Go Math Grade 6 Exercise 8.3: Understanding Percents Solutions

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 8 Percents

Page 53 Problem 1 Answer

Here the value is 22. Total number of students is 55.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents 1

22 students is 40% of 55.

Page 53 Problem 2 Answer

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents 2

red marbles is 40% of 60 marbles

Page 53 Problem 3 Answer

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents 3

15% of $9 is $1.35.

Go Math! Practice Fluency Workbook Grade 6 Chapter 8 Percents Exercise 8.3 Answer Key

Page 53 Problem 4 Answer

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents 4

12 is 6% of 200.

Go Math Grade 6 Exercise 8.3 Understanding Percents Solutions

Go Math Grade 6 Exercise 8.3: Understanding Percents Solutions  Page 53 Problem 5 Answer

Total number of messages Bethany sent is 60.

Percentage of messages she sent to her best friend is 15%.

The number of text messages Bethany sent to her best friend is 9.

Given, Bethany sent 60 text messages out of which 15% of messages were sent to her best friend.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents 5

Then the number of messages Bethany sent to her best friend is 9.

Page 53 Problem 6 Answer

Given: 27% of the people in a survey chose salads over a meat dish.

81 people chose salads.To find: the number of people in the survey

summary: Write the equation to find the percentage. Substitute the given values. Solve for the unknown.

Given that 27% of people in the survey chose salads over a meat dish.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents 6

The number of people who chose salad over meat dish is 81, the value of the part in the equation.

Substitute these values in the equation for percentage.

If the percentage and the number of people in a survey who chose salad over a meat fish are 27% and 81 respectively, then the total number of people in the survey is 300.

Page 53 Problem 7 Answer

Given: The sales tax on a $350 computer is $22.75.

To find: The sales tax rate. Summary: Given the part and the total. Substitute the values in the equation for percentage.

Simplify and find the percentage. Given the cost of the computer is $350.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents 7

The sale tax on it is $22.75. Substitute these values in the equation to If the sales tax on a computer of cost $350 is $22.75, then the sales tax rate of the product is 6.5%.

Go Math Grade 6 Exercise 8.3: Understanding Percents Solutions Page 53 Problem 8 Answer

Given: The circle graph

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents 8 1

To find: The number of voters of age 18−29 if the total number of voters is 6000.

Summary: Identify the percentage of the given age group from the figure. Substitute the value and the total in the equation for the percentage.

Solve for the unknown. The percentage of voters of age 18−29 is 10%.

The total number of people who voted for the election is 6000.

Substitute these values in the equation for percentage and solve.

The figure below shows the age group and the percentage of voters.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents 8 2

If the total number of voters is 6000, the number of voters of age 18−29 is 600.

Page 53 Problem 9 Answer

Given: The cycle graph,

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents 9 1

To find: If the total number of voters is 12,000, find the number of voters of age 50−64.

Summary: Identify the percentage of age group voters from the figure and the total. Substitute the values in the equation and solve.

Given the total number of voters is 12,000. The percentage of voters of age 50−64 is 35%.

percentage=value/total×100

value=percentage×total/100

​Substituting the values, we get, ​value=35×12000/100 = 4200; which is the number of voters of the age 50−64.

The circle graph is given below,

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents 9 2

If the total number of voters is 12,000 the number of voters of age 50−64 is 4200.

Page 53 Problem 10 Answer

Given: The circle graph is

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents 10 1

To find: The number of voters of age over 65 if the total number of voters is 596.

Summary: Find the percentage of voters of the given age group from the figure and the total is also given. Substitute the values in the equation and solve.

The total number of voters is 596. The percentage of voters of age over 60=25%

​Percentage=value/total×100

value=percentage×total/100

​Substitute the values in the equation, ​value=25×596/100=149; which is the number of voters of age over 60.

The circle graph is given below, If the total number of voters is 596, then the number of voters of age over 60 is 149.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents 10 2

Page 53 Problem 11 Answer

From the below given circle graph,

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents 11 1

The percentage of voters of age 18−29=10%.

The percentage of voters of age 30−49=30%.

The number of people of age group 18−29 who voted is 45.

The percentage of voters of age 30−49 is three tomes the percentage of voters of age 18−29.

So in order to find the number of people of age 30−49 who voted, we need to multiply the number of people of age 18−29 who voted with three.

That is, 45×3=135. This is how Sahil calculated the answer, and he is correct.

Given the circle graph,

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents 11 2

Knowing that the number of people of age 18−29 who voted is 45, Sahil said the number of people of age 30−49 who voted is 135.

He is correct. He did it by multiplying three to the number just like their percentages are got

Go Math Grade 6 Exercise 8.3 Understanding Percents Answers

Go Math Grade 6 Exercise 8.3: Understanding Percents Solutions Page 54 Exercise 1 Answer

Here the value is 14. If 14 is some percent of 25, then 14 is the part.

In the problem to find what percentage of 25 is 14, the part is 14.

Page 54 Exercise 2 Answer

In the question statement, ” What percent of 25 is 14″, The part or the value is 14 The total is 25.

In the problem statement, “what percent of 25 is 14”, the total is 25.

Page 54 Exercise 3 Answer

Given: Some percent of 25 is 14

To find: Percentage of 14 part of 25

By using the formula, first, we divide 14 by 25 and then multiply it by 100 to get the percentage.

Divide 14 by 25,

=14/25

=0.56

Multiply it by 100,

=0.56⋅100

=56%

In a statement, what percent of 25 is 14. We got the percentage which is, 56%

Page 54 Exercise 4 Answer

Given: __% of 25 is 14

To find: Write and solve the equation.

We put 56 % in the fill-in-the-blank and solve the equation.

We put 56% in the fill-in-the-blank,

=56 %of 25is 14We check,

=56% of 25

=56/100 ⋅25

=0.56⋅25

=14

We solved 56% of 25 is 14.

Page 54 Exercise 5 Answer

Given: 80% of some number is 16

To write what is the part. If 80% of some number is 16, then 16 is the part.

In the problem to find 80 percentage of some number is 16, the part is 16.

Page 54 Exercise 6 Answer

Given: 80% of some number is 16

To find :The total number whose 80% is 16

By using the formula, we divide the 16 by 80% to get the total number.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents e6

In the statement, 80% of what number is 16, what number or total came as 20.

Go Math Grade 6 Exercise 8.3: Understanding Percents Solutions Page 54 Exercise 7 Answer

Given: 80% of some number is 16

We need to write the percent In the statement, 80% of some number is 16, 80 is the percentage.

In the given statement of question , 80 is the percentage.

Page 54 Exercise 8 Answer

Given:80% of ___ is 16

To find: Write and solve the equation.

We put 20 in the fill-in-the-blank and solve it.

Given,80% of _______ is 16.

We put 20 in the fill-in-the-blank, = 80% of 20 is 16 We check,

=80 % of 20

=0.8⋅20

=16

Page 54 Exercise 9 Answer

Given that What percent of 20 is 11

As we have all the required values we need, Now we can put them in a simple mathematical formula as below

What percent of 20 is 11, just enter 11÷20×100 and you will get your answer which is55

Y=11/20×100/100=55/100

Percentage Calculator:11 is what percent of 20=55.

Page 54 Exercise 10 Answer

Given that 18 is 45 of what number?

As we have all the required values we need, Now we can put them in a simple mathematical formula as below:

You can easily calculate 18 is 45 percent of what number by, simply enter 18×100÷45 and you will get your answer which is 40

Solutions For Go Math Grade 6 Exercise 8.3 Understanding Percents

Go Math Grade 6 Exercise 8.3: Understanding Percents Solutions Page 54 Exercise 11 Answer

Given that 15 is what percent of 5

By simply dividing equation 1 by equation 2 and taking note of the fact that both the LHS (left-hand side) of both equations have the same unit percent we have 5 is what percent of 15=33.33

Page 54 Exercise 12 Answer

Given that 75 of what number is 105/75 of 140 is105.

You can prove this easily, divide 105 by 75 to find 1 percent (1.4) then simply multiply by100

Percentage: What is 75

percent of 105=78.75

Go Math Grade 6 Understanding Percents Exercise 8.3 Key

Go Math Answer Key

 

Go Math Grade 6 Exercise 8.2: Understanding Percents Solutions

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 8 Percents

Page 51 Problem 1 Answer

Given: The percent was given as 0.17

Find: Here, Write each decimal as a percent.

Solution: To convert a decimal to a percent, move the decimal point two places to the right and then write the % symbol after the number.

Therefore: 0.17=17%

The result of converting a decimal to a percent,17%

Page 51 Problem 2 Answer

The given decimal number is 0.56 .We have to write the given decimal number in percentage form.

Multiply by 100 to convert a number from decimal to percent then add a percent sign %.

We have to multiply the given decimal number by 100 to convert it into percent and then add the percent sign.

​⇒0.56×100=56/100×100=56%

0.56 is written as 56%.

Go Math Grade 6 Exercise 8.2: Understanding Percents Solutions Page 51 Problem 3 Answer

The given decimal number is 0.04. We have to write the given decimal number in percentage form.

Multiply by 100 to convert a number from decimal to percent then add a percent sign %.

We have to multiply the given decimal number by 100 to convert it into percent and then add the percent sign.

⇒0.04×100=4/100×100=4%

​0.04 is written as 4%.

Go Math! Practice Fluency Workbook Grade 6 Chapter 8 Percents Exercise 8.2 Answer Key

Go Math Grade 6 Exercise 8.2 Understanding Percents Answers

Page 51 Problem 4 Answer

The given decimal number is 0.7.We have to write the given decimal number in percentage form.

Multiply by 100 to convert a number from decimal to percent then add a percent sign %.

We have to multiply the given decimal number by 100 to convert it into percent and then add the percent sign.

​⇒0.7×100=7/10×100 = 70%​.

0.7 is written as 70%.

Page 51 Problem 5 Answer

The given decimal number is 0.025. We have to write the given decimal number in percentage form.

Multiply by 100 to convert a number from decimal to percent then add a percent sign %.

We have to multiply the given decimal number by 100 to convert it into percent and then add the percent sign.

​⇒0.025×100=25/1000×100

=25/10

=2.5%

0.025 is written as 2.5%.

Go Math Grade 6 Exercise 8.2: Understanding Percents Solutions Page 51 Problem 6 Answer

The given decimal number is 0.803. We have to write the given decimal number in percentage form.

Multiply by 100 to convert a number from decimal to percent then add a percent sign %.

We have to multiply the given decimal number by 100 to convert it into percent and then add the percent sign.

⇒0.803×100=803/1000×100=803/10=80.3%

0.803 is written as 80.3%.

Page 51 Problem 7 Answer

The given decimal number is 1.3. We have to write the given decimal number in percentage form.

Multiply by 100 to convert a number from decimal to percent then add a percent sign %.

We have to multiply the given decimal number by 100 to convert it into percent and then add the percent sign.

​⇒1.3×100 = 13/10×100.

=13×10

=130%

1.3 is written as 130%.

Page 51 Problem 8 Answer

The given decimal number is 2.10. We have to write the given decimal number in percentage form.

Multiply by 100 to convert a number from decimal to percent then add a percent sign %.

We have to multiply the given decimal number by 100 to convert it into percent and then add the percent sign.

​⇒2.1×100=21/10×100

=21×10

=210%

2.1 is written as 210%.

Solutions For Go Math Grade 6 Exercise 8.2 Understanding Percents

Go Math Grade 6 Exercise 8.2: Understanding Percents Solutions Page 51 Problem 9 Answer

The given fraction is 13/50. We have to write the given decimal number in percentage form.

Multiply by 100 to convert a fraction to percent then add a percent sign %.

We have to multiply the given fraction by 100 to convert it into percent and then add the percent sign.

​⇒13/50×100=13×2

=26%

13/50 is written as 26%.

Page 51 Problem 10 Answer

The given fraction is 3/5. We have to write the given decimal number in percentage form.

Multiply by 100 to convert a fraction to percent then add a percent sign %.

We have to multiply the given fraction by 100 to convert it into percent and then add the percent sign.

⇒3/5×100=3×20

=60%

3/5 is written as 60%.

Page 51 Problem 11 Answer

The given fraction is 3/20. We have to write the given decimal number in percentage form.

Multiply by 100 to convert a fraction to percent then add a percent sign %.

We have to multiply the given fraction by 100 to convert it into percent and then add the percent sign.

​⇒3/20×100=3×5

=15%

3/20 is written as 15%.

Go Math Grade 6 Exercise 8.2: Understanding Percents Solutions Page 51 Problem 12 Answer

The given fraction is 127/100.We have to write the given decimal number in percentage form.

Multiply by 100 to convert a fraction to percent then add a percent sign %.

We have to multiply the given fraction by 100 to convert it into percent and then add the percent sign.

⇒127/100×100=127%.

127/100 is written as 127%.

Page 51 Problem 13 Answer

The given fraction is 5/8. We have to write the given decimal number in percentage form.

Multiply by 100 to convert a fraction to percent then add a percent sign %.

We have to multiply the given fraction by 100 to convert it into percent and then add the percent sign.

​⇒5/8×100

=0.625×100

=62.5%

5/8 is written as 62.5%.

Page 51 Problem 14 Answer

The given fraction is 45/90. We have to write the given decimal number in percentage form.

Multiply by 100 to convert a fraction to percent then add a percent sign %.

We have to multiply the given fraction by 100 to convert it into percent and then add the percent sign.

​⇒45/90×100

=45/9×10

=5×10

=50%

45/90 is written as 50%.

Go Math Grade 6 Understanding Percents Exercise 8.2 Key

Go Math Grade 6 Exercise 8.2: Understanding Percents Solutions Page 51 Problem 15 Answer

The given fraction is 7/5. We have to write the given decimal number in percentage form.

Multiply by 100 to convert a fraction to percent then add a percent sign %.

We have to multiply the given fraction by 100 to convert it into percent and then add the percent sign.

​⇒7/5×100

=7×20

=140%

7/5 is written as 140%.

Page 51 Problem 16 Answer

The given fraction is 19/25. We have to write the given decimal number in percentage form.

Multiply by 100 to convert a fraction to percent then add a percent sign %.

We have to multiply the given fraction by 100 to convert it into percent and then add the percent sign.

​⇒19/25×100

=19×4

=76%

19/25 is written as 76%.

Go Math Grade 6 Exercise 8.2: Understanding Percents Solutions Page 51 Problem 17 Answer

Three numbers are given: 0.3,19/50,22%. We have to arrange the given numbers in increasing order.

We can convert all the given numbers either as fractions or decimal numbers or as a percent.

The given first number is 0.3. To convert the decimal number in percent, we have to multiply it by 100 and then add percent sign, “%”.

​⇒0.3×100

=3/10×100

=3×10

=30%

So, 0.3 is 30%.

The second given number is 19/50.

To convert the fraction in percent, we have to multiply it by 100 and then add percent sign, “%”.

​⇒19/50×100

=19×2

=38%.

So, 19/50 is 38%.

The given third number is 22% which is already a percent. Now, on comparing all the three percents, we get

∴38%>30%>22%.

⇒19/50>0.3>22%.

The final order of the numbers from the least to greatest is 22%<0.3<19/50.

Page 51 Problem 18 Answer

Three numbers are given: 11%,1/8,2/25. We have to arrange the given numbers in increasing order.

We can convert all the given numbers either as fractions or decimal numbers or as a percent.

The given first number 11% is already a percent. So, we have to convert the remaining two fractions, 1/8&2/25 to percent now.

∴To convert the fraction in percent, we have to multiply it by 100 and then add percent sign, “%”.

⇒1/8×100

=25/2%

=12.5%

and ​2/25×100

=2×4

=8%.

Now, on comparing all the three percents, we get

∴8%<11%<12.5%,

⇒2/25<11%<1/8.

Page 51 Problem 19 Answer

Three numbers are given: 5/8,0.675,5%.

We have to arrange the given numbers in increasing order.We can convert all the given numbers either as fractions or decimal numbers or as a percent.

The first given number is 5/8. To convert the fraction in percent, we have to multiply it by 100 and then add the percent sign, “%”.

⇒5/8×100

=5×12.5

=62.5%.

So, 5/8 is 62.5%.

The given second number is 0.675. To convert the decimal number in percent, we have to multiply it by 100 and then add the percent sign, “%”.

​⇒0.675×100

=675/1000×100

=675/10%

=67.5%.

So, 0.675 is 67.5%.

The given third number is 5% which is already a percent. Now, on comparing all the three percents, we get

∴5%<62.5%<67.5%,

⇒5%<5/8<0.675.

The final order of the numbers from the least to greatest is 5%<5/8<0.675.

Go Math Grade 6 Exercise 8.2: Understanding Percents Solutions Page 51 Problem 20 Answer

Three numbers are given: 1.25,0.51,250% . We have to arrange the given numbers in increasing order.

We can convert all the given numbers either as fractions or decimal numbers or as a percent.

The first two given numbers are 1.25&0.51 which are decimal numbers.

To convert the decimal number into percent, we have to multiply it by 100 and then add the percent sign, “%”.

⇒1.25×100

=125/100×100

=125%

and

⇒0.51×100

=51/100×100

=51%.

The given third number is 250% which is already a percent. Now, on comparing all the three percents, we get

∴51%<125%<250%,

⇒0.51<1.25<250%.

The final order of the numbers from the least to greatest is 0.51<1.25<250%.

Page 51 Problem 21 Answer

Three numbers are given: 350/100,0.351,27%. We have to arrange the given numbers in increasing order.

We can convert all the given numbers either as fractions or decimal numbers or as percent.

The first given number is 350/100.

To convert the fraction in percent, we have to multiply it by 100 and then add the percent sign, “%”.

⇒350/100×100=350%

So, 350/100 is 350%.

The given second number is 0.351.

To convert the decimal number into percent, we have to multiply it by 100 and then add the percent sign, “%”.

⇒0.351×100

=351/1000×100

=351/10%

=35.1%.

So, 0.351/ is 35.1%.

The given third number is 27% which is already a percent. Now, on comparing all the three percents, we get

∴27%<35.1%<350%,

⇒27%<0.351<350/100.

The final order of the numbers from the least to greatest is 27%<0.351<350/100.

Page 51 Problem 22 Answer

Given: – There are three given numbers 4/8,0.05,51%. We have to order these number from the least to greatest.

We can order the numbers by simplifying the fractions and percentages until it becomes a number with a decimal point which makes comparison easier.

Simplifying the first fraction 4/8
=2⋅2/2⋅2⋅2

=1/2

=0.5

The second number is already in its decimal representation. There is no further simplification possible.

The third number is in its percentage form. Using the definition of percentage, we can express the number as 51%=51/100

=0.51

Now, comparing the three numbers, we have 0.51>0.5>0.05

This implies 51%>4/8>0.05

The final order of the numbers from the least to greatest is 0.05<4/8<51%.

Go Math Grade 6 Exercise 8.2: Understanding Percents Solutions Page 51 Problem 23 Answer

Given: – During one hour, 6 out of 25 cars were travelling above the speed limit.

We have to find what percent of the cars were traveling above the speed limit.

Multiplying the numerator and denominator with a number that can make the denominator equal to 100.

The corresponding numerator is the required percentage.

The fraction form for the cars travelling above the speed limit is 6/25.

Multiplying the numerator and denominator with 4, we get

=6⋅4/25⋅4

=24/100

The numerator of the fraction with denominator 100 is the required percentage. 6/25=24%

The percentage of the cars that were travelling above the speed limit is 24%.

Page 51 Problem 24 Answer

Given: – At Oaknoll School, 90 out of 270 students own computers.

We have to find the percentage of students who do not own computers.

Multiplying the numerator and denominator with a number that can make the denominator equal to 100.

The corresponding numerator is the required percentage.

The number of students who do not use computers is 270−90=180.

The fraction form for the number of students who don’t use the computers is 180/270.

Simplifying the fraction, we get 90⋅2/90⋅3=2/3

The numerator of the fraction with denominator 100 is the required percentage. After rounding off to the nearest tenth, the percentage will be 2/3=66.7%

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents 24

The percent of students at Oaknoll School that do not own computers (rounded to the nearest tenth) is 66.7%.

Page 52 Exercise 1 Answer

Given: – The given number in decimal form is 0.34.

We have to find the percentage form of this decimal.

By appropriately shifting the decimal point, we can express a decimal in its percentage form.

The given number has its decimal point in between 0 and 3.

After shifting the decimal point two places to the right, the number becomes 034.0

Now adding the percentage symbol, the final value in percentage form is 34%.

Writing the decimal as a percent, we get 34%.

Page 52 Exercise 2 Answer

Given: – The given number in decimal form is 0.06.

We have to find the percentage form of this decimal.

By appropriately shifting the decimal point, we can express a decimal in its percentage form.

The given number has its decimal point in between 0 and 0.

After shifting the decimal point two places to the right, the number becomes 006.0 (or) 6.

Now adding the percentage symbol, the final value in percentage form is 6%.

Writing the decimal as a percent, we get 6%.

Detailed Solutions For Go Math Grade 6 Exercise 8.2 Understanding Percents

Go Math Grade 6 Exercise 8.2: Understanding Percents Solutions Page 52 Exercise 3 Answer

Given: – The given number in decimal form is 0.93.

We have to find the percentage form of this decimal.

By appropriately shifting the decimal point, we can express a decimal in its percentage form.

The given number has its decimal point in between 0 and 9.

After shifting the decimal point two places to the right, the number becomes 093.0 (or) 93.

Now adding the percentage symbol, the final value in percentage form is 93%.

Writing the decimal as a percent, we get 93%.

Page 52 Exercise 4 Answer

Given: – The given number in decimal form is 0.57.

We have to find the percentage form of this decimal.

By appropriately shifting the decimal point, we can express a decimal in its percentage form.

The given number has its decimal point in between 0 and 5.

After shifting the decimal point two places to the right, the number becomes 057.0 (or) 57.

Now adding the percentage symbol, the final value in percentage form is 57 %.

Writing the decimal as a percent, we get 57%.

Page 52 Exercise 5 Answer

Given: – The given number in decimal form is 0.8.

We have to find the percentage form of this decimal.

By appropriately shifting the decimal point, we can express a decimal in its percentage form.

The given number has its decimal point in between 0 and 8.

After shifting the decimal point two places to the right, the number becomes 080.0 (or) 80.

Now adding the percentage symbol, the final value in percentage form is 80%.

Writing the decimal as a percent, we get 80%.

Go Math Grade 6 Exercise 8.2: Understanding Percents Solutions Page 52 Exercise 6 Answer

Given: – The given number in decimal form is 0.734.

We have to find the percentage form of this decimal.

By appropriately shifting the decimal point, we can express a decimal in its percentage form.

The given number has its decimal point in between 0 and 7.

After shifting the decimal point two places to the right, the number becomes 073.4 (or) 73.4.

Now adding the percentage symbol, the final value in percentage form is 73.4 %.

Writing the decimal as a percent, we get 73.4%.

Page 52 Exercise 7 Answer

Given: – The given number in decimal form is 0.082.

We have to find the percentage form of this decimal.

By appropriately shifting the decimal point, we can express a decimal in its percentage form.

The given number has its decimal point in between 0 and 0.

After shifting the decimal point two places to the right, the number becomes 008.2 (or) 8.2.

Now adding the percentage symbol, the final value in percentage form is 8.2%

Writing the decimal as a percent, we get 8.2%.

Page 52 Exercise 8 Answer

Given: – The given number in decimal form is 0.225.

We have to find the percentage form of this decimal.

By appropriately shifting the decimal point, we can express a decimal in its percentage form.

The given number has its decimal point in between 0 and 2.

After shifting the decimal point two places to the right, the number becomes 022.5 (or) 22.5.

Now adding the percentage symbol, the final value in percentage form is 22.5%

Writing the decimal as a percent, we get 22.5%.

Go Math Grade 6 Exercise 8.2: Understanding Percents Solutions Page 52 Exercise 9 Answer

Given: – The given number in decimal form is 0.604.

We have to find the percentage form of this decimal.

By appropriately shifting the decimal point, we can express a decimal in its percentage form.

The given number has its decimal point in between 0 and 6.

After shifting the decimal point two places to the right, the number becomes 060.4 (or) 60.4.

Now adding the percentage symbol, the final value in percentage form is 60.4%.

Writing the decimal as a percent, we get 60.4%.

Page 52 Exercise 10 Answer

Given: – The given number in decimal form is 0.09.

We have to find the percentage form of this decimal.

By appropriately shifting the decimal point, we can express a decimal in its percentage form.

The given number has its decimal point in between 0 and 0.

After shifting the decimal point two places to the right, the number becomes 009.0 (or) 9.

Now adding the percentage symbol, the final value in percentage form is 9%.

Writing the decimal as a percent, we get 9%.

Page 52 Exercise 11 Answer

Given: – The given number in decimal form is 0.518.

We have to find the percentage form of this decimal.

By appropriately shifting the decimal point, we can express a decimal in its percentage form.

The given number has its decimal point in between 0 and 5.

After shifting the decimal point two places to the right, the number becomes 051.8 (or) 51.8.

Now adding the percentage symbol, the final value in percentage form is 51.8 %.

Writing the decimal as a percent, we get 51.8%.

Page 52 Exercise 12 Answer

Given: – The given number in decimal form is 1.03.

We have to find the percentage form of this decimal.

By appropriately shifting the decimal point, we can express a decimal in its percentage form.

The given number has its decimal point in between 1 and 0.

After shifting the decimal point two places to the right, the number becomes 103.0 (or) 103.

Now adding the percentage symbol, the final value in percentage form is 103%.

Writing the decimal as a percent, we get 103%.

Go Math Grade 6 Exercise 8.2: Understanding Percents Solutions Page 52 Exercise 13 Answer

Given: – The given fraction is 3/10.

We have to express the fraction as a percentage.

Multiplying the numerator and denominator with a number that can make the denominator equal to 100.

The corresponding numerator is the required percentage.

Multiply the numerator and the denominator with 10.

3⋅10/10⋅10=30/100

The numerator of the fraction with denominator 100 is the required percentage.

3/10=30%

Writing the fraction as a percent, we get 30%.

Page 52 Exercise 14 Answer

Given: – The given fraction is 2/50.

We have to express the fraction as a percentage.

Multiplying the numerator and denominator with a number that can make the denominator equal to 100.

The corresponding numerator is the required percentage.

Multiply the numerator and denominator with 2.

2⋅2/50⋅2=4/100

The numerator of the fraction with denominator 100 is the required percentage.

4/100=4%

Writing the fraction as a percentage, we get 4%.

Page 52 Exercise 15 Answer

Given: – The given fraction is 7/20.

We have to express the fraction as a percentage.

Multiplying the numerator and denominator with a number that can make the denominator equal to 100.

The corresponding numerator is the required percentage.

Multiply the numerator and denominator by 5.

7⋅5/20⋅5=35100

The numerator of the fraction with denominator 100 is the required percentage.

7/20=35%

Writing the fraction as a percentage, we get 35%.

Practice Problems For Go Math Grade 6 Exercise 8.2 Understanding Percents

Go Math Grade 6 Exercise 8.2: Understanding Percents Solutions Page 52 Exercise 16 Answer

Given: – The given fraction is 1/5.

We have to express the fraction as a percentage.

Multiplying the numerator and denominator with a number that can make the denominator equal to 100.

The corresponding numerator is the required percentage.

Multiply the numerator and denominator with 20.

1⋅20/5⋅20=20/100

The numerator of the fraction with denominator 100 is the required percentage.

1/5=20%

Writing the fraction as a percentage, we get 20%.

Page 52 Exercise 17 Answer

Given: – The given fraction is 1/8.

We have to express the fraction as a percentage.

Multiplying the numerator and denominator with a number that can make the denominator equal to 100.

The corresponding numerator is the required percentage.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents17

The numerator of the fraction with denominator 100 is the required percentage.

1/8=12.5%

Writing the fraction as a percentage, we get 12.5%.

Page 52 Exercise 18 Answer

Given: The fraction is 3/25

To find: Express the fraction as a percent.Summary: To convert a fraction into a percent, we divide the numerator by the denominator.

Multiply the decimal with 100. Another way is to multiply the numerator and the denominator with some number so that the denominator becomes 100.

The corresponding numerator is the required percentage.

Multiply the numerator and the denominator with 4.

3×4/25×4=12/100

The numerator of the fraction with denominator 100 is the required percentage.

3/25=12%

We can write the fraction 3/25 as 12%.

Go Math Grade 6 Exercise 8.2: Understanding Percents Solutions Page 52 Exercise 19 Answer

Given: The fraction is 3/4

To find: Express the fraction as a percent.Summary: To convert a fraction into a percent, we divide the numerator by the denominator.

Multiply the decimal thus obtained with 100.

Divide the numerator by the denominator.

3/4=0.75

Multiply the decimal with 100.

​0.75×100=75

3/4=75%

​We can write the fraction 3/4 as 75%.

Page 52 Exercise 20 Answer

Given: The fraction is 23/50.

To find: Express the fraction as a percent.

Summary: We can convert a fraction into a percent, by first dividing the numerator by the denominator and then multiplying the decimal with 100.

Another way is to multiply the numerator and the denominator with some number so that the denominator becomes 100, and the numerator is the percentage.

Multiply the numerator and the denominator with 2.

23×2/50×2=46/100

The numerator of the fraction with denominator 100 is the required percentage.

23/50=46%

Converting the fraction into percentage we get, 23/50

=46%.

Page 52 Exercise 21 Answer

Given: The fraction is 11/20.

To find: Express the fraction as a percent.

Summary: To convert a fraction into a percent, we divide the numerator by the denominator.

Multiply 100 to the decimal. Or else, multiply the numerator and the denominator with some number so that the denominator becomes 100.

And the corresponding numerator is the required percentage.

Multiply the numerator and the denominator with 5.

11×5/20×5=55/100.

The numerator of the fraction with denominator as 100 is the percentage.

11/20=55%.

Go Math Grade 6 Exercise 8.2: Understanding Percents Solutions Page 52 Exercise 22 Answer

Given: The fraction is 43/50

To find: Express the fraction as a percent.

Summary: First we divide the numerator by the denominator. Multiply the decimal with 100.

Another way is to multiply the numerator and the denominator with some number so that the denominator becomes 100.

Then the corresponding numerator becomes the required percentage.

Multiply the numerator and the denominator with 2.

43×2/50×2=86/100.

The numerator of the fraction with denominator as 100 is the required percentage.

43/50=86%.

Converting the fraction into percentage, we get, 43/50=86%.

Page 52 Exercise 23 Answer

Given: The fraction is 24/25

To find: Express the fraction as a percent.Summary: In order to convert a fraction into a percent, divide the numerator by the denominator.

Multiply the decimal with 100. Alternate way is to multiply the numerator and the denominator with some number so that the denominator becomes 100, so that the corresponding numerator is the required percentage.

Multiply both numerator and the denominator with 4.

24×4/25×4=96/100.

The required percentage is the numerator of the fraction with denominator as 100.

24/25=96%.

Converting the fraction to the percentage we get,

24/25=96%.

Step-By-Step Solutions For Go Math Grade 6 Understanding Percents Exercise 8.2

Go Math Grade 6 Exercise 8.2: Understanding Percents Solutions Page 52 Exercise 24 Answer

Given: The fraction is 7/8.

To find: Express the fraction as a percent.

Summary: To convert a fraction into a percent, first we divide the numerator by the denominator.

Then multiply the decimal with 100. Another way is to multiply the numerator and the denominator with some number so that the denominator becomes 100.

The corresponding numerator is the required percentage.

Divide the numerator by the denominator.

7/8=0.875

Multiply the decimal with 100.

0.875×100=87.5

7/8=87.5%.

We can write the given fraction 7/8 as 87.5%

Go Math Answer Key

 

Go Math Grade 6 Exercise 8.1: Understanding Percents Solutions

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 8 Percents

Page 49 Problem 1 Answer

Given:- We have30%

To Find:- Write each percent as a fraction in simplest form and as a decimal to the nearest hundredth.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents 1

30% can be written in simplest form as 3/10 and in decimal form will be 0.3.

Page 49 Problem 2 Answer

Given:- We have 42%.

To Find:- Write each percent as a fraction in simplest form and as a decimal to the nearest hundredth.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents 2

42% can be written in the simplest form as 21/50 and in the decimal form it will be 0.42.

Go Math! Practice Fluency Workbook Grade 6 Chapter 8 Percents Exercise 8.1 Answer Key

Go Math Grade 6 Exercise 8.1 Understanding Percents Solutions

Go Math Grade 6 Exercise 8.1: Understanding Percents Solutions Page 49 Problem 3 Answer

Given:- We have 18%.

To Find:- Write each percent as a fraction in simplest form and as a decimal to the nearest hundredth.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents 3

18% can be written in simplest form as 9/50 and in the decimal form will be 0.18.

Page 49 Problem 4 Answer

Given:- We have 35%.

To Find:- Write each percent as a fraction in simplest form and as a decimal to the nearest hundredth.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents 4

35% can be written in the simplest form as 7/20 and in decimal 0.35.

Page 49 Problem 5 Answer

Given;- We have 100 %.

To Find:- Write each percent as a fraction in simplest form and as a decimal to the nearest hundredth.

The simplest form of 100% is1/1 and the decimal form is1.00

Go Math Grade 6 Exercise 8.1: Understanding Percents Solutions Page 49 Problem 6 Answer

Given:- We have 29 %.

To Find:- Write each percent as a fraction in simplest form and as a decimal to the nearest hundredth.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents 6

The simplest form of 29%  is 29/100 and in decimal, it will be represented as 0.29.

Page 49 Problem 7 Answer

Given:- We have 56%

To Find:- Write each percent as a fraction in simplest form and as a decimal to the nearest hundredth.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents 7

Simplest form of 56% is 14/25 and in the decimal form 0.56.

Page 49 Problem 8 Answer

Given: There is a given equation 66×2/3%

Find: Write each percent as a fraction in simplest form and as a decimal to the nearest hundredth. Approach: Here, A percent is a portion of 100

so to write a percent as a fraction, Write the percent in the numerator and 100 in the denominator.

A percent is a portion of 100 so to write a percent as a fraction, write the percent in the numerator and 100 in the denominator. Then reduce if possible:

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents 8 1

To write 66×2/3% as a decimal, we need to convert 2/3 to decimal form by dividing 2 and 3.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents 8 2

Since we need to round to the nearest hundredth, divide up to 3 decimal places:

Rounding to the nearest hundredth then gives 66×2/3%≈0.67.

Go Math Grade 6 Exercise 8.1 Understanding Percents Answers

Page 49 Problem 9 Answer

Given: The given equation is 25%

Find: Here, percent as a fraction in simplest form and as a decimal to the nearest hundredth.

Approach: So to write a percent as a fraction, write the percent in the numerator and 100 in the denominator.

A percent is a portion of 100 so to write a percent as a fraction, write the percent in the numerator and 100 in the denominator. Then reduce if possible:

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents 9

Since 25/100 = 0.25, then 25% as a decimal is 0.25 1/4=​0.25

Go Math Grade 6 Exercise 8.1: Understanding Percents Solutions Page 49 Problem 10 Answer

Given: The equation is 0.03

Find: Write each decimal or fraction as a percent. Approach: we need to first write the decimal as a fraction that has a denominator of 100.

The percent is then the numerator of the fraction,

A percent is a portion of 100 so to write a decimal as a percent, we need to first write the decimal as a fraction that has a denominator of 100.

The percent is then the numerator of the fraction: ​0.03=3/100=3%

The percent is then the numerator of the fraction,3%

Page 49 Problem 11 Answer

Given: The given percent is 0.92

Find: Write each decimal or fraction as a percent. Approach: A percent is a portion of 100

so to write a decimal as a percent, we need to first write the decimal as a fraction that has a denominator of 100.

A percent is a portion of 100 so to write a decimal as a percent, we need to first write the decimal as a fraction that has a denominator of 100.

The percent is then the numerator of the fraction: ​0.92=92/100=92%

The percent is then the numerator of the fraction,92%

Page 49 Problem 12 Answer

Given: The equation show percent 0.18
Find: Write each decimal or fraction as a percent. so to write a decimal as a percent approach: We need to first write the decimal as a fraction that has a denominator of 100.

The percent is then the numerator of the fraction. A percent is a portion of 100

so to write a decimal as a percent, we need to first write the decimal as a fraction that has a denominator of 100.

The percent is then the numerator of the fraction: ​0.18=18/100=18%

The percent is then the numerator of the fraction,18%

Page 49 Problem 13 Answer

Given: The equation given percent is 2/5

Find: Here, decimal or fraction as a percent. Approach: We first need to find an equivalent fraction that has a denominator of 100.

A percent is a portion of 100 so to write a fraction as a percent, we first need to find an equivalent fraction that has a denominator of 100.

The percent is then the numerator of the equivalent fraction:

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents 13

The percent is then the numerator of the equivalent fraction,40%

Solutions For Go Math Grade 6 Exercise 8.1 Understanding Percents

Go Math Grade 6 Exercise 8.1: Understanding Percents Solutions Page 49 Problem 14 Answer

Given: here equation shows the percent 23/25

Find: Write each decimal or fraction as a percent. Approach: write a fraction as a percent, we first need to find an equivalent fraction that has a denominator of 100.

A percent is a portion of 100 so to write a fraction as a percent, we first need to find an equivalent fraction that has a denominator of 100.

The percent is then the numerator of the equivalent fraction:

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents 14

The percent is then the numerator of the equivalent fraction,92%

Page 49 Problem 15 Answer

Given: Here we have given that the percent is 7/10

Find: Write each decimal or fraction as a percent. Approach: A percent is a portion of 100

so to write a fraction as a percent, We first need to find an equivalent fraction.

A percent is a portion of 100 so to write a fraction as a percent, we first need to find an equivalent fraction that has a denominator of 100.

The percent is then the numerator of the equivalent fraction:

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents 15

The percent is then the numerator of the equivalent fraction, 70%

Page 49 Problem 16 Answer

Given: Bradley completed 3/5 of his homework.

Find: What percent of his homework does he still need to complete? Approach: It is given that Bradley has completed 0/5 his homework and we need to find what percent he has left to complete.

A percent is a portion of 100 so to write a fraction as a percent, we first need to find an equivalent fraction that has a denominator of 100.

The percent is then the numerator of the equivalent fraction:

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents 16

Since he has completed 60%, then he has 100%−60%=40% left to complete. 40%

Go Math Grade 6 Exercise 8.1: Understanding Percents Solutions Page 49 Problem 17 Answer

Given: After reading a book for English class, 100 students were asked whether or not they enjoyed it.

Nine twenty-fifths of the class did not like the book.Find: How many students liked the book? Approach: It is given that 100 students were asked whether or not they enjoyed a book and nine twenty-fifths.

It is given that 100 students were asked whether or not they enjoyed a book and nine twenty-fifths, which is the fraction 9/25

said they did not like the book. We need to find the number of students who did like the book.

First, we can use the given fraction to find the number of students who did not like the book:

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents 17

36 students did not like the book, then 100−36=64 students did like the book.

Page 49 Problem 18 Answer

Given: At a concert, 20% of the people are wearing black dresses or suits, 1/4 are wearing navy, 0.35 are wearing brown, and the rest are wearing a variety of colors (other).

Find: Write the percent, fraction, and decimal for each color of clothing. Approach: It is given that 20% wore black, 1/4 wore navy, 0.35 wore brown, and the rest wore other colors.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents 18

It is given that 20% wore black, 1/4

wore navy, 0.35 wore brown, and the rest wore other colors. We need to write the percent, fraction, and decimal for each color of clothing. Black

Other To find the percent, we must first find the combined percent for black, navy, and brown. Adding the percentages gives 20%+25%+35%=80%.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents 18

The percent for others is then 100%−80%=20%.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents 18 2

Since Other and Black have the same percent, they must also have the same fraction and decimal. The fraction and decimal for Other is then 1/5 and 0.2.

Go Math Grade 6 Exercise 8.1: Understanding Percents Solutions Page 50 Exercise 1 Answer

Given: The percent gives as 43%

Find: Here, percent as a fraction in the simplest form. Approach: We can write a fraction with the percent as the numerator and 100 as the denominator.

A percent is a portion of 100 so to write a percent as a fraction, we can write a fraction with the percent as the numerator and 100 as the denominator.

Then we can reduce if possible: 43%=43/100

Simplest form of the percentage is 43/100

Page 50 Exercise 2 Answer

Given: Here, we have given percent of the equation 72% Find: Here, Each percent is a fraction in simplest form.

Approach: A percent is a portion of 100 so writing a percent as a fraction.

A percent is a portion of 100 so to write a percent as a fraction, we can write a fraction with the percent as the numerator and 100 as the denominator.

Then we can reduce if possible:

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents e2

Simplest form of the percentage is 18/25

Page 50 Exercise 3 Answer

Given: Here, we have given percent is 88%

Find: Write each percent as a fraction in the simplest form.Approach: write a fraction with the percent as the numerator and 100 as the denominator.

A percent is a portion of 100 so to write a percent as a fraction, we can write a fraction with the percent as the numerator and 100 as the denominator.

Then we can reduce if possible:

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents e3

Simplest form of the percentage is=22/25

Go Math Grade 6 Understanding Percents Exercise 8.1 Key

Go Math Grade 6 Exercise 8.1: Understanding Percents Solutions Page 50 Exercise 4 Answer

Given: We have the percent of the equation is 35%

Find: Write each percent as a fraction in the simplest form.

Approach: A percent is a portion of 100 so to write a percent as a fraction, we can write a fraction.

A percent is a portion of 100 so to write a percent as a fraction, we can write a fraction with the percent as the numerator and 100 as the denominator.

Then we can reduce if possible:

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 8 Percents e4

The simplest form of the percentage is 7/20

Page 50 Exercise 5 Answer

Given: The percent is given 64%

Find: Write each percent as a decimal. Approach: so to write a percent as a decimal, we can write a fraction with the percent as the numerator and 100 as the denominator.

A percent is a portion of 100 so to write a percent as a decimal, we can write a fraction with the percent as the numerator and 100 as the denominator.

Then write the fraction as a decimal: ​64%= 64/100 = 0.64

The simplest form of the percentage is 0.64

Page 50 Exercise 6 Answer

Given: Here, The given percent is 92%

Find: Here, Write each percent as a decimal. Approach: A percent is a portion of 100 so write a percent as a decimal.

A percent is a portion of 100 so to write a percent as a decimal, we can write a fraction with the percent as the numerator and 100 as the denominator.

Then write the fraction as a decimal: ​92% = 92/100 = 0.92

The simplest form of the percentage is 0.92

Go Math Grade 6 Exercise 8.1: Understanding Percents Solutions Page 50 Exercise 7 Answer

Given: Here, we have the percent is 73%

Find: Write each percent as a decimal. Approach: write a percent as a decimal, we can write a fraction with the percent as the numerator and 100 as the denominator.

A percent is a portion of 100 so to write a percent as a decimal, we can write a fraction with the percent as the numerator and 100 as the denominator.

Then write the fraction as a decimal: ​73% = 73/100 = 0.73

The simplest form of the percentage is 0.73

Page 50 Exercise 8 Answer

Given: Here, The percent is 33%

Find: Write each percent as a decimal. Approach: A percent is a portion of 100

so to write a percent as a decimal, We can write a fraction with the percent as the numerator and 100 as the denominator.

A percent is a portion of 100 so to write a percent as a decimal, we can write a fraction with the percent as the numerator and 100 as the denominator.

Then write the fraction as a decimal: ​33% = 33/100 = 0.33

The simplest form of the percentage is 0.33

Detailed Solutions For Go Math Grade 6 Exercise 8.1 Understanding Percents

Go Math Answer Key

 

Go Math Grade 6 Exercise 5.5: Operations with Decimals Solutions

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 5 Operations with Decimals

Page 33 Problem 1 Answer

Given: It is given that the four friends equally shared the cost of supplies for a picnic. The total cost of supplies is $12.40

To find: We have to find how much money each friend paid.We will find the money each person paid by dividing the total cost by four.

Firstly, we will remove the decimal point from the denominator of the division expression by multiplying both the denominator and numerator by 10.

We have, 12.4/4 = 124/40

The division of the given cost by four is

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals 1

Each friend has to pay $3.1

Page 33 Problem 2 Answer

Given: It is given that there are twenty people who are going to the movie by van. Each van seats eight people.

To find: We have to find the total number of vans required. We will divide twenty by eight to get the answer.

The division of the given data is

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals 2

As there can not be a half van. So, three vans should be available.

The total number of vans required for twenty people is 3

Go Math! Practice Fluency Workbook Grade 6 Chapter 5 Operations with Decimals Exercise 5.5 Answer Key

Page 33 Problem 3 Answer

Given: It is given that we need forty forks and they come in packs of six.To find: We have to find how many packs do we need to buy.We will divide forty by six to get the answer.

The division is

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals 3

As packs must be in whole numbers, so seven packs should be baught.

The number of packs we need to buy if we need forty forks is7

Go Math Grade 6 Exercise 5.5 Operations With Decimals Solutions

Go Math Grade 6 Exercise 5.5: Operations with Decimals SolutionsPage 33 Problem 4 Answer

Given: It is given that Kesha spent a total of$9.60 on new shoelaces and each pair costs $1.20

To find: We have to find how many pairs did she buy. We will divide the total cost by the cost of one pair.

The pairs of shoelaces that she bought are ​9.60/1.20=96/12 =8

The total number of pairs of shoelaces that Kesha bought is 8

Page 33 Problem 5 Answer

Given: It is given that one inch equals 1/4

hand and average Clydesdale is 17×1/5 hands tall. To find: We have to find the height of the horse in inches and in feet.

We will convert the height of Clydesdale in inches with the help of the given relation and then convert it into feet.

We will convert mixed fractions into improper fractions. The height of Clydesdale is 17×1/5=86/5

We will convert this height in hands into height in inches. We get, 86/5×1/4=4.3in

Now, the height in feet is 4.3×1/12=0.3583ft

The height of the horse in inches is 4.3in

The height of the horse in feet is 0.3583ft

Go Math Grade 6 Exercise 5.5: Operations with Decimals Solutions Page 33 Problem 6 Answer

Given: It is given that the banana bread recipe contains 3/4

cup of butter and one tablespoon equals 1/16 cup. To find: We have to find the number of tablespoons of butter needed.

We will divide the total cup of butter required by the butter that one tablespoon contains.

As we want to divide the fractions, we will rewrite the division as multiplication by reciprocating the divisor. We get, 3/4×16=12

The number of tablespoons needed is 12

Page 33 Problem 7 Answer

Given: It is given that Cindy earns $5.75 per hour. She works 12.4 hours per week. She worked50 weeks in a year.

To find: We have to find how much she earned in a year. We will find this by multiplying the pay per hour by the total number of hours per year.

We will first calculate the total number of hours in a year.

We get,12.4×50=620 The total income that she earned in a year is 5.75×620=3565

The total earnings of Cindy per year are$3565

Go Math Grade 6 Exercise 5.5 Operations With Decimals Answers

Go Math Grade 6 Exercise 5.5: Operations with Decimals Solutions Page 33 Problem 8 Answer

Given: Joey scored 9.4,9.7,9.9,9.8 and Carlos scored 9.5,9.2,9.7,9.6

To find who’s average is high. Using the method of average.

To find the average of Joey’s score average = sum of all the numbers / Total numbers

Now to find the average for Carlos’s score,

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals 8 1

The average score of Joey’s is 9.7 and the average score of Carlos’s is 9.5 from this it is clear that the average score of Joey’s is higher than the score of Carlos.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals 8 2

Go Math Grade 6 Exercise 5.5: Operations with Decimals Solutions Page 33 Problem 9 Answer

Given: granola recipe calls for2x1/3

cup of almonds. A bag of almonds contains 2 cups. Ali buys 5 bags of almonds. To find how many cups of almonds will he have left over.

Using the method of multiplication. It is given that a granola recipe calls for 21/3

cups of almonds, each bag of almonds contain2 cups, Ali will make 21/2 batches of granola, and he bought 5bags of almonds.

First, we need to find how many cups of almonds he needs by multiplying the number of cups in each batch and the number of batches he will make,

=2×1/2×2×1/3=35/6  write in mixed fractions,

=5×5/6

To find how many cups are left, we can then subtract the number of cups he bought and the total number of cups he needs,

=10−5×5/6 simplify, = 4×1/6

There are 4×1/6 cups of almonds are left over.

Next, we need to find how many cups he bought by multiplying the number of cups in each bag and the number of bags he bought. He the bought2×5=10 cups of almonds.

Solutions For Go Math Grade 6 Exercise 5.5 Operations With Decimals

Go Math Grade 6 Exercise 5.5: Operations with Decimals Solutions Page 33 Problem 10 Answer

Given: three pandas eat a total of 181×1/2 pounds of bamboo shoot each day.

The male panda eats 3 times as much as the baby and the female panda eats 2 times as much as the baby.

To find how many pounds does the female panda eats. Using the method of division.

It is given that 3 pandas at a zoo each a total of 181×1/2 pounds of bamboo each day, the male eats 3 times as much as the baby, and the female eats 2 times as much as the baby.

Using the given information, we then get the ratio, male: female: baby=3:2:1

To find how much the female panda eats, we first need to write the ratio of females: all 3 pandas.

Since3+2+1=6, then the ratio is,female: all 3 pandas = 2:6

Reducing this ratio by dividing both numbers by 2 gives: female: all3 pandas2:6=2/6 simplify, =1/3

Now we need to write an equivalent ratio that has 181×1/2 = 363/2 as the second number in the ratio.

The female panda eats 60×1/2 pounds of the bamboo shoot aday.

Go Math Grade 6 Exercise 5.5: Operations with Decimals Solutions Page 34 Exercise 1 Answer

Given: Jan has $37.50 and the cost of one ticket is $5.25

To find how many tickets can Jan buy with the amount he has.Using the method of division.

Jan has $37.50 and the cost of one ticket is $5.25  to find how many tickets can he buy with the amount he has, here we should divide the total amount by the cost of one ticket because we are going to find the number of tickets can buy with the total amount so division takes place here.

Now divide the total amount by the cost of one ticket,

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals e1

Jan can buy 7 tickets with the $37.05 money.

Go Math Grade 6 Exercise 5.5: Operations with Decimals Solutions Page 34 Exercise 2 Answer

Given: Jon has $45.00 and he plans to spend 4/3 of his money on sports equipment.To find how much will he spend.Using the method of multiplication.

It is given that Jon has $ 45 and plans to spend 4/5 of his money on sports equipment.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals e2

Since we know the total amount of money he has and what fraction of the total he will spend on sports equipment, we need to multiply to find how much he will spend.

Go Math Grade 6 Operations With Decimals Exercise 5.5 Key

Go Math Grade 6 Exercise 5.5: Operations with Decimals Solutions Page 34 Exercise 3 Answer

Given: Ricki has $ 76.8 feet of cable. She plans to cut it into 7 pieces.To find the length of each cable. Using the method of division.

It is given that Ricki has 76.8 feet of cable and will cut it into 7 pieces.

Since we know the total length and how many pieces it will be cut into, we need to divide to find the length of each piece.

Dividing 76.8 and 7 up to 3 decimal place gives = 76.8/7 divide, =10.971

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals e4

Rounding to the nearest hundredth gives 10.97 so each piece will be about 10.97ft.

The length of the cable is 10.97 ft long.

Go Math Grade 6 Exercise 5.5: Operations with Decimals Solutions Page 34 Exercise 4 Answer

Given: Roger has 2×1/2 cups of butter. The recipe for the loaf of bread requires 3/4 cup of butter.

To find how many loaves can Roger make. Using the method of division.

It is given that Rogger has2x1/2 cups of butter and each recipe for a loaf of bread calls for3/4 cups of butter.

Since we know the total amount of butter he has and how much is needed for each loaf, we need to divide to find how many loaves he can make.

Roger can make 3 loaves of bread with the butter.

Go Math Grade 6 Exercise 5.5 Detailed Solutions

Go Math Answer Key

 

Go Math! Grade 6 Exercise 5.2: Operations with Decimals Solutions

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 5 Operations with Decimals

Page 27 Problem 1 Answer

The given expression is 1.5+2.3.

We are sked to find the sum of the expression.

The answer can be found by adding the digits to the right of the decimal.

Then add the number to the left of the decimal.Add the extras( tens place digit) obtained from adding the values right to the decimal.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals 1

The sum of 1.5+2.3 is 3.8.

Go Math! Grade 6 Exercise 5.2: Operations with Decimals Solutions Page 27 Problem 2 Answer

The given expression is 8.9−5.1.

We are asked to find the difference of the expression. The answer can be found by finding the difference form the right side.

Go Math! Practice Fluency Workbook Grade 6 Chapter 5 Operations with Decimals Exercise 5.2 Answer Key

Find the difference of the numbers right to the decimal.Then find the difference between the number left to the decimal. This will give you the required answer.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals 2

The difference of the expression 8.9−5.1 is 3.8.

Page 27 Problem 3 Answer

The given expression is 2.5+1.3+4.1.

We are asked to find the sum of the expression. The answer can be found by adding the digits to the right of the decimal.

Then add the number to the left of the decimal. Add the extras( tens place digit) obtained from adding the values right to the decimal.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals 3

The sum of the expression 2.5+1.3+4.1 is 7.9.

Go Math Grade 6 Exercise 5.2 Operations With Decimals Solutions

Go Math! Grade 6 Exercise 5.2: Operations with Decimals Solutions Page 27 Problem 4 Answer

The given expression is 7.25+8.75. We are sked to find the sum of the expression. The answer can be found by adding the digits to the right of the decimal.

If you get a two digit value, write the ones digit down and take the tens digit value as the carrier.

Then add the number to the left of the decimal. Add the carrier( tens place digit) obtained from adding the values right to the decimal.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals 4

The sum of the expression 7.25+8.75 is 16.00.

Page 27 Problem 5 Answer

The given expression is 8.16−7.72. We are asked to find the difference of the expression.

The answer can be found by finding the difference form the right side. Find the difference of the numbers right to the decimal.

Then find the difference between the number left to the decimal. This will give you the required answer.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals 5

The difference of the expression 8.16−7.72 is 0.44.

Page 27 Problem 6 Answer

The given numbers are 3.3+4.5+2.6

We need to find the sum or difference of the given question.

We will first add the first two numbers and then with that result add the third number to get the result.

The first two numbers are: – 3.3+4.5=7.8

Now, the addition of the result and the third number is 7.8+2.6=10.4

The answer to the given numbers are 10.4

Go Math! Grade 6 Exercise 5.2: Operations with Decimals Solutions Page 27 Problem 7 Answer

The given numbers are8.9+3.05

We need to find the sum or difference of the given question.

We will first add decimals side of these two numbers and then add the left side of the decimal number.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals 7

The addition of the numbers are11.95

Page 27 Problem 8 Answer

The given numbers are10.64−8.8

We need to find the sum or difference of the given question.

We will first subtract the right side of the decimals of these two numbers and then subtract the left side of the decimal number.

The numbers are: – ​10.64−8.80=1.84

​The subtraction of numbers is1.84

Page 27 Problem 9 Answer

The given numbers are4.1+0.35+6.564

We are asked to find the sum of the expression. The answer can be found by adding the digits to the right of the decimal.

Then add the number to the left of the decimal. Add the extras( tens place digit) obtained from adding the values right to the decimal.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals 9

The addition of the numbers is11.014

Go Math Grade 6 Exercise 5.2 Operations With Decimals Answers

Go Math! Grade 6 Exercise 5.2: Operations with Decimals Solutions Page 27 Problem 10 Answer

Given: -Height of Marcus=1.5m

Height is 0.1 meter taller than Marcus.

Height is 0.2 meter than his sister.

We need to find the height of the father.

We will first find the height of his sister and then by using the height of his sister we will then find the height of the father.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals 10

The height of the father is1.8m

Page 27 Problem 11 Answer

Given: – Jennifer buy a baseball game=24.75

She spent on drinks and snacks=12.45

We need to find that how much money does she have leftover.

We will subtract the amount spent on drinks and snacks from the total buy to find the leftover money.

The total leftover money= Total buy on a baseball game− money spent on drinks and snacks ​=24.75−12.45 =12.30

The total leftover money is $12.30

Go Math! Grade 6 Exercise 5.2: Operations with Decimals Solutions Page 27 Problem 12 Answer

The given numbers are:- 7.089+2.?13=9.502

We need to find the missing digit.

We will use the concept of addition to find the missing digit in the given question.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals 12

The given expression is 7.089+2.?13=9.502​

It can be written as 2.?13=9.502−7.089

Thus we have2.?13=2.413

On comparing both the numbers, it can be concluded that the missing digit is 4

The missing digit is 4

Page 27 Problem 13 Answer

The given numbers are16.594−?.175=11.419

We need to find the missing digit.

We will use the concept of subtraction to find the missing digit.

The given numbers are 16.594−?.175=11.419

16.594−5.175=11.419

The missing digit is5

Solutions For Go Math Grade 6 Exercise 5.2 Operations With Decimals

Go Math! Grade 6 Exercise 5.2: Operations with Decimals Solutions Page 27 Problem 14 Answer

The given numbers are6.2?67+9.75=15.9867

We need to find the missing digit.

We will use the concept of addition to find the missing digit in the given question.

The numbers are: – ​6.2?67+9.75=15.9867

6.2367+9.75=15.9867

The missing digit is 3

Page 27 Problem 15 Answer

Given: – Cost of all three pizzas is $46.24

Cost of medium pizza is $15.75

Cost of large pizza is $17.50

We need to find the cost of small pizza.

We will first add both sizes of pizza i.e., medium and large size pizza, and then subtract the result from the total cost of the pizza to find the small size of pizza.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals 15

The cost of small pizza is $13.

Page 27 Problem 16 Answer

Given: – Length of three sheets of plywood is 6.85 feet long.

Length cut from one sheet is3.4 foot.

Length cut from second sheet is0.5 foot.

We need to find the feet of plywood left in all.

We will cut the length of sheet from first and second sheet as given in the question and then add all the length of plywood to find the overall length.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals 16

The total length of plywood left in all is13.65 feet.

Go Math! Grade 6 Exercise 5.2: Operations with Decimals Solutions Page 28 Exercise 1 Answer

The given numbers are2.6+1.15

We need to find the addition of the given numbers.

We will use the concept of addition to find out the solution to the questions.

The numbers are: – ​2.60 + 1.15 =3.75

The addition of the given two numbers is3.75

Page 28 Exercise 2 Answer

The given numbers are2.53−1.7=?

We need to find the solution to the given question.

We will subtract the second number from the first number from right to left direction subtracting the decimal part first.

The given numbers are: – ​2.53 −1.70 =0.83

The solution is0.83

Page 28 Exercise 3 Answer

The given numbers are4.3+1.4=?
We need to find the sum of the given numbers when you find the sum of two or more numbers, you use the symbol+.

The addition is the arithmetic operation of adding two numbers. An expression is a mathematical statement with terms, operands and operators.

The numbers are:- Tens Ones Tenths Hundredths Thousandths
​4.3
1.4
=5.7
​The sum of the numbers is5.7

Go Math Grade 6 Operations With Decimals Exercise 5.2 Key

Go Math! Grade 6 Exercise 5.2: Operations with Decimals Solutions Page 28 Exercise 4 Answer

The given numbers are 14.4−3.8=?

We need to find the solution to the given question.

We will subtract the second number from the first number from the right to left direction subtracting the decimal part first.

The numbers in a table are

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals e4

The subtraction of the given numbers is10.6

Page 28 Exercise 5 Answer

The given numbers are7.3+8.5=?

We need to find the addition of the given numbers.

We will use the concept of addition to find out the solution to the questions.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals e5

The solution of the question is15.8

Page 28 Exercise 6 Answer

The given numbers are12.34−6.9=?

We need to find the solution to the given question.

We will subtract the second number from the first number from the right to left direction subtracting the decimal part first.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals e6

The solution of the given numbers is5.4

Page 28 Exercise 7 Answer

The given numbers are4.3+1.4=?

We need to find the addition of the given numbers.

We will use the concept of addition to find out the solution to the questions.

The numbers are ​4.3 + 1.4 =5.7

The answer by rounding to the nearest whole number is6

The solution of the given numbers after rounding is6

Go Math! Grade 6 Exercise 5.2: Operations with Decimals Solutions Page 28 Exercise 8 Answer

The given numbers are14.4−3.8=?

We need to find the solution to the given question.

We will subtract the second number from the first number from the right to left direction subtracting the decimal part first.

The numbers are ​14.4 − 3.8 = 10.6

The nearest whole number to10.6 is 11

The answer of the question is11

Page 28 Exercise 9 Answer

The given numbers are7.3+8.5=?

We need to find the addition of the given numbers.

We will use the concept of addition to find out the solution to the questions.

The numbers are 7.3+8.5=15.8

The nearest whole number to15.8 is 16

The solution the given number is16

Go Math! Grade 6 Exercise 5.2: Operations with Decimals Solutions Page 28 Exercise 10 Answer

The given numbers are 12.34−6.9

We need to find the solution to the given question.

We will subtract the second number from the first number from the right to left direction subtracting the decimal part first.

The numbers are ​12.34−6.9 =5.4

The nearest whole number to 5.4 is5

Here, estimated answer is 5 and exact answer is 5.4

So, we can say that estimated answer is less than the exact answer .

On rounding the digit to the nearest whole number , we get 5. And on comparing estimate to the exact answers we must say that estimated answers is less than the exact answer i.e.5.4

Detailed Solutions For Go Math Grade 6 Exercise 5.2 Operations With Decimals

Go Math Answer Key

 

Go Math! Grade 6 Exercise 5.1: Operations with Decimals Solutions

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 5 Operations with Decimals

Page 25 Problem 1 Answer

The given expression is 585÷13.

We are asked to estimate the quotient by rounding off the dividend and the divisor to the largest possible value.

The answer can be found by rounding off the dividend and the divisor to the nearest value.

Here, 13 is the divisor and 585 dividend.

That would be 10&600 respectively. Now, find the value of 600/10

This will give you the required answer.

The expression is 585÷13. Here 585 is closer to 600

Rounding off the value 585, we get 600.

The value 13 is closer to ten. Rounding of 13, we have:10.

Thus, we have the quotient of 600÷10 as 60

The estimated quotient value of 585÷13 by rounding the dividend and the divisor to the largest place value is 60

Page 25 Problem 2 Answer

The given expression is 2,756÷53.

We are asked to find the quotient. The answer can be found by rounding off the dividend and the divisor to the nearest value.

Here, is the divisor and dividend are 53&2756 respectively.

When rounding off, we have 50&3000 Now, find the value of the quotient. This will give you the required answer.

The given expression is: 2,756÷53. The value 2756 is closer to 3000

Rounding the value 2756 to the largest place value, we have: 3000.

Similarly, 53 is closest to 50. Thus, the quotient value is 3000/50=60.

The estimated value of the quotient of 2,756÷53 by rounding the dividend and the divisor to the largest place value is 60

Go Math! Practice Fluency Workbook Grade 6 Chapter 5 Operations with Decimals Exercise 5.1 Answer Key

Go Math Grade 6 Exercise 5.1 Operations With Decimals Solutions

Go Math! Grade 6 Exercise 5.1: Operations with Decimals Solutions Page 25 Problem 3 Answer

The given expression is 22,528÷98.

We are asked to find the value of the quotient when the dividend and the divisor is rounded to the largest value.

The answer can be found by rounding off the dividend and the divisor to the nearest value.

Here, is the divisor and dividend are98&22528 respectively.

When rounding off, we have 100&20,000 Now, find the value of the quotient. This will give you the required answer.

The expression 22,528÷98. The value 22528 is closest to 20,000.

Similarly, the value 98 is closest to 100. Thus, the quotient value is 20,000/100 = 200.

The estimated value of the quotient of the expression 22,528÷98 by rounding the dividend and the divisor to the largest place value is 200.

Page 25 Problem 4 Answer

The given expression is 7,790÷210.

We are asked to find the value of the quotient when the dividend and the divisor is rounded to the largest value.

The answer can be found by rounding off the dividend and the divisor to the nearest value.

Here, is the divisor and dividend are 210&7790 respectively.

When rounding off, we have 200&8000.

Now, find the value of the quotient. This will give you the required answer.

The given expression is 7,790÷210

The value 7790 is closest to 8000.

Similarly, the value 210 is closest to 200. So, the quotient value of 8000/200 =40.

The estimated quotient of the expression 7,790÷210 by rounding the dividend and the divisor to the largest place value is 40.

Go Math! Grade 6 Exercise 5.1: Operations with Decimals Solutions Page 25 Problem 5 Answer

The given expression is 17,658÷360.

We are asked to find the value of the quotient when the dividend and the divisor is rounded to the largest value.

The answer can be found by rounding off the dividend and the divisor to the nearest value. Here, is the divisor and dividend are 360&17658respectively.

When rounding off, we have 400&20,000 Now, find the value of the quotient.

This will give you the required answer.

The expression is 17,658÷360. The value 17658 is closer to 20,000 than 10,000.

Similarly, the value 360 is closest to 400. Thus, the quotient value is: 20,000/400=50.

The estimated quotient value of 17,658÷360 by rounding the dividend and the divisor to the largest place value is 50.

Page 25 Problem 6 Answer

The given expression is 916÷320.

We are asked to find the value of the quotient when the dividend and the divisor is rounded to the largest value.

The answer can be found by rounding off the dividend and the divisor to the nearest value.Here, is the divisor and dividend are320&916 respectively.

When rounding off, we have 300&900 Now, find the value of the quotient.This will give you the required answer.

The expression is 916÷320.

The value 916 is closer to 900 than 1000. Similarly, the value 320 is closer to 300 than 400. Thus, the quotient value of 900/300 =3.

The estimated quotient value of the expression is916÷320 by rounding the dividend and the divisor to the largest place value is 3

Go Math Grade 6 Exercise 5.1 Operations With Decimals Answers

Go Math! Grade 6 Exercise 5.1: Operations with Decimals Solutions Page 25 Problem 7 Answer

The given expression is 1334÷29. We are asked to find the quotient of 1334÷29 by long division method.The answer can be found by writing the number as dq+r.

Then find the quotient for the value 1334 when divided by 29. Perform the operation by long division.

The expression is 1334÷29. 1334 can be written as 29×46+0.

That is, 1334 is divisible by 29. The quotient obtained is: 46

The quotient of 1334÷29 using long division method is 46:

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals 7

Page 25 Problem 8 Answer

The given expression is 20884÷92.

We are asked to find the quotient of the expression by long division method. The answer can be found by writing the number as dq+r.

Then find the quotient for the value 20884 when divided by 92. Perform the operation by long division.

The expression is 20884÷92. Write the expression 20884 in the form dq+r.

Thus, we have: 227×92+0. That is the quotient value is q=227.

The quotient value of 20884÷92 using long division method is:

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals 8

Page 25 Problem 9 Answer

The given expression is 18175÷25. We are asked to find the value of the quotient 18175÷25 using the long division method.

The answer can be found by found by writing the expression in the form dq+r.

Find the value of q when 18175 is divided by 25. Represent this by the long division method.

The expression is 18175÷25. It can be written as 727×25+0.

Hence, the remainder is 0 .So, the quotient value of 18175÷25 is 727.

The value of the quotient for the expression 18175÷25 using the long division method is 727.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals 9

Go Math! Grade 6 Exercise 5.1: Operations with Decimals Solutions Page 25 Problem 10 Answer

The given expression is 2902÷18. We are asked to find the quotient and the remainder value by the method of long division.

The answer can be found by writing the number as dq+r, where q,r are the quotient and remainder.

Find the quotient value of 2902 when divided by 18. Write the answer in the long division form.

The expression is 2902÷18. It can be written as 18×161+4. Here, the remainder is 4.

That is, the number is not perfectly divisible by 18. The quotient is: 161

The quotient and the remainder value of 2902÷18by long division method is 161 and 4 respectively.:

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals 10

Page 25 Problem 11 Answer

The given expression is 34680÷64.We are asked to find the remainder and the quotient of the expression.

The answer can be found by writing the number in dq+r form, where r,q are the remainder and the quotient respectively.

Find the quotient value of 34680 when divided by 64. Similarly, find the remainder. This will give you the required answer.

The expression is 34680÷64. The value 34680 can be written as 34680=64×541+56.

From the above expansion, we have: q=541 and The remainder is r=56.

The remainder and the quotient of the expression 34680÷64 are 56 and 541.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals 11

Solutions For Go Math Grade 6 Exercise 5.1 Operations With Decimals

Go Math! Grade 6 Exercise 5.1: Operations with Decimals Solutions Page 25 Problem 12 Answer

The given expression is 52245÷215. We are asked to find the remainder and the quotient of the expression.

The answer can be found by writing the value as dq+r, where r,q are the remainder and the quotient respectively.

From the expansion, write down the remainder and the quotient value.Reflect the value in the long division form.

The expression is: 52245÷215. Writing the expression 52245 as dq+r, we have:243×215+0.

Here, r=0 and The value of quotient is q=243. The remainder and the quotient of the expression 52245÷215 using the long division method are 0&243respectively.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals 12

Page 25 Problem 13 Answer

Given that there were 4050 students from 15 different school district. We are asked to find the average number of students in the museum.

The answer can be found by using the definition of average.Find the quotient of 4050/15.

This will give you the required answer. The number of students are 4050. Number of school district are: 15.

Thus, the average number of students from each school is: ​4050/15

15•270=4050

∴4050/15=270.

Hence, there were an average of 270 students from each school district.

The average number of students from each school district are 270.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals 13

Page 25 Problem 14 Answer

Given that, The Appalachian Trail is 2175 miles long and a hiker averages twelve miles each day.

We are asked to find how long will it take her to hike the length of the trail.

The answer can be found using the definition of average.Find the quotient value of 2175/12.

This would give the number of days. The total distance is 2175 miles.

Each day the hiker completes a distance of 12 miles.

Thus, the number of days taken for him to complete the journey is: 2175/12 = 181.25

That is, it would take 181 days and six hours. It would take 181.25 days for the hiker to hike the length of the trail.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals 14

Go Math Grade 6 Operations With Decimals Exercise 5.1 Key

Go Math! Grade 6 Exercise 5.1: Operations with Decimals Solutions Page 26 Exercise 1 Answer

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals e1

We are asked to fill these up and hence find the number of markers available for each table.The answer can be found by the process of division.

Divide the number 47/11 and find the quotient and the remainder, Bring down the remainder and the other remaining number.

Find the quotient when divided by eleven. Find the quotient. This will give you the required answer.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals e2

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals e3

She will put on 43 markers on each table:

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals e4

Go Math Answer Key

 

Go Math! Grade 6 Exercise 4.4: Operations with Fractions Solutions

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 4 Operations with Fractions

Page 23 Problem 1 Answer

Given the values of leftover apple and pumpkin pies.

To do: Find how much more apple pie than pumpkin pie is left.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 4 Operations with Fractions 1

1/2 apple pie is left than pumpkin pie.

Go Math! Grade 6 Exercise 4.4: Operations with Fractions Solutions Page 23 Problem 2 Answer

Given about the inches of angelfish.

To do: Find how much it was grown.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 4 Operations with Fractions 2

Angelfish has grown 5/6 inches.

Go Math! Practice Fluency Workbook Grade 6 Chapter 4 Operations with Fractions Exercise 4.4 Answer Key

Go Math Grade 6 Exercise 4.4 Operations With Fractions Solutions

Go Math! Grade 6 Exercise 4.4: Operations with Fractions Solutions Page 23 Problem 3 Answer

Given about the wrapping paper for a birthday present.

To do: Find how many pieces of 6 square-foot paper are needed to wrap 3 of these presents.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 4 Operations with Fractions 3

We need 2 pieces of 6 square-foot paper are needed to wrap 3 of these presents.

Go Math! Grade 6 Exercise 4.4: Operations with Fractions Solutions Page 23 Problem 4 Answer

Given a bicycle was rode 5×1/2 miles today and 6×1/4 miles yesterday

To do: find the difference in length between the two rides.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 4 Operations with Fractions 4

The difference is 3/4 miles. and the fraction of the longer side is 3/25.

Page 23 Problem 5 Answer

Given a survey by the state health department

To do: Find what fraction of the total pounds of fruit and vegetables do the pounds of fruits represent

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 4 Operations with Fractions 5

1005/2669 is the fraction of the total pounds of fruit and vegetables do the pounds of fruits represent

Go Math! Grade 6 Exercise 4.4: Operations with Fractions Solutions Page 23 Problem 6 Answer

Given the values of leftover apple and pumpkin pies.

To do: if tom ate 1/2 of leftovers, find how much pie in all did he eat.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 4 Operations with Fractions 6

Tom ate 3×1/12 pies.

Page 23 Problem 7 Answer

Given about the inches of angelfish.

To do: find how much has the angelfish grown in feet.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 4 Operations with Fractions 7

Angelfish has grown 5/72 feet.

Go Math Grade 6 Exercise 4.4 Operations With Fractions Answers

Go Math! Grade 6 Exercise 4.4: Operations with Fractions Solutions Page 24 Exercise 1 Answer

Given the amount of cheese, that deli was ordered.

To do: Find how much cheese was left for Wednesday.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 4 Operations with Fractions e1

For Wednesday 1×1/8 wheels of cheese were left.

Solutions For Go Math Grade 6 Exercise 4.4 Operations With Fractions

Go Math Answer Key

 

Go Math! Grade 6 Exercise 3.3: Rational Numbers Solutions

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 3 Rational Numbers

Page 15 Problem 1 Answer

Given:- 3/8

To Find:- To write each fraction as a decimal and to round to the nearest hundredth if necessary The given fraction is 3/8.

It can be converted into a decimal by dividing the numerator 3 by the denominator 8 using the long division method. The decimal value of 3/8 would be 0.375

The decimal value of the fraction 3/8 is0.375

Page 15 Problem 2 Answer

Given:- 7/5

To Find:- To write each fraction as a decimal and to round to the nearest hundredth if necessary The given fraction is 7/5.

Here, we can notice that the denominator is 5 , so we can easily convert the denominator in terms of 10 by multiplying both the numerator and denominator by 2.

Then, the fraction 7/5 would become 14/10. Therefore, the decimal value of the fraction 7/5 would be1.4

The decimal value of the fraction 7/5 is 1.4

Go Math! Grade 6 Exercise 3.3: Rational Numbers Solutions Page 15 Problem 3 Answer

Given:- 21/7

To Find:- To write each fraction as a decimal and to round to the nearest hundredth if necessary The given fraction is 21/7.

Here, we can notice that the denominator is7 which cannot be easily converted in terms of 10,100,1000….

So, we divide the numerator by the denominator to get 3.0 or 3 as the numerator 21 is fully divisible by 7.

The fraction 21/7 can be written as 3.0 or 3

Go Math Grade 6 Exercise 3.3 Rational Numbers Solutions

Go Math! Practice Fluency Workbook Grade 6 Chapter 3 Rational Numbers Exercise 3.3 Answer Key

Page 15 Problem 4 Answer

Given:- 5/3

To Find:- To write each fraction as a decimal and to round to the nearest hundredth if necessary The given fraction is 5/3.

Here, we can notice that the denominator is 3 which cannot be easily converted in terms of 10,100,1000….

So, we divide the numerator by the denominator to get 1.666which can be rounded to the nearest hundredth as 1.67.

The fraction 5/3 can be written in decimal form as 1.67

Page 15 Problem 5 Answer

Given:- 0.55

To Find:- To write each decimal as a fraction or mixed fraction in simplest form The given decimal is 0.55.

Since there are two digits after the decimal point, we can write 0.55 as 55/100.

Now, we can reduce the 55/100 in its simplest form by dividing both the numerator and denominator by 5 to get 11/20.

The decimal 0.55 can be written in the fractional form 11/20.

Go Math! Grade 6 Exercise 3.3: Rational Numbers Solutions Page 15 Problem 6 Answer

Given:- 10.6

To Find:- To write each decimal as a fraction or mixed fraction in simplest form The given decimal is 10.6.

Since there is one digit after the decimal point, we can write 10.6 as 106/10

This can be further simplified by dividing the numerator and denominator by 2 to get 53/5.

The decimal 10.6 can be written in the fractional form as 53/5.

Page 15 Problem 7 Answer

Given:- −7.08

To Find:- To write each decimal as a fraction or mixed fraction in simplest form The given decimal is 7.08.

Since there are two digits after the decimal point, we can write 7.08 as 708/100

This can be further reduced by dividing the numerator and denominator by 4 to get 177/25.

The decimal−7.08can be written in the fractional form−177/25

Page 15 Problem 8 Answer

Given:- 0.5,0.05,5/8

To Find:- To write the numbers in order from least to greatest

To write the given numbers in order from the least to the greatest, we should first change all the numbers into similar terms such as all the numbers into fractions or all the numbers into decimals.

Here, we can convert 5/8 into a decimal by dividing the numerator by the denominator to get 0.625.

So, the numbers would be 0.5,0.05,0.625.

So, the order from the least to the greatest will be 0.05,0.5,0.625.

Rewriting 0.625 as a fraction and then writing the numbers in order from the least to the greatest will be 0.05,0.5,5/8.

The order of the numbers from the least to the greatest will be0.05,0.5,5/8.

Go Math Grade 6 Exercise 3.3 Rational Numbers Answers

Page 15 Problem 9 Answer

Given:- 1.3,1×1/3 ,1.34

To Find:- To write the numbers in order from least to greatest

To write the given numbers in order from the least to the greatest, we should first change all the numbers into similar terms such as all the numbers into fractions or all the numbers into decimals.

Here, we can rewrite 1×1/3 as 4/3 which can be converted into a decimal by dividing the numerator by the denominator to get 1.33.

So, the numbers would be 1.3,1.33,1.34 in order from the least to the greatest.

The order of the numbers after rewriting 1.33 as a mixed fraction would be 1.3,1×1/3,1.34.

The order of the numbers from the least to the greatest would be 1.3,1×1/3,1.34.

Go Math! Grade 6 Exercise 3.3: Rational Numbers Solutions Page 15 Problem 10 Answer

Given:- 2.07,2×7/10 10 ,2.67,−2.67

To Find:- To write the numbers in order from least to greatest

To write the given numbers in order from the least to the greatest, we should first change all the numbers into similar terms such as all the numbers into fractions or all the numbers into decimals.

Here, we can rewrite 2×7/10 as 2×7/10 which can be converted to decimal as 2.7.

So, the numbers would be2.07,2.7,2.67,−2.67.

Arranging them in the order from the least to the greatest will be −2.67,2.07,2.67,2.7.

Rearranging the numbers in the order from the least to the greatest after rewriting 2.7 as 2×7/10 will be −2.67,2.07,2.67,2×7/10.

The order of the numbers from the least to the greatest would be −2.67,2.07,2.67,2×7/10.

Page 15 Problem 11 Answer

Given:- Out of 45 times at bat, Raul got 19 hits.
To Find:- Raul’s batting average as a decimal It is given that out of 45 times while batting, Raul got 19 hits. So, the average can be represented as 19/45.

Converting it to a decimal, the numerator should be divided by the denominator to get0.4222.

Therefore, Raul’s batting average is0.422.

Raul’s batting average in decimal is0.422

Page 15 Problem 12 Answer

Given:- Karen’s batting average was 0.444. She was at bat 45 times

To Find:- To determine how many hits did Karen get

We know that the batting average is defined as the ratio of the number of hits to the number of times at the bat.

Substituting the given values in the formula of batting average, we get,

⇒ \(\text { Batting average }=\frac{\text { number of hits }}{\text { number of times at bat }}\)

⇒ \(0.444=\frac{\text { number of hits }}{45}\)

Multiplying both sides of the equation by 45, we get, the number of hits=0.444×45 =19.98  which can be rounded to the nearest whole number as20

The number of hits Karen got was 20.

Page 15 Problem 13 Answer

Given:- To have batting averages over 0.500 how many hits in 45 times at bat would Raul and Karen need?

To Find:- To determine the number of hits Raul and Karen would need to have batting averages over 0.500

We know that the batting average is defined as the ratio of the number of hits to the number of times at the bat. Substituting the given values in the formula of batting average, we get,

⇒ \(Batting average= \frac{\text { number of hits }}{\text { number of times at bat }}\)

⇒ \( 0.500=\frac{\text { number of hits }}{45}\)

Multiplying both sides of the equation by 45, we get,

the number of hits=0.500×45 =22.5 which can be rounded to the nearest whole number23

The number of hits Raul and Karen would need to have a batting average above 0.500 would be 23.

Solutions For Go Math Grade 6 Exercise 3.3 Rational Numbers

Go Math! Grade 6 Exercise 3.3: Rational Numbers Solutions Page 15 Problem 14 Answer

we have to Solve the given question It is given that a car travels 65 miles per hour, but then travels 3/5 of this speed when going through construction. First, we need to write this fraction as a decimal.

When converting a fraction to a decimal: If the denominator is a factor of 10,100,1000,…, multiply the numerator and denominator by the same number to get a denominator that is a power of 10.

Then write the equivalent fraction as a decimal. If the denominator is not a factor of 10,100,1000,…, then divide the numerator by the denominator to find the decimal.

For3/5, the denominator of 5 is a factor of 10 so we can write an equivalent fraction to find the decimal:

⇒ \(\frac{3}{5}=\frac{3 \times 2}{5 \times 2}=\frac{6}{10}\)

⇒ \(\frac{6}{10} in words is ” six tenths”.\)

Since tenths mean one decimal place, then 6/10 =0.6

To find the speed, we can either multiply 65 and 3/5 or multiply 65 and 0.6. It is easier to multiply 65 and 3/5 since 5 is a factor of 65.

From the above step, we get a fraction as a decimal is 0.6 and 3/5 of the given speed of 39 miles per hour.

⇒ Multiplying then gives:\(65 \times \frac{3}{5}=65 \times \frac{1}{5} \times 3=\frac{65}{5} \times 3=13 \times 3=39 miles per hour\)

Page 15 Problem 15 Answer

Given: A city’s sales tax is 0.07. Write this decimal as a fraction and tell how many cents of tax are on each dollar. 0.07 in words is ‘ ‘seven hundredths” so we can write it as a fraction with 7 as the numerator and 100 as the denominator.

We then get 0.07=7/100.

Since there are 100 cents in a dollar, then the fraction means there are 7 cents of tax in each dollar. the tax is equal to 7 cents

Page 15 Problem 16 Answer

It is given that Norm has 373 sheets of paper left in a ream and each ream of paper initially has 500 sheets of paper. The portion of a ream that Norm has written as a fraction is then 373/500.

To find; A ream of paper contains 500 sheets of paper. Norm has 373 sheets of paper left from a dream. Express the portion of a ream Norm has as a fraction and as a decimal. _______________
When converting a fraction to a decimal:- If the denominator is a factor of 10,100,1000,…, multiply the numerator and denominator by the same number to get a denominator that is a power of 10. Then write the equivalent fraction as a decimal.

If the denominator is not a factor of 10,100,1000,…, then divide the numerator by the denominator to find the decimal.

⇒ \(For \frac{373}{500},\) the denominator of 500 is a factor of 1000 so we can write an equivalent fraction to find the decimal:

⇒ \(\frac{373}{500}=\frac{373 \times 2}{500 \times 2}=\frac{746}{1000}\)

⇒ \(\frac{746}{1000}\) in words is “seven hundred forty-six thousandths”.

Since thousandths mean three decimal places, then\( \frac{746}{1000}=0.746.\)

So, the portion written as a decimal number is 0.746

Go Math Grade 6 Rational Numbers Exercise 3.3 Key

Go Math! Grade 6 Exercise 3.3: Rational Numbers Solutions Page 16 Exercise 1 Answer

we have to write each decimal as a fraction or mixed number.0.61 in words is ” sixty-one hundredths” so we can write it as a fraction with 61 as the numerator and 100 as the denominator.

We then get 0.61=61/100

From the above step, we will get the answer 61/100

Page 16 Exercise 2 Answer

we have to write each decimal as a fraction or mixed number.3.43 in words is ‘ three and forty-three hundredths” so we can write it as a mixed number with 43 as the numerator and 100 as the denominator.

We then get

⇒  \(3.43=3 \frac{43}{100}\)

From the above step, we will get the answer 3×43/100

Page 16 Exercise 3 Answer

we have to write each decimal as a fraction or mixed number. 0.009 in words is “nine thousandths” so we can write it as a fraction with 9 as the numerator and 1000 as the denominator.

We then get\(0.009=\frac{9}{1000}\)

From the above step, we will get the answer 9/1000

Page 16 Exercise 4 Answer

we have to write each decimal as a fraction or mixed number.4.7 in words is ‘ ‘four and seven-tenths” so we can write it as a mixed number with 7 as the numerator and 10 as the denominator.

We then get \(4.7=4 \frac{7}{10}\)

From the above step, we will get the answer4x7/10

Page 16 Exercise 5 Answer

we have to write each decimal as a fraction or mixed number.1.5 in words is ” one and five tenths” so we can write it as a mixed number with 5 as the numerator and 10 as the denominator.

We then get \(1.5=1 \frac{5}{10}\)

Since 5 and 10 have a GCF of 5, we can reduce the fraction to:

⇒ \(1 \frac{5}{10}=1 \frac{5 \div 5}{10 \div 5}=11 \frac{1}{2}\)

From the above step, we will get the answer11x1/2

Go Math! Grade 6 Exercise 3.3: Rational Numbers Solutions Page 16 Exercise 6 Answer

we have to write each decimal as a fraction or mixed number. 0.13 in words is ” thirteen hundredths” so we can write it as a fraction with 13 as the numerator and 100 as the denominator.

We then get \(0.13=\frac{13}{100}\)

From the above step, we will get the answer13/100

Detailed Solutions For Go Math Grade 6 Exercise 3.3 Rational Numbers

Page 16 Exercise 7 Answer

we have to write each decimal as a fraction or mixed number 5.0002 in words is ” five and two thousandths” so we can write it as a mixed number with 2 as the numerator and 1000 as the denominator.

We then get \(5.002=5 \frac{2}{1000} .\)

Since 5 and 10 have a GCF of 2, we can reduce the fraction to:

⇒ \(5 \frac{2}{1000}=5 \frac{2 \div 2}{1000 \div 2}=5 \frac{1}{500}\)

From the above step, we will get the answer5x1/500

Page 16 Exercise 8 Answer

we have to write each decimal as a fraction or mixed number.

0.021 in words is ‘ ‘twenty-one thousandths” so we can write it as a fraction with 21 as the numerator and 1000 as the denominator.

We then get \(0.021=\frac{21}{1000}\)

From the above step, we will get the answer 21/1000

Practice Problems For Go Math Grade 6 Exercise 3.3 Rational Numbers

Go Math Answer Key

 

Go Math! Grade 6 Exercise 1.1 Solutions

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Exercise 1.1 Answer Key

Page 1 Problem 1 Answer

Here given that depositing of $85 in a bank account.
Now we have to represent this condition with positive or negative number.

So here given that deposit of money in bank ,we know that if you deposit some money in bank then our saving amount in bank account increasing so we can represent it by positive number.

Depositing of $ 85 in a bank account =+85

Page 1 Problem 2, Answer

Here given that Riding an elevator down 3 floor.
Now we have to represent this situation with Positive or negative number.

If we take riding of an elevator to up as positive then obviously the riding of an elevator down is negative.

So Riding an elevator down 3 floor is 3

 Page 1 Problem 3, Answer

Here given that the foundation of a house sinking 5 inches.
Now we have to present this situation by positive or negative numbers.

The foundation of a house sinking means a decrease of the height of the foundation.

So we can represent the foundation of a house sinking 5 inches =5

The foundation of a house sinking 5 inches =−5

Go Math! Practice Fluency Workbook Grade 6 Chapter 1 Integers Exercise 1.1 Answer Key

Go Math Practice Fluency Workbook Grade 6 Chapter 1 Integers Exercise 1.1 Answer Key

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Exercise 1.1 Answer Key Page 1 Problem 4, Answer

Here given that the temperature of 98° above zero —–
Now we have to represent it by positive or negative number

The temperature above zero degree is assumed as positive so

The temperature of 98above zero =+98°

So the temperature of 98° above zero is +98°

Page 1 Problem 5, Answer

Here given the number line

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1: Integers problem 5

We have to graph the −2 and its opposite number in number line.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1: Integersproblem 5 answer

The graph for −2 and its opposite number +2 is

Problem 6 Page 1 Answer

Here given the number +3

Now we have to write the opposite number of it and have to graph both.

So the opposite number of +3 is −3 And the graph is

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1Integers problem 6

Page 1  Problem 7, Answer

Here given that the number −5

Now we have to write the opposite number of the given number and have to draw the graph for both of them.

So the opposite number of −5 is +5 And the graph is

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 7

Go Math Grade 6 Practice Fluency Workbook Exercise 1.1 Integers solutions

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Exercise 1.1 Answer Key Page 1 Problem 8, Answer

Here given that the number +1

Now we have to write the opposite number of it and have to draw the graph.

So the opposite number of +1 is −1 And the graph is

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 8
Page 1 Problem 9, Answer

Here given that the average temperature in Fairbanks, Alaska, in February is 4°F below zero.
Now we have to write this temperature as an integer.

We know that the temperature below zero degree is represented by negative number.

So the average temperature of Fairbanks, Alaska, in February is 4°F below zero can be represented as 4F

So the temperature can be represented as −4° F.

Page 1 Problem 10, Answer

Here given that the average temperature in Fairbanks, Alaska, in November is 2° F above zero.
Now we have to write this temperature as an integer.

So the temperature above zero degree can be represented by positive number.

So the average temperature in Fairbanks, Alaska, in November is F above zero is +F

The average temperature in Fairbanks, Alaska, in November is 2° F above zero is +2° F

Page 1  Problem 11, Answer

Here given that the highest point in the state of Louisiana is Driskill Mountain. It rises 535 feet above sea level.
Now we have to represent this elevation as an integer.

We know that the elevation of a mountain is represented by positive number

So the elevation of Driskill Mountain is +535feet

The elevation of Driskill Mountain is +535feet

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Exercise 1.1 Answer Key Page 1 Problem 12, Answer

Here given that the lowest point in the state of Louisiana is New Orleans. The city’s elevation is 8 feet below sea level.
Now we have to represent the elevation of New Orleans as an integer.

If we take the elevation above sea as positive number than the city’s elevation below the sea level is represented by negative integer.

Now the elevation of New Orleans below the sea level is 8feet

The elevation of New Orleans is −8feet

Page 1 Problem 13, Answer

Here given that Death Valley, California, has the lowest elevation in the United States. Its elevation is 282 feet below sea level. Mount McKinley, Alaska, has the highest elevation in the United States. Its elevation is 20,320 feet above sea level

Now we have to write the elevation of two locations United States as an integer.

If we take the elevation above sea level as positive then the elevation below the sea level must be negative.

The elevation of Mount McKinley is 282 feet.

The elevation of is Alaska is +20320 feet

So The height of the two location of united states are as follows

The elevation of Mount McKinley is −282feet and The elevation of Alaska is +20320feet

Page 1 Problem 14, Answer

No there are no integers between zero and one . Because there are no whole number between them

No there are no integers between zero and one . Because there are no whole number between them.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1: Integers

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Exercise 1.1 Answer Key Page 2 Exercise 1 Answer

Here given that an increase of 3 points
Now we have to represent this by positive or negative numbers.

So an increase of 3 points is represented by +3

Page 2 Exercise 2, Answer

Here given that spending $10

Now we have to express this by positive or negative numbers.

So the spending of $10 can be represented by $(−10)

Page 2 Exercise 3, Answer

Here given that earning of $25

Now we have to express it by a positive or negative number.

So the earnings of $25 can be represented by $+25

Solutions for Go Math Grade 6 Practice Fluency Workbook Chapter 1 Exercise 1.1

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Exercise 1.1 Answer Key Page 2 Exercise 4, Answer

Here given that a loss of 5 yards.

Now we have to represent it by positive or negative number.

So the loss of 5 yards is −5yards

Page 2 Exercise 5,  Answer:

Here given the integer −1 ,now we have to write it’s opposite number and have to place them in number line

So the opposite number of −1 is +1

And the number line is

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 5 page 2

 

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Exercise 1.1 Answer Key Page 2 Exercise 6, Answer

Here given the number 9

Now we have to write it’s opposite number and have to present them in the number line

So the opposite number of 9 is −9

And the number line is

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 6 page 2
Page 2 Exercise 7, Answer

Given: 6 Write each integer and its opposite. Then graph them on the number line.

The graph of the given integer 6 and its opposite is

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 7 page 2

Go Math Grade 6 Integers Exercise 1.3 Key

Page 2 Exercise 8, Answer

Given: -5. Write each integer and its opposite. Then graph them on the number line.

The graph of the given integer -5 and its opposite 5 is

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 8 page 2

Go Math Grade 6 Integers Exercise 1.1 key from Practice Fluency Workbook

Go Math Answer Key

 

Go Math! Grade 6 Chapter 1 Integers Exercise 1.2 Answers

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Exercise 1.2 Answer Key

Page 3 Problem 1, Answer

Given: 10? -2 Use the number line to compare each pair of integers. Write < or >.ExplanationThe number line is

From the number line, 10 is to the right of -2. So, 10>−2.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 1 page 3

Thus, 10>−2.

Page 3 Problem 2, Answer

Given: 0? 3.

Use the number line to compare each pair of integers. Write < or >.ExplanationThe number line is

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 2

From the number line, 0 is to the left of 3.So, 0<3.

Page 3 Problem 3, Answer

Given: −5?0.

Use the number line to compare each pair of integers. Write < or >.ExplanationThe number line is
From the number line, -5 is to the left of 0.So, −5<0.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 3

Thus, −5<0.

Go Math Grade 6 Chapter 1 Integers Exercise 1.2 Answers

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Exercise 1.2 Answer Key Page 3 Problem 4, Answer

Given: −7?6.

Go Math! Practice Fluency Workbook Grade 6 Chapter 1 Integers Exercise 1.2 Answer Key

Use the number line to compare each pair of integers. Write < or >.ExplanationThe number line is

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 4

From the number line, -7 is to the left of 6.So, −7<6

Thus, −7<6.

Page 3 Problem 5 , Answer

Given: −6?−9.

Use the number line to compare each pair of integers. Write < or >.ExplanationThe number line is

From the number line, -6 is to the right of -9. So, −6>−9.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 5

Thus, −6>−9.

Page 3 Problem 6, Answer

Given: −8?−10.

Use the number line to compare each pair of integers. Write < or >.ExplanationThe number line is

From the number line, -8 is to the right of -10. So, −8>−10.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 6

Thus, −8>−10.

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Exercise 1.2 Answer Key Page 3 Problem 7, Answer

Given: 5,−2,6.

Order the integers in each set from least to greatest.Use the number line to order the integers.

List all the numbers in the order in which they appear from left to right. So, the integers in order from least to greatest are −2,5,6.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 7

Thus, the Order of Integers from least to greatest is −2,5,6.

Page 3 Problem 8, Answer

Given: 0,9,-3. Order the integers in each set from least to greatest.

List all the numbers in the order in which they appear from left to right. So, the integers in order from least to greatest are −3,0,9.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 8

Thus, the Order of Integers from least to greatest is −3,0,9.

Page 3 Problem 9, Answer

Given: -1,6,1.Order the integers in each set from least to greatest.Use the number line to order the integers.

List all the numbers in the order in which they appear from left to right. So, the integers in order from least to greatest are −1,1,6.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 9
Thus, the Order of Integers from least to greatest is −1,1,6.

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Exercise 1.2 Answer Key Page 3 Problem 10 , Answer

Given: -1,1,0.Order the integers in each set from greatest to least.

List all the numbers in the order in which they appear from right to left. So, the integers in order from greatest to least are 1,0,−1.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 10

Thus, the Order of Integers from greatest to least is 1,0,−1.

Page 3 Problem 11, Answer

Given: −12,2,1.

Order the integers in each set from greatest to least.

List all the numbers in the order in which they appear from right to left. So, the integers in order from greatest to least are 2,1,−12.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 11

Go Math Grade 6 Exercise 1.2 Integers Solutions

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Exercise 1.2 Answer Key Page 3 Problem 12, Answer

Given: −10,−12,−11.

Order the integers in each set from greatest to least.

List all the numbers in the order in which they appear from right to left. So, the integers in order from greatest to least are −10,−11,−12.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 12

Thus, the Order of Integers from greatest to least is −10,−11,−12.

Page 3 Problem 13, Answer

Given: 205,−20,−5,50.

Order the integers in each set from greatest to least.

List all the numbers in the order in which they appear from right to left.So, the integers in order from greatest to least are 205,50,−5,−20.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 13

Thus, the Order of Integers from greatest to least is 205,50,−5,−20.

Page 3 Problem, 14 Answer

Given: -78, -89, 78, 9.

Order the integers in each set from greatest to least.

List all the numbers in the order in which they appear from right to left. So, the integers in order from greatest to least are 78,9,−78,−89

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 14
Thus, the Order of Integers from greatest to least is 78,9,−78,−89.

Page 3 Problem 15, Answer

Given: -55, -2, -60, 0.Order the integers in each set from greatest to least.

List all the numbers in the order in which they appear from right to left. So, the integers in order from greatest to least are 0,−2,−55,−60.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 15

Thus, the Order of Integers from greatest to least is 0,−2,−55,−60.

Page 3 Problem 16, Answer

Given: 28, 8, -8, 0.

Order the integers in each set from greatest to least.

List all the numbers in the order in which they appear from right to left. So, the integers in order from greatest to least are 28,8,0,−8.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 16

Thus, the Order of Integers from greatest to least is 28,8,0,−8

Page 3 Problem 17, Answer

Given: 37, -37, -38, 38.

Order the integers in each set from greatest to least.

List all the numbers in the order in which they appear from right to left.So, the integers in order from greatest to least are 38,37,−37,−38.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 17
Thus, the Order of Integers from greatest to least is 38,37,−37,−38.

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Exercise 1.2 Answer Key Page 3 Problem 18, Answer

Given: -111, -1, 1, 11.

Order the integers in each set from greatest to least.

List all the numbers in the order in which they appear from right to left. So, the integers in order from greatest to least are11,1,−1,−111.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 18

Thus, the Order of Integers from greatest to least is 11,1,−1,−111

Solutions For Go Math Grade 6 Chapter 1 Exercise 1.2 Integers

Page 3 Problem 19, Answer

Given: Four friends went scuba diving today. Ali dove 70 feet, Tim went down 50 feet, Carl dove 65 feet, and Brenda reached 48 feet below sea level. Write the 4 friends’ names in order from the person whose depth was closest to the surface to the person whose depth was the farthest from the surface.

Here Ali dove 70 feet, Tim went down 50 feet, Carl dove 65 feet, and Brenda reached 48 feet below sea level. We know that 48<50<65<70.

So, from the given, the order from the person whose depth was closest to the surface to the person whose depth was the farthest from the surface is
Brenda, Tim, Carl, Ali.

The order from the person whose depth was closest to the surface to the person whose depth was the farthest from the surface is Brenda, Tim, Carl, Ali.

Page 3 Problem 20, Answer

Given: The temperatures on Monday and Tuesday were opposites.
The temperature on Wednesday was neither positive nor negative.
The temperature dropped below zero on Monday. To find: Write the 3 days in order from the highest to the lowest temperature.

We know that Monday and Tuesday were opposites and Monday dropped below zero. so, Tuesday had the highest temperature. Wednesday wasn’t positive or negative so it would be exactly zero. Monday dropped below zero so it had the lowest temperature. So, the order from the highest to the lowest temperature is Tuesday, Wednesday, Monday.

The 3 days in order from the highest to the lowest temperature are Tuesday, Wednesday, Monday.

Page 4 Exercise 1 Answer

Given: 1? −4. Use the number line to compare each pair of integers. Write < or >.

The number line is From the number line, 1 is to the right of -4. So, 1>−4.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers exercise 1

Thus, 1>−4.

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Exercise 1.2 Answer Key Page 4 Exercise 2 Answer

Given: −5? −2.

Use the number line to compare each pair of integers. Write < or >.ExplanationThe number line is
From the number line, -5 is to the left of -2. So, −5<−2.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers exercise 2

Thus, −5<−2.

Page 4 Exercise 3 Answer

Given: −3?2.

Use the number line to compare each pair of integers. Write < or >.ExplanationThe number line is
From the number line, -3 is to the left of 2. So, −3<2.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers exercise 3

Thus, −3<2.

Page 4 Exercise 4 Answer

Given: −1? −4.

Use the number line to compare each pair of integers. Write < or >.ExplanationThe number line is

From the number line, -1 is to the right of -4.So, −1>−4.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers exercise 4
Thus, −1>−4.

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Exercise 1.2 Answer Key Page 4 Exercise 5 Answer

Given: 5? 0.

Use the number line to compare each pair of integers. Write < or >.ExplanationThe number line is
From the number line, 5 is to the right of 0. So, 5>0

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers exercise 5

Thus, 5>0.

Page 4 Exercise 6 Answer

Given: −2?3.

Use the number line to compare each pair of integers. Write < or >.Explanation The number line is From the number line, -2 is to the left of 3. So, −2<3

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers exercise 6

Thus, −2<3

Page 4 Exercise 7 Answer

Given: −2,−5,1.

Order the integers from least to greatest.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers exercise 7

List all the numbers in the order in which they appear from left to right. So, the integers in order from least to greatest are −5,−2,−1.

Thus, the Order of Integers from least to greatest is −5,−2,−1.

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Exercise 1.2 Answer Key Page 4 Exercise 8 Answer

Given:0,−5,5
Order the integers from least to greatest.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers exercise 8

List all the numbers in the order in which they appear from left to right.So, the integers in order from least to greatest are−5,0,5.

Thus, the Order of Integers from least to greatest is−5,0,5.

Page 4 Exercise 9 Answer

Given:−5,2,−3
Order the integers from least to greatest. The given integers are−5,2,−3and the number line is

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers exercise 9

List all the numbers in the order in which they appear from left to right. So, the integers in order from least to greatest are -5,−3,2.

Hence the integers in order from least to greatest are−5,−3,2.

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Exercise 1.2 Answer Key Page 4 Exercise 10 Answer

Given:3,−1,−4
Order the integers from least to greatest. The given integers are3,−1,−4 and the number line is

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers exercise 10
List all the numbers in the order in which they appear from left to right.

So, the integers in order from least to greatest are−4,−1,3.Hence the integers in order from least to greatest are−4,−1,3

Page 4 Exercise 11 Answer

Given:3,−5,0
Order the integers from least to greatest.

The given integers are3,−5,0, and the number line is

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers exercise 11

List all the numbers in the order in which they appear from left to right.
So, the integers in order from least to greatest are−5,0,3

Hence the integers in order from least to greatest are−5,0,3

Page 4 Exercise 12 Answer

Given:−2,−4,1.

Order the integers from least to greatest.

The given integers are−2,−4,1 and the number line is

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers exercise 12

list all the numbers in the order in which they appear from left to right. So, the integers in order from least to greatest are−4,−2,1

Hence the integers in order from least to greatest are−4,−2,1.

Go Math Grade 6 Integers Exercise 1.2 Key

Go Math Answer Key