Distance, Rate, and Time Worksheets

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In math, distance, rate, and time are three important concepts you can use to solve many problems if you know the formula. Distance is the length of space traveled by a moving object or the length measured between two points. It is usually denoted by  d  in math problems.

The rate is the speed at which an object or person travels. It is usually denoted by  r  in equations. Time is the measured or measurable period during which an action, process, or condition exists or continues. In distance, rate, and time problems , time is measured as the fraction in which a particular distance is traveled. Time is usually denoted by  t  in equations.

Use these free, printable worksheets to help students learn and master these important math concepts. Each slide provides the student worksheet, followed by an identical worksheet that includes the answers for ease of grading. Each worksheet provides three distance, rate, and time problems for students to solve.

Worksheet No. 1

Print the PDF: Distance, Rate, and Time Worksheet No. 1

When solving distance problems, explain to students that they will use the formula:

or rate (speed) times time equals distance. For example, the first problem states:

The Prince David ship headed south at an average speed of 20 mph. Later the Prince Albert traveled north with an average speed of 20 mph. After the Prince David ship had traveled for eight hours, the ships were 280 miles apart. How many hours did the Prince David Ship Travel?

Students should find that the ship traveled for six hours.

Worksheet No. 2

Print the PDF: Distance, Rate, and Time Worksheet No. 2

If students are struggling, explain that to solve these problems, they will apply the formula that solves distance, rate, and time, which is  distance = rate x tim e. It is abbreviated as:

The formula can also be rearranged as:

r = d/t or  t = d/r

Let students know that there are many examples where you might use this formula in real life. For example, if you know the time and rate a person is traveling on a train, you can quickly calculate how far he traveled. And if you know the time and distance a passenger traveled on a plane, you could quickly figure the distance she traveled simply by reconfiguring the formula.

Worksheet No. 3

Print the PDF: Distance, Rate, Time Worksheet No. 3

On this worksheet, students will solve problems such as:

Two sisters Anna and Shay left the home at the same time. They headed out in opposite directions toward their destinations. Shay drove 50 mph faster than her sister Anna. Two hours later, they were 220 mph apart from each other. What was Anna’s average speed?

The students should find that Anna's average speed was 30 mph.

Worksheet No. 4

Print the PDF: Distance, Rate, Time Worksheet No. 4

Ryan left home and drove to his friend's house driving 28 mph. Warren left an hour after Ryan traveling at 35 mph hoping to catch up with Ryan. How long did Ryan drive before Warren caught up to him?

Students should find that Ryan drove for five hours before Warren caught up to him.

Worksheet No. 5

Print the PDF: Distance, Rate, and Time Worksheet No. 5

On this final worksheet, students will solve problems including:

Pam drove to the mall and back. It took one hour longer to go there than it did to come back home. The average speed she was traveling on the trip there was 32 mph. The average speed on the way back was 40 mph. How many hours did the trip there take?

They should find that Pam's trip took five hours.

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Convert Units of Length - Lesson 6.1

Convert Units of Capacity - Lesson 6.2

Convert Units of Weight and Mass - Lesson 6.3

Transform Units - Lesson 6.4

Problem Solving - Distance, Rate, and Time - Lesson 6.5

Fractions and Decimals - Lesson 2.1

Compare and order Fractions and Decimals - Lesson 2.2

Multiply Fractions - Lesson 2.3

Simplify Factors - Lesson 2.4

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Estimate Quotients - Lesson 2.6

Dividing Fractions - Lesson 2.7

Model Mixed Number Division - Lesson 2.8

Divide Mixed Numbers - Lesson 2.9

Problem Solving - Fraction Operations - Lesson 2.10

Understanding Positive and Negative Integers - Lesson 3.1

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Rational Numbers and the Number Line - Lesson 3.3

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Compare Absolute Value - Lesson 3.6

Rational Numbers and the Coordinate Plane - Lesson 3.7

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Distance on the Coordinate Plane - Lesson 3.9

Problem Solving - The Coordinate Plane - Lesson 3.10

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Area of Regular Polygons - Lesson 10.6

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Divide Multi-Digit Numbers - Lesson 1.1

Prime Factorization - Lesson 1.2

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Exponents - Lesson 7.1

Evaluating Expressions Involving Exponents - Lesson 7.2

Write Algebraic Expressions - Lesson 7.3

Identify Parts of Expressions - Lesson 7.4

Evaluate Algebraic Expressions and Formulas - Lesson 7.5

Use Algebraic Expressions - Lesson 7.6

Problem Solving - Combining Like Terms - Lesson 7.7

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Independent and Dependent Variables - Lesson 9.1

Equations and Tables - Lesson 9.2

Model Percents - Lesson 5.1

Write Percents as Fractions and Decimals - Lesson 5.2

Write Fractions and Decimals as Percents - Lesson 5.3

Percent of a Quantity - Lesson 5.4

Problem Solving - Percents - Lesson 5.5

Find the Whole From a Percent - Lesson 5.6

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Distance rate time problems

Distance rate time problems involve object moving at a constant rate and this is called uniform motion. The formula d = r × t is the formula to use to solve problems related to distance, rate, and time.

Examples showing how to solve distance rate time problems

It may be obvious that for this type of problem, the distance is the same since you left the same place (your house) and are going to the same location.

Suppose that you went to no other places, the distance then to go back to your house is again the same.

Therefore, when solving distance rate time problems involving opposite direction travel, you can add the distances to get the total distance.

Study also the distance rate time problem in the figure below carefully

Distance rate time problems

Mixture word problems

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Chapter 8: Rational Expressions

8.8 Rate Word Problems: Speed, Distance and Time

Distance, rate and time problems are a standard application of linear equations. When solving these problems, use the relationship rate (speed or velocity) times time equals distance .

[latex]r\cdot t=d[/latex]

For example, suppose a person were to travel 30 km/h for 4 h. To find the total distance, multiply rate times time or (30km/h)(4h) = 120 km.

The problems to be solved here will have a few more steps than described above. So to keep the information in the problem organized, use a table. An example of the basic structure of the table is below:

The third column, distance, will always be filled in by multiplying the rate and time columns together. If given a total distance of both persons or trips, put this information in the distance column. Now use this table to set up and solve the following examples.

Example 8.8.1

Joey and Natasha start from the same point and walk in opposite directions. Joey walks 2 km/h faster than Natasha. After 3 hours, they are 30 kilometres apart. How fast did each walk?

The distance travelled by both is 30 km. Therefore, the equation to be solved is:

[latex]\begin{array}{rrrrrrl} 3r&+&3(r&+&2)&=&30 \\ 3r&+&3r&+&6&=&30 \\ &&&-&6&&-6 \\ \hline &&&&\dfrac{6r}{6}&=&\dfrac{24}{6} \\ \\ &&&&r&=&4 \text{ km/h} \end{array}[/latex]

This means that Natasha walks at 4 km/h and Joey walks at 6 km/h.

Example 8.8.2

Nick and Chloe left their campsite by canoe and paddled downstream at an average speed of 12 km/h. They turned around and paddled back upstream at an average rate of 4 km/h. The total trip took 1 hour. After how much time did the campers turn around downstream?

The distance travelled downstream is the same distance that they travelled upstream. Therefore, the equation to be solved is:

[latex]\begin{array}{rrlll} 12(t)&=&4(1&-&t) \\ 12t&=&4&-&4t \\ +4t&&&+&4t \\ \hline \dfrac{16t}{16}&=&\dfrac{4}{16}&& \\ \\ t&=&0.25&& \end{array}[/latex]

This means the campers paddled downstream for 0.25 h and spent 0.75 h paddling back.

Example 8.8.3

Terry leaves his house riding a bike at 20 km/h. Sally leaves 6 h later on a scooter to catch up with him travelling at 80 km/h. How long will it take her to catch up with him?

The distance travelled by both is the same. Therefore, the equation to be solved is:

[latex]\begin{array}{rrrrr} 20(t)&=&80(t&-&6) \\ 20t&=&80t&-&480 \\ -80t&&-80t&& \\ \hline \dfrac{-60t}{-60}&=&\dfrac{-480}{-60}&& \\ \\ t&=&8&& \end{array}[/latex]

This means that Terry travels for 8 h and Sally only needs 2 h to catch up to him.

Example 8.8.4

On a 130-kilometre trip, a car travelled at an average speed of 55 km/h and then reduced its speed to 40 km/h for the remainder of the trip. The trip took 2.5 h. For how long did the car travel 40 km/h?

[latex]\begin{array}{rrrrrrr} 55(t)&+&40(2.5&-&t)&=&130 \\ 55t&+&100&-&40t&=&130 \\ &-&100&&&&-100 \\ \hline &&&&\dfrac{15t}{15}&=&\dfrac{30}{15} \\ \\ &&&&t&=&2 \end{array}[/latex]

This means that the time spent travelling at 40 km/h was 0.5 h.

Distance, time and rate problems have a few variations that mix the unknowns between distance, rate and time. They generally involve solving a problem that uses the combined distance travelled to equal some distance or a problem in which the distances travelled by both parties is the same. These distance, rate and time problems will be revisited later on in this textbook where quadratic solutions are required to solve them.

For Questions 1 to 8, find the equations needed to solve the problems. Do not solve.

  • A is 60 kilometres from B. An automobile at A starts for B at the rate of 20 km/h at the same time that an automobile at B starts for A at the rate of 25 km/h. How long will it be before the automobiles meet?
  • Two automobiles are 276 kilometres apart and start to travel toward each other at the same time. They travel at rates differing by 5 km/h. If they meet after 6 h, find the rate of each.
  • Two trains starting at the same station head in opposite directions. They travel at the rates of 25 and 40 km/h, respectively. If they start at the same time, how soon will they be 195 kilometres apart?
  • Two bike messengers, Jerry and Susan, ride in opposite directions. If Jerry rides at the rate of 20 km/h, at what rate must Susan ride if they are 150 kilometres apart in 5 hours?
  • A passenger and a freight train start toward each other at the same time from two points 300 kilometres apart. If the rate of the passenger train exceeds the rate of the freight train by 15 km/h, and they meet after 4 hours, what must the rate of each be?
  • Two automobiles started travelling in opposite directions at the same time from the same point. Their rates were 25 and 35 km/h, respectively. After how many hours were they 180 kilometres apart?
  • A man having ten hours at his disposal made an excursion by bike, riding out at the rate of 10 km/h and returning on foot at the rate of 3 km/h. Find the distance he rode.
  • A man walks at the rate of 4 km/h. How far can he walk into the country and ride back on a trolley that travels at the rate of 20 km/h, if he must be back home 3 hours from the time he started?

Solve Questions 9 to 22.

  • A boy rides away from home in an automobile at the rate of 28 km/h and walks back at the rate of 4 km/h. The round trip requires 2 hours. How far does he ride?
  • A motorboat leaves a harbour and travels at an average speed of 15 km/h toward an island. The average speed on the return trip was 10 km/h. How far was the island from the harbour if the trip took a total of 5 hours?
  • A family drove to a resort at an average speed of 30 km/h and later returned over the same road at an average speed of 50 km/h. Find the distance to the resort if the total driving time was 8 hours.
  • As part of his flight training, a student pilot was required to fly to an airport and then return. The average speed to the airport was 90 km/h, and the average speed returning was 120 km/h. Find the distance between the two airports if the total flying time was 7 hours.
  • Sam starts travelling at 4 km/h from a campsite 2 hours ahead of Sue, who travels 6 km/h in the same direction. How many hours will it take for Sue to catch up to Sam?
  • A man travels 5 km/h. After travelling for 6 hours, another man starts at the same place as the first man did, following at the rate of 8 km/h. When will the second man overtake the first?
  • A motorboat leaves a harbour and travels at an average speed of 8 km/h toward a small island. Two hours later, a cabin cruiser leaves the same harbour and travels at an average speed of 16 km/h toward the same island. How many hours after the cabin cruiser leaves will it be alongside the motorboat?
  • A long distance runner started on a course, running at an average speed of 6 km/h. One hour later, a second runner began the same course at an average speed of 8 km/h. How long after the second runner started will they overtake the first runner?
  • Two men are travelling in opposite directions at the rate of 20 and 30 km/h at the same time and from the same place. In how many hours will they be 300 kilometres apart?
  • Two trains start at the same time from the same place and travel in opposite directions. If the rate of one is 6 km/h more than the rate of the other and they are 168 kilometres apart at the end of 4 hours, what is the rate of each?
  • Two cyclists start from the same point and ride in opposite directions. One cyclist rides twice as fast as the other. In three hours, they are 72 kilometres apart. Find the rate of each cyclist.
  • Two small planes start from the same point and fly in opposite directions. The first plane is flying 25 km/h slower than the second plane. In two hours, the planes are 430 kilometres apart. Find the rate of each plane.
  • On a 130-kilometre trip, a car travelled at an average speed of 55 km/h and then reduced its speed to 40 km/h for the remainder of the trip. The trip took a total of 2.5 hours. For how long did the car travel at 40 km/h?
  • Running at an average rate of 8 m/s, a sprinter ran to the end of a track and then jogged back to the starting point at an average of 3 m/s. The sprinter took 55 s to run to the end of the track and jog back. Find the length of the track.

Answer Key 8.8

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Go Math! 6 Common Core Edition, Grade: 6 Publisher: Houghton Mifflin Harcourt

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Distance, Rate, and Time

Our free downloadable PDF Distance, Rate and Time lesson plan introduces the formulas for solving distance, rate, and time equations. This hands-on lesson allows students to utilize resources when applying the formula, and leverage group learning as students exchange papers for review.

Included with this lesson are some adjustments or additions that you can make if you’d like, found in the “Options for Lesson” section of the Classroom Procedure page. One of the optional additions to this lesson is to have your students create additional distance, time, and rate problems for other students to solve.

Description

Additional information, what our distance, rate, and time lesson plan includes.

Lesson Objectives and Overview: Distance, Rate, and Time lesson plan introduces the formulas for solving distance, rate, and time equations. This hands-on lesson allows students to utilize resources when applying the formula, and leverage group learning as students exchange papers for review. At the end of the lesson, students will be able to identify and use the formula for distance, rate, and time. This lesson is for students in 4th grade, 5th grade, and 6th grade.

Classroom Procedure

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

Options for Lesson

Included with this lesson is an “Options for Lesson” section that lists a number of suggestions for activities to add to the lesson or substitutions for the ones already in the lesson. If you’d like to adjust the lesson activity, you can have students work alone. For an additional activity, you could take your students outdoors to participate in several timed events, like running, skipping, and hopping, and have them estimate the distance they could travel in one hour, one day, and so on. Finally, you could have your students create additional distance, time, and rate problems for other students to solve.

Teacher Notes

The teacher notes page includes a paragraph with additional guidelines and things to think about as you begin to plan your lesson. This page also includes lines that you can use to add your own notes as you’re preparing for this lesson.

DISTANCE, RATE, AND TIME LESSON PLAN CONTENT PAGES

What is distance, rate, and time.

The Distance, Rate, and Time lesson plan includes two content pages. When you’re on a road trip, you might wonder when you’re going to reach your destination. The answer to that question depends on two things: the rate of speed you’re traveling at and the distance you need to travel. If you know the the distance, rate (speed), and time, you can answer your question!

Distance is the amount of space or length between two points or places. We usually measure it in inches, feet, miles, meters, or kilometers. Rate is a ratio that compares to quantities of different units, like speed (miles per hour) or cost (dollars per pound). In this lesson, when we talk about rate, we’re talking about the speed of an object. Time measures a sequence of events in seconds, minutes, hours, days, weeks, months, or years.

We use distance, rate, and time formulas often without even being aware of them! The lesson offers a few examples. Weather forecasters use it to predict when a hurricane or storm may be in the area. Airlines and airports use it to plan their take-off and landing schedules. Schools use it to make sure their busses pick students up on time. Delivery companies use it to figure out how many packages their drivers can deliver. There are many more examples as well!

The Formula

You can write the formula three ways, depending on which variable you need to find: Distance = Rate x Time or d = rt , Rate = Distance ÷ Time or r =d/t , and Time = Distance ÷ Rate or t = d/r.

The lesson shows each of these equations as parts of a triangle. Visualizing the formula in this way can help you remember if you need to multiply or divide. In the triangles where the measures are next to each other, you multiply. When they’re on top of each other, you divide.

Using the Formula

The first step in using the formula is to determine what measure you’re trying to find: distance, speed, or time. Once you know that, you can use the right version of the formula, replacing the variables with the figures. The lesson walks students through an example of finding each measure—distance, speed, and time.

Using this formula is as simple as multiplying or dividing. However, it is important to pay attention to the unit that the question is asking for. A problem might give you information in minutes, for example, but want the answer in hours. In that case, you’d have to convert the minutes into hours. Therefore, it’s very important to read each problem thoroughly before solving. And always remember to ask yourself if your answer makes sense!

DISTANCE, RATE, AND TIME LESSON PLAN WORKSHEETS

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

CHART ACTIVITY WORKSHEET

For the activity worksheet, students will complete the chart shown on the worksheet by using the formula from the lesson to solve word problems.

QUESTIONS PRACTICE WORKSHEET

The practice worksheet asks students to answer 12 questions about the lesson content, testing their understanding of the lesson.

DISTANCE, RATE, AND TIME HOMEWORK ASSIGNMENT

For the homework assignment, students will create six word problems for another student in the class to solve. Two of these problems should be solved for distance, two for speed, and two for time. Each student will also create an answer key for their problems.

Worksheet Answer Keys

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

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CCSS Math Answers

Go Math Grade 6 Answer Key Chapter 6 Convert Units of Length

In order to solve real-life mathematical problems, students must understand how the information is related, and how to convert the units. You can learn the concepts only when you start from the basics. Download Free Pdf of Go Math Grade 6 Answer Key Chapter 6 Convert Units of length to practice the exercise and homework problems. We have provided the solutions for all the questions in the HMH Go Math Grade 6 Answer Key Chapter 6 Convert Units of length.

Go Math Grade 6 Chapter 6 Convert Units of Length Answer Key

The topics covered in this chapter are Convert units of length, capacity, convert units of weight and mass, transform units, distance, rate and time formulas. This is the easiest and important among all the chapters in the 6th standard. You can score the maximum marks in the exams with the help of our HMH Go Math Grade 6 Answer Key Chapter 6 Convert Units of Length. Tap the links which are provided according to the topics and kickstart your preparation.

Lesson 1: Convert Units of Length

  • Convert Units of Length – Page No. 317
  • Convert Units of Length – Page No. 318
  • Convert Units of Length – Page No. 319
  • Convert Units of Length Lesson Check – Page No. 320

Lesson 2: Convert Units of Capacity

  • Convert Units of Capacity – Page No. 323
  • Convert Units of Capacity – Page No. 324
  • Convert Units of Capacity – Page No. 325
  • Convert Units of Capacity Lesson Check – Page No. 326

Lesson 3: Convert Units of Weight and Mass

  • Convert Units of Weight and Mass – Page No. 329
  • Convert Units of Weight and Mass – Page No. 330
  • Convert Units of Weight and Mass – Page No. 331
  • Convert Units of Weight and Mass Lesson Check – Page No. 332

Mid-Chapter Checkpoint

  • Mid-Chapter Checkpoint – Page No. 333
  • Mid-Chapter Checkpoint – Page No. 334

Lesson 4: Transform Units

  • Transform Units – Page No. 337
  • Transform Units – Page No. 338
  • Transform Units – Page No. 339
  • Transform Units Lesson Check – Page No. 340

Lesson 5: Problem Solving • Distance, Rate, and Time Formulas

  • Distance, Rate, and Time Formulas – Page No. 343
  • Distance, Rate, and Time Formulas – Page No. 344
  • Distance, Rate, and Time Formulas – Page No. 345
  • Distance, Rate, and Time Formulas Lesson Check – Page No. 346

Chapter 6 Review/Test

  • Review/Test – Page No. 347
  • Review/Test – Page No. 348
  • Review/Test – Page No. 349
  • Review/Test – Page No. 350
  • Review/Test – Page No. 351
  • Review/Test – Page No. 352

Share and Show – Page No. 317

Convert to the given unit.

Question 1. 3 miles = ? yards _______ yd

Answer: 5280 yd

Explanation: 3 miles = ? yards 1 yard = 3 feet 1 mile = 5280 feet So, 3 miles = 3 x 5280 feet = 15,840 feet 3 feet = 1 yard Then, 15,840 feet = 15,840 ÷ 3 = 5280 yards So, 3 miles = 5280 yards

Question 2. 43 dm = ? hm _______ hm

Answer: 0.043 hm

Explanation: 43 dm= ?hm 10 decimeters = 1 meter 1 hectometer = 100 meters 1 meter = 10 decimeter 100 meters = 10×100 decimeters = 1000 decimeters So 1 hectometer = 1000 decimeters Then, 43 decimeters = 43/1000 = 0.043 hectometers So, 43 dm = 0.043 hm

Question 3. 9 yd = ? in. _______ inches

Answer: 324 inches

Explanation: 9 yd= ? in. 1 yard = 36 inches So 9 yards = 9×36 = 324 inches 9 yards = 324 inches

Question 4. 72 ft = 24 yd _______ yd

Answer: 24 yd

Explanation: 72 ft = 24 yd 1 yard = 3 feet So, 1 feet = 1/3 yard Then, 72 feet = 72/3 yard So, 72 feet = 24 yards

Question 5. 7,500 mm = ? dm _______ dm

Answer: 75 dm

Explanation: 7,500 mm = ?dm 1000 millimeters = 1 meter 10 decimeters = 1 meter So, 1000 millimeters = 10 decimeters Then 1 millimeter = 10/1000 decimeter = 1/100 decimeters So 7500 millimeters = 7500/100 decimeters Then 7500 mm = 75 dm

On Your Own

Question 6. Rohan used 9 yards of ribbon to wrap gifts. How many inches of ribbon did he use? _______ inches

Explanation: As per the given data, Rohan used 9 yards of ribbon to wrap gifts 1 yard = 36 inches So, 9 yards = 9×36 = 324 inches So, Rohan used 324 inches ribbon to wrap gifts

Question 7. One species of frog can grow to a maximum length of 12.4 millimeters. What is the maximum length of this frog species in centimeters? _______ cm

Answer: 1.24 cm

Explanation: One species of frog can grow to a maximum length of 12.4 millimeters. From the given information One species of frog can grow to a maximum length of 12.4 millimeters 1000 millimeters = 1 meter 100 centimeters = 1 meter So, 1000 millimeters = 100 centimeters 1 millimeter = 100/1000 centimeters = 1/10 centimeters So, 12.4 millimeters = 12.4/10 centimeters = 1.24 centimeters 12.4 millimeters = 1.24 centimeters

Question 8. The height of the Empire State Building measured to the top of the lightning rod is approximately 443.1 meters. What is this height in hectometers? _______ hectometers

Answer: 4.431 hectometers

Explanation: The height of the Empire State Building measured to the top of the lightning rod is approximately 443.1 meters. 443.1 meters in hectometers 1 hectometer = 100 meters Then, 1 meter = 1/100 hectometers So, 443.1 meters = 443.1/100 hectometers 443.1 meters = 4.431 hectometers

Question 9. A snail moves at a speed of 2.5 feet per minute. How many yards will the snail have moved in half of an hour? _______ yards

Answer: 25 yards

Explanation: From the given information A snail moves at a speed of 2.5 feet per minute 1 hour = 60 minutes 1 minute = 2.5 feet speed 60 minutes = 60×2.5 feet = 150 feet 1 yard = 3 feet So 1 feet = 1/3 yards Then, 150 feet = 150/3 yards = 50 yards per hour For half of an hour, a snail moves 25 yards

Practice: Copy and Solve Compare. Write <, >, or =.

Question 10. 32 feet _______ 11 yards

Answer: 32 feet < 11 yards

Explanation: 32 feet _______ 11 yards 1 yard = 3 feet So, 11 yards = 11×3 = 33 feet So, 32 feet < 11 yards

Question 11. 537 cm _______ 5.37 m

Answer: 537 cm = 5.37 m

Explanation: 537 cm _______ 5.37 m 100 centimeters = 1 meter 1 centimeter = 0.01 meter So, 537 centimeters = 537×0.01 meters That is 537 centimeters = 5.37 meters

Question 12. 75 inches _______ 6 feet

Answer: 75 inches > 6 feet

Explanation: 75 inches _______ 6 feet 1 foot = 12 inches 6 feet = 6×12 = 72 inches So, 75 inches > 6 feet

Problem Solving + Applications – Page No. 318

What’s the Error?

Question 13. The Redwood National Park is home to some of the largest trees in the world. Hyperion is the tallest tree in the park, with a height of approximately 379 feet. Tom wants to find the height of the tree in yards. Tom converted the height this way : 3 feet = 1 yard conversion factor: \(\frac{3 \mathrm{ft}}{1 \mathrm{yd}}\) \(\frac{379 \mathrm{ft}}{1} \times \frac{3 \mathrm{ft}}{1 \mathrm{yd}}\) = 1,137 yd Find and describe Tom’s error. Show how to correctly convert from 379 feet to yards. Explain how you knew Tom’s answer was incorrect. Type below: ____________

Answer: conversion factor: 3ft1yd 379ft1 × 3ft1yd = 1,137 yd We need to divide the 379 feet with 3 to get the height of the Hyperion tree, but tom multiplies the 379 with 3 and that is the error part 1 yard = 3 feet 1 feet = 1/3 yards So, 379 feet = 379/3 yards = 126.3 yards So, the height of the Hyperion tree is 126.3 yards

Question 14. Choose <, >, or =. 14a. 12 yards Ο 432 inches 14b. 321 cm Ο 32.1 m 12 yards _______ 432 inches 321 cm _______ 32.1 m

Answer: 14a. 12 yards Ο 432 inches 14b. 321 cm Ο 32.1 m 12 yards = 432 inches 321 cm < 32.1 m

Explanation: 14a. 12 yards Ο 432 inches 1 yard = 36 inches 12 yards = 12×36 = 432 inches So, 12 yards = 432 inches 14b. 321 cm Ο 32.1 m 100 centimeters = 1 meter 1 centimeter = 0.01 meter 321 centimeters = 321×0.01 meters = 3.21 meters 3.21 < 32.1 So, 321 centimeters < 32.1 meters

Convert Units of Length – Page No. 319

Question 1. 42 ft = ? yd _______ yd

Answer: 14yd

Explanation: 42 ft= ?yd 3 feet = 1 yard 1 feet = 1/3 yard So, 42 feet = 42/3 = 14 yard So, 42 feet = 14 yards

Question 2. 2,350 m = ? km _______ km

Answer: 2.350 km

Explanation: 2,350 m = ? km 1 kilometer = 1000 meters 1 meter = 1/1000 kilometers Then, 2350 meters = 2350/1000 kilometers 2350 meters = 2.350 kilometers

Question 3. 18 ft = ? in. _______ inches

Answer: 216 inches

Explanation: 18 ft= ? in 1 foot = 12 inches 18 feet = 12×18 = 216 inches 18 feet = 216 inches

Question 4. 289 m = ? dm _______ dm

Answer: 2890 dm

Explanation: 289 m = ?dm 10 decimeters = 1 meter 289 meters = 289×10 decimeters So, 289 meters = 2890 decimeters

Question 5. 5 mi = ? yd _______ yd

Answer: 8,800 yd

Explanation: 1. 5 mi = ? yd 1 mile = 1760 yards 5 miles = 5×1760 = 8800 yards 5 mi = 8,800 yards

Question 6. 35 mm = ? cm _______ cm

Answer: 3.5 cm

Explanation: 35 mm = ? cm 1000 millimeters = 1 meter 100 centimeters = 1 meter So, 1000 millimeters = 100 centimeters 1 millimeter = 100/1000 centimeters Then, 35 millimeters = 35×100/1000 centimeters = 3.5 centimeters 35 millimeters = 3.5 centimeters

Compare. Write <, >, or =.

Question 7. 1.9 dm _______ 1,900 mm

Answer: 1.9 dm < 1,900 mm

Explanation: 1.9 dm _______ 1,900 mm 10 decimeters = 1 meter 1000 millimeters = 1 meter So, 10 decimeters = 1000 millimeters 1 decimeter = 100 millimeters 1.9 decimeters = 1.9 x 100 = 190 millimeters So, 1.9 decimeters = 190 millimeters So, 1.9 dm < 1900 mm

Question 8. 12 ft _______ 4 yd

Answer: 12 ft  = 4 yd

Explanation: 12 ft _______ 4 yd 3 feet = 1 yard 3×4 feet = 12 feet = 1×4 = 4 yard So, 12 feet = 4 yards

Question 9. 56 cm _______ 56,000 km

Answer: 56 cm < 56,000 km

Explanation: 56 cm _______ 56,000 km 100 centimeters = 1 meter 1 kilometer = 1000 meters 0.01 kilometer = 1 meter So, 100 centimeters = 0.01 kilometers 1 centimeter = 0.01/100 kilometers 56 centimeters = 56 x 0.01/100 kilometers =0.0056 kilometers So, 56 cm < 56,000 km

Question 10. 98 in. _______ 8 ft

Answer: 98 in. > 8 ft

Explanation: 98 in. _______ 8 ft 1 foot = 12 inches 8 feet = 8×12 = 96 inches So, 98 in > 8 feet

Question 11. 64 cm _______ 630 mm

Answer: 64 cm  > 630 mm

Explanation: 64 cm _______ 630 mm 1000 millimeters = 1 meter 100 centimeters = 1 meter So, 100 centimeters = 1000 millimeters 1 centimeter = 10 millimeters so, 64 centimeters = 64×10 millimeters = 640 millimeters then, 64 cm > 630 mm

Question 12. 2 mi _______ 10,560 ft

Answer: 2 mi  = 10,560 ft

Explanation: 1 mi _______ 10,560 ft 1 mile = 5280 feet so, 2 miles = 2×5280 = 10560 feet then, 2 miles = 10,560 feet

Question 13. The giant swallowtail is the largest butterfly in the United States. Its wingspan can be as large as 16 centimeters. What is the maximum wingspan in millimeters? _______ mm

Answer: 160 mm

Explanation: The giant swallowtail is the largest butterfly in the United States. Its wingspan can be as large as 16 centimeters. 100 centimeters = 1 meter 1000 millimeters = 1 meter So, 100 centimeters = 1000 millimeters 1 centimeters = 10 millimeters then 16 centimeters = 16×10 millimeters = 160 millimeters So, giant swallowtail wingspan is 160 millimeters large

Question 14. The 102nd floor of the Sears Tower in Chicago is the highest occupied floor. It is 1,431 feet above the ground. How many yards above the ground is the 102nd floor? _______ yd

Answer: 477 yd

Explanation: The 102nd floor of the Sears Tower in Chicago is the highest occupied floor. It is 1,431 feet above the ground. 3 feet = 1 yard 1 feet = 1/3 yard Then, 1431 feet = 1431/3 yard = 477 yards So, the height of the 102nd floor from the ground = 477 yards

Question 15. Explain why units can be simplified first when measurements are multiplied. Type below: ____________

Answer: Units can be simplified first, because if (60 min)/(1 hr) = 1, then I can multiply any measurement by that fraction and not change its value.

Lesson Check – Page No. 320

Question 1. Justin rides his bicycle 2.5 kilometers to school. Luke walks 1,950 meters to school. How much farther does Justin ride to school than Luke walks to school? _______ meters

Answer: 550 meters

Explanation: Justin rides his bicycle 2.5 kilometers to school. Luke walks 1,950 meters to school. 1 kilometer = 1000 meters Then, 2.5 kilometers = 2.5 x 1000 = 2500 meters So, Justin rides his bicycle 2500 meters and Luke walks 1950 meters 2500 – 1950 = 550 meters So, Justin rides more 550 meters than Luke to school

Question 2. The length of a room is 10 \(\frac{1}{2}\) feet. What is the length of the room in inches? _______ inches

Answer: 126 inches

Explanation: 1 feet = 12 inches 10 1/2 feet = ? 10 1/2 = 21/2 21/2 × 12 = 21 × 6 = 126 126 inches

Spiral Review

Go Math Grade 6 Answer Key Chapter 6 Convert Units of Length img 1

Answer: 4 miles

Explanation: Each unit on the map represents 1 mile The distance between the campground and the waterfall is 4 units that is 4 miles

Question 4. On a field trip, 2 vans can carry 32 students. How many students can go on a field trip when there are 6 vans? _______ students

Answer: 96 students

Explanation: On a field trip, 2 vans can carry 32 students So, 1 van can carry the students = 32/2 = 16 students Then, students can go in 6 vans = 6×16 = 96 students

Question 5. According to a 2008 survey, \(\frac{29}{50}\) of all teens have sent at least one text message in their lives. What percent of teens have sent a text message? _______ %

Answer: 58%

Explanation: From the given information According to a 2008 survey 29/50 of all teens have sent at least one text message in their lives Percent of teens have sent a text message = 29/50 x 100 = 58% So, 58% of teens have sent text messages

Question 6. Of the students in Ms. Danver’s class, 6 walk to school. This represents 30% of her students. How many students are in Ms. Danver’s class? _______ students

Answer: 20 students

Explanation: Of the students in Ms. Danver’s class, 6 walk to school It represents 30% of her students That is 30% = 6 students Then 100% = (100×6)/30 = 20 Total number of students in Ms. Danver’s class = 20 students

Share and Show – Page No. 323

Question 1. 5 quarts = ? cups _______ cups

Answer: 20 cups

Explanation: 5 quarts = ? cups 4cups = 1 quart So, 5 quarts = 5×4 = 20 cups 5 quarts = 20 cups

Question 2. 6.7 liters = ? hectoliters _______ hectoliters

Answer: 0.067 hectoliters

Explanation: 1.7 liters = ? hectoliters 1 hectoliter= 100 liters 1 liter = 1/100 hectoliters 6.7 liters = 6.7/100 hectoliters = 0.067 hectoliters

Question 3. 5.3 kL = ? L _______ L

Answer: 5300 L

Explanation: 5.3 kL= ? L 1 Kiloliter = 1000 liters Then, 5.3 kiloliters = 5.3 x 1000 = 5300 liters So, 5.3 kL = 5300 L

Question 4. 36 qt = ? gal _______ gal

Answer: 9 gal

Explanation: 36 qt = ? gal 4 quarts = 1 gallon So, 36 qts = 9×4 quarts = 9×1 gallons So, 36 qt = 9 gallons

Question 5. 5,000 mL = ? cL _______ cL

Answer: 500 cL

Explanation: 5,000 mL = ?cL 1000 milliliters = 1 liter 100 centiliters = 1 liter So, 1000 milliliters = 100 centiliters Then, 5000 milliliters = 5×100 centiliters = 500 centiliters 5000 milliliters = 500 centiliters

Question 6. It takes 41 gallons of water for a washing machine to wash a load of laundry. How many quarts of water does it take to wash one load? _______ quarts

Answer: 164 quarts

Explanation: It takes 41 gallons of water for a washing machine to wash a load of laundry. 41 gallons of water is required for a washing machine to wash a load of laundry 1 gallon = 4 quarts Then, 41 gallons = 41×4 quarts = 164 quarts 164 quarts of water us required for a washing machine to wash a load of laundry

Question 7. Sam squeezed 237 milliliters of juice from 4 oranges. How many liters of juice did Sam squeeze? _______ L

Answer: 0.237 L

Explanation: Sam squeezed 237 milliliters of juice from 4 oranges 1000 liliters = 1 liter 1 milliliter = 1/1000 liter 237 milliliters = 237/1000 liters 237 milliliters = 0.237 liters

Question 8. Reason Quantitatively A bottle contains 3.78 liters of water. Without calculating, determine whether there are more or less than 3.78 deciliters of water in the bottle. Explain your reasoning Type below: ____________

Answer: Reason Quantitatively A bottle contains 3.78 liters of water 1 liter = 10 deciliters Then 3.78 liters = 3.78×10 = 37.8 deciliters So, bottle contains more than 3.78 deciliters of water

Question 9. Tonya has a 1-quart, a 2-quart, and a 3-quart bowl. A recipe asks for 16 ounces of milk. If Tonya is going to triple the recipe, what is the smallest bowl that will hold the milk? The _______ bowl

Answer: The 3 quarts bowl

Explanation: Tonya has a 1-quart, a 2-quart, and a 3-quart bowl A recipe asks for 16 ounces of milk If Tonya triples the recipe, then 1 quart = 3, 2 quart = 6, 3 quart = 9 The smallest bowl is 3 quarts

Question 10. 700,000 L _______ 70 kL

Answer: 700,000 L > 70 kL

Explanation: 700,000 L _______ 70 kL 1 kiloliter = 1000 liters Then, 70 kiloliters = 70×1000 liters = 70,000 liters So, 700,000 liters > 70 kiloliters

Question 11. 6 gal _______ 30 qt

Answer: 6 gal < 30 qt

Explanation: 6 gal _______ 30 qt 4 quarts = 1 gallon So, 6 gallons = 6×4 = 24 quarts So, 6 gallons < 30 quarts

Question 12. 54 kL _______ 540,000 dL

Answer: 54 kL  = 540,000 dL

Explanation: 54 kL _______ 540,000 dL 1 kiloliter = 1000 liters 1 liter = 10 deciliters Then, 1000 liters = 10×1000 = 10,000 deciliters So, 1 kiloliter = 10,000 deciliters Then, 54 kiloliters = 54×10,000 = 540,000 deciliters So, 54 kL = 540,000 dL

Question 13. 10 pt _______ 5 qt

Answer: 10 pt  = 5 qt

Explanation: 10 pt _______ 5 qt 1 pints = 1 quart then, 10 pints = 2×5 pints = 1×5 quart = 5 quarts So, 10 pints = 5 quarts

Question 14. 500 mL _______ 50 L

Answer: 500 mL  < 50 L

Explanation: 500 mL _______ 50 L 1000 milliliters = 1 liter Then, 1000/2 milliliters = 500 milliliters = ½ liters= 0.5 liters So, 500 mL < 50 L

Question 15. 14 c _______ 4 qt

Answer: 14 c  < 4 qt

Explanation:

14 c _______ 4 qt 4 cups = 1 quart 1 cup = ¼ quart Then, 14 cups = 14/4 quarts = 3.5 quarts So, 14 cups < 4 quarts

Unlock the Problem – Page No. 324

Question 16. Jeffrey is loading cases of bottled water onto a freight elevator. There are 24 one-pint bottles in each case. The maximum weight that the elevator can carry is 1,000 pounds. If 1 gallon of water weighs 8.35 pounds, what is the maximum number of full cases Jeffrey can load onto the elevator? a. What do you need to find? Type below: ____________

Answer: the maximum number of full cases Jeffrey can load onto the elevator

Question 16. b. How can you find the weight of 1 case of bottled water? What is the weight? Type below: ____________

Answer: Using one-pint bottles and 1 gallon of water weighs 8.35 pounds information

Question 16. c. How can you find the number of cases that Jeffrey can load onto the elevator? Type below: ____________

Answer: 1 US liquid pint is equivalent to 0.125 US liquid gallons. So, 24 one-pint bottles is equivalent to (24 × 0.125) =3 gallons. Therefore, one full case of bottled water is equal to 3 gallons. Now, 1 gallon is equal to 8.35 pounds And hence, 3 gallons is equal to (8.35 × 3) = 25.05 pounds.

Question 16. d. What is the maximum number of full cases Jeffrey can load onto the elevator? _______ cases

Answer: 39 cases

Explanation: 1 US liquid pint is equivalent to 0.125 US liquid gallons. So, 24 one-pint bottles is equivalent to (24 × 0.125) =3 gallons. Therefore, one full case of bottled water is equal to 3 gallons. Now, 1 gallon is equal to 8.35 pounds And hence, 3 gallons is equal to (8.35 × 3) = 25.05 pounds. If the maximum weight that the elevator can carry is 1000 pounds, then the maximum number of cases of bottled water that the elevator can carry is ≈ 39 We can not take the number as 40, because then the total weight will become more than 1000 pounds which is not allowed.

Question 17. Monica put 1 liter, 1 deciliter, 1 centiliter, and 1 milliliter of water into a bowl. How many milliliters of water did she put in the bowl? _______ milliliters

Answer: 1111 milliliters

Explanation: Monica put 1 liter, 1 deciliter, 1 centiliter, and 1 milliliter of water into a bowl 1 liter = 1000 milliliters 1 liter = 10 deciliters so, 10 deciliters = 1000 milliliters then, 1 deciliter = 100 milliliters 1 liter = 100 centiliters So, 100 centiliters = 1000 milliliters Then, 1 centiliter = 10 milliliters 1 liter + 1 deciliter + 1 centiliter + 1 milliliter = 1000 milliliters + 100 milliliters + 10 milliliters + 1 milliliter = 1111 milliliters Monica filled the bowl with 1111 milliliters of water

Question 18. Select the conversions that are equivalent to 235 liters. Mark all that apply. Options: a. 235,000 milliliters b. 0.235 milliliters c. 235,000 kiloliters d. 0.235 kiloliters

Answer: a. 235,000 milliliters

Explanation: a. 235,000 milliliters 1000 milliliters = 1 liter Then, 235×1000 milliliters = 1×235 liters = 235 liters So, 235,000 milliliters are equivalent to 235 liters

Convert Units of Capacity – Page No. 325

Question 1. 7 gallons = ? quarts _______ quarts

Answer: 28 quarts

Explanation: 6 gallons = ? quarts 4 quarts = 1 gallon then, 7 gallons = 4×7 = 28 quarts

Question 2. 5.1 liters = ? kiloliters _______ kiloliters

Answer: 0.0051 kiloliters

Explanation: 5.1 liters = ? kiloliters 1 kiloliter = 1000 liters So, 1 liter = 1/1000 kiloliter Then, 5.1 liters = 5.1/1000 kiloliters 5.1 liters = 0.0051 kiloliters

Question 3. 20 qt = ? gal _______ gal

Answer: 5 gal

Explanation: 20 t = ? gal 4 quarts = 1 gallon Then, 4×5 quarts = 1×5 gallons That is 20 quarts = 5 gallons

Question 4. 40 L = ? mL _______ mL

Answer: 40,000 mL

Explanation: 40 L = ? mL 1000 milliliters = 1 liter Then, 40 liters = 40×1000 milliliters = 40,000 milliliters 40 L = 40,000 mL

Question 5. 33 pt = ? qt ? pt _______ qt _______ pt

Answer: 33/2 quarts = 16.5 quarts

Explanation: 33 pt= ?qt ? pt 1 pints = 1 quart 1 pint = ½ quart then, 33 pint = 33/2 quarts = 16.5 quarts

Question 6. 29 cL = ? daL _______ daL

Answer: 0.029 daL

Explanation: 29 cL = ? daL 100 centiliters = 1 liter 1 dekaliter = 10 liters So, 1 liter = 1/10 dekaliters Then, 100 centiliters = 1/10 dekaliters 1 centiliter = 1/1000 dekaliters then, 29 centiliters = 29/1000 dekaliters = 0.029 dekaliters 29 cL = 0.029 daL

Question 7. 7.7 kL = ? cL _______ cL

Answer: 7,70,000 cL

Explanation: 6.7 kL = ? cL 1 kiloliter = 1000 liters 100 centiliters = 1 liter So, 1000 liters = 100×1000 centiliters = 1,00,000 centiliters Then, 1 kiloliter = 1,00,000 centiliters Then, 7.7 kiloliters = 7.7 x 1,00,000 centiliters = 7,70,000 centiliters

Question 8. 24 fl oz = ? pt ? c _______ pt _______ c

Answer: 3/2 pt and 3 cups

Explanation: 24 floz= ?pt ? c 6 fluids ounces = 1 cup then, 24 fluid ounces = 8×3 = 1×3 cups = 3 cups 1 cups = 1 pint then, 1 cup = ½ pint then, 3 cups = 3/2 pint so, 24 fluids ounces = 3/2 pint and 3 cups

Problem Solving

Question 9. A bottle contains 3.5 liters of water. A second bottle contains 3,750 milliliters of water. How many more milliliters are in the larger bottle than in the smaller bottle? _______ mL

Answer: 250 mL

Explanation: A bottle contains 3.5 liters of water. A second bottle contains 3,750 milliliters of water. A bottle contains 3.5 liters of water A second bottle contains 3,750 milliliters of water 1000 milliliters = 1 liter Then, 3.5 liters = 3.5×1000 = 3500 milliliters So, 3750 – 3500 = 250 milliliters 250 milliliters of water is more than in the larger bottle than the smaller bottle

Question 10. Arnie’s car used 100 cups of gasoline during a drive. He paid $3.12 per gallon for gas. How much did the gas cost? $ _______

Answer: $19.5

Explanation: Arnie’s car used 100 cups of gasoline during a drive. He paid $3.12 per gallon for gas. Arnie’s car used 100 cups of gasoline during a drive He paid $3.12 per gallon for gas 1 gallon = 4 quarts 1 quart = 4 cups then, 4 quarts = 4×4 cups = 16 cups So, 1 gallon = 16 cups Then, 1 cup = 1/16 gallons Then, 100 cups = 100/16 gallons = 6.25 gallons Total gas cost = $3.12 x 6.25 = $19.5

Question 11. Explain how units of length and capacity are similar in the metric system. Type below: ____________

Answer: In the metric system, The unit of length is a meter (m) and the unit of capacity is the liter (L)

Lesson Check – Page No. 326

Question 1. Gina filled a tub with 25 quarts of water. What is this amount in gallons and quarts? _______ gallons _______ quart

Answer: 6 gallons and 1 quart

Explanation: Gina filled a tub with 25 quarts of water 4quarts = 1 gallon 1 quart = ¼ gallon 25 quarts = 25/4 gallon = 6 gallons and 1 quart Gina filled a tub with 6 gallons and 1 quart

Question 2. Four horses are pulling a wagon. Each horse drinks 45,000 milliliters of water each day. How many liters of water will the horses drink in 5 days? _______ liters

Answer: 900 liters

Explanation: Four horses are pulling a wagon Each horse drinks 45,000 milliliters of water each day Then, four horses drinks 4×45,000 milliliters = 1,80,000 1000 milliliters = 1 liter Then, 180×1000 = 1,80,000 milliliters = 180 liters 180 x 5 = 900 liters Horses drink 900 liters of water in 5 days

Go Math Grade 6 Answer Key Chapter 6 Convert Units of Length img 2

Answer: 7 kilometers

Explanation: The map shows Henry’s town. Each unit represents 1 kilometer. After school, Henry walks to the library. Each unit represents 1 kilometer After school, Henry walks to the library Distance between school and library = 7 kilometers So, henry walks 7 kilometers from school to library

Question 4. An elevator travels 117 feet in 6.5 seconds. What is the elevator’s speed as a unit rate? _______ feet per second

Answer: 18 feet per second

Explanation: An elevator travels 117 feet in 6.5 seconds. The elevator’s speed as a unit rate = 117/6.5 = 18 feet per second

Question 5. Julie’s MP3 player contains 860 songs. If 20% of the songs are rap songs and 15% of the songs are R&B songs, how many of the songs are other types of songs? _______ songs

Answer: 559 songs

Explanation: Julie’s MP3 player contains 860 songs 20% of the songs are rap songs = 860×20/100 = 172 15% of the songs are R & B songs = 860×15/100 = 129 Other types of songs = 860 – 172-129 = 559

Question 6. How many kilometers are equivalent to 3,570 meters? _______ kilometers

Answer: 3.57 kilometers

Explanation: 1 kilometer = 1000 meters then,1 meter = 1/1000 kilometer So, 3570 meters = 3570/1000 kilometer 3570 meters = 3.57 kilometers

Share and Show – Page No. 329

Question 1. 9 pounds = ? ounces _______ ounces

Answer: 144 ounces

Explanation: 6 pounds = ? ounces 1 pound = 16 ounces then, 9 pounds = 9×16 ounces = 144 ounces

Question 2. 3.77 grams = ? dekagram _______ dekagram

Answer: 0.377 dekagram

Explanation: 3.77 grams = ? dekagram 1 dekagram = 10 grams 1 gram = 1/10 dekagram Then, 3.77 grams = 3.77/10 dekagram = 0.377 dekagram So, 3.77 grams = 0.377 dekagram

Question 3. Amanda’s computer weighs 56 ounces. How many pounds does it weigh? _______ pounds

Answer: 3.5 pounds

Explanation: Amanda’s computer weighs 56 ounces 1 pound = 16 ounces then, 1 ounce = 1/16 pound So, 56 ounces = 56/16 pounds = 3.5 pounds

Question 4. A honeybee can carry 40 mg of nectar. How many grams of nectar can a honeybee carry? _______ grams

Answer: 0.04 grams

Explanation: A honeybee can carry 40 mg of nectar. 1000 milligrams = 1 gram 1 milligram = 1/1000 grams Then, 40 milligrams = 40/1000 grams = 0.04 grams So, the honeybee can carry 0.04 grams of nectar

Question 5. 4 lb = ? oz _______ oz

Answer: 64 oz

Explanation: 4lb = ?oz 1 pound (lb) = 16 ounces then, 4 pounds = 4×16 ounces = 64 ounces

Question 6. 7.13 g = ? cg _______ cg

Answer: 713 cg

Explanation: 7.13g = ? cg 100 centigrams = 1 gram Then, 7.13 grams = 100×7.13 = 713 centigrams So, 7.13 grams = 713 centigrams

Question 7. 3 T = ? lb _______ lb

Answer: 6000 lb

Explanation: 3T = ?lb 1 ton = 2000 pounds (lb) then, 3 tons = 3×2000 = 6000 pounds (lb)

Question 8. The African Goliath frog can weigh up to 7 pounds. How many ounces can the Goliath frog weigh? _______ ounces

Answer: 112 ounces

Explanation: The African Goliath frog can weigh up to 7 pounds. 1 pound = 16 ounces 7 pounds = 7×16 = 112 pounds So, the goliath frog can weigh up to 112 pounds

Question 9. The mass of a standard hockey puck must be at least 156 grams. What is the minimum mass of 8 hockey pucks in kilograms? _______ kg

Answer: 1.248 kg

Explanation: The mass of a standard hockey puck must be at least 156 grams. 1 kilogram = 1000 grams 1 gram = 1/1000 kilogram then, 156 grams = 156/1000 kilograms = 0.156 kilograms mass of a hockey puck is 0.156 kilograms then, the mass of 8 hockey pucks is 8×0.156 = 1.248 kilograms

Question 10. 250 lb _______ 0.25 T

Answer: 250 lb < 0.25 T

Explanation: 250 lb_______ 0.25 T 1 ton = 2000 pounds(lb) then, 0.25 tons =0.25×2000 = 500 pounds = 500lb So, 250 lb < 0.25 T

Question 11. 65.3 hg _______ 653 dag

Answer: 65.3 hg = 653 dag

Explanation: 65.3 hg _______ 653 dag 1 hectogram = 100 grams Then, 65.3 hectograms = 65.3×100 = 6530 grams 1 dekagram = 10 grams then, 653 dekagram = 6530 grams So, 65.3 hectogram = 653 dekagram

Question 12. 5 T _______ 5,000 lb

Answer: 5 T  > 5,000 lb

Explanation: 5 T _______ 5,000 lb 1 ton = 2000 pounds (lb) 5 tons = 5×2000 lb = 10,000 lb Then, 5 T > 5000 lb

Question 13. Masses of precious stones are measured in carats, where 1 carat = 200 milligrams. What is the mass of a 50-dg diamond in carats? _______ carats

Answer: 25 carats

Explanation: Masses of precious stones are measured in carats, where 1 carat = 200 milligrams. 1 carat = 200 milligrams 6 decigrams = 1 gram 1000 milligrams = 1 gram So, 10 decigrams = 1000 milligrams Then, 1 decigram = 100 milligram 2 decigrams = 200 milligrams = 1 carat then, 50 decigrams = 2×25 decigrams = 25×200 milligrams = 25 carats

Problem Solving + Applications – Page No. 330

Go Math Grade 6 Answer Key Chapter 6 Convert Units of Length img 3

Question 14. Express the weight range for bowling balls in pounds. _______ lb

Answer: 16 lb

Explanation: Weight range for bowling balls = 160 to 256 ounces 1 pound = 16 ounces So, 1 ounce = 1/16 pounds Then, 160 ounces = 160/16 pounds = 10 pounds 256 ounces = 256/16 pounds = 16 pounds So, the weight range for bowling balls is 10 to 16 pounds

Question 15. How many more pounds does the heaviest soccer ball weigh than the heaviest baseball? Round your answer to the nearest hundredth. _______ lb

Answer: 0.68 lb

Explanation: Heaviest soccer ball weight = 16 ounces 1 pound = 16 ounces Heaviest baseball weight = 5.25 ounces 1 pound = 16 ounces 1 ounce = 1/16 pounds then, 5.25 ounces = 5.25/16 = 0.32 pounds difference between soccer ball and baseball weight = 1 – 0.32 = 0.68 pounds So, the soccer ball weight is 0.68 pounds more than the weight of the baseball.

Question 16. A manufacturer produces 3 tons of baseballs per day and packs them in cartons of 24 baseballs each. If all of the balls are the minimum allowable weight, how many cartons of balls does the company produce each day? _______ cartons

Answer: 800 cartons

Explanation: 3 tons = 6000 lbs. Base ball = 5 ounces 16 ounces in 1 pound 6000 × 16 = 96,000 96,000/5 = 19,200 19,200/24 = 800

Question 17. Communicate Explain how you could use mental math to estimate the number of soccer balls it would take to produce a total weight of 1 ton. Type below: ____________

Answer: Soccer balls range 14 to 16 ounces 1 ton = 2000 pounds then, 1 pound = 1/2000 tons 1 pound = 16 ounces So, 16 ounces = 1/2000 tons = 0.0005 tons 1 ounce = 1/32000 tons then, 14 ounces = 14/32000 tons =0.0004375 tons So, the range of soccer balls is 0.0005 to 0.0004375 tons

Question 18. The Wilson family’s newborn baby weighs 84 ounces. Choose the numbers to show the baby’s weight in pounds and ounces. _______ pounds and _______ ounces

Answer: 5 pounds and 4 ounces

Explanation: The Wilson family’s newborn baby weighs 84 ounces 1 pound = 16 ounces then, 1 ounce = 1/16 pounds So, 84 ounces = 84/16 pounds = 5 pounds and 4 ounces

Convert Units of Weight and Mass – Page No. 331

Question 1. 5 pounds = ? ounces _______ ounces

Answer: 80 ounces

Explanation: 5 pounds = ? ounces 1 pound = 16 ounces Then, 5 pounds = 5×16 = 80 ounces So, 5 pounds = 80 ounces

Question 2. 2.36 grams = ? hectograms _______ hectograms

Answer: 0.0236 hectograms

Explanation: 1.36 grams = ? hectograms 1 hectogram = 100 grams 1 gram = 1/100 hectograms then, 2.36 grams = 2.36/100 hectograms = 0.0236 hectograms So, 2.36 grams = 0.0236 hectograms

Question 3. 30 g = ? dg _______ dg

Answer: 300 dg

Explanation: 29 g = ? dg 10 decigrams = 1 gram then, 30 grams = 30×10 decigrams = 300 decigrams 30 grams = 300 decigrams

Question 4. 17.2 hg = ? g _______ g

Answer: 1720 g

Explanation: 17.2 hg = ? g 1 hectogram = 100 grams Then, 17.2 hectograms = 17.2×100 = 1720 grams So, 17.2 hectograms = 1720 grams

Question 5. 400 lb = ? T _______ T

Answer: 0.2 T

Explanation: 1. 400 lb = ? T 1 ton = 2000 pounds (lb) 400 lb = 2000/5 pounds (lb) = 1/5 tons So, 400 lb = 0.2 tons

Question 6. 38,600 mg = ? dag _______ dag

Answer: 3.86 dag

Explanation: 38,600 mg = ? dag 1000 milligrams = 1 gram 1 dekagram = 10 grams So, 1 gram = 1/10 dekagram Then, 1000 milligrams = 1/10 dekagrams 1 milligram = 1/10,000 dekagrams So, 38,600 milligrams = 38,600/10,000 = 3.86 dekagrams 38,600 milligrams = 3.86 dekagrams

Question 7. 87 oz = ? lb ? oz _______ pounds _______ ounces

Answer: 5 pounds and 7 ounces

Explanation: 87 oz = ? lb ? oz 1 pound = 16 ounces 1 ounce = 1/16 pounds then, 87 ounces = 87/16 pounds 87 ounces = 5 pounds and 7 ounces

Question 8. 0.65 T = ? lb _______ lb

Answer: 1300 lb

Explanation: 0.65 T = ?lb 1 ton = 2000 pounds Then, 0.65 tons = 0.65×2000 = 1300 pounds 0.65 T = 1300 lb

Question 9. Maggie bought 52 ounces of swordfish selling for $6.92 per pound. What was the total cost? $ _______

Answer: $22.49

Explanation: Maggie bought 52 ounces of swordfish selling for $6.92 per pound. Maggie bought 52 ounces of swordfish selling for $6.92 per pound 1 pound = 16 ounces 1 ounce = 1/16 pounds then, 52 ounces = 52/16 pounds = 3.25 pounds 1 pound cost = $6.92 then, 3.25 pounds cost = $6.92 x 3.25 = $22.49 So, the cost for swordfish is $22.49

Question 10. Three bunches of grapes have masses of 1,000 centigrams, 1,000 decigrams, and 1,000 grams, respectively. What is the total combined mass of the grapes in kilograms? _______ kg

Answer: 1.11 kg

Explanation: Three bunches of grapes have masses of 1,000 centigrams, 1,000 decigrams, and 1,000 grams, respectively. Three bunches of grapes have masses of 1,000 centigrams, 1,000 decigrams, and 1,000 grams 100 centigrams = 1 gram then, 1000 centigrams = 10×100 centigrams = 10 grams 1 kilogram = 1000 grams So, 1 gram = 1/1000 kilograms Then, 10 grams = 10/1000 = 1/100 kilograms = 0.01 kilograms 10 decigrams = 1 gram then, 100×10 decigrams = 100×1 gram = 100 grams 1000 grams = 1 kilogram Then, 100 grams = 1/10 kilograms = 0.1 kilograms 1000 grams = 1 kilogram Total weight of the grapes = 1 + 0.1 + 0.01 = 1.11 kilograms

Question 11. Explain how you would find the number of ounces in 0.25T. Type below: ____________

Answer: number of ounces in 0.25T 1 ton = 2000 pounds then, 1 pound = 1/2000 tons 1 pound = 16 ounces so, 16 ounces = 1/2000 tons then, 1 ton = 16×2000 ounces = 32000 ounces So, 0.25 tons = 0.25×32000 ounces = 8000 ounces 8000 ounces = 0.25 T

Lesson Check – Page No. 332

Question 1. The mass of Denise’s rock sample is 684 grams. The mass of Pauline’s rock sample is 29,510 centigrams. How much greater is the mass of Denise’s sample than Pauline’s sample? _______ centigrams

Answer: 38900 centigrams

Explanation: The mass of Denise’s rock sample is 684 grams The mass of Pauline’s rock sample is 29,510 centigrams 100 centigrams = 1 gram 1 centigram = 1/100 gram then, 29,510 centigrams = 29,510/100 grams = 295.1 grams So, the mass of Pauline’s rock sample is 295.1 grams By comparing Denise’s rock sample with Pauline’s rock sample 684 – 295 = 389 The mass of Denise’s rock sample is 389 grams more than the mass of Pauline’s rock sample 389 grams = 38900 centigrams

Question 2. A sign at the entrance to a bridge reads: Maximum allowable weight 2.25 tons. Jason’s truck weighs 2,150 pounds. How much additional weight can he carry? _______ pounds

Answer: 2,350 pounds

Explanation: A sign at the entrance to a bridge reads: Maximum allowable weight 2.25 tons Jason’s truck weighs 2,150 pounds 1 ton = 2000 pounds then, 2.25 tons = 2.25×2000 = 4500 pounds So, maximum allowable weight = 4500 pounds 4500 – 2150 = 2350 So, Jason can carry an additionally 2350 pounds’ weight

Question 3. There are 23 students in a math class. Twelve of them are boys. What is the ratio of girls to total number of students? Type below: ____________

Answer: 11 : 23

Explanation: There are 23 students in a math class. Twelve of them are boys. Number of students in a math class = 23 Number of boys in a class = 12 Number of girls in a class = 23-12 = 11 Then, the ratio of girls to the total number of students = 11/23

Question 4. Miguel hiked 3 miles in 54 minutes. At this rate, how long will it take him to hike 5 miles? _______ minutes

Answer: 90 minutes

Explanation: Miguel hiked 3 miles in 54 minutes. Then, time for 5 miles = 5×54/3 = 90 minutes So, Miguel hikes 5 miles in 90 minutes

Question 5. Marco borrowed $150 from his brother. He has paid back 30% so far. How much money does Marco still owe his brother? $ _______

Answer: $60

Explanation: Marco borrowed $150 from his brother He has paid back 30% of amount = 30/100 (150) = $45 Remaining amount = 150 -45 = 60 So, still $60 amount Marco need to give his brother

Question 6. How many milliliters are equivalent to 2.7 liters? _______ milliliters

Answer: 2,700 milliliters

Explanation: 2.7 liters 1000 milliliters = 1 liter Then, 2.7 liters = 2.7 x 1000 = 2700 milliliters So, 2,700 milliliters are equivalent to 2.7 liters

Mid-Chapter Checkpoint – Vocabulary – Page No. 333

Go Math Grade 6 Answer Key Chapter 6 Convert Units of Length img 4

Question 1. A _____ is a rate in which the two quantities are equal, but use different units. Type below: ____________

Answer: Conversion factor

Question 2. _____ is the amount a container can hold. Type below: ____________

Answer: Capacity

Concepts and Skills

Convert units to solve.

Question 3. A professional football field is 160 feet wide. What is the width of the field in yards? _____ \(\frac{□}{□}\) yd

Answer: 53\(\frac{1}{3}\) yd

Explanation: A professional football field is 160 feet wide 3feet = 1 yard Then, 160 feet = 160/3 = 53.33 So, the width of football field is 53.33 yards 160/3 = 53 1/3

Question 4. Julia drinks 8 cups of water per day. How many quarts of water does she drink per day? _____ quarts

Answer: 2 quarts

Explanation: Julia drinks 8 cups of water per day. 4 cups = 1 quart Then, 8 cups = 8/4 = 2 quarts So, Julia drinks 2 quarts of water per day

Question 5. The mass of Hinto’s math book is 4,458 grams. What is the mass of 4 math books in kilograms? _____ kilograms

Answer: 17.832 kilograms

Explanation: The mass of Hinto’s math book is 4,458 grams 1kilogram = 1000 grams Then, 4,458 grams = 4,458/1000 = 4.458 kilograms Then, the mass of 4 math books = 4×4.458 = 17.832 kilograms The mass of 4 math books is 17.832 kilograms

Question 6. Turning off the water while brushing your teeth saves 379 centiliters of water. How many liters of water can you save if you turn off the water the next 3 times you brush your teeth? _____ liters

Answer: 11.37 liters

Explanation: Turning off the water while brushing your teeth saves 379 centiliters of water 100centiliters = 1 liter Then, 379 centiliters = 379/100 = 3.79 liters if you turn off the water the next 3 times = 3×3.79 liters = 11.37 liters So, you can save 11.37 liters of water when you turn off the water for 3 times

Question 7. 34.2 mm = ? cm _____ cm

Answer: 3.42 cm

Explanation: 34.2 mm = ? cm 1000 millimeters = 1 meter 100centimeters = 1 meter so, 1000 millimeters = 100 centimeters then, 10 millimeters = 1 centimeter then, 34.2 millimeters = 34.2/10 = 3.42 centimeters So, 34.2 mm = 3.42 cm

Question 8. 42 in. = ? ft _____ \(\frac{□}{□}\) ft

Answer: 3\(\frac{1}{2}\) ft

Explanation: 41 in. = ? ft 12 inches = 1 foot then, 42 inches = 42/12 = 3.5 feet So, 42 in = 3.5 ft 42/12 = 3 1/2

Question 9. 1.4 km = ? hm _____ hm

Answer: 140 hm

Explanation: 1.4 km = ? hm 1 kilometer = 1000 meters 1 hectometer = 100 meters So, 1 meter = 0.001 kilometers 1 meter = 0.01 hectometers Now, 0.001 kilometer = 0.01 hectometer That is 0.1 kilometer = 1 hectometer Then, 1.4 kilometer = 1.4/0.1 = 140 hectometers So, 1.4 km = 140 hm

Question 10. 4 gal = ? qt _____ qt

Answer: 16 qt

Explanation: 4gal = ?qt 1gallon = 4 quarts Then, 4 gallons = 4×4 = 16 quarts So, 4 gal = 16 qt

Question 11. 53 dL = ? daL _____ daL

Answer: 0.53 daL

Explanation: 53 dL = ? daL 10deciliters = 1 liter 1 dekaliter = 10 liters that is 0.1 dekaliters = 1 liter So, 10 dL = 0.1 daL Then, 53 dL = 53×0.1/10 =0.53 daL So, 53 dL = 0.53 daL

Question 12. 28 c = ? pt _____ pt

Answer: 14 pt

Explanation: 28 c = ?pt 1 cups = 1pint then, 28 cups = 28/2 = 14 pints So, 28 c = 14 pt

Page No. 334

Question 13. Trenton’s laptop is 32 centimeters wide. What is the width of the laptop in decimeters? _____ dm

Answer: 3.2 dm

Explanation: Trenton’s laptop is 32 centimeters wide. 100 centimeters = 1 meter 10decimeters = 1 meter So, 100 centimeters = 10 decimeters Then, 32 centimeters = 32×10/100 = 3.2 decimeters So, the width of the laptop is 3.2 decimeters

Question 14. A truck is carrying 8 cars weighing an average of 4,500 pounds each. What is the total weight in tons of the cars on the truck? _____ tons

Answer: 18 tons

Explanation: A truck is carrying 8 cars weighing an average of 4,500 pounds each. So, total weight = 8 x 4500 pounds = 36,000 pounds 2000 pounds = 1 ton Then, 36,000 pounds = 36,000 / 2000 = 18 tons So, total weight of the cars in truck is 18 tons

Question 15. Ben’s living room is a rectangle measuring 10 yards by 168 inches. By how many feet does the length of the room exceed the width? _____ feet

Answer: 16 feet

Explanation: Ben’s living room is a rectangle measuring 10 yards by 168 inches. 12inches = 1 foot Then, 168 inches = 168/12 = 14 feet 1 yard = 3 feet then, 10 yards = 10×3 = 30 feet 30-14 = 16 feet So, the length of the room exceeds 16 feet in width

Question 16. Jessie served 13 pints of orange juice at her party. How many quarts of orange juice did she serve? _____ quarts

Answer: 6.5 quarts

Explanation: Jessie served 13 pints of orange juice at her party 1 pints = 1 quart then, 13 pints = 13/2 = 6.5 quarts So, Jessie served 6.5 quarts of orange juice at her party

Question 17. Kaylah’s cell phone has a mass of 50,000 centigrams. What is the mass of her phone in grams? _____ grams

Answer: 500 grams

Explanation: Kaylah’s cell phone has a mass of 50,000 centigrams 100 centigrams = 1 gram then, 50,000 centigrams = 50,000/100 = 500 grams So, the mass of Kaylah’s phone is 500 grams

Share and Show – Page No. 337

Question 1. A dripping faucet leaks 12 gallons of water per day. How many gallons does the faucet leak in 6 days? _____ gallons

Answer: 72 gallons

Explanation: A dripping faucet leaks 12 gallons of water per day Then, faucet leaks how many gallons of water per 6 days = 12 x 6 = 72 gallons

Question 2. Bananas sell for $0.44 per pound. How much will 7 pounds of bananas cost? $ _____

Answer: $3.08

Explanation: Bananas sell for $0.44 per pound 1 pound banana cost is $0.44 then, 7 pounds bananas cost is = 7 x 0.44 = $3.08

Question 3. Grizzly Park is a rectangular park with an area of 24 square miles. The park is 3 miles wide. What is its length in miles? _____ miles

Answer: 8 miles

Explanation: Grizzly Park is a rectangular park with an area of 24 square miles The park is 3 miles wide Rectangular park area = length x breadth That is 24 = 3 x b So, breadth = 8 miles The rectangular park length is 8 miles

Multiply or divide the quantities.

Question 4. \(\frac{24 \mathrm{kg}}{1 \mathrm{min}}\) × 15 min _____ kg

Answer: 6 kg

Explanation: 24kg1min × 15 min 24 kg / 1min x 15 min 60 min = 1 hour Then, 15 min = 15/60 = ¼ hours 24 kg x 1/ 4 = 6 kg

Question 5. 216 sq cm÷8 cm _____ cm

Answer: 27 cm

Explanation: 216 sq cm ÷ 8 cm 216 sq cm/ 8 cm = 27 cm

Question 6. \(\frac{17 \mathrm{L}}{1 \mathrm{hr}}\) × 9 hr _____ L

Answer: 153 L

Explanation: 17L1hr x 9 hr 17L/1hr x 9 hr = 153 L

Question 7. The rectangular rug in Marcia’s living room measures 12 feet by 108 inches. What is the rug’s area in square feet? _____ square feet

Answer: 108 square feet

Explanation: The rectangular rug in Marcia’s living room measures 12 feet by 108 inches 1 foot = 12 inches 108 inches = 108/12 = 9 feet 12 x 9 = 108 square feet Are of rug is 108 square feet

Question 8. Make Sense of Problems A box-making machine makes cardboard boxes at a rate of 72 boxes per minute. How many minutes does it take to make 360 boxes? _____ minutes

Answer: 5 minutes

Explanation: A box-making machine makes cardboard boxes at a rate of 72 boxes per minute Then, time for 360 boxes = 360/72 = 5 minutes So, it takes 5 minutes’ time to make 360 boxes

Question 9. The area of an Olympic-size swimming pool is 1,250 square meters. The length of the pool is 5,000 centimeters. Select True or False for each statement. 9a. The length of the pool is 50 meters. 9b. The width of the pool is 25 meters. 9c. The area of the pool is 1.25 square kilometers 9a. ____________ 9b. ____________ 9c. ____________

Answer: 9a. True 9b. True 9c. True

Explanation: The area of an Olympic-size swimming pool is 1,250 square meters The length of the pool is 5,000 centimeters 100centimeters = 1meter Then, 5000 centimeters = 5000/100 = 50 meters Areas of the swimming pool = length x width 1250 square meters = 50 length x 25 width Then, width = 25 meters 1000 meters = 1 kilometer then, 1250 square meters = 1250/1000 = 1.25 square meters

Make Predictions – Page No. 338

A prediction is a guess about something in the future. A prediction is more likely to be accurate if it is based on facts and logical reasoning.

Go Math Grade 6 Answer Key Chapter 6 Convert Units of Length img 5

Question 10. An average of 19,230 people tour the Hoover Dam each week. Predict the number of people touring the dam in a year. _____ people

Answer: 999,960 people

Explanation: An average of 19,230 people tour the Hoover Dam each week Number of weeks per year = 52 Then, total number of people tour the hoover dam in the year = 52 x 19, 230 = 999,960 So, 999,960 people touring the hoover dam per year

Question 11. The Hoover Dam generates an average of about 11,506,000 kilowatt-hours of electricity per day. Predict the number of kilowatt-hours generated in 7 weeks. _____ kilowatt-hours

Answer: 563,794 kilowatt-hours

Explanation: The Hoover Dam generates an average of about 11,506,000 kilowatt-hours of electricity per day 1 week = 7 days 7weeks = 7 × 7 = 49 days Then, Hoover Dam generated electricity per 7 weeks = 49 × 11,506,000 = 563,794,000 So, the total number of kilowatt-hours generated in 7 weeks by the Hoover Dam is 563,794,000

Transform Units – Page No. 339

Question 1. \(\frac{62 \mathrm{g}}{1 \mathrm{day}}\) × 4 days _____ g

Answer: 248 g

Explanation: 62g1day × 4 days 62 g÷1 day × 4 days Then, 62 g × 4 = 248 g

Question 2. 322 sq yd ÷ 23 yd _____ yd

Answer: 14 yd

Explanation: 322 sqyd ÷ 23 yd 322 sqyd / 23 yd = 14 sq

Question 3. \(\frac{128 \mathrm{kg}}{1 \mathrm{hr}}\) × 10 hr _____ kg

Answer: 1,280 kg

Explanation: 128kg1hr × 10 hr 128 kg/1hr * 10hr So, 1,280 kg

Question 4. 136 sq km ÷ 8 km _____ km

Answer: 17 km

Explanation: 136 sq km ÷ 8 km 136 sq km / 8 km 136 sq / 8 = 17

Question 5. \(\frac{88 \mathrm{lb}}{1 \mathrm{day}}\) × 12 days _____ lb

Answer: 1,056 lb

Explanation: 88lb1day × 12 days 88lb / 1 day × 12days That is 88lb × 12 = 1,056 lb

Question 6. 154 sq mm ÷ 11 mm _____ mm

Answer: 14  mm

Explanation: 154 sq mm ÷ 11 mm 154 sq / 11 = 14

Question 7. \(\frac{\$ 150}{1 \mathrm{sq} \mathrm{ft}}\) × 20 sq ft $ _____

Answer: $30,020 sqft

Explanation: $1501sqft × 20 sqft Multiplication of 1501 and 20 is 30,020 That is $1501sqft x 20 sqft = $30,020 sqft

Question 8. 234 sq ft÷18 ft _____ ft

Answer: 13 ft

Explanation: 234 sq ft÷18 ft 234 sq / 18 = 13

Question 9. Green grapes are on sale for $2.50 a pound. How much will 9 pounds cost? $ _____

Answer: $22.5

Explanation: Green grapes are on sale for $2.50 a pound 1 pound = $2.50 then, 9 pounds cost = 9*$2.50 = $22.5 green grapes cost for 9 pounds is $22.5

Question 10. A car travels 32 miles for each gallon of gas. How many gallons of gas does it need to travel 192 miles? _____ gallons

Answer: 6 gallons

Explanation: A car travels 32 miles for each gallon of gas Then, 192 miles is = 192/ 32 = 6 gallons of gas So, total 6 gallons of gas is required to travel 192 miles

Question 11. Write and solve a problem in which you have to transform units. Use the rate 45 people per hour in your question. Type below: ____________

Answer: A fast-food restaurant is trying to find out how many customers they had in the last 3 hours, and they know they get 45 people per hour. How many customers were served in the last 3 hours? The answer is 135 people.

Lesson Check – Page No. 340

Question 1. A rectangular parking lot has an area of 682 square yards. The lot is 22 yards wide. What is the length of the parking lot? _____ yards

Answer: 31 yards

Explanation: A rectangular parking lot has an area of 682 square yards Width of the parking lot = 22 yards wide Area = length *width 682 square yards= length * 22 yards wide So, length = 682 square yards / 22 yards = 31 yards Then, length of the parking lot = 31 yards

Question 2. A machine assembles 44 key chains per hour. How many key chains does the machine assemble in 11 hours? _____ key chains

Answer: 484 key chains

Explanation: A machine assembles 44 key chains per hour Then, the machine assembles key chains per 11 hours = 11*44 = 484 key chains So, the machine assembles totally 484 key chains in 11 hours

Question 3. Three of these ratios are equivalent to \(\frac{8}{20}\). Which one is NOT equivalent? \(\frac{2}{5} \quad \frac{12}{24} \quad \frac{16}{40} \quad \frac{40}{100}\) \(\frac{□}{□}\)

Answer: \(\frac{8}{20}\)

Explanation: The below mentioned ratios are equivalent to 8/20 i. 2/5 Multiply the numerator and denominator with 4 That is (2*4)/(5*4) = 8/20 ii. 12/24 Divide the numerator and denominator with 6 That is (12÷6)/(24÷6) = 2/4 Now, multiply the numerator and denominator with 4 That is (2*4)/(4*4) = 8/16 So, 12/14 is not equal to 8/20 iii. 16/40 Divide the numerator and denominator with 2 That is, (16÷2)/(40÷2) = 8/20 iv. 40/100 Divide the numerator and denominator with 5 That is (40÷5)/(100÷5) = 8/20

Go Math Grade 6 Answer Key Chapter 6 Convert Units of Length img 6

Answer: $80

Explanation: Total number of days worked = 7 Total earned money = 560 dollars 560 / 7 = 80 dollars per day

Question 5. Megan answered 18 questions correctly on a test. That is 75% of the total number of questions. How many questions were on the test? _____ questions

Answer: 24 questions

Explanation: Megan answered 18 questions correctly That is 75% of the total number of questions = 18 Then, 100% of the questions = 18*100/75 = 24 So, the total number of questions on the test = 24 questions

Share and Show – Page No. 343

Question 1. Mariana runs at a rate of 180 meters per minute. How far does she run in 5 minutes? _____ meters

Answer: 900 meters

Explanation: Mariana runs at a rate of 180 meters per minute Then, Mariana runs per 5 minutes = 5×180 = 900 meters So, Mariana runs 900 meters per 5 minutes

Question 2. What if Mariana runs for 20 minutes at the same speed? How many kilometers will she run? _____ kilometers

Answer: 3.6 kilometers

Explanation: From the given information Marians runs at a rate of 180 meters per minute Then the speed of Mariana = 180/1 = 180 meters per minute If Mariana runs 20 minutes then the covered distance = 20×180 = 3600 meters 1000 meters = 1 kilometer Then, 3600 meters = 3600/1000 = 3.6 kilometers So, Mariana runs 3.6 kilometers in 20 minutes

Question 3. A car traveled 130 miles in 2 hours. How fast did the car travel? _____ miles per hour

Answer: 65 miles per hour

Explanation: A car travelled 130 miles in 2 hours Then the speed of the car = Distance/Time That is, Speed of the car = 130 miles/ 2 hours = 65 miles per hour So, the car travels 65 miles per hour

Question 4. A subway car travels at a rate of 32 feet per second. How far does it travel in 16 seconds? _____ feet

Answer: 512 feet

Explanation: A subway car travels at a rate of 32 feet per second 1 second = 32 feet then, 16 seconds = 16 x 32/1 = 512 feet So, a subway car travels 512 feet per 16 seconds

Question 5. A garden snail travels at a rate of 2.6 feet per minute. At this rate, how long will it take for the snail to travel 65 feet? _____ minutes

Answer: 25 minutes

Explanation: A garden snail travels at a rate of 2.6 feet per minute So, 2.6 feet = 1 minute Then, 65 feet = 65/2.6 = 650/26 = 25 minutes So, the snail travels 65 feet in 25 minutes

Question 6. A squirrel can run at a maximum speed of 12 miles per hour. At this rate, how many seconds will it take the squirrel to run 3 miles? _____ seconds

Answer: 900 seconds

Explanation: A squirrel can run at a maximum speed of 12 miles per hour 1 hour = 3600 seconds So, the squirrel can run 12 miles in 3600 seconds Then, the squirrel can run 3 miles in 3×3600/12 = 900 seconds So, the squirrel can take 900 seconds of time to run 3 miles

Question 7. A cyclist rides 8 miles in 32 minutes. What is the speed of the cyclist in miles per hour? _____ miles per hour

Answer: 15 miles per hour

Explanation: A cyclist rides 8 miles in 32 minutes 32minutes = 8 miles Then, 60 minutes = 60×8/32 = 15 miles So, a cyclist rides 15 miles in 60 minutes that is one hour So, the speed of the cyclist per hour = 15 miles/ 1 = 15 miles per hour

Share and Show – Page No. 344

Question 8. A pilot flies 441 kilometers in 31.5 minutes. What is the speed of the airplane? _____ kilometers per minute

Answer: 14 kilometers per minute

Explanation: From the given information A pilot flies 441 kilometers in 31.5 minutes Speed = Distance / Time Here, distance = 441 kilometers Time = 31.5 minutes Speed of the airplane = 441/31.5 = 4410/315 = 14 kilometers per minute

Question 9. Chris spent half of his money on a pair of headphones. Then he spent half of his remaining money on CDs. Finally, he spent his remaining $12.75 on a book. How much money did Chris have to begin with? $ _____

Answer: $51

Explanation: Total money with the Chris= x amount Chris spent half of his money on a pair of headphones = x/2 Then he spent half of his remaining money on CDs = x/4 Finally, he spent his remaining $12.75 on a book So, total amount x = x/2+x/4+$12.75 $12.75 = (x-x/2-x/4) = (4x-2x-x)/4 $12.75 = x/4 Then, x = $12.75×4 = $51 So, Chris have to begin with $51

Question 10. André and Yazmeen leave at the same time and travel 75 miles to a fair. André drives 11 miles in 12 minutes. Yazmeen drives 26 miles in 24 minutes. If they continue at the same rates, who will arrive at the fair first? Explain. ____________

Answer: André and Yazmeen leave at the same time and travel 75 miles to a fair André drives 11 miles in 12 minutes So, Andre can reach 75 miles in = 75×12/11 That is, Andre can travel 75 miles in 81 minutes Yazmeen drives 26 miles in 24 minutes So, Yazmeen can reach 75 miles in = 75×24/26 = 69 minutes That means, Yazmeen can reach 75 miles in 69 minutes So, Yazmeen can reach the fair first

Question 11. Make Arguments Bonnie says that if she drives at an average rate of 40 miles per hour, it will take her about 2 hours to drive 20 miles across town. Does Bonnie’s statement make sense? Explain. ____________

Answer: Make Arguments Bonnie says that if she drives at an average rate of 40 miles per hour, it will take her about 2 hours to drive 20 miles across town Speed of the Bonnie = 40 miles per hour Then, Bonnie can cover the distance in 2 hours = 2×40 = 80 miles So, Bonnie statement is wrong

Question 12. Claire says that if she runs at an average rate of 6 miles per hour, it will take her about 2 hours to run 18 miles. Do you agree or disagree with Claire? Use numbers and words to support your answer. ____________

Answer: Claire says that if she runs at an average rate of 6 miles per hour, it will take her about 2 hours to run 18 miles Claire runs in 1 hour = 6 miles Then, Claire runs in 2 hours = 2×6 = 12 miles So, the Claire statement is wrong

Problem Solving Distance, Rate, and Time Formulas – Page No. 345

Read each problem and solve.

Question 1. A downhill skier is traveling at a rate of 0.5 mile per minute. How far will the skier travel in 18 minutes? _____ miles

Answer: 9 miles

Explanation: A downhill skier is traveling at a rate of 0.5 miles per minute 1 minute = 0.5 mile then, 18 minutes = 18×0.5 = 9 miles So, the skier travel 9 miles in 18 minutes

Question 2. How long will it take a seal swimming at a speed of 8 miles per hour to travel 52 miles? _____ hours

Answer: 6.5 hours

Explanation: A seal swimming at a speed of 8 miles per hour Then,52 miles = 52/8 = 6.5 hours So, A seal swimming can travel 52 miles in 6.5 hours

Question 3. A dragonfly traveled at a rate of 35 miles per hour for 2.5 hours. What distance did the dragonfly travel? _____ miles

Answer: 87.5 miles

Explanation: A dragonfly traveled at a rate of 35 miles per hour for 2.5 hours That means, 1 hour = 35 miles Then, 2.5 hours = 2.5×35 = 87.5 miles So, a dragonfly travels 87.5 miles in 2.5 hours

Question 4. A race car travels 1,212 kilometers in 4 hours. What is the car’s rate of speed? _____ kilometers per hour

Answer: 303 kilometers per hour

Explanation: A race car travels 1,212 kilometers in 4 hours Speed = Distance/ Time Here, distance = 1212 kilometers Time = 4 hours Then, Speed of the race car = 1212/4 = 303 kilometers per hour

Question 5. Kim and Jay leave at the same time to travel 25 miles to the beach. Kim drives 9 miles in 12 minutes. Jay drives 10 miles in 15 minutes. If they both continue at the same rate, who will arrive at the beach first? ____________

Answer: Kim reaches the beach first

Explanation: Kim and Jay leave at the same time to travel 25 miles to the beach Kim drives 9 miles in 12 minutes Then, Kim travels 25 miles in = 25×12/9 = 33 minutes Jay drives 10 miles in 15 minutes Then, Jay travels 25 miles in = 25×15/10 = 37.5 minutes So, Kim reaches the beach first

Question 6. Describe the location of the variable d in the formulas involving rate, time, and distance. Type below: ____________

Answer: Formula Distance = Rate x Time Distance (d) = Rate x Time

Lesson Check – Page No. 346

Question 1. Mark cycled 25 miles at a rate of 10 miles per hour. How long did it take Mark to cycle 25 miles? _____ hours

Answer: 2.5 hours

Explanation: Mark cycled 25 miles at a rate of 10 miles per hour That means, 10 miles = 1 hour Then, 25 miles = 25/10 =2.5 hours So, Mark take 2.5 hours to cycle 25 miles

Question 2. Joy ran 13 miles in 3 \(\frac{1}{4}\) hours. What was her average rate? _____ miles per hour

Answer: 4 miles per hour

Explanation: Joy ran 13 miles in 3 ¼ hours 3 ¼ = 13/4 = 3.25 hours Then, the average rate of the Joy = 13/3.25 hours = 4 miles per hour

Question 3. Write two ratios that are equivalent to \(\frac{9}{12}\). Type below: ____________

Answer: 3/4 and 18/24

Explanation: Equivalent ratios of 9/12 is 3/4 and 18/24 Multiply the numerator and denominator of ¾ with 3 That is 3×3/4×3 = 9/12 Divide the numerator and denominator of 18/24 with 2 That is (18/2)/(24/2) = 9/12

Question 4. In the Chang family’s budget, 0.6% of the expenses are for internet service. What fraction of the family’s expenses is for internet service? Write the fraction in simplest form. \(\frac{□}{□}\)

Answer: \(\frac{3}{500}\)

Explanation: In the Chang family’s budget, 0.6% of the expenses are for internet service That is 0.6% = 0.6/100 = 6/1000 = 3/500 So, 3/500 part of the family’s expenses is used for internet service

Question 5. How many meters are equivalent to 357 centimeters? _____ meters

Answer: 3.57 meters

Explanation: 357 ntimeters 100centimeters = 1 meter Then, 357 centimeters = 357/100 = 3.57 meters 3.57 meters is equivalent to 357 centimeters

Question 6. What is the product of the two quantities shown below? \(\frac{60 \mathrm{mi}}{1 \mathrm{hr}}\) × 12 hr _____ miles

Answer: 720 miles

Explanation: 60 mi/1hr x 12 hr That is, 60 milesx12 = 720 miles So, the equivalent quantity of 60mi/1hr x 12hr is 720 miles

Chapter 6 Review/Test – Page No. 347

Question 1. A construction crew needs to remove 2.5 tons of river rock during the construction of new office buildings The weight of the rocks is ____________ pounds

Answer: The weight of the rocks is 5,000 pounds

Explanation: A construction crew needs to remove 2.5 tons of river rock during the construction of new office buildings 1 Ton = 2000 pounds Then, 2.5 Tons = 2.5×2000 = 25×200 = 5000 pounds So, the weight of the rocks is 5000 pounds

Question 2. Select the conversions that are equivalent to 10 yards. Mark all that apply Options: a. 20 feet b. 240 inches c. 30 feet d. 360 inches

Answer: c. 30 feet d. 360 inches

Explanation: a. 20 feet 3feet = 1 yard Then, 20 feet = 20/3 yard b. 240 inches 36 inches = 1 yard Then, 240 inches = 240/36 = 6 yards c. 30 feet 3feet = 1 yard Then, 30 feet =30/3 = 10 yards d. 360inches 36 inches = 1 yard Then, 360 inches = 360/36 = 10 yards So, 30 feet and 360 inches are equivalent to 10 yards

Question 3. Meredith runs at a rate of 190 meters per minute. Use the formula d=r×t to find how far she runs in 6 minutes. _____ meters

Answer: 1,140 meters

Explanation: Meredith runs at a rate of 190 meters per minute Formula d = r x t Here, d= 190 meters, t = 1 minute Then, r = 190/1 = 190 meters per minute Now, t = 6 minutes and r = 190 meters per minute Then d = 190 x 6 = 1,140 meters

Go Math Grade 6 Answer Key Chapter 6 Convert Units of Length img 7

Answer: D = RxT Alisha 36 = Rx3 Then, Rate of Alisha =36/3 =12 miles per hour Jose 39 = Rx3 Then, Rate of Jose = 39/3 = 13 miles per hour Raul 40 = Rx4 Then, Rate of Raul = 40/4 = 10 miles per hour Ruthie 22= Rx2 Then, Rate of Ruthie = 22/2 = 11 miles per hour

Page No. 348

Question 5. For numbers 5a–5c, choose <, >, or =. 5a. 5 kilometers Ο 5,000 meters 5b. 254 centiliters Ο 25.4 liters 5c. 6 kilogram Ο 600 gram 5 kilometers _____ 5,000 meters 254 centiliters _____ 25.4 liters 6 kilogram _____ 600 gram

Answer: 5a. 5 kilometers Ο 5,000 meters 5b. 254 centiliters Ο 25.4 liters 5c. 6 kilogram Ο 600 gram 5 kilometers = 5,000 meters 254 centiliters < 25.4 liters 6 kilogram > 600 gram

Explanation: a. 5 kilometers —— 5000 meters 1 kilometer = 1000 meters then,5 kilometers 5×1000 = 5000 meters So, 5 kilometers = 5000 meters b. 254 centiliters ——25.4 liters 100centiliters = 1 liter Then, 254 centiliters = 254/100 = 2.54 liters So, 254 centiliters < 25.4 liters c. 6 kilograms —– 600 grams 1kilogram = 1000 grams Then, 6 kilograms = 6000 grams So, 6 kilograms > 600 grams

Question 6. A recipe calls for 16 fluid ounces of light whipping cream. If Anthony has 1 pint of whipping cream in his refrigerator, does he have enough for the recipe? Explain your answer using numbers and words. ____________

Answer: A recipe calls for 16 fluid ounces of light whipping cream 8 fluid ounces = 1 cup So, 16 fluid ounces = 2 cups = 1 pint If Anthony has 1 pint of whipping cream in his refrigerator, then it is enough for the recipe

Question 7. For numbers 7a–7d, choose <, >, or =. 7a. 43 feet Ο 15 yards 7b. 10 pints Ο 5 quarts 7c. 5 tons Ο 5,000 pounds 7d. 6 miles Ο 600 yards 43 feet _____ 15 yards 10 pints _____ 5 quarts 5 tons _____ 5,000 pounds 6 miles _____ 600 yards

Answer: 43 feet < 15 yards 10 pints = 5 quarts 5 tons > 5,000 pounds 6 miles > 600 yards

Explanation: a. 43 feet —- 15 yards 3feet = 1 yard Then, 43 feet = 43/3 = 14.3 yards So, 43 feet < 15 yards b. 10 pints —- 5 quarts 1 pints = 1 quart then, 10 pints = 10/2 = 5 quarts So, 10 pints = 5 quarts c. 5 tons —– 5000 pounds 1 ton = 2000 pounds then, 5 tons = 5×2000 = 10,000 pounds So, 5 tons > 5000 pounds d. 6 miles —- 600 yards 1 mile =1760 yards then, 6 miles = 6×1760 = 10,560yards So, 6 miles > 600 yards

Question 8. The distance from Caleb’s house to the school is 1.5 miles, and the distance from Ashlee’s house to the school is 3,520 feet. Who lives closer to the school, Caleb or Ashlee? Use numbers and words to support your answer. ____________

Answer: There are 5280 feet in one mile. So, you need to change the miles to feet. 1.5 x 5280 = 7920. 7920 > 3520 So, Ashley lives closer.

Page No. 349

Go Math Grade 6 Answer Key Chapter 6 Convert Units of Length img 8

Answer: 7.4 kilograms, 7.4 centigrams, 7.4 decigrams

Go Math Grade 6 Answer Key Chapter 6 Convert Units of Length img 9

Answer: 840 beats

Explanation: An elephant’s heart beats 28 times per minute Then, elephant’s heart beats in 30 minutes = 28 x 30 = 840 So, an elephant’s heartbeat is 840 times in 30 minutes

Question 11. The length of a rectangular football field, including both end zones, is 120 yards. The area of the field is 57,600 square feet. For numbers 11a–11d, select True or False for each statement. 11a. The width of the field is 480 yards. 11b. The length of the field is 360 feet. 11c. The width of the field is 160 feet. 11d. The area of the field is 6,400 square yards. 11a. ____________ 11b. ____________ 11c. ____________ 11d. ____________

Answer: 11a. True 11b. True 11c. False 11d. False

Explanation: The length of a rectangular football field, including both end zones, is 120 yards The area of the field is 57,600 square feet That is length x width = 57,600 square feet Here, length = 120 yards Then, width = 57,600/120 = 480 yards 11a. True 11b. 1 yard = 3 feet Then, 120 yards = 120×3 = 360 feet True 11c. 480 yards = 480×3 = 1440 False 11d. 6400 square yards 3 feet = 1 yard then, 57,600 square feet = 57,600/3 = 19,200 square yards False

Question 12. Harry received a package for his birthday. The package weighed 357,000 centigrams. Select the conversions that are equivalent to 357,000 centigrams. Mark all that apply. Options: a. 3.57 kilograms b. 357 dekagrams c. 3,570 grams d. 3,570,000 decigrams

Answer: a. 3.57 kilograms b. 357 dekagrams c. 3,570 grams

Explanation: Harry received a package for his birthday The package weighed 357,000 centigrams 100centigrams = 1 gram Then, 357,000 centigrams = 357,000/100 = 3570 grams 1000 grams = 1 kilogram Then, 3570 grams = 3570/1000 = 3.57 kilograms 10grams = 1 dekagram Then, 3570 grams = 3570/10 = 357 dekagrams 1gram = 10 decigrams Then, 3570 grams = 35700 decigrams Options a, b and c are true

Page No. 350

Question 13. Mr. Martin wrote the following problem on the board. Juanita’s car has a gas mileage of 21 miles per gallon. How many miles can Juanita travel on 7 gallons of gas? Alex used the expression \(\frac{21 \text { miles }}{\text { 1 gallon }} \times \frac{1}{7 \text { gallons }}\) to find the answer. Explain Alex’s mistake. Type below: ____________

Answer: Juanita’s car has a gas mileage of 21 miles per gallon Juanita traveled miles on 7 gallons of gas = 21×7 = 147 miles But, Alex used the expression 21 miles 1 gallon ×17 gallons In the place of 7 gallons, Alex used 17 gallons

Question 14. Mr. Chen filled his son’s wading pool with 20 gallons of water. 20 gallons is equivalent to ____________ quarts.

Answer: 80 quarts

Explanation: Mr. Chen filled his son’s wading pool with 20 gallons of water 1gallon = 4 quarts Then, 20 gallons = 20×4 = 80 quarts So, 20 gallons is equivalent to 80 quarts

Go Math Grade 6 Answer Key Chapter 6 Convert Units of Length img 10

Answer: 0.411, 41.1, 4.11

Explanation: Nadia has a can of vegetables with a mass of 411 grams 1000 grams = 1 kilogram Then, 411 grams = 411/1000 = 0.411 kilograms 100grams = 1 hectogram Then, 411 grams = 411/100 = 4.11 hectograms 10grams = 1 dekagram Then, 411 grams = 411/10 = 41.1 dekagram

Question 16. Steve is driving 440 miles to visit the Grand Canyon. He drives at an average rate of 55 miles per hour. Explain how you can find the amount of time it will take Steve to get to the Grand Canyon. Type below: ____________

Answer: Steve is driving 440 miles to visit the Grand Canyon He drives at an average rate of 55 miles per hour Then, 440 miles = 440/55 = 8 hours So, Steve can take 8 hours of time to visit the Grand Canyon

Page No. 351

Question 17. Lucy walks one time around the lake. She walks for 1.5 hours at an average rate of 3 miles per hour. What is the distance, in miles, around the lake? _____ miles

Answer: 4.5 miles

Explanation: Lucy walks one time around the lake She walks for 1.5 hours at an average rate of 3 miles per hour 1 hour = 3 miles Then, 1.5 hours = 1.5×3 = 4.5 miles So, Luke walks 4.5 miles around the lake

Go Math Grade 6 Answer Key Chapter 6 Convert Units of Length img 11

Answer: Derrick says that the width could also be written as 22 feet 1yard = 3 feet 20 yards = 60 feet So, we cannot write 20 yards 2 feet as 22 feet

Question 19. Part B The cost to repave the parking lot is $2 per square foot. Explain how much it would cost to repave the parking lot. Type below: ____________

Answer: The cost to repave the parking lot is $2 per square foot Parking lot area =20 yards 2 feet x 30 yards 1yard = 3 feet Then, 20 yards = 20×3 = 60 feet 30 yards = 30×3 = 90 feet so, Parking lot area = 62 feet x 90 feet = 5580 feet 1 square foot cost = $2 then, 5580 feet cost = 2×5580 = $11,160

Page No. 352

Go Math Grade 6 Answer Key Chapter 6 Convert Units of Length img 12

Answer: Horse weight in Tons = 0.5 T Trailer weight in Tons = 1.25 T 1 ton = 2000 pounds then, 0.5 T = 0.5×2000 = 1000 pounds then, 1.25 T = 1.25×2000 = 2,500 pounds

Question 19. Part B Jake’s truck can tow a maximum weight of 5,000 pounds. What is the maximum number of horses he can take in his trailer at one time without going over the maximum weight his truck can tow? Use numbers and words to support your answer. Type below: ____________

Answer: Max. No of Horses = (Max weight truck can tow)/(Average weight of one horse) The weight of a horse is not given in the question .Thus, we assume the average weight of one horse, to be equal to 1000 pounds, Max. No of Horses = 5000 pounds/ 1000 pounds Max. No of Horses = 5

Question 20. A rectangular room measures 13 feet by 132 inches. Tonya said the area of the room is 1,716 square feet. Explain her mistake, then find the area in square feet. Type below: ____________

Answer: A rectangular room measures 13 feet by 132 inches =13 feetx132 inches Tonya said the area of the room is 1,716 square feet Area of the rectangular room = 13 feet x 132 inches 12inches = 1 foot Then, 132 inches = 132/12 = 11 feet So, the area of the rectangular room = 13 feet x 12 feet = 156 feet So, Tonya’s answer is wrong

Conclusion:

Get free access to Download Go Math Grade 6 Answer Key Chapter 6 Convert Units of Length from here. I wish the information prevailed in Go Math Grade 6 Answer Key is beneficial for all the students. We have given the step by step explanation for each and every problem here. So refer to our Go Math 6th Grade Solution Key Chapter 6 Convert Units of Length. If you have any doubts regarding the tops you can ask your doubts in the below mentioned comment section.

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Appendix A: Applications

Using the distance, rate, and time formula, learning outcomes.

  • Use the problem-solving method to solve problems using the distance, rate, and time formula

One formula you’ll use often in algebra and in everyday life is the formula for distance traveled by an object moving at a constant speed. The basic idea is probably already familiar to you. Do you know what distance you traveled if you drove at a steady rate of [latex]60[/latex] miles per hour for [latex]2[/latex] hours? (This might happen if you use your car’s cruise control while driving on the Interstate.) If you said [latex]120[/latex] miles, you already know how to use this formula!

The math to calculate the distance might look like this:

[latex]\begin{array}{}\\ \text{distance}=\left(\Large\frac{60\text{ miles}}{1\text{ hour}}\normalsize\right)\left(2\text{ hours}\right)\hfill \\ \text{distance}=120\text{ miles}\hfill \end{array}[/latex]

In general, the formula relating distance, rate, and time is

[latex]\text{distance}\text{=}\text{rate}\cdot \text{time}[/latex]

Distance, Rate, and Time

For an object moving at a uniform (constant) rate, the distance traveled, the elapsed time, and the rate are related by the formula

[latex]d=rt[/latex]

where [latex]d=[/latex] distance, [latex]r=[/latex] rate, and [latex]t=[/latex] time.

Notice that the units we used above for the rate were miles per hour, which we can write as a ratio [latex]\Large\frac{miles}{hour}[/latex]. Then when we multiplied by the time, in hours, the common units “hour” divided out. The answer was in miles.

Jamal rides his bike at a uniform rate of [latex]12[/latex] miles per hour for [latex]3\Large\frac{1}{2}[/latex] hours. How much distance has he traveled?

In the following video we provide another example of how to solve for distance given rate and time.

Rey is planning to drive from his house in San Diego to visit his grandmother in Sacramento, a distance of [latex]520[/latex] miles. If he can drive at a steady rate of [latex]65[/latex] miles per hour, how many hours will the trip take?

Show Solution

In the following video we show another example of how to find rate given distance and time.

  • Question ID 145550, 145553,145619,145620. Authored by : Lumen Learning. License : CC BY: Attribution
  • Ex: Find the Rate Given Distance and Time. Authored by : James Sousa (Mathispower4u.com). Located at : https://youtu.be/3rYh32ErDaE . License : CC BY: Attribution
  • Example: Solve a Problem using Distance = Rate x Time. Authored by : James Sousa (Mathispower4u.com). Located at : https://youtu.be/lMO1L_CvH4Y . License : CC BY: Attribution
  • Prealgebra. Provided by : OpenStax. License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/contents/[email protected]

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IMAGES

  1. Rate Time Distance Problems

    problem solving distance rate and time formulas lesson 6.5 answers

  2. How to Solve a Distance, Rate & Time Problem Using a Rational Equation

    problem solving distance rate and time formulas lesson 6.5 answers

  3. 6.5: Problem Solving

    problem solving distance rate and time formulas lesson 6.5 answers

  4. 6.5 Distance, Rate, & Time Formulas P117 1

    problem solving distance rate and time formulas lesson 6.5 answers

  5. Lesson 6.5: Problem Solving: Distance, Rate, And Time Formulas

    problem solving distance rate and time formulas lesson 6.5 answers

  6. How to Solve Distance Rate Time Problems

    problem solving distance rate and time formulas lesson 6.5 answers

VIDEO

  1. 13. Class-5 । Mathematics । Unit-6 । Distance ।

  2. Speed distance time formulas #mathstricks #mathshorts

  3. 9.6 Applications, Rate times Time = Distance

  4. Speed, Distance and Time Math Steps 5 Introduction Video

  5. Distance and Time Formulas #shorts #mathematics #shortvideo# Speed

  6. DRT04 Solving Motion Word Problems Part 3 (Distance, Rate, Time Problems)

COMMENTS

  1. Distance, Rate, and Time Formulas

    This lesson uses the formula d = r x t, and variations to solve problems.

  2. Lesson 6.5 Problem solving Distance , rate and time formulas , Go math

    About Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How YouTube works Test new features NFL Sunday Ticket Press Copyright ...

  3. 6th Grade Go Math lesson 6.5 problems distance, rate, time formulas

    About Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How YouTube works Test new features NFL Sunday Ticket Press Copyright ...

  4. Distance Rate and Time Worksheets with Answers

    In math, distance, rate, and time are three important concepts you can use to solve many problems if you know the formula. Distance is the length of space traveled by a moving object or the length measured between two points. It is usually denoted by d in math problems.. The rate is the speed at which an object or person travels. It is usually denoted by r in equations.

  5. PDF Name Lesson 6.5 Problem Solving • Distance, Rate,

    A dragonfly traveled at a rate of 35 miles per hour for 2.5 hours. What distance did the dragonfly travel? 4. A race car travels 1,212 kilometers in 4 hours. What is the car's rate of speed? 5. Kim and Jay leave at the same time to travel 25 miles to the beach. Kim drives 9 miles in 12 minutes. Jay drives 10 miles in 15 minutes.

  6. Sixth Grade Math

    Problem Solving - Distance, Rate, and Time - Lesson 6.5. Fractions and Decimals - Lesson 2.1 ... Lesson 3.9. Problem Solving - The Coordinate Plane - Lesson 3.10. Solutions of Equations - Lesson 8.1. Writing Equations - Lesson 8.2. Model and Solve Addition Equations - Lesson 8.3. Solve Addition & Subtraction Equations - Lesson 8.4. Sixth Grade ...

  7. Using the Distance, Rate, and Time Formula

    Use the problem-solving method to solve problems using the distance, rate, and time formula; ... the common units "hour" divided out. The answer was in miles. example Jamal rides his bike at a uniform rate of . 12 12 12. miles per hour for .

  8. Solving problems with the formula for distance, rate, and time

    Before you can use the distance, rate, and time formula, D=RT, you need to make sure that your units for the distance and time are the same units as your rate. If they aren't, you'll need to change them so you're working with the same units.

  9. Using the Distance, Rate, and Time Formula

    Distance, Rate, and Time. For an object moving at a uniform (constant) rate, the distance traveled, the elapsed time, and the rate are related by the formula. d= rt d = r t. where d = d = distance, r = r = rate, and t= t = time. Notice that the units we used above for the rate were miles per hour, which we can write as a ratio miles hour m i l ...

  10. Distance Rate Time Problems

    30. t. 30 × t. The total distance you and your wife travel is 45t + 30t = 75t and this must be equal to 225. 75 × t = 225. Since 75 times 3 = 225, t = 3. Therefore, you will be 225 miles apart after 3 hours. The above are great examples of distance rate time problems. Try to do some now in your math textbook.

  11. Problems Involving Formulas I

    For example, if we want to find the time it will take to accrue a specific amount of interest, we can solve for t t using division: I = P ⋅ r ⋅ t I P ⋅r = P ⋅r⋅t P ⋅r t = I r⋅t I = P ⋅ r ⋅ t I P ⋅ r = P ⋅ r ⋅ t P ⋅ r t = I r ⋅ t. Below is a table showing the result of solving for each individual variable in the formula.

  12. 8.8 Rate Word Problems: Speed, Distance and Time

    Distance, rate and time problems are a standard application of linear equations. When solving these problems, use the relationship rate (speed or velocity) times time equals distance. r⋅t = d r ⋅ t = d. For example, suppose a person were to travel 30 km/h for 4 h. To find the total distance, multiply rate times time or (30km/h) (4h) = 120 km.

  13. PDF 1. d r t d 0.5 mi 1 min 18 min d 9 miles

    Problem Solving • Distance, Rate, and Time Formulas Chapter 6 P107 Lesson 6.5 PROBLEM SOLVING Read each problem and solve. 1. A downhill skier is traveling at a rate of 0.5 mile per minute. How far will the skier travel in 18 minutes? ... Kim and Jay leave at the same time to travel 25 miles to the beach. Kim drives 9 miles in

  14. Go Math! 6 Common Core Edition answers & resources

    Go Math! 6 Common Core Edition grade 6 workbook & answers help online. Grade: 6, Title: Go Math! 6 Common Core Edition, Publisher: Houghton Mifflin Harcourt, ISBN: 547587783 ... Lesson 5: Problem Solving: Apply the Greatest Common Factor. apps. videocam. create. Lesson 6: Add and Subtract Decimals ... Problem Solving: Distance, Rate, and Time ...

  15. PDF New York Common Core Mathematics Curriculum Grade 6 Modules

    Lesson 6.5 Problem Solving • Distance, Rate, and Time Formulas 6.RP.3d Module 1: Topic C Lesson 7.1 Exponents 6.EE.1 Module 4: Topic B Lesson 7.2 Evaluate Expressions Involving Exponents 6.EE.1 Module 4: Topic B Lesson 7.3 Write Algebraic Expressions 6.EE.2a Module 4: Topic D Topic F

  16. PDF 6 Letter capacity

    Problem Solving • Distance, Rate, and Time Formulas Chapter 6 P107 Lesson 6.5 PROBLEM SOLVING Read each problem and solve. 1. A downhill skier is traveling at a rate of 0.5 mile per minute. How far will the skier travel in 18 minutes? 2. How long will it take a seal swimming at a speed of 8 miles per hour to travel 52 miles? 3.

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    Lesson 6.5: Problem Solving: Distance, Rate, And Time Formulas by Melani Fulwider - January 22, 2014

  18. GO MATH Grade 6 Chapter 6 Printable Handouts & Activities

    Printable review worksheets for each lesson. New math problems to solve. Cut and glue vocabulary cards. Printable math vocabulary puzzles. Answer Key included! BONUS: Conversion Table Chart. Chapter 6 Lessons: Lesson 6.1: Convert Units of Length (CCSS.6.RP.3D) Lesson 6.2: Convert Units of Capacity (CCSS.6.RP.3D)

  19. Distance, Rate, and Time, Free PDF Download

    The Formula. You can write the formula three ways, depending on which variable you need to find: Distance = Rate x Time or d = rt, Rate = Distance ÷ Time or r =d/t, and Time = Distance ÷ Rate or t = d/r. The lesson shows each of these equations as parts of a triangle. Visualizing the formula in this way can help you remember if you need to ...

  20. Distance Rate Time Problems Teaching Resources

    This product includes 2 worksheets on solving word problems relating to distance, time, and rate. Each sheet has 8 problems to solve including 6 single-step problems and 2 multi-step problems.I've also included a paper for your students to use as scrap as well as an answer key so you don't have to work through all the problems :)If you are using the Go Math series, this product accompanies ...

  21. Go Math Grade 6 Answer Key Chapter 6 Convert Units of Length

    Problem Solving Distance, Rate, and Time Formulas - Page No. 345. Read each problem and solve. Question 1. A downhill skier is traveling at a rate of 0.5 mile per minute. How far will the skier travel in 18 minutes? _____ miles. Answer: 9 miles. Explanation: A downhill skier is traveling at a rate of 0.5 miles per minute 1 minute = 0.5 mile

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  23. Using the Distance, Rate, and Time Formula

    Distance, Rate, and Time. For an object moving at a uniform (constant) rate, the distance traveled, the elapsed time, and the rate are related by the formula. d= rt d = r t. where d = d = distance, r = r = rate, and t= t = time. Notice that the units we used above for the rate were miles per hour, which we can write as a ratio miles hour m i l ...