade-6-Answer-Key-Lesson-6.4-Subtracting-Fractions-with-Like-Denominators-Exercises-Subtract-Fractions-10

Question 11.
McGraw Hill Math Grade 6 Chapter 6 Lesson 6.4 Answer Key Subtracting Fractions with Like Denominators 3
Answer:
Difference between \(\frac{5}{7}\) and \(\frac{2}{7}\), we get \(\frac{3}{7}\)

Explanation:
\(\frac{5}{7}\) – \(\frac{2}{7}\) =
McGraw-Hill-Math-Grade-6-Answer-Key-Lesson-6.4-Subtracting-Fractions-with-Like-Denominators-Exercises-Subtract-Fractions-11

Question 12.
McGraw Hill Math Grade 6 Chapter 6 Lesson 6.4 Answer Key Subtracting Fractions with Like Denominators 4
Answer:
Difference between \(\frac{10}{17}\) and \(\frac{7}{17}\), we get \(\frac{3}{17}\)

Explanation:
\(\frac{10}{17}\) – \(\frac{7}{17}\) =
McGraw-Hill-Math-Grade-6-Answer-Key-Lesson-6.4-Subtracting-Fractions-with-Like-Denominators-Exercises-Subtract-Fractions-12

Question 13.
McGraw Hill Math Grade 6 Chapter 6 Lesson 6.4 Answer Key Subtracting Fractions with Like Denominators 5
Answer:
Difference between \(\frac{3}{4}\) and \(\frac{2}{4}\), we get \(\frac{1}{4}\)

Explanation:
\(\frac{3}{4}\) – \(\frac{2}{4}\) =
McGraw-Hill-Math-Grade-6-Answer-Key-Lesson-6.4-Subtracting-Fractions-with-Like-Denominators-Exercises-Subtract-Fractions-13

Question 14.
McGraw Hill Math Grade 6 Chapter 6 Lesson 6.4 Answer Key Subtracting Fractions with Like Denominators 6
Answer:
Difference between \(\frac{14}{15}\) and \(\frac{4}{15}\), we get \(\frac{2}{3}\)

Explanation:
\(\frac{14}{15}\) – \(\frac{4}{15}\) =
McGraw-Hill-Math-Grade-6-Answer-Key-Lesson-6.4-Subtracting-Fractions-with-Like-Denominators-Exercises-Subtract-Fractions-14

Question 15.
McGraw Hill Math Grade 6 Chapter 6 Lesson 6.4 Answer Key Subtracting Fractions with Like Denominators 7
Answer:
Difference between \(\frac{7}{12}\) and \(\frac{5}{12}\), we get \(\frac{1}{6}\)

Explanation:
\(\frac{7}{12}\) – \(\frac{5}{12}\) =
McGraw-Hill-Math-Grade-6-Answer-Key-Lesson-6.4-Subtracting-Fractions-with-Like-Denominators-Exercises-Subtract-Fractions-15

Question 16.
McGraw Hill Math Grade 6 Chapter 6 Lesson 6.4 Answer Key Subtracting Fractions with Like Denominators 8
Answer:
Difference between \(\frac{6}{7}\) and \(\frac{2}{7}\), we get \(\frac{4}{7}\)

Explanation:
\(\frac{6}{7}\) – \(\frac{2}{7}\) =
McGraw-Hill-Math-Grade-6-Answer-Key-Lesson-6.4-Subtracting-Fractions-with-Like-Denominators-Exercises-Subtract-Fractions-16

Question 17.
McGraw Hill Math Grade 6 Chapter 6 Lesson 6.4 Answer Key Subtracting Fractions with Like Denominators 9
Answer:
Difference between \(\frac{7}{8}\) and \(\frac{5}{8}\), we get \(\frac{1}{4}\)

Explanation:
\(\frac{7}{8}\) – \(\frac{5}{8}\) =
McGraw-Hill-Math-Grade-6-Answer-Key-Lesson-6.4-Subtracting-Fractions-with-Like-Denominators-Exercises-Subtract-Fractions-17

Question 18.
McGraw Hill Math Grade 6 Chapter 6 Lesson 6.4 Answer Key Subtracting Fractions with Like Denominators 10
Answer:
Difference between \(\frac{5}{13}\) and \(\frac{2}{13}\), we get \(\frac{3}{13}\)

Explanation:
\(\frac{5}{13}\) and \(\frac{2}{13}\) =
McGraw-Hill-Math-Grade-6-Answer-Key-Lesson-6.4-Subtracting-Fractions-with-Like-Denominators-Exercises-Subtract-Fractions-18

McGraw Hill Math Grade 6 Lesson 6.4 Answer Key Subtracting Fractions with Like Denominators Read More »

McGraw Hill Math Grade 6 Unit Test Lessons 24-25 Answer Key

Practice questions available in McGraw Hill Math Grade 6 Answer Key PDF Unit Test Lessons 24-25 will engage students and is a great way of informal assessment.

McGraw-Hill Math Grade 6 Unit Test Lessons 24-25 Answer Key

Question 1.
Look at the chart of Walter’s Dog House sales of hot dogs.
McGraw Hill Math Grade 6 Unit Test Lessons 24-25 Answer Key 1
On what day were the most hot dogs sold? _______________
The fewest? _______________
About how many more hot dogs were sold on Sunday than Saturday? _______________
Answer:
The most hot dogs sold on Sunday.
The fewest hot dogs sold on Wednesday and Friday.
The number of hot dogs sold on Saturday are 80.
The number of hot dogs sold on Sunday are 100.
100 – 80 = 20
20 more hot dogs are sold on Sunday than Saturday.

Question 2.
Look at the line graph of school dance ticket sales.
McGraw Hill Math Grade 6 Unit Test Lessons 24-25 Answer Key 2
In what month were the most tickets sold? _______________
About how many tickets were sold that month? _______________
In what month were the fewest tickets sold? _______________
Answer:
Most of the tickets are sold in May month.
In may month 110 tickets are sold.
In March month fewest tickets are sold.

Question 3.
Look at the double-line graph of teenage employment rates.
McGraw Hill Math Grade 6 Unit Test Lessons 24-25 Answer Key 3
At which age are a higher percentage of British teenagers employed than Canadian teenagers? _______________
Based on this line graph, would you say it is easier for a fourteen-year-old to find a job in Canada or Great Britain? _______________
Answer:
A higher percentage of British teenagers employed than Canadian teenagers at age 17.
Based on the above line graph, we can say it is easier for a fourteen-year-old to find a job in Canada.

Question 4.
Look at the circle graph of favorite breakfast foods.
McGraw Hill Math Grade 6 Unit Test Lessons 24-25 Answer Key 4
What is the favorite breakfast food as voted on by students? _______________
Do more students prefer cereal than meat and eggs combined? _______________
If so, by what percentage? _______________
Answer:
The students voted the favorite breakfast food as Cereal.
Yes, more students prefer cereal than meat and eggs combined.
30% of students prefer Cereal.
28% of students prefer meat and egg combined.
13% + 15% = 28%
30% – 28% = 2%
2% of students prefer Cereal than meat and egg combined.

Question 5.
Bird Watchers of America was conducting its annual census for birds of prey. They recorded the following number of daily sightings for a ten-day period in October: 6, 8,10,10,14, 9, 8,10, 6, 5
What is the average number (mean) of sightings per day? _______________
What is the median of the sample of daily sightings? _______________
What is the mode of the sample of daily sightings? _______________
Answer:
The given set of data is 6, 8,10,10,14, 9, 8,10, 6, 5.
Mean = Sum of all the numbers/Total number of numbers.
Sum of given set of data is as below,
6 + 8 + 10 + 10 + 14 + 9 + 8 + 10 + 6 + 5 = 86
The total number of numbers in the given set of data is 10.
Mean = 86/10
Mean = 8.6
The average number of sightings per day is 8.6.

The given set of data is 6, 8,10,10,14, 9, 8,10, 6, 5.
The median of a given set of data is the middle number in the set.
To calculate the median first we have to arrange the data either in ascending order or descending order.
The given set of data in ascending order is 5, 6, 6, 8, 8, 9, 10,10, 10, 14.
If the number of addends are even then the median is equal to average of the two middle ones.
(8 + 9)/2 = 17/2 = 8.5
The median of the sample of daily sightings is 8.5.

The number that occurs the most in the given set of numbers is called a mode.
We can have more than one mode or no mode at all.
In the given series 6, 8,10,10,14, 9, 8,10, 6, 5.
The number 6 are 2.
The number 8 are 2.
The number 10 are 3.
The number 14 is 1.
The number 9 is 1.
The number 5 is 1.
The mode of the sample of daily sightings is 10.

Question 6.
Freda has been appointed the official scorekeeper for the school bowling team. She recorded the following scores for her teammates over the past two weeks:
McGraw Hill Math Grade 6 Unit Test Lessons 24-25 Answer Key 5
Marissa: 55, 59, 63, 71, 45
Jeremy: 65, 64, 46, 56, 49
Freda: 71, 78, 81, 65, 72
Charmaine: 56, 59, 60, 65, 76
Create a stem-and-leaf plot using the data.
Answer:
McGraw-Hill-Math-Grade-6-Unit-Test-Lessons-24-25-Answer-Key-5
Explanation:
The recorded scores of bowling team over the past two weeks are arranged from least to greatest as 45, 46, 49, 55, 56, 56, 59, 59, 60, 63, 64, 65, 65, 65, 71, 71, 72, 76, 78, 81.
In this stem and leaf plot, the stems are the scores in tens and the leaves are the ones of the scores. The stem and leaf plot are arranged using the above data.

Question 7.
Using the raw data for the bowling scores in Exercise 6, create a box-and-whisker plot of the data.
What is the range of the data? _______________
What is the median of the data? _______________
Answer:
McGraw-Hill-Math-Grade-6-Unit-Test-Lessons-24-25-Answer-Key-5(1)
The above box-and-whisker plot allows us to look at data to tell where most of the numbers lie.
The raw data for the bowling score is 45, 46, 49, 55, 56, 56, 59, 59, 60, 63, 64, 65, 65, 65, 71, 71, 72, 76, 78, 81.
The lower extreme is the lowest number in the given data which is 45.
The higher extreme is the highest number in the given data which is 81.
The range of the data is equal to the difference between the higher extreme and lower extreme.
81 – 45 = 36
The range of the data is 36.
The median of a given set of data is the middle number in the set.
If the number of addends are even then the median is equal to average of the two middle ones.
(63 + 64)/2 = 63.5
The median of the data is 63.5.
The lower quartile is the median of the numbers below the median.
(49 + 55)/2 = 52
The upper quartile is the median of the numbers above the median.
(71+ 72) = 71.5

Question 8.
Eddie gathered stones from a nearby stream. He collected 12 stones in all and he noticed that 8 were black and 4 were white before he put them in a bag. Draw a tree diagram to show the possible outcomes if Eddie were to pick a stone from the bag, record the color, and return the stone to the bag and pick again.
What is the probability that Eddie will pick a black stone on the first pick? _______________
What is the probability that Eddie will pick a white stone on the first pick? _______________
Answer:
McGraw-Hill-Math-Grade-6-Unit-Test-Lessons-24-25-Answer-Key-5(3)
Total number of stones = 12
Total number of black stones = 8
Total number of white stones = 4
Probability of picking a black stone from the bag = 8/12 = 2/3
Probability of picking a white stone from the bag = 4/12 = 1/3

Question 9.
Priscilla recorded the arrival of the members of her book club. She noted that there were 25 members in attendance. 16 brought dessert and 15 brought beverages. Draw a Venn Diagram and record the data.
How many in the book group brought dessert and beverages? _______________
What percentage brought beverages only? _______________
Answer:
McGraw-Hill-Math-Grade-6-Unit-Test-Lessons-24-25-Answer-Key-5(2)
Out of 25 members 16 brought dessert.
So, members who brought only beverages = 25 – 16 = 9

Out of 25 members 15 brought beverages.
So, members who brought only desserts = 25 – 15 = 10

Members who brought both dessert and beverages = 25 – (9 + 10) = 6
So, 6 people brought both dessert and beverages.
Percentage of members who brought beverages only = (9/25) x 100 = 36%

Question 10.
Wallace and 15 other students each put their names on a piece of paper and put them in a cap.
If there are 9 girls in the group, what is the probability that a girl’s name will be picked from the cap? _______________
What is the probability that Wallace’s name will not be chosen? _______________
Answer:
Total number of students are 16.
1 + 15 = 16
There are 9 girls in the group.
Probability = The number of favorable outcomes/ The total number of possible outcomes
P = 9/16
The probability that a girl’s name will be picked from the cap is 9/16.
Except Wallace 15 other students each put their names on a piece of paper and put them in a cap.
The probability that Wallace’s name will not be chosen is 15/16.

Question 11.
Sue randomly throws darts at a wall covered with 100 balloons, 40 of which are red, 25 blue, 20 green, 10 yellow, and 5 orange.
What is the probability that Sue will pop a red or green balloon? _______________
What is the probability that Sue will pop an orange balloon? _______________
What is the probability that Sue will not pop a blue balloon? _______________
Answer:
Total number of balloons are 100.
Red color balloons are 40.
Blue color balloons are 25.
Green color balloons are 20.
Yellow color balloons are 10.
Orange color balloons are 5.
Probability of red or green balloons = The number of favorable outcomes/ The total number of possible outcomes
P = (40 + 20)/100
P = 60/100
The probability that Sue will pop a red or green balloon is 60/100 or 3/5.
Probability of orange balloons = The number of favorable outcomes/ The total number of possible outcomes
P = 5/100
The probability that Sue will pop an orange balloon is 5/100 or 1/20.
Probability of not blue balloons = The number of favorable outcomes/ The total number of possible outcomes
Except blue balloons add all balloons.
40 + 20 + 10 + 5 = 75
P = 75/100
The probability that Sue will not pop a blue balloon is 75/100 or 3/4.

McGraw Hill Math Grade 6 Unit Test Lessons 24-25 Answer Key Read More »

McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key

Practice questions available in McGraw Hill Math Grade 6 Answer Key PDF Unit Test Lessons 21-23 will engage students and is a great way of informal assessment.

McGraw-Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key

Identify each angle as obtuse, acute, or right.

Question 1.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 1
Answer:
obtuse angle
Explanation:
In the above image we can observe an angle 128 degrees. The angle which measures greater than 90 degrees and less than 180 degrees is called as obtuse angle. So, the above angle is obtuse angle.

Question 2.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 2
Answer:
Right angle
Explanation:
In the above image we can observe an angle 90 degrees. The angle which measures exactly 90 degrees is called as right angle. So, the above angle is right angle.

Question 3.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 3
Answer:
Acute angle
Explanation:
In the above image we can observe an angle 55 degrees. The angle which measures less than 90 degrees is called as acute angle. So, the above angle is acute angle.

Question 4.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 4
Answer:
Acute angle
Explanation:
In the above image we can observe an angle 35 degrees. The angle which measures less than 90 degrees is called as acute angle. So, the above angle is acute angle.

Question 5.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 5
Answer:
obtuse angle
Explanation:
In the above image we can observe an angle 101 degrees. The angle which measures greater than 90 degrees and less than 180 degrees is called as obtuse angle. So, the above angle is obtuse angle.

Identify each pair of angles as supplementary or complementary, and explain why.

Question 6.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 6
Answer:
The measure of angle ∠DAC is 24°.
The measure of angle ∠CAB is 56°.
∠DAC + ∠CAB = 24° + 56° = 80°
The supplementary angles are two angles that form a line and their sum will be 180°.
The complementary angles are two angles that form a right angle and their sum will be 90°.
So, the above pair of angles are neither supplementary nor complementary..

Question 7.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 7
Answer:
The measure of angle ∠CAB is 56°.
The measure of angle ∠DAC is 124°.
∠CAB + ∠DAC = 56° + 124° = 180°
The supplementary angles are two angles that form a line and their sum will be 180°.
The above pair of angles are called as supplementary angles.

Question 8.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 8
Answer:
The measure of angle ∠CAB is 66°.
The measure of angle ∠DAC is 24°.
∠CAB + ∠DAC = 66° + 24° = 90°
The complementary angles are two angles that form a right angle and their sum will be 90°.
The above pair of angles are called as complementary angles.

Identify each triangle as scalene, equilateral, or isosceles.

Question 9.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 9
Answer:
Isosceles Triangle
Explanation:
An isosceles triangle has two sides of equal length. In the above triangle we can observe two sides having same length and one side having different length. So, the above triangle is isosceles triangle.

Question 10.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 10
Answer:
Isosceles Triangle
Explanation:
An isosceles triangle has two sides of equal length. In the above triangle we can observe two sides having same length and one side having different length. So, the above triangle is isosceles triangle.

Question 11.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 11
Answer:
Equilateral Triangle
Explanation:
An equilateral triangle has three sides of equal length. In the above image we can observe the triangle having three sides with a same length. So, the above triangle is equilateral triangle.

Question 12.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 12
Answer:
Scalene Triangle
Explanation:
A scalene triangle has no sides of equal length. In the above image we can observe the triangle having three sides with a different lengths. So, the above triangle is scalene triangle.

Question 13.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 13
Answer:
Equilateral Triangle
Explanation:
An equilateral triangle has three sides of equal length. In the above image we can observe the triangle having three sides with a same length. So, the above triangle is equilateral triangle.

Question 14.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 14
Answer:
Scalene Triangle
Explanation:
A scalene triangle has no sides of equal length. In the above image we can observe the triangle having three sides with a different lengths. So, the above triangle is scalene triangle.

Identify each triangle as obtuse, right, or acute.

Question 15.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 15
Answer:
Obtuse triangle
Explanation:
In the above triangle we can observe one obtuse angle. The angle which measures greater than 90 degrees and less than 180 degrees is called as obtuse angle. So, the above triangle is obtuse angle triangle.

Question 16.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 16
Answer:
Obtuse triangle
Explanation:
In the above triangle we can observe one obtuse angle. The angle which measures greater than 90 degrees and less than 180 degrees is called as obtuse angle. So, the above triangle is obtuse angle triangle.

Question 17.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 17
Answer:
Right angle Triangle
Explanation:
In the above triangle we can observe the angle 90 degrees. The angle which measures exactly 90 degrees is called as right angle. So, the above triangle is right angle triangle.

Question 18.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 18
Answer:
Right angle Triangle
Explanation:
In the above triangle we can observe the angle 90 degrees. The angle which measures exactly 90 degrees is called as right angle. So, the above triangle is right angle triangle.

Question 19.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 19
Answer:
Acute triangle
Explanation:
In the above triangle we can observe all the angles are less than 90 degrees. The angle which measures less than 90 degrees is called as acute angle. So, the above triangle is acute angle triangle.

Question 20.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 20
Answer:
Acute triangle
Explanation:
In the above triangle we can observe all the angles are less than 90 degrees. The angle which measures less than 90 degrees is called as acute angle. So, the above triangle is acute angle triangle.

Question 21.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 21
Answer:
Obtuse triangle
Explanation:
In the above triangle we can observe one obtuse angle. The angle which measures greater than 90 degrees and less than 180 degrees is called as obtuse angle. So, the above triangle is obtuse angle triangle.

Question 22.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 22
Answer:
Right angle Triangle
Explanation:
In the above triangle we can observe the angle 90 degrees. The angle which measures exactly 90 degrees is called as right angle. So, the above triangle is right angle triangle.

Question 23.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 23
Answer:
Acute triangle
Explanation:
In the above triangle we can observe all the angles are less than 90 degrees. The angle which measures less than 90 degrees is called as acute angle. So, the above triangle is acute angle triangle.

Question 24.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 24
Answer:
Acute triangle
Explanation:
In the above triangle we can observe all the angles are less than 90 degrees. The angle which measures less than 90 degrees is called as acute angle. So, the above triangle is acute angle triangle.

Question 25.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 25
Answer:
Acute triangle
Explanation:
In the above triangle we can observe all the angles are less than 90 degrees. The angle which measures less than 90 degrees is called as acute angle. So, the above triangle is acute angle triangle.

Question 26.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 26
Answer:
Obtuse triangle
Explanation:
In the above triangle we can observe one obtuse angle. The angle which measures greater than 90 degrees and less than 180 degrees is called as obtuse angle. So, the above triangle is obtuse angle triangle.

Identity the following quadrilaterals.

Question 27.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 27
Answer:
The above quadrilateral is Rectangle.
Explanation:
A rectangle is a quadrilateral having four sides. The opposite sides of a rectangle are parallel and equal in length. All the interior angles of rectangle are equal to 90 degrees.

Question 28.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 28
Answer:
The above quadrilateral is Rhombus.
Explanation:
A rhombus is a quadrilateral having four equal sides. The opposite sides of rhombus are parallel to each other.

Question 29.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 29
Answer:
The above quadrilateral is Trapezoid.
Explanation:
A trapezoid has four sides and having one pair of parallel sides.

Question 30.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 30
Answer:
The above quadrilateral is Kite.
Explanation:
A kite is a quadrilateral having four sides. Two angles in a kite are equal. Two touching sides are equal in length and the other two touching sides are equal in length.

McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 31

Question 31.
Name the center point.
Answer:
The center point of a circle is A.

Question 32.
Which segments are chords?
Answer:
A chord is a line segment that has both end points on the circumference.
The line segments BD, DC, CB are chords.

Question 33.
Which segment is the diameter?
Answer:
The line segment FG is the diameter.

Question 34.
Which segments are radii?
Answer:
The line segments AG, AF, AD are radii.
A line segment which starts at a circle’s origin and extends to its circumference is called as radius. In circle all radii are equal in length.

Identify each figure and fill in the information requested.

Question 35.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 32
Figure _______________
Base is _______________
Number of faces _______________
Number of edges _______________
Number of vertices _______________
Answer:
Figure : Cube
Base is : Square
Number of faces : 6
Number of edges : 12
Number of vertices : 8
Explanation:
In the above figure we can observe cube. A cube is a three dimensional solid figure having 6 faces, 12 edges, and 8 vertices. The base of the cube is a square.

Question 36.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 33
Figure _______________
Base is _______________
Number of faces _______________
Number of edges _______________
Number of vertices _______________
Answer:
Figure : Rectangular prism 
Base is : Rectangle
Number of faces : 6
Number of edges: 12
Number of vertices : 8 
Explanation:
In the above figure we can observe rectangular prism. A rectangular prism is a three dimensional solid figure having 6 faces, 12 edges, and 8 vertices. The base of the rectangular prism is rectangle.

Question 37.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 34
Figure _______________
Base is _______________
Number of faces _______________
Number of edges _______________
Number of vertices _______________
Answer:
Figure : Rectangular pyramid
Base is : Rectangle
Number of faces : 5
Number of edges: 8
Number of vertices : 5
Explanation:
In the above figure we can observe rectangular pyramid. A rectangular pyramid is a three dimensional solid figure having 5 faces, 8 edges, and 5 vertices. The base of the rectangular pyramid is rectangle.

Question 38.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 35
Figure _______________
Base is _______________
Number of faces _______________
Number of edges _______________
Number of vertices _______________
Answer:
Figure : Triangular prism 
Base is : Rectangle
Number of faces : 5
Number of edges: 9
Number of vertices : 6 
Explanation:
In the above figure we can observe triangular prism. A triangular prism is a three dimensional solid figure having 5 faces, 9 edges, and 6 vertices. The base of the triangular prism is rectangle.

Question 39.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 36
Figure _______________
Base is _______________
Number of faces _______________
Number of edges _______________
Number of vertices _______________
Answer:
Figure : Triangular pyramid
Base is : Triangle
Number of faces : 4
Number of edges: 6
Number of vertices : 4
Explanation:
In the above figure we can observe triangular pyramid. A triangular pyramid is a three dimensional solid figure having 4 faces, 6 edges, and 4 vertices. The base of the triangular pyramid is triangle.

Question 40.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 37
Figure _______________
Base is _______________
Number of faces _______________
Number of edges _______________
Number of vertices _______________
Answer:
Figure : Cone
Base is : Circle
Number of faces : 2
Number of edges: 0
Number of vertices : 1
Explanation:
In the above figure we can observe cone. A cone is a three dimensional solid figure having 2 faces, 0 edges, and 1 vertex. The base of the cone is circle.

Question 41.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 38
Figure _______________
Base is _______________
Number of faces _______________
Number of edges _______________
Number of vertices _______________
Answer:
Figure : Cylinder
Base is : Circle
Number of faces : 3
Number of edges: 0
Number of vertices : 0
Explanation:
In the above figure we can observe cylinder. A cylinder is a three dimensional solid figure having 3 faces, 0 edges, and 0 vertices. The base of the cylinder is circle.

Find the surface area of each figure.

Question 42.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 39
Answer:

Question 43.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 40
Answer:

Question 44.
Carolyn wants to paint a cube and is trying to figure out the surface area so she knows how much paint to buy. If her cube has sides that measure 4.5 inches, what is the surface area?
Answer:
Given sides of the cube is 4.5 inches.
A cube has 6 square faces.
Surface area of the cube (A) = 6 x s x s
A = 6 x 4.5 in x 4.5 in
A = 121.5 square inches
The surface area of the cube is equal to 121.5 square inches.

Question 45.
Plot the coordinates (1, 1), (6, 1), (1, 6), and (6, 6). Write in the length of each side.
McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key 41
The figure is a _______________
The perimeter is _______________
The area is _______________
Answer:
McGraw-Hill-Math-Grade-6-Unit-Test-Lessons-21-23-Answer-Key-41
After plotting the given coordinates on the grid, we can observe square in the above figure.
Side of a square s = 6
Perimeter of a square = 4 x s
= 4 x 6
= 24
The perimeter of the square is equal to 24.
Area of square = s x s
= 6 x 6
= 36
The area of the square is equal to 36.

McGraw Hill Math Grade 6 Unit Test Lessons 21-23 Answer Key Read More »

McGraw Hill Math Grade 6 Unit Test Lessons 18-20 Answer Key

Practice questions available in McGraw Hill Math Grade 6 Answer Key PDF Unit Test Lessons 18-20 will engage students and is a great way of informal assessment.

McGraw-Hill Math Grade 6 Unit Test Lessons 18-20 Answer Key

Solve

Question 1.
Wilfredo is replacing the trim around all of the windows in his house. The perimeter of each window measures 124 inches and he has 9 windows. How many feet of trim does Wilfredo need to replace?
Answer:

Question 2.
Ainsley wants to fill a fish tank with water for her new fish. The fish tank holds 176 quarts and she is adding the water with a one-gallon container. How many gallon containers will she need to fill the tank? ___________________________
Answer:
The fish tank holds 176 quarts. Ainsley is adding the water with a one – gallon container.
We know that 1 gallon is equal to 4 quarts.
1 gallon = 4 quarts
? gallons = 176 quarts
(176 x 1)/4 = ?
176/4 = 44 gallon containers
Ainsley need 44 gallon containers to fill the tank.

Question 3.
Tom is weighing potted plants for shipping to customers. The first plant weighs 1670 ounces, the second plant weighs 100 \(\frac{1}{8}\) pounds, and third plant weighs one-twentieth of a ton.
Which plant weighs the most? _________________________________ ____________
Which plants weighs the least? ____________________________________________________
Answer:
The first plant weighs 1670 ounces.
The second plant weighs 100 \(\frac{1}{8}\) pounds.
The third plant weighs one-twentieth of a ton.
Convert the second and third plant weight into ounces because the first plant weight is in ounces.
Second plant weighs 100 \(\frac{1}{8}\) pounds.
(800 + 1)/8 = 801/8 pounds
1 pound = 16 ounces
801/8 pounds = ? ounces
16 x 801/8 = 1,602 ounces
The second plant weighs 1,602 ounces.
The third plant weighs one-twentieth of a ton i.e. 1/20 of a ton.
1 ton = 2,000 pounds
1/20 ton = ? pounds
2,000 x 1/20 = 100 pounds
To convert 100 pounds to ounces we need to multiply 100 with 16 the product is equal to 1,600 ounces.
100 x 16 = 1,600 ounces.
The third plant weighs 1,600 ounces.
The first plant weighs the most.
The third plant weighs the least.

Question 4.
Sally wants to build a fence in her garden to keep the rabbits out, and she needs to know how much fencing to buy. The garden has an irregular shape with sides of 15 \(\frac{1}{4}\) feet, 660 inches, 12 yards, one foot, \(\frac{1}{160}\) of a mile, and 6 feet. How much fence material does she need (in feet)?
Answer:
The garden has an irregular shape with sides of 15 \(\frac{1}{4}\) feet, 660 inches, 12 yards, one foot, \(\frac{1}{160}\) of a mile, and 6 feet.
Convert all sides of an irregular shape of a garden into customary unit feet.
Converting 660 inches into feet :
We know that,
1 foot = 12 inches
? feet = 660 inches
(660 x 1)/12 = 55 feet
The side 660 inches is converted into feet. So, 660 inches is equal to 55 feet.
Converting 12 yards into feet :
We know that,
1 yard = 3 feet
12 yards = ? feet
(12 x 3)/1 = 36 feet
The side 12 yards is converted into feet. So, 12 yards is equal to 36 feet.
Converting \(\frac{1}{160}\) of a mile into feet :
We know that,
1 mile = 1,760 yards
\(\frac{1}{160}\) mile = ? yards
1760 x \(\frac{1}{160}\) = 11 yards
We know that,
1 yard = 3 feet
11 yards = ? feet
11 x 3 = 33 feet
The side \(\frac{1}{160}\) mile is converted into feet. So, \(\frac{1}{160}\) of a mile is equal to 33 feet.
Add all sides of the garden.
15 \(\frac{1}{4}\) feet + 55 feet + 36 feet + 1 foot + 33 feet + 6 feet = 146 \(\frac{1}{4}\) feet 
Sally need 146 \(\frac{1}{4}\) feet fence material.

Question 5.
What is the area of a rectangle with a length of 20 ft and a width of 144 inches? ____________________________
Answer:
Given
length = 20 feet
width = 144 inches
Area of a rectangle (A) = l x w
Convert 144 inches into feet.
We know that,
1 foot = 12 inches
? feet = 144 inches
(144 x 1)/12 = 12 feet
So, width is equal to 12 feet.
A = 20 feet x 12 feet
A = 240 square feet
Area of a rectangle is equal to 240 square feet.

Question 6.
What is the area of a right triangle with lengths of 3 feet, 4 feet, and 60 inches? ____________________________________________
Answer:
Area of a right triangle (A) = 1/2 x length(also called as base) x height
A = 1/2 x 3 feet x 4 feet
A = 6 square feet
Area of a right triangle is equal to 6 square feet.

Question 7.
What is the volume of a rectangular box with sides of 24 inches, and 1\(\frac{1}{2}\) feet, and a height of 18 inches? _________________________ cubic feet
Answer:
Given length = 24 inches
Width = 1\(\frac{1}{2}\) feet = \(\frac{3}{2}\) feet
height = 18 inches
Volume of a rectangular box (V) = Length x width x height
Convert length and height to feet.
2 feet = 24 inches
1\(\frac{1}{2}\) feet = 18 inches
V = 2 feet x \(\frac{3}{2}\) feet x \(\frac{3}{2}\) feet
V = \(\frac{9}{2}\) cubic feet
V = 4.5 cubic feet
The volume of a rectangular box is equal to 4.5 cubic feet.

Question 8.
A modern spacecraft travels 4.9 miles per second to maintain enough speed to stay in orbit around the earth. How far does the spacecraft travel
in a minute? _______________________________
in an hour? _______________________________
in a day? _______________________________
Answer:
A modern spacecraft travels 4.9 miles per second to maintain enough speed to stay in orbit around the earth.
1 second = 4.9 miles
First we have to calculate how far the spacecraft travel in a minute.
We know that 1 minute is equal to 60 seconds.
Multiply 60 with 4.9 the product is equal to 294 miles
60 x 4.9 = 294 miles
The spacecraft travels 294 miles in a minute.
Second we have to calculate how far the spacecraft travel in a hour.
We know that 1 minute is equal to 60 seconds and 1 hour is equal to 60 minutes.
Multiply 60, and 60 with 4.9 the product is equal to 17,640 miles
60 x 60 x 4.9 = 17,640 miles
The spacecraft travels 17,640 miles in an hour.
Third we have to calculate how far the spacecraft travel in a day.
We know that 1 minute is equal to 60 seconds and 1 hour is equal to 60 minutes and 1 day is equal to 24 hours.
Multiply 24, 60, and 60 with 4.9 the product is equal to 423,360 miles
24 x 60 x 60 x 4.9 = 423,360 miles
The spacecraft travels 423,360 miles in a day.

Question 9.
If the earth is 24,000 miles in circumference, how much time would it take the spacecraft in Exercise 8 to orbit the earth once?
How many times a day would the spacecraft orbit the earth? _______________________________
How many times in a week? _______________________________
How many times in a year? _______________________________
Answer:

Question 10.
The world record for the high jump is 2.45 meters. The world record for the pole vault is 6.14 meters. How many centimeters higher is the pole vault record than the high jump record?
Answer:
The world record for the high jump is 2.45 meters.
The world record for the pole vault is 6.14 meters.
Subtract 2.45 meters from 6.14 meters the difference is equal to 3.69 meters
6.14 – 2.45 = 3.69 meters
We know that 1 meter is equal to 100 centimeters.
Multiply 3.69 with 100 the product is equal to 369 centimeters.
3.69 x 100 = 369 cm
The pole vault record is 369 cm higher than the high jump record.

Question 11.
Stacy is measuring fabric for her grandmother, who is going to make a rectangular banner for a parade float. The banner will be 4.5 meters in length and 1,350 millimeters in width. How much fabric is needed for the banner?_______________ sq cm
Answer:
The length of the banner (l) = 4.5 meters
Convert 4.5 meters into centimeters.
We know that,
1 meter = 100 cm
4.5 meters = ? cm
100 x 4.5 = 450 cm
The length of the banner (l) = 450 centimeters
The width of the banner (w) = 1,350 millimeters
Convert 1,350 millimeters into centimeters.
We know that,
1 cm = 10 mm
?cm = 1,350 mm
1,350/10 = 135 cm
The width of the banner (w) = 135 centimeters 
Area of the rectangle (A) = l x w
A = 450 cm x 135 cm
A = 60,750 Square centimeters
The fabric needed for the banner is 60,750 square centimeters.

Question 12.
Heidi has three cans of cooking oil to recycle. One can holds 1,456 milliliters of oil, the second can holds 23.4 centiliters of oil, and the third can holds 4.5 liters of oil. How much cooking oil is Heidi going to recycle? ______________ liters
Answer:
One can holds 1,456 milliliters of oil.
Convert milliliters into liters.
We know that,
1 liter = 1,000 milliliters
? liters = 1,456 milliliters
(1,456 x 1)/1,000 = 1.456 liters
One can holds 1.456 liters of oil.
The second can holds 23.4 centiliters of oil.
Convert centiliters into liters.
We know that,
1 liter = 100 liters
? liters = 23.4 centiliters
(23.4 x 1)/100 = 0.234 liters
The second can holds 0.234 liters of oil.
The third can holds 4.5 liters of oil.
Add all three cans of oil.
1.456 liters + 0.234 liters + 4.5 liters = 6.19 liters
Heidi going to recycle 6.19 liters of cooking oil.

Question 13.
What is the area of a triangle with sides of 6 cm, 6 cm, a base of 9 cm, and a height of 4 cm? _______________________________ sq cm
Answer:
Given
sides = 6 cm, 6 cm
Base (b) = 9 cm
Height (h) = 4 cm
Area of a triangle (A) = 1/2 x b x h
A = 1/2 x 9 cm x 4 cm
A = 18 square centimeters
Area of the triangle is equal to 18 square centimeters.

Question 14.
Jerold walked around the rectangular school gymnasium, which measures 56 meters by 6,500 centimeters. How many meters did Jerold walk?
Answer:
Given
length (l) = 56 meters
Breadth (b) = 6,500 centimeters
Convert breadth into meters.
1 meter = 100 cm
? meters = 6,500 cm
6,500/100 = 65 meters
breadth = 65 meters
Perimeter of a rectangle = 2(l + b)
= 2 (56 + 65)
= 2 (121)
= 242 meters
Jerold walk 242 meters.

Question 15.
How much corn feed can fit into a rectangular bin that measures 4.6 meters in width, 14.5 meters in length, and 5.5 meters high?
_______________________________ cu meters
Answer:
Given
length = 14.5 meters
Width = 4.6 meters
Height = 5.5 meters
Volume of rectangular bin (V) = l x w x h
V = 14.5 x 4.6 x 5.5
V = 366.85 cubic meters
366.85 cubic meters of corn feed can fit into a rectangular bit.

Question 16.
The average player on the basketball team is 6 feet 4 inches tall. About how tall is that in centimeters? __________________
Answer:
The average player on the basketball team is 6 feet 4 inches tall.
Convert 6 feet into inches.
1 feet = 12 inches
6 feet = ? inches
12 x 6 = 72 inches
Add 72 inches to 4 inches the sum is equal to 76 inches.
72 + 4 = 76 inches
Convert 76 inches to centimeters.
We know that
1 inch = 2.54 cm
76 inches = ? cm
76 x 2.54 = 193.04 cm
The average player on the basketball team is 193.04 cm tall.

Question 17.
The average US car has a gas tank that holds 65 liters of gasoline. How much is that in gallons?
Answer:
The average US car has a gas tank that holds 65 liters of gasoline.
We know that,
1 gallon = 3.785 liters
? gallons = 65 liters
65/3.785 = 17.17 = 17.2 gallons
The gas tank holds 17.2 gallons of gasoline.

Question 18.
Paulie was looking for lawn mowing jobs. He surveyed the people in his neighborhood and found out that the average lawn measured 54 feet by 30 feet.
What is the total area in square feet? _________________
In square yards? _________________
About how much is the total area in square meters? _________________
Answer:
The average lawn is measured 54 feet by 30 feet.
Area = length x width
A = 54 feet x 30 feet
A = 1,620 square feet
The total area of the lawn is 1,620 square feet.

The average lawn is measured 54 feet by 30 feet.
Convert 54 feet to yards.
1 yard = 3 feet
? yards = 54 feet
54/3 = 18 yards
Length = 18 yards
Convert 30 feet into yards
1 yard =  3 feet
? yards = 30 feet
30/3 = 10 yards
Width = 10 yards
Area = length x width
A = 18 yards x 10 yards
A = 180 square yards
The total area of the lawn is 180 square yards.

Question 19.
Normal body temperature is 98.6° F. What is that in Celsius?
Answer:
Given, Normal body temperature is 98.6° F.
To convert Fahrenheit to Celsius the formula is (F – 32) x 5/9 = C
(98.6 – 32) x 5/9 = C
66.6 x 5/9 = C
37 = C
37° C
Normal body temperature is 37° C.

Question 20.
The oil company suggests that people set the temperature in their homes at between 20 and 22 degrees Celsius during the winter. What is that range in Fahrenheit?
Answer:
If the temperature is 20°C in their homes during the winter. To convert the temperature 20°C to Fahrenheit the formula is C x 9/5 + 32 = F
20 x 9/5 + 32 = F
36 + 32 = F
68 = F
68° F
If the temperature is 22°C in their homes during the winter. To convert the temperature 22°C to Fahrenheit the formula is C x 9/5 + 32 = F
22 x 9/5 + 32 = F
71.6 = F
72° F
The range in Fahrenheit is 68° F – 72° F.

Question 21.
What is the volume of the rectangular solid?
McGraw Hill Math Grade 6 Unit Test Lessons 18-20 Answer Key 1
Answer:
Given,
Length (l) = 5 inches
Width (w) = 4 inches
height (h) = 6 inches
Volume of a rectangular solid (V) = l x w x h
V = 5 x 4 x 6
V = 120 cubic inches
The volume of a rectangular solid is 120 cubic inches.

Question 22.
What is the volume of the triangular solid?
McGraw Hill Math Grade 6 Unit Test Lessons 18-20 Answer Key 2
Answer:
Base area (b) = length x width
b = 9 inches x 5 inches
b 45 square inches
Volume of triangular solid (v) = 1/2 x base area x height
v = 1/2 x 45 square inches x 12 inches
v = 540/2 cubic inches
V = 270 cubic inches
Volume of triangular solid is equal to 270 cubic inches.

Question 23.
In track and field, the standard middle distance event is the 1,500 meters.
About how many feet is 1,500 meters? ______________
About how much farther beyond 1,500 meters would someone have to run in order to reach a mile? ______________ feet
about ______________ meters
Answer:

Question 24.
The distance of a flight from New York to Phoenix, Arizona is about 2,400 miles. The average ground speed for a commercial airliner is about 850 kilometers per hour. About how much time will it take to fly from New York to Phoenix?
______________ hours
Answer:
The distance of a flight from New York to Phoenix, Arizona is about 2,400 miles.
Convert 2,400 miles into kilometers.
1 mile = 1.6 km
2,400 miles = ? km
2,400 x 1.6 = 3,840 km
The distance of a flight from New York to Phoenix, Arizona is about 3,840 km.
1 hour = 850 km
? hours = 3,840 km
3,840/850 = 4.5 hours
The flight takes 4.5 hours to fly from New York to Phoenix.

McGraw Hill Math Grade 6 Unit Test Lessons 18-20 Answer Key Read More »

McGraw Hill Math Grade 5 Chapter 4 Lesson 4 Answer Key Strategies for Multiplying Decimals

All the solutions provided in McGraw Hill Math Grade 5 Answer Key PDF Chapter 4 Lesson 4 Strategies for Multiplying Decimals are as per the latest syllabus guidelines.

McGraw-Hill Math Grade 5 Answer Key Chapter 4 Lesson 4 Strategies for Multiplying Decimals

Solve.

Estimate and multiply to complete the table. You can use the grids to help.

McGraw Hill Math Grade 5 Chapter 4 Lesson 4 Answer Key Strategies for Multiplying Decimals 1

Question 1.
0.2 × 0.2 =
Estimate: 0.04
Product: 0.04
Answer:
0.2
×0.2
0.04
By multiplying 0.2 with 0.2 we get 0.04
Estimate: 0.04
Product: 0.04

Question 2.
0.4 × 0.7 =
Estimate:
Product:
Answer:
0.4
×0.7
0.28
By multiplying 0.4 with 0.7 we get 0.28
Estimate: 0.3
Product: 0.28

Question 3.
0.3 × 0.8 =
Estimate:
Product:
Answer:
0.3
×0.8
0.24
By multiplying 0.3 with 0.8 we get 0.24
Estimate: 0.20
Product: 0.24

Question 4.
0.6 × 0.5 =
Estimate:
Product:
Answer:
0.6
×0.5
0.30
By multiplying 0.6 with 0.5 we get 0.30
Estimate: 0.3
Product: 0.30

Question 5.
0.9 × 0.8 =
Estimate:
Product:
Answer:
0.9
×0.8
0.72
By multiplying 0.9 with 0.8 we get 0.72
Estimate: 0.7
Product: 0.72

Question 6.
0.2 × 0.02 =
Estimate:
Product:
Answer:
0.2
×0.02
0.004
By multiplying 0.2 with 0.02 we get 0.04
Estimate: 0.004
Product: 0.004

McGraw Hill Math Grade 5 Chapter 4 Lesson 4 Answer Key Strategies for Multiplying Decimals Read More »

McGraw Hill Math Grade 6 Lesson 7.3 Answer Key Multiplying Fractions and Mixed Numbers: Reducing

Practice questions available in McGraw Hill Math Grade 6 Answer Key PDF Lesson 7.3 Multiplying Fractions and Mixed Numbers: Reducing will engage students and is a great way of informal assessment.

McGraw-Hill Math Grade 6 Answer Key Lesson 7.3 Multiplying Fractions and Mixed Numbers: Reducing

Exercises Multiply, Reduce

Question 1.
1\(\frac{1}{2}\) × \(\frac{4}{5}\)
Answer:
1\(\frac{1}{2}\) × \(\frac{4}{5}\) = \(\frac{6}{5}\)

Explanation:
1\(\frac{1}{2}\) × \(\frac{4}{5}\)
= {[(1 × 2) + 1] ÷ 2} × \(\frac{4}{5}\)
= [(2 + 1) ÷ 2] × \(\frac{4}{5}\)
= \(\frac{3}{2}\) × \(\frac{4}{5}\)
= \(\frac{3}{1}\) × \(\frac{2}{5}\)
= \(\frac{6}{5}\)

Question 2.
\(\frac{2}{9}\) × 4\(\frac{1}{4}\)
Answer:
\(\frac{2}{9}\) × 4\(\frac{1}{4}\) = \(\frac{17}{18}\)

Explanation:
\(\frac{2}{9}\) × 4\(\frac{1}{4}\)
= \(\frac{2}{9}\) ×{[(4 × 4) + 1] ÷ 4}
= \(\frac{2}{9}\) × [(16 + 1) ÷ 4]
= \(\frac{2}{9}\) × \(\frac{17}{4}\)
= \(\frac{1}{9}\) × \(\frac{17}{2}\)
= \(\frac{17}{18}\)

Question 3.
\(\frac{3}{4}\) × 2\(\frac{1}{7}\)
Answer:
\(\frac{3}{4}\) × 2\(\frac{1}{7}\) = \(\frac{45}{28}\)

Explanation:
\(\frac{3}{4}\) × 2\(\frac{1}{7}\)
= \(\frac{3}{4}\) × {[(2 × 7) + 1] ÷ 7}
= \(\frac{3}{4}\) × [(14 + 1) ÷ 7]
= \(\frac{3}{4}\) × \(\frac{15}{7}\)
= \(\frac{45}{28}\)

Question 4.
5\(\frac{1}{3}\) × \(\frac{1}{2}\)
Answer:
5\(\frac{1}{3}\) × \(\frac{1}{2}\) = \(\frac{8}{3}\)

Explanation:
5\(\frac{1}{3}\) × \(\frac{1}{2}\)
= {[(5 × 3) + 1] ÷ 3} × \(\frac{1}{2}\)
= [(15 + 1) ÷ 3] × \(\frac{1}{2}\)
= \(\frac{16}{3}\) × \(\frac{1}{2}\)
= \(\frac{8}{3}\) × \(\frac{1}{1}\)
= \(\frac{8}{3}\)

Question 5.
\(\frac{8}{9}\) × 3\(\frac{1}{5}\)
Answer:
\(\frac{8}{9}\) × 3\(\frac{1}{5}\) = \(\frac{128}{45}\)

Explanation:
\(\frac{8}{9}\) × 3\(\frac{1}{5}\)
= \(\frac{8}{9}\) × {[(3 × 5) + 1] ÷ 5}
= \(\frac{8}{9}\) × [(15 + 1) ÷ 5]
= \(\frac{8}{9}\) × \(\frac{16}{5}\)
= \(\frac{128}{45}\)

Question 6.
1\(\frac{2}{3}\) × \(\frac{2}{7}\)
Answer:
1\(\frac{2}{3}\) × \(\frac{2}{7}\) = \(\frac{10}{21}\)

Explanation:
1\(\frac{2}{3}\) × \(\frac{2}{7}\)
= {[(1 × 3) + 2] ÷ 3} × \(\frac{2}{7}\)
= [(3 + 2) ÷ 3] × \(\frac{2}{7}\)
= \(\frac{5}{3}\) × \(\frac{2}{7}\)
= \(\frac{10}{21}\)

Question 7.
5\(\frac{1}{4}\) × \(\frac{2}{3}\)
Answer:
5\(\frac{1}{4}\) × \(\frac{2}{3}\) = \(\frac{7}{2}\)

Explanation:
5\(\frac{1}{4}\) × \(\frac{2}{3}\)
= {[(5 × 4) + 1] ÷ 4} × \(\frac{2}{3}\)
= [(20 + 1) ÷ 4] × \(\frac{2}{3}\)
= \(\frac{21}{4}\) × \(\frac{2}{3}\)
= \(\frac{21}{2}\) × \(\frac{1}{3}\)
= \(\frac{7}{2}\) × \(\frac{1}{1}\)
= \(\frac{7}{2}\)

Question 8.
7\(\frac{1}{8}\) × \(\frac{1}{4}\)
Answer:
7\(\frac{1}{8}\) × \(\frac{1}{4}\) = \(\frac{57}{32}\)

Explanation:
7\(\frac{1}{8}\) × \(\frac{1}{4}\)
= {[(7 × 8) + 1] ÷ 8} × \(\frac{1}{4}\)
= [(56 + 1) ÷ 8] × \(\frac{1}{4}\)
= \(\frac{57}{8}\) × \(\frac{1}{4}\)
= \(\frac{57}{32}\)

Question 9.
1\(\frac{1}{2}\) × \(\frac{8}{9}\)
Answer:
1\(\frac{1}{2}\) × \(\frac{8}{9}\) = \(\frac{4}{3}\)

Explanation:
1\(\frac{1}{2}\) × \(\frac{8}{9}\)
= {[(1 × 2) + 1] ÷ 2 × \(\frac{8}{9}\)
= [(2 + 1) ÷ 2]× \(\frac{8}{9}\)
= \(\frac{3}{2}\) × \(\frac{8}{9}\)
= \(\frac{3}{1}\) × \(\frac{4}{9}\)
= \(\frac{1}{1}\) × \(\frac{4}{3}\)
= \(\frac{4}{3}\)

Question 10.
\(\frac{3}{7}\) × 5\(\frac{1}{7}\)
Answer:
\(\frac{3}{7}\) × 5\(\frac{1}{7}\) = \(\frac{108}{49}\)

Explanation:
\(\frac{3}{7}\) × 5\(\frac{1}{7}\)
= \(\frac{3}{7}\) × {[(5 × 7) + 1] ÷ 7}
= \(\frac{3}{7}\) × [(35 + 1) ÷ 7
= \(\frac{3}{7}\) × \(\frac{36}{7}\)
= \(\frac{108}{49}\)

Question 11.
\(\frac{1}{3}\) × 5\(\frac{2}{3}\)
Answer:
\(\frac{1}{3}\) × 5\(\frac{2}{3}\) = \(\frac{17}{9}\)

Explanation:
\(\frac{1}{3}\) × 5\(\frac{2}{3}\)
= \(\frac{1}{3}\) × {[(5 × 3) + 2] ÷ 3}
= \(\frac{1}{3}\) × [(15 + 2) ÷ 3]
= \(\frac{1}{3}\) × \(\frac{17}{3}\)
= \(\frac{17}{9}\)

Question 12.
\(\frac{5}{6}\) × 1\(\frac{1}{5}\)
Answer:
\(\frac{5}{6}\) × 1\(\frac{1}{5}\) = 1.

Explanation:
\(\frac{5}{6}\) × 1\(\frac{1}{5}\)
= \(\frac{5}{6}\) × {[(1 × 5) + 1] ÷ 5}
= \(\frac{5}{6}\) × [(5 + 1) ÷ 5]
= \(\frac{5}{6}\) × \(\frac{6}{5}\)
= 1.

Question 13.
\(\frac{3}{4}\) × 5\(\frac{1}{3}\)
Answer:
\(\frac{3}{4}\) × 5\(\frac{1}{3}\) = 4.

Explanation:
\(\frac{3}{4}\) × 5\(\frac{1}{3}\)
= \(\frac{3}{4}\) ×{[(5 × 3) + 1] ÷ 3}
= \(\frac{3}{4}\) × [(15 + 1) ÷ 3]
= \(\frac{3}{4}\) × \(\frac{16}{3}\)
= \(\frac{1}{4}\) × \(\frac{16}{1}\)
= \(\frac{1}{1}\) × \(\frac{4}{1}\)
= \(\frac{4}{1}\)
= 4.

Question 14.
3\(\frac{1}{4}\) × \(\frac{3}{7}\)
Answer:
3\(\frac{1}{4}\) × \(\frac{3}{7}\) = \(\frac{39}{28}\)

Explanation:
3\(\frac{1}{4}\) × \(\frac{3}{7}\)
= {[(3 × 4) + 1] ÷ 4} × \(\frac{3}{7}\)
= [(12 + 1) ÷ 4] × \(\frac{3}{7}\)
= \(\frac{13}{4}\) × \(\frac{3}{7}\)
= \(\frac{39}{28}\)

Question 15.
5\(\frac{2}{11}\) × \(\frac{22}{23}\)
Answer:
5\(\frac{2}{11}\) × \(\frac{22}{23}\) = \(\frac{114}{23}\)

Explanation:
5\(\frac{2}{11}\) × \(\frac{22}{23}\)
= {[(5 × 11) + 2] ÷ 11}× \(\frac{22}{23}\)
= [(55 + 2) ÷ 11] × \(\frac{22}{23}\)
= \(\frac{57}{11}\) × \(\frac{22}{23}\)
= \(\frac{57}{1}\) × \(\frac{2}{23}\)
= \(\frac{114}{23}\)

Exercises Multiply, Reduce
Question 1.
1\(\frac{1}{4}\) × 2\(\frac{2}{3}\)
Answer:
1\(\frac{1}{4}\) × 2\(\frac{2}{3}\) = \(\frac{10}{3}\)

Explanation:
1\(\frac{1}{4}\) × 2\(\frac{2}{3}\)
= {[(1× 4) + 1] ÷ 4} × {[(2× 3) + 2] ÷ 3}
= [(4 + 1) ÷ 4] × [(6 + 2) ÷ 3]
= \(\frac{5}{4}\) × \(\frac{8}{3}\)
= \(\frac{5}{1}\) × \(\frac{2}{3}\)
= \(\frac{10}{3}\)

Question 2.
2\(\frac{1}{5}\) × 5\(\frac{1}{4}\)
Answer:
2\(\frac{1}{5}\) × 5\(\frac{1}{4}\) = \(\frac{231}{15}\)

Explanation:
2\(\frac{1}{5}\) × 5\(\frac{1}{4}\)
= {[(2× 5) + 1] ÷ 5} × {[(5× 4) + 1] ÷ 3}
= [(10 + 1) ÷ 5] × [(20 + 1) ÷ 3]
= \(\frac{11}{5}\) × \(\frac{21}{3}\)
= \(\frac{231}{15}\)

Question 3.
5\(\frac{1}{8}\) × 2\(\frac{2}{3}\)
Answer:
5\(\frac{1}{8}\) × 2\(\frac{2}{3}\) = \(\frac{41}{2}\)

Explanation:
5\(\frac{1}{8}\) × 2\(\frac{2}{3}\)
= {[(5 × 8) + 1] ÷ 8} × {[(2 × 3) + 2] ÷ 3}
= [(40 + 1) ÷ 8] × [(6 + 2) ÷ 3]
= \(\frac{41}{8}\) × \(\frac{12}{3}\)
= \(\frac{41}{2}\) × \(\frac{3}{3}\)
= \(\frac{41}{2}\) × \(\frac{1}{1}\)
= \(\frac{41}{2}\)

Question 4.
2\(\frac{3}{7}\) × 2\(\frac{3}{4}\)
Answer:
2\(\frac{3}{7}\) × 2\(\frac{3}{4}\) = \(\frac{187}{28}\)

Explanation:
2\(\frac{3}{7}\) × 2\(\frac{3}{4}\)
= {[(2 × 7) + 3] ÷ 7} × {[(2 × 4) + 3] ÷ 4}
= [(14 + 3) ÷ 7] × [(8 + 3) ÷ 4]
= \(\frac{17}{7}\) × \(\frac{11}{4}\)
= \(\frac{187}{28}\)

Question 5.
2\(\frac{1}{7}\) × 1\(\frac{1}{3}\)
Answer:
2\(\frac{1}{7}\) × 1\(\frac{1}{3}\) = \(\frac{20}{7}\)

Explanation:
2\(\frac{1}{7}\) × 1\(\frac{1}{3}\)
= {[(2 × 7) + 1] ÷ 7} × {[(1 × 3) + 1] ÷ 3}
= [(14 + 1) ÷ 7] × [(3 + 1) ÷ 3]
= \(\frac{15}{7}\) × \(\frac{4}{3}\)
= \(\frac{5}{7}\) × \(\frac{4}{1}\)
= \(\frac{20}{7}\)

Question 6.
2\(\frac{1}{2}\) × 3\(\frac{2}{3}\)
Answer:
2\(\frac{1}{2}\) × 3\(\frac{2}{3}\) = \(\frac{55}{6}\)

Explanation:
2\(\frac{1}{2}\) × 3\(\frac{2}{3}\)
= {[(2 × 2) + 1] ÷ 2} × {[(3 × 3) + 2] ÷ 3}
= [(4 + 1) ÷ 2] × [(9 + 2) ÷ 3]
= \(\frac{5}{2}\) × \(\frac{11}{3}\)
= \(\frac{55}{6}\)

Question 7.
3\(\frac{1}{3}\) × 1\(\frac{2}{3}\)
Answer:
3\(\frac{1}{3}\) × 1\(\frac{2}{3}\) = \(\frac{50}{9}\)

Explanation:
3\(\frac{1}{3}\) × 1\(\frac{2}{3}\)
= {[(3 × 3) + 1] ÷ 3} × {[(1 × 3) + 2] ÷ 3}
= [(9 + 1) ÷ 3] × [(3 + 2) ÷ 3]
= \(\frac{10}{3}\) × \(\frac{5}{3}\)
= \(\frac{50}{9}\)

Question 8.
3\(\frac{4}{5}\) × 1\(\frac{1}{2}\)
Answer:
3\(\frac{4}{5}\) × 1\(\frac{1}{2}\) = \(\frac{57}{10}\)

Explanation:
3\(\frac{4}{5}\) × 1\(\frac{1}{2}\)
= {[(3 × 5) + 4] ÷ 5} × {[(1 × 2) + 1] ÷ 2}
= [(15 + 4) ÷ 5] × [(2 + 1) ÷ 2]
= \(\frac{19}{5}\) × \(\frac{3}{2}\)
= \(\frac{57}{10}\)

Question 9.
3\(\frac{1}{2}\) × 2\(\frac{1}{7}\)
Answer:
3\(\frac{1}{2}\) × 2\(\frac{1}{7}\) = \(\frac{15}{2}\)

Explanation:
3\(\frac{1}{2}\) × 2\(\frac{1}{7}\)
= {[(3 × 2) + 1] ÷ 2} × {[(2 × 7) + 1] ÷ 7}
= [(6 + 1) ÷ 2] × [(14 + 1) ÷ 7]
= \(\frac{7}{2}\) × \(\frac{15}{7}\)
= \(\frac{1}{2}\) × \(\frac{15}{1}\)
= \(\frac{15}{2}\)

Question 10.
4\(\frac{3}{5}\) × 2\(\frac{1}{3}\)
Answer:
4\(\frac{3}{5}\) × 2\(\frac{1}{3}\) = \(\frac{91}{15}\)

Explanation:
4\(\frac{3}{5}\) × 2\(\frac{1}{3}\)
= {[(4 × 5) + 3] ÷ 5} × {[(2 × 3) + 1] ÷ 3}
= [(10 + 3) ÷ 5] × [(6 + 1) ÷ 3]
= \(\frac{13}{5}\) × \(\frac{7}{3}\)
= \(\frac{91}{15}\)

Question 11.
4\(\frac{1}{2}\) × 3\(\frac{1}{3}\)
Answer:
4\(\frac{1}{2}\) × 3\(\frac{1}{3}\) = 15.

Explanation:
4\(\frac{1}{2}\) × 3\(\frac{1}{3}\)
= {[(4 × 2) + 1] ÷ 2} × {[(3 × 3) + 1] ÷ 3}
= [(8 + 1) ÷ 2] × [(9 + 1) ÷ 3]
= \(\frac{9}{2}\) × \(\frac{10}{3}\)
= \(\frac{3}{2}\) × \(\frac{10}{1}\)
= \(\frac{3}{1}\) × \(\frac{5}{1}\)
= \(\frac{15}{1}\)
= 15.

Question 12.
5\(\frac{2}{3}\) × 2\(\frac{1}{16}\)
Answer:
5\(\frac{2}{3}\) × 2\(\frac{1}{16}\) = \(\frac{187}{16}\)

Explanation:
5\(\frac{2}{3}\) × 2\(\frac{1}{16}\)
= {[(5 × 3) + 2] ÷ 3} × {[(2 × 16) + 1] ÷ 16}
= [(15 + 2) ÷ 3] × [(32 + 1) ÷ 16]
= \(\frac{17}{3}\) × \(\frac{33}{16}\)
= \(\frac{17}{1}\) × \(\frac{11}{16}\)
= \(\frac{187}{16}\)

Question 13.
1\(\frac{1}{2}\) × 1\(\frac{1}{8}\)
Answer:
1\(\frac{1}{2}\) × 1\(\frac{1}{8}\) = \(\frac{27}{16}\)

Explanation:
1\(\frac{1}{2}\) × 1\(\frac{1}{8}\)
= {[(1 × 2) + 1] ÷ 2} × {[(1 × 8) + 1] ÷ 8}
= [(2 + 1) ÷ 2] × [(8 + 1) ÷ 8]
= \(\frac{3}{2}\) × \(\frac{9}{8}\)
= \(\frac{27}{16}\)

Question 14.
3\(\frac{1}{4}\) × 2\(\frac{3}{5}\)
Answer:
3\(\frac{1}{4}\) × 2\(\frac{3}{5}\) = \(\frac{169}{20}\)

Explanation:
3\(\frac{1}{4}\) × 2\(\frac{3}{5}\)
= {[(3 × 4) + 1] ÷ 4} × {[(2 × 5) + 3] ÷ 5}
= [(12 + 1) ÷ 4] × [(10 + 3) ÷ 5]
= \(\frac{13}{4}\) × \(\frac{13}{5}\)
= \(\frac{169}{20}\)

Question 15.
5\(\frac{1}{2}\) × 1\(\frac{7}{8}\)
Answer:
5\(\frac{1}{2}\) × 1\(\frac{7}{8}\) = \(\frac{165}{16}\)

Explanation:
5\(\frac{1}{2}\) × 1\(\frac{7}{8}\)
= {[(5 × 2) + 1] ÷ 2} × {[(1 × 8) + 7] ÷ 8}
= [(10 + 1) ÷ 2] × [(8 + 7) ÷ 8]
= \(\frac{11}{2}\) × \(\frac{15}{8}\)
= \(\frac{165}{16}\)

McGraw Hill Math Grade 6 Lesson 7.3 Answer Key Multiplying Fractions and Mixed Numbers: Reducing Read More »

McGraw Hill Math Grade 6 Lesson 7.2 Answer Key Multiplying Fractions: Reciprocals

Practice questions available in McGraw Hill Math Grade 6 Answer Key PDF Lesson 7.2 Multiplying Fractions: Reciprocals will engage students and is a great way of informal assessment.

McGraw-Hill Math Grade 6 Answer Key Lesson 7.2 Multiplying Fractions: Reciprocals

Exercises Multiply

Question 1.
\(\frac{1}{2}\) × \(\frac{1}{2}\)
Answer:
Multiplying \(\frac{1}{2}\) by \(\frac{1}{2}\), we get the product \(\frac{1}{4}\)

Explanation:
\(\frac{1}{2}\) × \(\frac{1}{2}\)
= \(\frac{1}{4}\)

Question 2.
\(\frac{2}{3}\) × \(\frac{6}{7}\)
Answer:
Multiplying \(\frac{2}{3}\) by \(\frac{6}{7}\), we get the product \(\frac{4}{7}\)

Explanation:
\(\frac{2}{3}\) × \(\frac{6}{7}\)
= \(\frac{2}{1}\) × \(\frac{2}{7}\)
= \(\frac{4}{7}\)

Question 3.
\(\frac{5}{9}\) × \(\frac{3}{11}\)
Answer:
Multiplying \(\frac{5}{9}\) by \(\frac{3}{11}\) , we get the product \(\frac{5}{33}\)

Explanation:
\(\frac{5}{9}\) × \(\frac{3}{11}\)
= \(\frac{5}{3}\) × \(\frac{1}{11}\)
= \(\frac{5}{33}\)

Question 4.
\(\frac{10}{13}\) × \(\frac{1}{3}\)
Answer:
Multiplying \(\frac{10}{13}\) by \(\frac{1}{3}\) , we get the product \(\frac{10}{39}\)

Explanation:
\(\frac{10}{13}\) × \(\frac{1}{3}\)
= \(\frac{10}{39}\)

Question 5.
\(\frac{3}{11}\) × \(\frac{11}{3}\)
Answer:
Multiplying \(\frac{3}{11}\) by \(\frac{11}{3}\) , we get the product 1.

Explanation:
\(\frac{3}{11}\) × \(\frac{11}{3}\)
= 1.

Question 6.
\(\frac{7}{3}\) × \(\frac{3}{11}\)
Answer:
Multiplying \(\frac{7}{3}\) by \(\frac{3}{11}\), we get the product \(\frac{7}{11}\)

Explanation:
\(\frac{7}{3}\) × \(\frac{3}{11}\)
= \(\frac{7}{1}\) × \(\frac{1}{11}\)
= \(\frac{7}{11}\)

Question 7.
\(\frac{4}{5}\) × \(\frac{7}{8}\)
Answer:
Multiplying \(\frac{4}{5}\) by \(\frac{7}{8}\), we get the product \(\frac{7}{10}\)

Explanation:
\(\frac{4}{5}\) × \(\frac{7}{8}\)
= \(\frac{1}{5}\) × \(\frac{7}{2}\)
= \(\frac{7}{10}\)

Question 8.
\(\frac{3}{4}\) × \(\frac{4}{3}\)
Answer:
Multiplying \(\frac{3}{4}\) by \(\frac{4}{3}\) , we get the product 1.

Explanation:
\(\frac{3}{4}\) × \(\frac{4}{3}\)
= 1.

Question 9.
\(\frac{17}{27}\) × \(\frac{5}{3}\)
Answer:
Multiplying \(\frac{17}{27}\) by \(\frac{5}{3}\), we get the product \(\frac{85}{81}\)

Explanation:
\(\frac{17}{27}\) × \(\frac{5}{3}\)
= \(\frac{85}{81}\)

Question 10.
\(\frac{6}{10}\) × \(\frac{6}{11}\)
Answer:
Multiplying \(\frac{6}{10}\) by \(\frac{6}{11}\), we get the product \(\frac{18}{55}\)

Explanation:
\(\frac{6}{10}\) × \(\frac{6}{11}\)
= \(\frac{6}{5}\) × \(\frac{3}{11}\)
= \(\frac{18}{55}\)

Question 11.
\(\frac{13}{14}\) × \(\frac{2}{3}\)
Answer:
Multiplying \(\frac{13}{14}\) by \(\frac{2}{3}\), we get the product \(\frac{13}{21}\)

Explanation:
\(\frac{13}{14}\) × \(\frac{2}{3}\)
= \(\frac{13}{7}\) × \(\frac{1}{3}\)
= \(\frac{13}{21}\)

Question 12.
\(\frac{4}{5}\) × \(\frac{4}{5}\)
Answer:
Multiplying \(\frac{4}{5}\) by \(\frac{4}{5}\), we get the product \(\frac{16}{25}\)

Explanation:
\(\frac{4}{5}\) × \(\frac{4}{5}\)
= \(\frac{16}{25}\)

Question 13.
\(\frac{133}{145}\) × \(\frac{145}{133}\)
Answer:
Multiplying \(\frac{133}{145}\) by \(\frac{145}{133}\), we get the product 1.

Explanation:
\(\frac{133}{145}\) × \(\frac{145}{133}\)
= 1.

Question 14.
\(\frac{8}{19}\) × \(\frac{5}{7}\)
Answer:
Multiplying \(\frac{8}{19}\) by \(\frac{5}{7}\), we get the product \(\frac{40}{133}\)

Explanation:
\(\frac{8}{19}\) × \(\frac{5}{7}\)
= \(\frac{40}{133}\)

Question 15.
\(\frac{11}{13}\) × \(\frac{9}{14}\)
Answer:
Multiplying \(\frac{11}{13}\) by \(\frac{9}{14}\), we get the product \(\frac{99}{182}\)

Explanation:
\(\frac{11}{13}\) × \(\frac{9}{14}\)
= \(\frac{99}{182}\)

Question 16.
\(\frac{32}{33}\) × \(\frac{1}{2}\)
Answer:
Multiplying \(\frac{32}{33}\) by \(\frac{1}{2}\), we get the product \(\frac{16}{33}\)

Explanation:
\(\frac{32}{33}\) × \(\frac{1}{2}\)
= \(\frac{16}{33}\) × \(\frac{1}{1}\)
= \(\frac{16}{33}\)

Question 17.
\(\frac{1}{5}\) × \(\frac{1}{7}\)
Answer:
Multiplying \(\frac{1}{5}\) by \(\frac{1}{7}\), we get the product \(\frac{1}{35}\)

Explanation:
\(\frac{1}{5}\) × \(\frac{1}{7}\)
= \(\frac{1}{35}\)

Question 18.
\(\frac{5}{6}\) × \(\frac{5}{9}\)
Answer:
Multiplying \(\frac{5}{6}\) by \(\frac{5}{9}\), we get the product \(\frac{25}{54}\)

Explanation:
\(\frac{5}{6}\) × \(\frac{5}{9}\)
= \(\frac{25}{54}\)

Question 19.
\(\frac{2}{3}\) × \(\frac{33}{34}\)
Answer:
Multiplying \(\frac{2}{3}\) by \(\frac{33}{34}\), we get the product \(\frac{11}{17}\)

Explanation:
\(\frac{2}{3}\) × \(\frac{33}{34}\)
= \(\frac{1}{1}\) × \(\frac{11}{17}\)
= \(\frac{11}{17}\)

Question 20.
\(\frac{45}{47}\) × \(\frac{47}{45}\)
Answer:
Multiplying \(\frac{45}{47}\) by \(\frac{47}{45}\), we get the product 1.

Explanation:
\(\frac{45}{47}\) × \(\frac{47}{45}\)
= 1.

McGraw Hill Math Grade 6 Lesson 7.2 Answer Key Multiplying Fractions: Reciprocals Read More »

McGraw Hill Math Grade 6 Lesson 7.1 Answer Key Multiplying Fractions and Whole Numbers

Practice questions available in McGraw Hill Math Grade 6 Answer Key PDF Lesson 7.1 Multiplying Fractions and Whole Numbers will engage students and is a great way of informal assessment.

McGraw-Hill Math Grade 6 Answer Key Lesson 7.1 Multiplying Fractions and Whole Numbers

Exercises Estimate

Question 1.
5 × \(\frac{3}{4}\)
Answer:
5 × \(\frac{3}{4}\) = 3\(\frac{3}{4}\)

Explanation:
5 × \(\frac{3}{4}\)
= \(\frac{15}{4}\)
= 3\(\frac{3}{4}\)

Question 2.
4 × \(\frac{2}{7}\)
Answer:
4 × \(\frac{2}{7}\) = 1\(\frac{1}{7}\)

Explanation:
4 × \(\frac{2}{7}\)
= \(\frac{8}{7}\)
= 1\(\frac{1}{7}\)

Question 3.
21 × \(\frac{5}{8}\)
Answer:
21 × \(\frac{5}{8}\) = 13\(\frac{1}{8}\)

Explanation:
21 × \(\frac{5}{8}\)
= \(\frac{105}{8}\)
= 13\(\frac{1}{8}\)

Question 4.
11 × \(\frac{2}{9}\)
Answer:
11 × \(\frac{2}{9}\) = 2\(\frac{4}{9}\)

Explanation:
11 × \(\frac{2}{9}\)
= \(\frac{22}{9}\)
= 2\(\frac{4}{9}\)

Question 5.
4 × \(\frac{3}{11}\)
Answer:
4 × \(\frac{3}{11}\) = 1\(\frac{1}{11}\)

Explanation:
4 × \(\frac{3}{11}\)
= \(\frac{12}{11}\)
= 1\(\frac{1}{11}\)

Question 6.
13 × \(\frac{13}{14}\)
Answer:
13 × \(\frac{13}{14}\) = 12\(\frac{1}{14}\)

Explanation:
13 × \(\frac{13}{14}\)
= \(\frac{169}{14}\)
= 12\(\frac{1}{14}\)

Question 7.
21 × \(\frac{10}{23}\)
Answer:
21 × \(\frac{10}{23}\) = 9 \(\frac{3}{23}\)

Explanation:
21 × \(\frac{10}{23}\)
= \(\frac{210}{23}\)
= 9 \(\frac{3}{23}\)

Question 8.
7 × \(\frac{2}{3}\)
Answer:
7 × \(\frac{2}{3}\) = 4 \(\frac{2}{3}\)

Explanation:
7 × \(\frac{2}{3}\)
= \(\frac{14}{3}\)
= 4 \(\frac{2}{3}\)

Question 9.
12 × \(\frac{7}{19}\)
Answer:
12 × \(\frac{7}{19}\) = 4\(\frac{8}{19}\)

Explanation:
12 × \(\frac{7}{19}\)
= \(\frac{84}{19}\)
= 4\(\frac{8}{19}\)

Question 10.
14 × \(\frac{3}{5}\)
Answer:
14 × \(\frac{3}{5}\) = 8\(\frac{2}{5}\)

Explanation:
14 × \(\frac{3}{5}\)
=\(\frac{42}{5}\)
= 8\(\frac{2}{5}\)

Question 11.
14 × \(\frac{11}{13}\)
Answer:
14 × \(\frac{11}{13}\) = 11\(\frac{11}{13}\)

Explanation:
14 × \(\frac{11}{13}\)
= \(\frac{154}{13}\)
= 11\(\frac{11}{13}\)

Question 12.
13 × \(\frac{4}{17}\)
Answer:
13 × \(\frac{4}{17}\) = 3\(\frac{1}{17}\)

Explanation:
13 × \(\frac{4}{17}\)
= \(\frac{52}{17}\)
= 3\(\frac{1}{17}\)

Question 13.
10 × \(\frac{9}{23}\)
Answer:
10 × \(\frac{9}{23}\) = 3\(\frac{21}{23}\)

Explanation:
10 × \(\frac{9}{23}\)
= \(\frac{90}{23}\)
= 3\(\frac{21}{23}\)

Question 14.
13 × \(\frac{21}{22}\)
Answer:
13 × \(\frac{21}{22}\) = 12\(\frac{9}{22}\)

Explanation:
13 × \(\frac{21}{22}\)
= \(\frac{273}{22}\)
= 12\(\frac{9}{22}\)

Question 15.
5 × \(\frac{14}{27}\)
Answer:
5 × \(\frac{14}{27}\) = 2\(\frac{16}{27}\)

Explanation:
5 × \(\frac{14}{27}\)
= \(\frac{70}{27}\)
= 2\(\frac{16}{27}\)

Question 16.
Chet found a pair of sunglasses that he would like to buy. They normally cost $43.00, but this week they are on sale for only \(\frac{7}{8}\) of the usual price. How much money will Chet need to buy the sunglasses?
Answer:
Amount of money Chet needs to buy the sunglasses = $37\(\frac{5}{8}\)

Explanation:
Cost of a pair of sunglasses usually = $43.00
Cost of a pair of sunglasses this week on sale for only of the usual price = \(\frac{7}{8}\)
Amount of money Chet needs to buy the sunglasses = Cost of a pair of sunglasses usually × Cost of a pair of sunglasses this week on sale for only of the usual price
= 43 × \(\frac{7}{8}\)
= \(\frac{301}{8}\)
= $37\(\frac{5}{8}\)

Question 17.
Ashlee is riding her bike to the beach. If the beach is 17 miles from her home, and she has already traveled \(\frac{3}{5}\) of the way, how much farther does she have to cycle?
Answer:
Number of miles more she needs to cycle = 10\(\frac{1}{5}\)

Explanation:
Number of miles from her home the beach = 17.
Number of miles already travelled of the way = \(\frac{3}{5}\)
Number of miles more she needs to cycle = Number of miles from her home the beach × Number of miles already travelled of the way
= 17 × \(\frac{3}{5}\)
= \(\frac{51}{5}\)
= 10\(\frac{1}{5}\)

McGraw Hill Math Grade 6 Lesson 7.1 Answer Key Multiplying Fractions and Whole Numbers Read More »

McGraw Hill Math Grade 5 Chapter 4 Lesson 5 Answer Key Relating Multiplication and Division of Decimals

All the solutions provided in McGraw Hill Math Grade 5 Answer Key PDF Chapter 4 Lesson 5 Relating Multiplication and Division of Decimals are as per the latest syllabus guidelines.

McGraw-Hill Math Grade 5 Answer Key Chapter 4 Lesson 5 Relating Multiplication and Division of Decimals

Solve.

Solve each problem. Then list the related decimal multiplication and division facts.

Question 1.
2.4 ÷ 8 = 0.3
2.4 ÷ 0.3 = 8
0.3 × 8 = 2.4
8 × 0.3 = 2.4
Answer:
Given the expression is 2.4 ÷ 8 which is
2.4 ÷ 8 = 0.3,
2.4 ÷ 0.3 = 8,
0.3 × 8 = 2.4,
8 × 0.3 = 2.4

Question 2.
6 ÷ 0.2 = __________________
___________________________
Answer:
Given the expression is 6 ÷ 0.2 which is
6 ÷ 0.2 = 30,
30 ÷ 6 = 5,
30 × 0.2 = 6,
0.2 × 30 = 6.

Question 3.
2.2 ÷ 10 = __________________
___________________________
Answer:
Given the expression is 2.2 ÷ 10 which is
2.2 ÷ 10 = 0.22,
0.22 × 10 = 2.2,
10 × 0.22 = 2.2.

Question 4.
2 ÷ 0.1 = __________________
___________________________
Answer:
Given the expression is 2 ÷ 0.1 which is
2 ÷ 0.1 = 20,
20 × 0.1 = 2,
0.1 × 20 = 2.

Question 5.
8 ÷ 0.4 = __________________
___________________________
Answer:
Given the expression is 8 ÷ 0.4 which is
8 ÷ 0.4 = 20,
20 × 0.4 = 8,
0.4 × 20 = 8.

Question 6.
1.6 ÷ 5 = __________________
___________________________
Answer:
Given the expression is 1.6 ÷ 5 which is
1.6 ÷ 5 = 0.32,
0.32 × 5 = 1.6,
5 × 0.32 = 1.6.

Question 7.
When a whole number is divided by a decimal, is the quotient greater than or less than the dividend?
Answer:
The quotient is greater than the dividend

Explanation:
When a whole number is divided by a decimal, the quotient is greater than the dividend. For example, if we divide 25 by 0.5 which is
25 ÷ 0.5 = 50.

McGraw Hill Math Grade 5 Chapter 4 Lesson 5 Answer Key Relating Multiplication and Division of Decimals Read More »

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