# Everyday Math Grade 4 Answers Unit 7 Multiplication of a Fraction by a Whole Number; Measurement

## Everyday Mathematics 4th Grade Answer Key Unit 7 Multiplication of a Fraction by a Whole Number; Measurement

Liquid Measures

Find at least one container that holds each of the amounts listed below.
Describe each container and record all the measurements on the label.

Question 1.  Explanation :
Containers with quantities near to about 1 gallon are written .

Question 2.  Explanation :
Containers with quantities near to about 1 Quart are written .

Question 3.  Explanation :
Containers with quantities near to about 1 pint are written .

Question 4.  Explanation :
Containers with quantities near to about 1 cup are written .

Complete.

Question 5.
2 quarts = pints
2 quarts = 4 pints
Explanation :
1 quart = 2 pints
2 quarts = ? pints
multiply  with 2 to get the value in pints.
2 quarts = 2 × 2 = 4 pints .

Question 6.
3 gallons = cups
3 gallons = 48 cups
Explanation :
1 gallons = 16 cups .
3 gallons = ? cups
multiply with 16 to get the value in cups
3 gallons = 3 × 16 = 48 cups .

Question 7.
__ pints = 4 cups
Explanation :
1 pint = 2 cups .
? pints = 4 cups
Divide cups with 4 we get value in pints .
4 ÷ 4 cups = 1 pint .

Question 8.
___ quarts = 12 cups
3 quarts = 12 cups
Explanation :
1 quart = 4 cups
? quarts = 12 cups
Divide the cups with 4 to get the value in quarts .
12 ÷ 4 cups = 3 quarts .

Question 9.
6 pints = __ quarts
6 pints = 3 quarts
Explanation :
1 pints = 0.5 quarts
Divide the pint value with 2 to get value in quarts .
6 ÷ 2 pints = 3 quarts

Question 10.
___ quarts = 2$$\frac{1}{2}$$ gallons
10 quarts = 2$$\frac{1}{2}$$ gallons
Explanation :
1 quart = 0.25 gallons
Multiply the gallons with 4 to get the value in quarts .
$$\frac{5}{2}$$ × 4 = 10 quarts .

Practice

Question 11.
273 ∗ 2 = ___
273 ∗ 2 = 546
Explanation : Question 12.
385 ∗ 4 = ___
385 ∗ 4 = 1,540
Explanation : Question 13.
__ = 886 ∗ 5
4,430 = 886 ∗ 5
Explanation : Question 14.
__ = 98 ∗ 38
3,724 = 98 ∗ 38
Explanation : Sugar in Drinks

Use the information in the table to solve the number stories. In the space below each problem, use pictures or equations to show what you did to find your answers. Sources: National Institutes of Health and California Department of Public Health

Question 1.
Carmen drinks one 12-ounce can of orange soda every day. How much sugar is that in 1 week? cup(s)
Quantity of sugar content in one 12-ounce can of orange soda = $$\frac{1}{4}$$
Quantity of sugar content in 7 cups of 12-ounce can of orange soda = 7 × $$\frac{1}{4}$$ = $$\frac{7}{4}$$ = 1$$\frac{3}{4}$$ .
Therefore, Quantity of sugar content in 7 cups of 12-ounce can of orange soda =1$$\frac{3}{4}$$  cups

Question 2.
If you drink one 12-ounce glass of cranberry juice cocktail every morning, how much sugar will that be in 2 weeks? cup(s)
Number of days = 2 weeks = 2 × 7 = 14 days .
Quantity of sugar content in one 12-ounce glass of cranberry juice cocktail = $$\frac{1}{4}$$
Quantity of sugar content in 7 cups of 12-ounce glass of cranberry juice cocktail = 14 × $$\frac{1}{4}$$ = $$\frac{7}{2}$$ = 3$$\frac{1}{2}$$ .
Therefore, Quantity of sugar content in 7 cups of 12-ounce glass of cranberry juice cocktail = 3$$\frac{1}{2}$$ .

Question 3.
Mike drinks three 12-ounce servings of sweet tea per day.
a. How much sugar is he drinking in his tea in 1 day?
___ cup(s)
b. In 5 days? cup(s)
a.
Quantity of sugar in one 12-ounce servings of sweet tea = $$\frac{1}{6}$$
Quantity of sugar in three 12-ounce servings of sweet tea = 3 × $$\frac{1}{6}$$ = latex]\frac{1}{2}[/latex]
Therefore. Quantity of sugar in three 12-ounce servings of sweet tea per day = = latex]\frac{1}{2}[/latex] cups .
b.
Quantity of sugar in three 12-ounce servings of sweet tea per day = latex]\frac{1}{2}[/latex] cup
Quantity of sugar in three 12-ounce servings of sweet tea per 5 days = 5 × latex]\frac{1}{2}[/latex] = latex]\frac{5}{2}[/latex] = 2 latex]\frac{1}{2}[/latex] cups .
Therefore, Quantity of sugar in three 12-ounce servings of sweet tea per 5 days = 2$$\frac{1}{2}$$ cups .

Practice

Question 4.
951 ∗ 4 = _____5
951 ∗ 4 = 3,804
Explanation : Question 5.
650 ∗ 5 = ___
650 ∗ 5 = 3,250
Explanation : Question 6.
425 ∗ 7 = _____
425 ∗ 7 = 2,975
Explanation : Question 7.
3,684 ∗ 6 =
3,684 ∗ 6 = 22,104
Explanation : Multiplying Unit Fractions

Question 1. Multiplication equation:
What is the fourth multiple of $$\frac{1}{5}$$?
The fourth multiple of $$\frac{1}{5}$$ = 4 × $$\frac{1}{5}$$ = $$\frac{4}{5}$$
Explanation :
To find the fourth multiple of given number multiply 4 with the given number , we get the multiple .

Question 2. Multiplication equation: ___
What is the third multiple of $$\frac{1}{10}$$?
The third multiple of $$\frac{1}{10}$$ = 3 × $$\frac{1}{10}$$ = $$\frac{3}{10}$$
Explanation :
To find the third multiple of given number multiply 3 with the given number , we get the multiple .

Question 3.
Dmitri fixed a snack for 5 friends. Each friend got $$\frac{1}{2}$$ of an avocado. How many avocados did Dmitri use?
Multiplication equation: ___
Number of friends = 5.
Fraction of snack given to each friend = $$\frac{1}{2}$$ of an avocado
Number of avocados given to friend’s or used = Number of friends × $$\frac{1}{2}$$ = 5 × $$\frac{1}{2}$$ = $$\frac{5}{2}$$ = 2$$\frac{1}{2}$$ avocados .
Therefore, Number of avocados Dmitri used = 2$$\frac{1}{2}$$ avocados .

Question 4.
Juanita made 3 protein shakes. All together, she used 1 cup of protein powder to make them. Each had the same amount.
How many cups of protein powder are in each shake?
Multiplication equation: ___
Number of protein shakes = 3
Quantity of protein used for 3 protein shake  = 1 cup
Quantity of protein used for 3 protein shakes =  1 cup ÷ 3 = $$\frac{1}{3}$$ .
Therefore, Quantity of protein used for each protein shake = $$\frac{1}{3}$$ cup .

Practice

Question 5.
$$\frac{1}{2}$$ + $$\frac{1}{2}$$ + $$\frac{1}{2}$$ =_____
$$\frac{1}{2}$$ + $$\frac{1}{2}$$ + $$\frac{1}{2}$$ = 1$$\frac{1}{2}$$
Explanation :
$$\frac{1}{2}$$ + $$\frac{1}{2}$$ + $$\frac{1}{2}$$ = $$\frac{3}{2}$$ = 1 $$\frac{1}{2}$$

Question 6.
$$\frac{2}{3}$$ + $$\frac{2}{3}$$ + $$\frac{1}{3}$$ = ____
$$\frac{2}{3}$$ + $$\frac{2}{3}$$ + $$\frac{1}{3}$$ = $$\frac{1}{2}$$
Explanation :
$$\frac{2}{3}$$ + $$\frac{2}{3}$$ + $$\frac{1}{3}$$ = $$\frac{6}{3}$$ = $$\frac{1}{2}$$

Question 7.
$$\frac{9}{10}$$ – $$\frac{4}{10}$$ = ___
$$\frac{9}{10}$$ – $$\frac{4}{10}$$ = $$\frac{1}{2}$$
Explanation :
$$\frac{9}{10}$$ – $$\frac{4}{10}$$ = $$\frac{5}{10}$$ = $$\frac{1}{2}$$

Question 8.
$$\frac{8}{12}$$ – $$\frac{5}{12}$$ = ___
$$\frac{8}{12}$$ – $$\frac{5}{12}$$ = $$\frac{1}{4}$$
Explanation :
$$\frac{8}{12}$$ – $$\frac{5}{12}$$ = $$\frac{3}{12}$$ = $$\frac{1}{12}$$

Multiplying Fractions by Whole Numbers

Solve the problems below.

Question 1.
5 ∗ $$\frac{1}{5}$$ = ___
Draw a picture.
5 ∗ $$\frac{1}{5}$$ = 1
Explanation : Question 2.
3 ∗ $$\frac{4}{9}$$ = __
Draw a picture.
3 ∗ $$\frac{4}{9}$$ = $$\frac{4}{3}$$ = 1$$\frac{1}{3}$$
Explanation : Question 3.
6 ∗ $$\frac{3}{6}$$ = __
Draw a picture.
6 ∗ $$\frac{3}{6}$$ = $$\frac{3}{1}$$ = 3
Explanation : Write a multiplication equation to represent the problem and then solve.

Question 4.
Rahsaan needs to make 5 batches of granola bars. A batch calls for $$\frac{1}{2}$$ cup of honey. How much honey does he need? Equation: ___
Number of batches = 5
Quantity of honey used for 1 batch = $$\frac{1}{2}$$ cup
Quantity of Honey used for 5 batches = 5 ×  $$\frac{1}{2}$$ cup = $$\frac{5}{2}$$ = 2 $$\frac{1}{2}$$ cups
Therefore, Quantity of Honey used for 5 batches = 2 $$\frac{1}{2}$$ cups .

Question 5.
a. Joe swims $$\frac{6}{10}$$ of a mile 5 days per week. How far does he swim every week?
Equation: ___
b. How far would he swim if he swam every day of the week?
Equation: ____
a.
Number of days per week = 5 days .
Fraction of distance covered in a swim = $$\frac{6}{10}$$ of a mile
Total Distance covered in swimming for 5 days = 5 ×  $$\frac{6}{10}$$ =  $$\frac{6}{2}$$ = 3 miles Therefore, Total Distance covered in swimming for 5 days = 3 miles .
b.
Number of days = 7 days .
Fraction of distance covered in a swim = $$\frac{6}{10}$$ of a mile
Total Distance covered in swimming for 7 days = 7 ×  $$\frac{6}{10}$$ =  $$\frac{42}{10}$$ = 4.2 miles
Therefore, Total Distance covered in swimming for 7 days = 4.2 miles .

Practice

Question 6.
653 ∗ 3 = ___
653 ∗ 3 = 1,959
Explanation : Question 7.
262 ∗ 8 = ___
262 ∗ 8 = 2,096
Explanation : Question 8.
357 ∗ 9 = ___
357 ∗ 9 = 3,213
Explanation : Question 9.
7,376 ∗ 2 = __
7,376 ∗ 2 = 14,752
Explanation : Multiplying Mixed Numbers by Whole Numbers

Solve.

Question 1.
Michelle’s grandmother sent her 5 small gifts for her fifth birthday. Each one weighed 1$$\frac{1}{2}$$ pounds. How much did the gifts weigh all together?
Number model with unknown: ___
Between what two whole numbers is this? __ and ___
How many ounces did the gifts weigh? __ ounces
Weight of each gift = 1$$\frac{1}{2}$$ =  $$\frac{3}{2}$$ pounds
Total weight of 5 gifts = 5 ×  $$\frac{3}{2}$$ =  $$\frac{15}{2}$$ = 7$$\frac{1}{2}$$ pounds
Therefore, Total weight of 5 gifts = 7$$\frac{1}{2}$$ pounds .
7$$\frac{1}{2}$$ lies between 7 and 8 whole numbers .

Question 2.
Rochelle bought 4 pieces of ribbon to finish a project. Each piece was 1$$\frac{5}{12}$$ yards long. What is the combined length of the ribbon she bought?
Number model with unknown: ____
Between what two whole numbers is this? __ and ___
How many feet is this? __ feet
Number of Ribbons = 4
Length of each ribbon = 1$$\frac{5}{12}$$ yards = $$\frac{17}{12}$$ yards
Total length of 5 Ribbons = 4 × $$\frac{17}{12}$$  = $$\frac{17}{3}$$ = 5$$\frac{2}{3}$$ yards .
Therefore, Total length of 5 Ribbons = 5$$\frac{2}{3}$$ yards .
5$$\frac{2}{3}$$ lies between 5 and 6 whole numbers .

Question 3.
3 ∗ 4$$\frac{5}{6}$$ = __
Between what two whole numbers is this? __ and __
3 ∗ 4$$\frac{5}{6}$$ = 14$$\frac{1}{2}$$
Explanation :
3 ∗ 4$$\frac{5}{6}$$ = 3 ∗ $$\frac{29}{6}$$ = $$\frac{29}{2}$$ = 14$$\frac{1}{2}$$
14$$\frac{1}{2}$$ lies between 14 and 15 whole numbers .

Question 4.
6 ∗ 7$$\frac{3}{8}$$ = ___
Between what two whole numbers is this? __ and ___
6 ∗ 7$$\frac{3}{8}$$ = 44$$\frac{1}{4}$$
Explanation :
6 ∗ 7$$\frac{3}{8}$$ = 6 ∗ $$\frac{59}{8}$$ = 3 ∗ $$\frac{59}{4}$$ = $$\frac{177}{4}$$ = 44$$\frac{1}{4}$$ .
44$$\frac{1}{4}$$ lies between 44 and 45 whole numbers .

Practice

Question 5.
$$\frac{3}{4}$$ + $$\frac{2}{4}$$ + $$\frac{1}{4}$$ = __
$$\frac{3}{4}$$ + $$\frac{2}{4}$$ + $$\frac{1}{4}$$ = 1$$\frac{1}{2}$$
Explanation :
$$\frac{3}{4}$$ + $$\frac{2}{4}$$ + $$\frac{1}{4}$$ = $$\frac{6}{4}$$ = $$\frac{3}{2}$$ =1$$\frac{1}{2}$$ .

Question 6.
$$\frac{4}{8}$$ + $$\frac{3}{8}$$ + $$\frac{2}{8}$$ = ___
$$\frac{4}{8}$$ + $$\frac{3}{8}$$ + $$\frac{2}{8}$$ = 1$$\frac{1}{8}$$
Explanation :
$$\frac{4}{8}$$ + $$\frac{3}{8}$$ + $$\frac{2}{8}$$ = $$\frac{9}{8}$$ = 1$$\frac{1}{8}$$

Question 7.
$$\frac{5}{6}$$ – $$\frac{2}{6}$$ = ___
$$\frac{5}{6}$$ – $$\frac{2}{6}$$ = $$\frac{1}{2}$$
Explanation :
$$\frac{5}{6}$$ – $$\frac{2}{6}$$ = $$\frac{3}{6}$$ = $$\frac{1}{2}$$

Question 8.
$$\frac{88}{100}$$ – $$\frac{57}{100}$$ = ___
$$\frac{88}{100}$$ – $$\frac{57}{100}$$ = $$\frac{31}{100}$$

Mr. Chou makes fruit salad that he sells in his store. Today he plans to make a fruit salad with 8 pears, 2 cups of grapes, and 4 pints of strawberries. Use the weights below to solve the problems.

• A medium pear weighs about $$\frac{3}{8}$$ lb.
• A cup of grapes weighs about $$\frac{2}{8}$$ lb.
• A pint of strawberries weighs about $$\frac{5}{8}$$ lb.

Question 1.
Write a multiplication sentence to show how much the pears weigh. ___
Total Weight of 8 pears = 3 pounds .
Explanation :
Number of pears = 8
Weight of each pear = $$\frac{3}{8}$$ lb.
Total Weight of 8 pears = Number of pears × Weight of each pear = 8 ×  $$\frac{3}{8}$$ = 3 lbs .
1 lbs = 1 pound
3 lbs = 3 pounds.
Therefore, Total Weight of 8 pears = 3 pounds .

Question 2.
Write a multiplication sentence to show how much the grapes weigh. ___
Total Weight of 8 pears = 2 pounds .
Explanation :
Number of cups of grapes = 2 cups
Weight of each cup = $$\frac{2}{8}$$ lb.
Total Weight of 2 cups of grapes = Number of cups of grapes × Weight of each cup of grapes = 2 ×  $$\frac{2}{8}$$ = 2 lbs .
1 lbs = 1 pound
2 lbs = 2 pounds.
Therefore, Total Weight of 2 cups of grapes = 2 pounds .

Question 3.
Write a multiplication sentence to show how much the strawberries weigh. ____
Total Weight of 4 pints of strawberries = 2$$\frac{1}{2}$$ pounds .
Explanation :
Number of pints of strawberries = 4
Weight of each pint of strawberries = $$\frac{5}{8}$$ lb.
Total Weight of 4 pints of strawberries = Number of pints of strawberries × Weight of each pint of strawberries = 4 × $$\frac{5}{8}$$ =  $$\frac{5}{2}$$ lbs  = 2$$\frac{1}{2}$$ lbs
1 lbs = 1 pound
2$$\frac{1}{2}$$ lbs = 2$$\frac{1}{2}$$ pounds.
Therefore, Total Weight of 4 pints of strawberries = 2$$\frac{1}{2}$$ pounds .

Question 4.
How much does Mr. Chou’s salad weigh in all? Show your work.
Total weight of salad = 7$$\frac{1}{2}$$ pounds.
Explanation :
Total Weight of 8 pears = 3 pounds .
Total Weight of 2 cups of grapes = 2 pounds .
Total Weight of 4 pints of strawberries = 2$$\frac{1}{2}$$ pounds .
Total Weight of salad = 3 + 2 + 2$$\frac{1}{2}$$ = 5 + 2$$\frac{1}{2}$$ = 5 + 2 + $$\frac{1}{2}$$= 7 + $$\frac{1}{2}$$ = 7$$\frac{1}{2}$$ pounds.
Therefore, Total weight of salad = 7$$\frac{1}{2}$$ pounds.

Division Number Stories

Question 1.
Robert and Jason want to buy a group ticket package for football games. Package A costs $276 and includes 2 tickets for each of 6 games. Package B costs$336 and includes 2 tickets for each of 8 games. Which package charges more per ticket? How much more per ticket?
Package ___ charges $___ more per ticket. Answer: Cost of Package A =$276
Number of tickets = 2 tickets for each of 6 games = 2 × 6 = 12 tickets
Cost of each ticket in Package A = $276 ÷ 12 = 23$ .
Cost of Package B = $336 Number of tickets = 2 tickets for each of 8 games = 2 × 8 = 16 tickets Cost of each ticket in Package A =$336 ÷ 16 = 21$. Package A costs more per ticket . More charge per ticket = 23$ – 21$= 2$ .

Question 2.
Rebecca wants to put 544 pennies in a coin-collection book. The blue book fits 9 pennies per page. The red book fits 7 pennies per page. How many more pages would she need if she used the red book rather than the blue one?

The red book will take __ more pages than the blue book.
What did you do with any remainders you found?
Total Number of Pennies = 544 pennies
Number of pennies in one page of blue book = 9 pennies per page
Number of pages required for 544 pennies in Blue book= 544 ÷ 9 = 60$$\frac{4}{9}$$ pages .
Number of pennies in one page of Red book = 7 pennies per page
Number of pages required for 544 pennies in Red book = 544 ÷ 7 = 77$$\frac{5}{7}$$ pages .
Number of More pages Red will take than Blue book = 77$$\frac{5}{7}$$ – 60$$\frac{4}{9}$$ = 77 + $$\frac{5}{7}$$ – 60 – $$\frac{4}{9}$$ = 17 + $$\frac{45}{63}$$ – $$\frac{28}{63}$$ = 17 + $$\frac{17}{63}$$ = 17$$\frac{17}{63}$$ .

Practice

Question 3.
754 ∗ 6 = ___
754 ∗ 6 = 4,524
Explanation : Question 4.
906 ∗ 2 =
906 ∗ 2 = 1,812
Explanation : Question 5.
__ = 831 ∗ 7
5,817 = 831 ∗ 7
Explanation : Question 6.
__ = 84 ∗ 29
2,436 = 84 ∗ 29
Explanation : More Division Home Measurement Number Stories

Read each number story. Use the information to write a number model with an unknown and then solves.

Question 1.
Kelly is in charge of bringing water for her softball game. The 8 members of the team have matching team water bottles that hold 500 mL. Kelly buys 5 liters of water at the store. If she fills all the bottles, how many milliliters of water will Kelly have left?
Number model with unknown: ___
Number of members = 8
Quantity of each water bottle = 500 ml
Quantity of 8 water bottles = 8 × 500 ml = 4000 ml
1 liter = 1000ml
5 liters = 5000ml
Quantity of water Kelly bought = 5 liters = 5000 milliliters
Quantity of water left = 5000 – 4000 = 1000 milliliters .
Therefore, Quantity of water left = 1000 milliliters

Question 2.
The distance around all the bases in softball is 72 meters. If Kelly hits 2 home runs and runs around the bases twice, how many millimeters will she run?
Number model with unknown: _____
Distance of the bases = 72 meters .
Number of runs = 2
Distance covered in running = 2 × 72 meters = 144 meters .
1 meter = 1000 millimeters
144 meters = 144 × 1000 = 144000 millimeters .
Therefore, Distance covered in running =1,44,000 millimeters .

Question 3.
In women’s softball the pitcher stands about 13 meters from the batter’s box. In men’s softball the pitcher stands about 1,400 centimeters from the batter’s box. About how many more centimeters is it from the men’s pitcher to the batter’s box than from the women’s pitcher to the batter’s box?
Number model with unknown: _____
Distance between the pitcher stands and the batter’s box in Women’s soft ball = 13 meters = 1300 centimeters .
as, 1meter = 100 centimeters .
Distance between the pitcher stands and the batter’s box in Men’s soft ball = 1400 centimeters
More distance from the men’s pitcher to the batter’s box than from the women’s pitcher to the batter’s box = 1400 – 1300 = 100 centimeters .
Therefore, More distance from the men’s pitcher to the batter’s box than from the women’s pitcher to the batter’s box = 100 centimeters .

Question 4.
The 6 games Kelly’s team played took a total of 7 hours.
a. How many minutes total did they play softball?
Number model with unknown: ____
b. If each game lasted the same amount of time, how many minutes did each one last?
Number model with unknown: ______
a.
Number of games = 6
As, 1 hour = 60 minutes
7 hours = 7 × 60 = 420 minutes .
Time taken to play 6 games = 7 hours . = 420 minutes .
b.
Number of games = 6
Total Time for 6 games = 420 minutes.
Time for each game = 420 ÷ 6 = 70 minutes = 1 hour 10 minutes.
Therefore , Each game last for 70 minutes .

Practice

Question 5.
1$$\frac{3}{6}$$ + 2$$\frac{1}{6}$$ = ___
1$$\frac{3}{6}$$ + 2$$\frac{1}{6}$$ = 3$$\frac{2}{3}$$
1$$\frac{3}{6}$$ + 2$$\frac{1}{6}$$ = $$\frac{9}{6}$$ + $$\frac{13}{6}$$ = $$\frac{22}{6}$$ = $$\frac{11}{3}$$ = 3$$\frac{2}{3}$$

Question 6.
4$$\frac{3}{5}$$ + 5$$\frac{4}{5}$$ = ___
4$$\frac{3}{5}$$ + 5$$\frac{4}{5}$$ = 10$$\frac{2}{5}$$
Explanation :
4$$\frac{3}{5}$$ + 5$$\frac{4}{5}$$ = $$\frac{23}{5}$$ + $$\frac{29}{5}$$ = $$\frac{52{5}$$ = 10$$\frac{2}{5}$$

Question 7.
7$$\frac{5}{12}$$ – 2$$\frac{3}{12}$$ = ___
7$$\frac{5}{12}$$ – 2$$\frac{3}{12}$$ = 5$$\frac{1}{6}$$
Explanation :
7$$\frac{5}{12}$$ – 2$$\frac{3}{12}$$ = $$\frac{89}{12}$$ – $$\frac{27}{12}$$ = $$\frac{62}{12}$$ = $$\frac{31}{6}$$ = 5$$\frac{1}{6}$$ .

Question 8.
6$$\frac{1}{3}$$ – 2$$\frac{2}{3}$$ = ___
6$$\frac{1}{3}$$ – 2$$\frac{2}{3}$$ = 3$$\frac{2}{3}$$
Explanation :
6$$\frac{1}{3}$$ – 2$$\frac{2}{3}$$ = $$\frac{19}{3}$$ – $$\frac{8}{3}$$ = $$\frac{11}{3}$$ = 3$$\frac{2}{3}$$ .

Perimeter Patterns

Alice was making squares out of toothpicks. She noticed a pattern involving the length of one side and the perimeter of the square. Complete the table and then answer the questions that follow.  Question 1.
What rule describes the relationship between the length of one side and the perimeter of a square?
The relationship between the length of one side and the perimeter of a square is perimeter is 4 times the length of the side .

Question 2.
What would be the perimeter of a square with a side length of 25 toothpicks? ___ toothpicks
100 toothpicks
Explanation :
length of the side = 25 toothpicks
perimeter = 4 times the length of a side = 4 ( 25) = 100 toothpicks .

Question 3.
What would be the side length of a square with a perimeter of 500 toothpicks? __ toothpicks
length of a side = 125 toothpicks .
Explanation :
perimeter = 500 toothpicks
perimeter = 4 times the length of a side
500 = 4 × length of a side
length of a side  = 500 ÷ 4 = 125 toothpicks .

Question 4.
Describe at least two other patterns you notice in the table ____
The two patterns are
Perimeter is 4 times the length of the side .
All the Values of perimeter are multiples of 4 .

Practice

Question 5.
753 ÷ 3 = ___
753 ÷ 3 = 251
Explanation : Question 6.
__ = 386 ÷ 2
193 = 386 ÷ 2
Explanation : Question 7.
283 ÷ 9 → __
283 ÷ 9 → 31 + 4 Remainder
Explanation : Question 8.
505 ÷ 6 → ___
505 ÷ 6 → 84 + 1 Remainder
Explanation : Fitness Challenge

Use the information in the table below to solve the number stories.
During Marcy School’s 2-week challenge, each student who meets a goal wins a prize. Question 1.
Tony will run $$\frac{1}{2}$$ mile after school each day. Will he win a prize?
a. Distance run in 1 week: mile(s)
b. In 2 weeks: mile(s)
a.
Distance run by Tony each day = $$\frac{1}{2}$$ mile
Number of days for 1 week = 5 days .
Distance run in 1 week = 5 × $$\frac{1}{2}$$ = $$\frac{5}{2}$$ = 2$$\frac{1}{2}$$ miles .
b.
Distance run by Tony each day = $$\frac{1}{2}$$ mile
Number of days in 2 weeks = 2 × 5 = 10 days
Distance run in 2 weeks = 10 × $$\frac{1}{2}$$ = 5 miles .
To win a prize distance run should be 4 miles .
In 2 weeks tony runs 5 miles that means 5 > 4 . So, Tony will win a prize .

Question 2.
Three times a week, Tina walks $$\frac{3}{10}$$ mile from school to the library, studies for 1 hour,
and then walks $$\frac{4}{10}$$ mile home. How much more will she need to walk to win a prize?
____ mile(s)
Number of days in a week = 3 days .
Distance walk by Tina from school to library  = $$\frac{3}{10}$$ mile
Distance walk by Tina from library to home =  $$\frac{4}{10}$$ mile
Total Distance Tina walks for a day  = $$\frac{3}{10}$$ + $$\frac{4}{10}$$ mile = $$\frac{7}{10}$$ mile .
Total Distance Tina walks for 3 days or 1 week = 3 × $$\frac{7}{10}$$ = $$\frac{21}{10}$$ miles = 2.1 miles .
Total Distance Tina walks for 2 weeks = 2 × 2.1 = 4.2 miles .
To win a prize distance walk should be 6 miles .
2.1 < 6 miles that means she doesnot win a prize .
Therefore, Tina doesn’t receive a prize .

Practice

Question 3.
642 ÷ 2 = ___
642 ÷ 2 = 321
Explanation : Question 4.
386 / 9 → ____
386 / 9 → 42   + 8 Remainder
Explanation : Question 5.
739 / 5 → ___
739 / 5 → 147  + 4 Remainder .
Explanation : Question 6.  = 207  + 1 Remainder .
Explanation : Fractions and Mixed Numbers

Solve. Draw a picture or show how you solved the problem.

Question 1.
5 ∗ $$\frac{3}{2}$$ =
5 ∗ $$\frac{3}{2}$$ = 7$$\frac{1}{2}$$
Explanation :
5 ∗ $$\frac{3}{2}$$ = $$\frac{15}{2}$$ = 7$$\frac{1}{2}$$ Question 2.
__ = 4$$\frac{2}{6}$$ – 2$$\frac{4}{6}$$
1$$\frac{2}{3}$$ = 4$$\frac{2}{6}$$ – 2$$\frac{4}{6}$$
Explanation :
4$$\frac{2}{6}$$ – 2$$\frac{4}{6}$$ = $$\frac{26}{6}$$ – $$\frac{16}{6}$$ = $$\frac{10}{6}$$ = $$\frac{5}{3}$$ = 1$$\frac{2}{3}$$ .

Question 3.
5$$\frac{7}{8}$$ + 3$$\frac{1}{8}$$ = ___
5$$\frac{7}{8}$$ + 3$$\frac{1}{8}$$ =
Explanation :
5$$\frac{7}{8}$$ + 3$$\frac{1}{8}$$ = $$\frac{47}{8}$$ + $$\frac{25}{8}$$ = $$\frac{72}{8}$$ = 9 inches .

Question 4.
___ = 3 ∗ 4$$\frac{1}{4}$$
12$$\frac{3}{4}$$ = 3 ∗ 4$$\frac{1}{4}$$
Explanation :
3 ∗ 4$$\frac{1}{4}$$ = 3 ∗ $$\frac{17}{4}$$ = $$\frac{51}{4}$$ = 12$$\frac{3}{4}$$

Question 5.
The combined weight of an assortment of fruit is 8$$\frac{3}{4}$$ pounds. When the fruit is on a tray, the tray weighs 10$$\frac{1}{4}$$ pounds. How many pounds does the tray weigh when empty? __ pound(s)
How many ounces does the tray weigh when empty? __ ounce(s)
Weight of the assortment of fruit = 8$$\frac{3}{4}$$ = $$\frac{35}{4}$$ pounds
Weight of the fruit on tray = 10$$\frac{1}{4}$$ = $$\frac{41}{4}$$ pounds .
Weight of the tray = $$\frac{41}{4}$$ – $$\frac{35}{4}$$ = $$\frac{6}{4}$$ = $$\frac{3}{2}$$ = 1$$\frac{1}{2}$$ pounds.
Therefore, Weight of the tray = 1$$\frac{1}{2}$$ pounds

Question 6.
(3 ∗ 2$$\frac{2}{3}$$) + (2 ∗ 4$$\frac{1}{3}$$) = ___
(3 ∗ 2$$\frac{2}{3}$$) + (2 ∗ 4$$\frac{1}{3}$$) = 16 $$\frac{2}{3}$$
Explanation :
(3 ∗ 2$$\frac{2}{3}$$) + (2 ∗ 4$$\frac{1}{3}$$) = (3 ∗ $$\frac{8}{3}$$) + (2 ∗ $$\frac{13}{3}$$) = $$\frac{24}{3}$$ + $$\frac{26}{3}$$ = $$\frac{50}{3}$$ = 16 $$\frac{2}{3}$$ .

Practice

Question 7.  = 116 + 2 Remainder
Explanation : Question 8.  = 138   + 4 Remainder
Explanation : Question 9.  = 42  + 1 Remainder
Explanation : Question 10.  = 64  + 6 Remainder
Explanation : Shopping for Bargains

Solve each number story and show how you solved the problems.

Question 1.
Phil wants to buy some Creepy Creature erasers that cost $1.05 each. If he buys 5 or more, the price is$0.79 each. If he decides to buy 7 erasers, how much will he spend?
Answer: $___ Cost each Creepy eraser =$1.05 each
Cost of each Creepy Eraser if buyed more than 5 = $0.79 each Cost of 7 creepy eraser = 7 ×$0.79 = 7 × $$\frac{79}{100}$$ = $$\frac{553}{100}$$ = 5.53$. Therefore, Cost of 7 creepy erasers = 5.53$ .

Question 2.
Mrs. Katz bought 3 pounds of apples and a muffin for snacks. The apples cost $2.59 per pound if you buy less than 3 pounds and$2.12 per pound if you buy 3 or more pounds. The muffin cost $1.95. How much did she spend? Answer:$ _____
3 pounds of apples and a muffin for snacks
Katz brought 3 pounds so, cost = $2.12 per pound Cost of apples =$2.12 per pound
Cost of 3 pounds of apples = 3 × $2.12 = 6.36$
Cost of muffin = $1.95 Total cost spent = 6.36$ + $1.95 =$ 8.31 .
Therefore, Total cost spent = $8.31 . Try This Question 3. Mrs. Katz paid with a$10 bill. How much change did she get back?
Answer: $___ Money given =$ 10 .
Total cost spent = $8.31 . Change given = 10 – 8.31 = 1.69$

Practice

Fill in the blanks with >, <, or =.

Question 4.
0.55 __ 0.65
0.55 < 0.65
Explanation :
When comparing decimals, start in the tenths place. The decimal with the biggest value there is greater. If they are the same, move to the hundredths place and compare these values. If the values are still the same keep moving to the right until you find one that is greater or until you find that they are equal.

Question 5.
0.3 __ 0.30
0.3 = 0.30
Explanation :
When comparing decimals, start in the tenths place. The decimal with the biggest value there is greater. If they are the same, move to the hundredths place and compare these values. If the values are still the same keep moving to the right until you find one that is greater or until you find that they are equal.

Question 6.
0.72 __ 0.8
0.72 < 0.8
Explanation :
When comparing decimals, start in the tenths place. The decimal with the biggest value there is greater. If they are the same, move to the hundredths place and compare these values. If the values are still the same keep moving to the right until you find one that is greater or until you find that they are equal.

Question 7.
0.4 __ 0.31
0.4 > 0.31
Explanation :
When comparing decimals, start in the tenths place. The decimal with the biggest value there is greater. If they are the same, move to the hundredths place and compare these values. If the values are still the same keep moving to the right until you find one that is greater or until you find that they are equal.

Pencil Lengths

At the beginning of the year Mrs. Kerry gave each student in her class a new pencil with “Welcome to 4th Grade” written on it. A month later the class measured their pencils to the nearest $$\frac{1}{8}$$ inch.
Pencil Lengths to the Nearest $$\frac{1}{8}$$ inch Plot the data set on the line plot.  Use the completed line plot to answer these questions.

Question 1.
How many students have a pencil that is shorter than 2$$\frac{7}{8}$$ inches? ___ students
Number of students shorter than 2$$\frac{7}{8}$$ inches = 13 students .
Explanation : Number of students have 1$$\frac{7}{8}$$ inches = 1
Number of students have 2 inches = 1
Number of students have 2$$\frac{1}{8}$$ inches = 1
Number of students have 2$$\frac{3}{8}$$ inches = 2
Number of students have 2$$\frac{4}{8}$$ inches = 3
Number of students have 2$$\frac{5}{8}$$ inches = 3
Number of students have 2$$\frac{6}{8}$$ inches = 2
Total students = 1+1+1+ 2+3+3+2 = 13 students .

Question 2.
What is the most common pencil length? __ inches
The most common pencil length is 2$$\frac{7}{8}$$ as it is used by 5 students .

Question 3.
a. How many pencils are less than 2$$\frac{2}{8}$$ inches long? ___ pencils
b. What is their combined length? ___ inches
a.
Number of students have 1$$\frac{7}{8}$$ inches pencil = 1 pencil
Number of students have 2 inches pencil = 1
Number of students have 2$$\frac{1}{8}$$ inches pencil = 1
Total pencils = 1+1+1 = 3 pencils .
b.
Combined length = 1$$\frac{7}{8}$$ + 2 + 2$$\frac{1}{8}$$ = 1 + $$\frac{7}{8}$$  + 2 + 2+ $$\frac{1}{8}$$ = 5 + $$\frac{8}{8}$$ = 5 + 1 = 6 inches.

Question 4.
a. How many pencils are between 2$$\frac{7}{8}$$ and 3$$\frac{2}{8}$$ inches long? ___ pencils
b. What is their combined length? inches
a.
Number of pencils have 3 inches = 1
Number of pencils have 3$$\frac{1}{8}$$ inches = 2
Total number of pencils between 2$$\frac{7}{8}$$ and 3$$\frac{2}{8}$$ inches long = 1 + 2 = 3 pencils .
b.
Combined lengths = 3 + 3$$\frac{1}{8}$$ = 3 + $$\frac{25}{8}$$ = $$\frac{24}{8}$$ + $$\frac{25}{8}$$ = $$\frac{49}{8}$$ = 6$$\frac{1}{8}$$ inches .

Question 5.
a. How long is the longest pencil? ___inches
b. How long is the shortest pencil? ___ inches
c. What is the combined length of the longest and shortest pencils? __ inches
d. What is the difference in length of the longest and shortest pencils? __ inches
a.  Longest pencil = 3$$\frac{4}{8}$$ inches .
b. Shortest pencil = 1$$\frac{7}{8}$$ inches .
c. Combined length of the longest and shortest pencils = 3$$\frac{4}{8}$$ + 1$$\frac{7}{8}$$ = $$\frac{27}{8}$$ + $$\frac{15}{8}$$ = $$\frac{42}{8}$$ = 6 inches .
d. Difference in length of the longest and shortest pencils = = 3$$\frac{4}{8}$$ – 1$$\frac{7}{8}$$ = $$\frac{27}{8}$$ – $$\frac{15}{8}$$ = $$\frac{12}{8}$$ = $$\frac{3}{2}$$ inches.

Practice

Question 6.
2$$\frac{1}{4}$$ + 5$$\frac{2}{4}$$ = ___
2$$\frac{1}{4}$$ + 5$$\frac{2}{4}$$ =
Explanation :
2$$\frac{1}{4}$$ + 5$$\frac{2}{4}$$ = $$\frac{9}{4}$$ + $$\frac{22}{4}$$ = $$\frac{31}{4}$$ = 7$$\frac{3}{4}$$ .

Question 7.
8$$\frac{5}{10}$$ + 3$$\frac{7}{10}$$ = __
8$$\frac{5}{10}$$ + 3$$\frac{7}{10}$$ = 12$$\frac{1}{5}$$
Explanation :
8$$\frac{5}{10}$$ + 3$$\frac{7}{10}$$ = $$\frac{85}{10}$$ + $$\frac{37}{10}$$ = $$\frac{122}{10}$$ = $$\frac{61}{5}$$ = 12$$\frac{1}{5}$$

Question 8.
3$$\frac{7}{8}$$ – 1$$\frac{3}{8}$$ = ___
3$$\frac{7}{8}$$ – 1$$\frac{3}{8}$$ = 2latex]\frac{1}{2}[/latex]
3$$\frac{7}{8}$$ – 1$$\frac{3}{8}$$ = $$\frac{31}{8}$$ – $$\frac{11}{8}$$ = latex]\frac{20}{8}[/latex] = latex]\frac{5}{2}[/latex] = 2latex]\frac{1}{2}[/latex] .
7$$\frac{41}{8}$$ – 3$$\frac{51}{100}$$ = __
7$$\frac{41}{8}$$ – 3$$\frac{51}{100}$$ = 8$$\frac{123}{200}$$
7$$\frac{41}{8}$$ – 3$$\frac{51}{100}$$ = $$\frac{41}{8}$$ – 3$$\frac{51}{100}$$ = $$\frac{97}{8}$$ – $$\frac{351}{100}$$ = $$\frac{2425}{200}$$ – $$\frac{702}{200}$$ = $$\frac{1723}{200}$$ = 8$$\frac{123}{200}$$ .