## Everyday Mathematics 4th Grade Answer Key Unit 8 Fraction Operations; Applications

### Everyday Math Grade 4 Home Link 8.1 Answer Key

**Multistep Number Stories**

The fourth-grade students in Mr. Kennedy’s class are investigating energy and motion. Students worked in teams to build two machines: a car that is propelled by a mousetrap and a boat that is propelled by balloons. Today the teams are competing to see which cars and boats go farthest.

Each car or boat gets 3 trials. The total distance from all 3 trials is used to determine which car or boat went farthest. Solve the number stories to help Mr. Kennedy’s class compare the machines made by various teams.

Question .1

Team A’s car went 173 cm on the first trial, 206 cm on the second trial, and 245 cm on the third trial. Team B’s car went 217 cm on each of the three trials.

Which car went the farthest overall? ____

How much farther did it go? ____

Answer:

Distance travelled by Team A car in first trial = 173 cm

Distance travelled by Team A car in second trial = 206 cm

Distance travelled by Team A car in third trial = 245 cm

Total Distance travelled by Team A car 3 trials = 173 + 206 + 245 = 624 cm

Distance travelled by Team B car in each trial = 217 cm

Total Distance travelled by Team B car 3 trials = 217 + 217 + 217 = 651 cm

651 cm > 624 cm

Team B car went the farthest overall .

Distance More covered by Team B car than Team A car = 651 – 624 = 27 cm .

Therefore the farther distance it went is 27 cms

Question 2.

Team A’s boat went 130 cm in all. Team B’s boat went the same distance on all 3 trials and lost to Team A’s boat by 7 cm.

How far did Team B’s boat go on each trial? ____

Answer:

Distance travelled by Team A’s boat in all trials= 130 cm .

Distance travelled by Team B’s boat in each trial = x

Distance travelled by Team B’s boat in all trials = 3x cms .

Team B is lost by 7 cms that means distance travelled by Team B + 7cms = Distance travelled by Team A .

3x + 7 = 130

3x = 130 – 7

3x = 123

x = 41 cms

Therefore, Distance travelled by Team B’s boat in each trial = 41 cms .

Question 3.

Team D’s car went the same distance on each of its trials. Team C’s car went exactly 1 cm farther in each trial than Team D’s car. Team C’s car went 543 cm in all.

How far did Team D’s cargo on each trial? ___

Answer :

Distance travelled by Team D car in each trial = x

Distance travelled by Team D car in 3 trials = 3x

Distance travelled by Team C car in each trial = 1 cm farther in each trial than Team D’s car. = x + 1.

Distance travelled by Team C car in 3 trials = 3 ( x + 1 ) = 3x + 3 .

Distance travelled by Team C car in 3 trials = 543 .

3x + 3 = 543 .

3x = 543 – 3

3x = 540 .

x = 180 cms.

Therefore, Distance travelled by Team D car in each trial = x = 180 cms .

**Practice**

Question 4.

5,624 ÷ 8 = ____

Answer:

5,624 ÷ 8 = 703

Explanation :

Question 5.

8,500 ÷ 3 = ___

Answer :

8,500 ÷ 3 = 2833 remainder 1

Explanation :

Question 6.

Answer:

9,207 ÷ 4 = 2301 remainder 3

Explanation :

Question 7.

Answer:

3,578 ÷ 5 = 715 remainder 3

Explanation :

### Everyday Math Grade 4 Home Link 8.2 Answer Key

**Finding Unknown Angle Measures**

Find the missing angle measures. For each problem, write an equation with a letter for the unknown to show how you found your answer.

Question 1.

f = ___

Equation: ____

Answer:

f = 165 °

Equation :

82 ° + 83 ° = f

165 ° = f .

Question 2.

p = ___

Equation: ____

Answer:

The given figure is circle . the Total angle = 360°

the total angle = 210 ° + p = 360 °

Equation:

210 ° + p = 360 °

p = 360 ° – 210 °

p = 150°

Question 3.

w = ___

Equation: ____

Answer:

The angle formed in the above figure is right angle = 90 °

Equation :

3 ° + w = 90 °

w = 90 ° – 3 °

w = 87 °

Question 4.

l = ___

Equation: ____

Answer:

The angle forward in a straight line.

Total angle = 180 ° .

Equation :

I + 75 ° = 180 °

I = 180 ° – 75 °

I = 105 °

Question 5.

s = ___

Equation: ____

Answer:

The angle forward in a straight line.

Total angle = 180 ° .

Equation :

s + 43 ° = 180 °

s = 180 ° – 43 °

s = 137 °

Question 6.

i = ___

Equation: ____

Answer:

The given figure is circle . the Total angle = 360°

the total angle = 120 ° + i = 360 °

Equation:

120 ° + i = 360 °

i = 360 ° – 120 °

i = 240°

**Practice**

Question 7.

\(\frac{1}{3}\) + \(\frac{2}{3}\) + \(\frac{2}{3}\) = ___

Answer:

\(\frac{1}{3}\) + \(\frac{2}{3}\) + \(\frac{2}{3}\)

All denominators are equal so add directly .

= \(\frac{5}{3}\) = 1\(\frac{2}{3}\)

Question 8.

\(\frac{1}{4}\) + \(\frac{3}{4}\) + \(\frac{3}{4}\) = ____

Answer:

\(\frac{1}{4}\) + \(\frac{3}{4}\) + \(\frac{3}{4}\) = \(\frac{7}{4}\) = 1\(\frac{3}{4}\)

Question 9.

\(\frac{4}{5}\) + \(\frac{4}{5}\) + \(\frac{3}{5}\) = ____

Answer:

\(\frac{4}{5}\) + \(\frac{4}{5}\) + \(\frac{3}{5}\) = \(\frac{11}{5}\) = 2\(\frac{1}{5}\)

Question 10.

\(\frac{5}{12}\) + \(\frac{3}{12}\) + \(\frac{7}{12}\) = ___

Answer:

\(\frac{5}{12}\) + \(\frac{3}{12}\) + \(\frac{7}{12}\) = \(\frac{15}{12}\) = 1\(\frac{3}{12}\)

### Everyday Math Grade 4 Home Link 8.3 Answer Key

**Finding Pattern-Block Measures**

Molly is using pattern blocks to find angle measures of other pattern blocks. She knows that the measure of the small angle of a white rhombus is 30°.

Question 1.

Molly fills an angle of the green triangle with the small angles of white rhombuses. What is the measure of the triangle’s angle? Explain how you know.

Answer:

Given Angle of small angles of white Rhombus = 30°

Angle formed in the above figure is by two small white rhombuses.

Angle formed = 30 ° + 30 ° = 60 ° .

Therefore, Angle Measure = 60 °

Question 2.

Molly fills a red trapezoid’s large angle with angles of the green triangle. What is the measure of the red trapezoid’s large angle? Explain how you know.

Answer :

Angle of the green triangle = 60 ° .

Angle formed by two green triangles .

Angle Measure of the red trapezoid’s large angle = 60 ° + 60 ° = 120 °

Therefore, Angle Measure of the red trapezoid’s large angle = 120 °

**Practice**

Question 3.

5,588 ∗ 3 = ____

Answer:

5,588 ∗ 3 = 16764

Explanation :

Step I: Arrange the numbers vertically.

Step II: First multiply the digit at the ones place by 3.

8 × 3 = 24 = 2 ten + 4 ones

Write 4 in the ones column and carry 1 ten to tens place.

Step III: Multiply the digit at the tens place by 3.

8 × 3 = 24 tens

24 + 2 (carried) = 26 tens = 2 hundreds + 6 ten

Write 6 in ten place and carry 2 hundreds to hundreds place.

Step IV: Multiply the digit at hundreds place by 3.

5 hundred × 3 = 15 hundreds

15 hundreds + 2 hundreds (carried) = 1 thousands + 7 hundreds

Write 7 in hundreds place and carry 1 thousands to thousands place.

Step V: Multiply the digit at thousands place by 3.

5 thousands × 3 = 15 thousands

Write 5 in thousands place and carry 1 ten thousands to ten thousands place.

Question 4.

9,037 ∗ 5 = ___

Answer:

9,037 ∗ 5 = 45,185

Explanation :

Step I: Arrange the numbers vertically.

Step II: First multiply the digit at the ones place by 5.

7 × 5 = 35 = 3 ten + 5 ones

Write 5 in the ones column and carry 3 ten to tens place.

Step III: Multiply the digit at the tens place by 5.

3 × 5 = 15 tens

15 + 3 (carried) = 18 tens = 1 hundreds + 8 ten

Write 8 in ten place and carry 1 hundreds to hundreds place.

Step IV: Multiply the digit at hundreds place by 5

0 hundred × 5 = 0 hundreds

0 hundreds + 1 hundreds (carried) = 1 hundreds

Write 1 in hundreds place .

Step V: Multiply the digit at thousands place by 5.

9 thousands × 5 = 45 thousands

Write 5 in thousands place and carry 4 ten thousands to ten thousands place.

Question 5.

52 ∗ 94 = ___

Answer:

52 ∗ 94 = 4,888

Explanation :

Step I: Arrange the numbers vertically.

Step II: Multiply 52 by 4 ones

52 × 4 = 208

Step III: Multiply 52 by 9 tens

52 × 90 = 4680

Step IV: Add

208 + 4680 = 4888

Question 6.

83 ∗ 77 = ___

Answer:

83 ∗ 77 = 6,391

Explanation :

Step I: Arrange the numbers vertically.

Step II: Multiply 83 by 7 ones

83 × 7 = 581

Step III: Multiply 83 by 7 tens

83 × 70 = 5810

Step IV: Add

581 + 5810 = 6391

### Everyday Math Grade 4 Home Link 8.4 Answer Key

**Line Symmetry**

Use a straightedge to draw the lines of symmetry on each shape.

Question 1.

Draw 2 lines of symmetry.

Answer:

Explanation :

Something is symmetrical when it is the same on both sides. A shape has symmetry if a central dividing line (a mirror line) can be drawn on it, to show that both sides of the shape are exactly the same.

Question 2.

Draw 6 lines of symmetry

Answer:

6 symmetrical lines are drawn from centre and 6 identical shapes are formed .

Explanation :

Something is symmetrical when it is the same on both sides. A shape has symmetry if a central dividing line (a mirror line) can be drawn on it, to show that both sides of the shape are exactly the same.

Question 3.

Draw 1 line of symmetry.

Answer:

1 symmetrical lines is drawn and 2 identical shapes are formed .

Explanation :

Something is symmetrical when it is the same on both sides. A shape has symmetry if a central dividing line (a mirror line) can be drawn on it, to show that both sides of the shape are exactly the same.

Question 4.

Draw 3 lines of symmetry.

Answer:

3 symmetrical lines is drawn from the centre and 3 identical shapes are formed .

Explanation :

Something is symmetrical when it is the same on both sides. A shape has symmetry if a central dividing line (a mirror line) can be drawn on it, to show that both sides of the shape are exactly the same.

Question 5.

How many lines of symmetry does this shape have? ___

Draw the line(s) of symmetry.

Answer:

Number of lines of symmetry for above shape = 1.

1 symmetrical lines is drawn and 2 identical shapes are formed .

Explanation :

Something is symmetrical when it is the same on both sides. A shape has symmetry if a central dividing line (a mirror line) can be drawn on it, to show that both sides of the shape are exactly the same.

Question 6.

Draw your own shape. Show the lines of symmetry. Be sure your shape includes at least 1 right angle.

Answer:

2 symmetrical lines is drawn and 4 identical shapes are formed .

Explanation :

Something is symmetrical when it is the same on both sides. A shape has symmetry if a central dividing line (a mirror line) can be drawn on it, to show that both sides of the shape are exactly the same.

**Practice**

Question 7.

6 ∗ \(\frac{5}{6}\) = ___

Answer:

6 ∗ \(\frac{5}{6}\) = 5

Question 8.

3 ∗ \(\frac{3}{8}\) = ___

Answer :

3 ∗ \(\frac{3}{8}\) = \(\frac{9}{8}\) = 1\(\frac{1}{8}\)

Question 9.

4 ∗ \(\frac{7}{10}\) = ___

Answer:

4 ∗ \(\frac{7}{10}\) = 2 ∗ \(\frac{7}{5}\) = \(\frac{14}{5}\) = 2\(\frac{4}{5}\)

Question 10.

6 * \(\frac{4}{12}\) = ___

Answer:

6 * \(\frac{4}{12}\) = 1 * \(\frac{4}{2}\) = \(\frac{2}{1}\) = 2

### Everyday Math Grade 4 Home Link 8.5 Answer Key

**Designing a Bookcase**

Nicholas is building a bookcase. To help with the design, he measured the height

of each of his books to the nearest \(\frac{1}{8}\) inch. His measurements are given below.

6\(\frac{1}{2}\), 9\(\frac{1}{4}\), 7\(\frac{1}{8}\), 7\(\frac{1}{2}\), 8, 6\(\frac{7}{8}\), 9\(\frac{1}{4}\), 9\(\frac{1}{4}\), 9\(\frac{1}{4}\), 9\(\frac{1}{4}\), 9\(\frac{1}{4}\), 8\(\frac{1}{4}\), 8, 8\(\frac{1}{4}\), 8\(\frac{3}{8}\)

6\(\frac{1}{2}\), 7\(\frac{1}{8}\), 9, 6\(\frac{7}{8}\), 7\(\frac{1}{2}\), 8, 8\(\frac{1}{4}\), 9\(\frac{1}{4}\), 6\(\frac{7}{8}\), 6\(\frac{7}{8}\), 8\(\frac{1}{4}\), 8\(\frac{1}{4}\), 8\(\frac{1}{4}\)

Plot the data set on the line plot below

Use the completed line plot to answer the questions below.

Answer :

Question 1.

What is the difference in height between the tallest and shortest books? __ in.

Answer:

Height of tallest book = 9\(\frac{3}{8}\) inches

Height of the shortest book = 6\(\frac{1}{2}\) inches

The difference in height between the tallest and shortest books = 9\(\frac{3}{8}\) – 6\(\frac{1}{2}\) = \(\frac{75}{8}\)– \(\frac{13}{2}\) =\(\frac{75}{8}\)– \(\frac{52}{8}\) = \(\frac{23}{8}\)= 2\(\frac{7}{8}\) .

Therefore, the difference in height between the tallest and shortest books = 2\(\frac{7}{8}\) inches

Question 2.

Nicholas wants the space between the shelves to be \(\frac{7}{8}\) inch taller than his tallest book.

a. How far apart should he make the shelves? ___ in.

b. If the thickness of the wood he uses for the shelves is \(\frac{5}{8}\) inch, what will be the total height of each shelf? (Hint: The total height is the thickness of one piece of wood plus the distance between shelves.) __ in.

Answer :

a. Height of tallest book = 9\(\frac{3}{8}\) = \(\frac{75}{8}\) inches

the space between the shelves to be \(\frac{7}{8}\) inch

Height of the shelves to apart = \(\frac{75}{8}\) + \(\frac{7}{8}\) = \(\frac{82}{8}\) = \(\frac{41}{4}\) inches .

**Practice**

Question 3.

8,207 ÷ 7 → ___

Answer:

8,207 ÷ 7 → 1172 + 3 remainder

Explanation :

Question 4.

7,109 ÷ 8 → ___

Answer:

7,109 ÷ 8 → 888 + 5 remainder

Explanation :

### Everyday Math Grade 4 Home Link 8.6 Answer Key

**Perimeters and Missing Measures**

Use a formula to find the perimeter of each rectangle. Show your work in the space provided.

Question 1.

Perimeter: ___ yd

Answer:

Length of the Rectangle = 3\(\frac{3}{6}\) = \(\frac{21}{6}\) yd

Width of the rectangle = \(\frac{1}{6}\) yd

Perimeter of a Rectangle = 2 ( length + width ) = 2 ( \(\frac{21}{6}\) + \(\frac{1}{6}\) ) = 2 ( \(\frac{22}{6}\) ) = \(\frac{22}{3}\) yds .

Therefore, Perimeter of a Rectangle =\(\frac{22}{3}\) yds .

Question 2.

Perimeter: __ ft

Answer:

Length of the Rectangle = 5\(\frac{1}{12}\) = \(\frac{61}{12}\) ft

Width of the rectangle = 4\(\frac{11}{12}\) ft = \(\frac{59}{12}\) ft

Perimeter of a Rectangle = 2 ( length + width ) = 2 ( \(\frac{61}{12}\) + \(\frac{59}{12}\) ) = 2 ( \(\frac{120}{12}\) ) = 2 ( 10 ) = 20 ft .

Therefore, Perimeter of a Rectangle = 20 ft .

Question 3.

Answer:

Perimeter of a Rectangle = \(\frac{74}{100}\) kms

Length of the Rectangle = \(\frac{25}{100}\) km

Width of the Rectangle = W

Perimeter of a Rectangle = 2 ( length + width )

=> \(\frac{74}{100}\) = 2 ( \(\frac{25}{100}\) + W )

=> \(\frac{74}{100}\) = \(\frac{50}{100}\) + 2W

=> \(\frac{74}{100}\) – \(\frac{50}{100}\) = 2W

=> \(\frac{24}{100}\) = 2W

=> W = \(\frac{24}{200}\)

=> W = \(\frac{3}{25}\) km .

Therefore,Width of a Rectangle = \(\frac{3}{25}\) km

Question 4.

Answer:

Perimeter of a Rectangle = 10 in

Length of the Rectangle = 4\(\frac{3}{8}\) = \(\frac{35}{8}\) in

Width of the Rectangle = W

Perimeter of a Rectangle = 2 ( length + width )

=> 10 = 2 ( \(\frac{35}{8}\) + W )

=> 10 = \(\frac{35}{8}\)+ 2W

=> \(\frac{80}{8}\) – \(\frac{35}{8}\) = 2W

=> \(\frac{45}{8}\) = 2W

=> W = \(\frac{45}{16}\) = 2\(\frac{13}{16}\) in

Therefore,Width of a Rectangle = 2\(\frac{13}{16}\) in

Question 5.

Answer:

Perimeter of a Rectangle = 12\(\frac{8}{10}\) = \(\frac{128}{10}\) cm

Length of the Rectangle = 4\(\frac{1}{10}\) = \(\frac{41}{10}\) cm

Width of the Rectangle = W

Perimeter of a Rectangle = 2 ( length + width )

=> \(\frac{128}{10}\) = 2 ( \(\frac{41}{10}\) + W )

=> \(\frac{128}{10}\) = \(\frac{82}{10}\) + 2W

=> \(\frac{128}{10}\) – \(\frac{82}{10}[/latex = 2W

=> [latex]\frac{46}{10}\) = 2W

=> W = \(\frac{46}{20}\)

=> W = \(\frac{23}{10}\) cm .

Therefore,Width of a Rectangle = \(\frac{23}{10}\) cm

**Try this**

Question 6.

Answer:

Perimeter of a Rectangle = 16\(\frac{1}{2}\) = \(\frac{33}{2}\) ft

Width of the rectangle = 3\(\frac{1}{2}\) ft = \(\frac{7}{2}\) ft

Length of the Rectangle = L

Perimeter of a Rectangle = 2 ( length + width )

=> \(\frac{33}{2}\) = 2 ( L + \(\frac{7}{2}\) )

=> \(\frac{33}{2}\) = 2L + \(\frac{14}{2}\)

=> \(\frac{33}{2}\) – \(\frac{14}{2}\) = 2L

=> \(\frac{19}{2}\) = 2L

=> L = \(\frac{19}{4}\) = 4\(\frac{3}{4}\) ft

Therefore, Perimeter of a Rectangle = 4\(\frac{3}{4}\) ft .

**Practice**

Question 7.

2 ∗ \(\frac{2}{3}\) = ___

Answer:

2 ∗ \(\frac{2}{3}\) = 1 \(\frac{1}{3}\)

Explanation :

2 ∗ \(\frac{2}{3}\) = \(\frac{4}{3}\) =1 \(\frac{1}{3}\)

Question 8.

5 ∗ \(\frac{3}{4}\) = __

Answer:

5 ∗ \(\frac{3}{4}\) = 3\(\frac{3}{4}\)

Explanation :

5 ∗ \(\frac{3}{4}\) = \(\frac{15}{4}\) = 3\(\frac{3}{4}\)

Question 9.

9 ∗ \(\frac{4}{5}\) = __

Answer:

9 ∗ \(\frac{4}{5}\) = 7\(\frac{1}{5}\)

Explanation :

9 ∗ \(\frac{4}{5}\) = \(\frac{36}{5}\) = 7\(\frac{1}{5}\)

Question 10.

8 ∗ \(\frac{6}{12}\) = ___

Answer:

8 ∗ \(\frac{6}{12}\) = 4

Explanation :

8 ∗ \(\frac{6}{12}\) = 8 ∗ \(\frac{1}{2}\) = 4 ∗ \(\frac{1}{1}\) = 4

### Everyday Math Grade 4 Home Link 8.7 Answer Key

**Decimal Number Stories**

Solve each number story. Write your answer as a decimal. Show how you found your answer.

Question 1.

An Olympic men’s shot put weighs 7.26 kilograms. An Olympic women’s shot put weighs 4 kilograms. How much more does the men’s shot put weigh than the women’s shot put?

__ kilograms

Answer:

Weight of the Olympic men’s shot put = 7.26 kilograms

Weight of the Olympic women’s shot put = 4 kilograms

More Weight of men’s shot put weigh than the women’s shot put = 7.26 – 4.00 = 3.26 kilograms .

Therefore, More Weight of men’s shot put weigh than the women’s shot put = 3.26 kilograms .

Question 2.

The recipe for homemade glue calls for 0.5 liter of skim milk, 0.09 liter of vinegar, and 0.06 liter of water. When you combine the ingredients, how much liquid will you have?

__ liter

Answer:

Quantity of skim milk = 0.5 liter

Quantity of vinegar = 0.09 liter

Quantity of water = 0.06 liter

Quantity of liquid formed = 0.5 + 0.09 + 0.06 = 0.65 liters .

Therefore, Quantity of liquid formed = 0.65 liters .

Question 3.

Ben cut a piece of string 11.4 cm long. Then he cut 3.6 cm off of it. How long is the string now?

__ cm

Answer:

Length of the string = 11.4 cm long

Quantity of Length cut off = 3.6 cm

Quantity of length of the string left off = 11.4 – 3.6 = 7.8 cm .

Therefore, length of the string now = 7.8 cm .

**Try This**

Question 4.

What is the answer to Problem 3 in millimeters? __ millimeters

Answer:

length of string now = 7.8 cm

1cm = 10 m

multiply cm value into 10 to get the quantity in millimeters .

7.8 cm = 78 millimeters .

Therefore, length of string now = 7.8 cm = 78 millimeters .

**Practice**

Question 5.

3,579 ∗ 4 = __

Answer:

3,579 ∗ 4 = 14,316

Explanation :

Step I: Arrange the numbers vertically.

Step II: First multiply the digit at the ones place by 4.

9 × 4 = 36 = 3 ten + 6 ones

Write 6 in the ones column and carry 3 ten to tens place.

Step III: Multiply the digit at the tens place by 4.

7 × 4 = 28 tens

28 + 3 (carried) = 31 tens = 3 hundreds + 1 ten

Write 1 in ten place and carry 3 hundreds to hundreds place.

Step IV: Multiply the digit at hundreds place by 4

5 hundred × 4 = 20 hundreds

20 hundreds + 3 hundreds (carried) = 23 hundreds = 2 thousands + 3 hundreds

Write 3 in hundreds place carry 2 thousands to thousands place.

Step V: Multiply the digit at thousands place by 4.

3 thousands × 4 = 12 thousands

12 + 2 (carried) = 14 thousands = 1 ten thousands + 4 thousands

Write 4 in thousands place and carry 1 ten thousands to ten thousands place.

Question 6.

2,904 ∗ 6 = __

Answer:

2,904 ∗ 6 = 17,424

Question 7.

36 ∗ 56 = ___

Answer:

36 ∗ 56 = 2,116

Step I: Arrange the numbers vertically.

Step II: Multiply 36 by 6 ones

36 × 6 = 216

Step III: Multiply 36 by 50 tens

36 × 50 = 1800

Step IV: Add

216 + 1800 = 2116

Question 8.

47 ∗ 72 = __

Answer:

Step I: Arrange the numbers vertically.

Step II: Multiply 47 by 2 ones

47 × 2 = 94

Step III: Multiply 36 by 50 tens

47 × 70 = 3290

Step IV: Add

94 + 3290 = 3384

### Everyday Math Grade 4 Home Link 8.8 Answer Key

**Area and Perimeter**

Solve the problems below.

Question 1.

The Murphy family bought two rectangular dog beds for their pets. Fluffy’s bed was 3 feet by 1\(\frac{9}{12}\) feet. Pete’s bed was 4 feet by 2\(\frac{4}{12}\) feet.

a. How much more area does Pete’s bed have than Fluffy’s?

Answer: ____ square feet

Length of the Fluffy’s bed = 3 feet

Width of Fluffy’s bed = 1\(\frac{9}{12}\) = \(\frac{21}{12}\) feet .

Area of the fluffy’s bed = length × Width = 3 × \(\frac{21}{12}\) = \(\frac{63}{12}\) feet .

Length of the pete’s bed = 4 feet

Width of pete’s bed = 2\(\frac{4}{12}\) = \(\frac{28}{12}\) feet .

Area of the pete’s bed = length × Width = 4 × \(\frac{28}{12}\) = \(\frac{112}{12}\) feet .

Quantity of more Area of Pete’s bed have than Fluffy’s = \(\frac{112}{12}\) – \(\frac{63}{12}\) = \(\frac{49}{12}\) feet .

Therefore, Quantity of more Area of Pete’s bed have than Fluffy’s = \(\frac{49}{12}\) feet

b. What is the perimeter of Pete’s bed?

Answer: ___ feet

Length of the pete’s bed = 4 feet

Width of pete’s bed = 2\(\frac{4}{12}\) = \(\frac{28}{12}\) feet .

Perimeter of the pete’s bed = 2 ( l + w ) = 2 ( 4 + \(\frac{28}{12}\) ) = 2 ( \(\frac{48}{12}\) + \(\frac{28}{12}\) ) = 2 ( \(\frac{76}{12}\) ) = \(\frac{76}{6}\) = \(\frac{38}{3}\) feet .

Therefore, Perimeter of the pete’s bed = \(\frac{38}{3}\) feet

Question 2.

The Cho family bought two rectangular cat beds for their cats. George’s bed is 2 feet by 1\(\frac{2}{12}\) feet. Sammie’s bed is 2 feet by 1\(\frac{7}{12}\) feet.

a. What is the total area of these two beds? Answer: __ square feet

b. What is the perimeter of George’s bed? Answer: ___ feet

Answer :

a .

Length of George’s bed = 2 feet

Width of George’s bed = 1\(\frac{2}{12}\) = \(\frac{14}{12}\) Feet .

Area of George’s bed = length × Width = 2 × \(\frac{14}{12}\) = \(\frac{28}{12}\) square Feet .

Length of Sammie’s bed = 2 feet

Width of Sammie’s bed = 1\(\frac{7}{12}\) = \(\frac{19}{12}\) Feet .

Area of Sammie’s bed = length × Width = 2 × \(\frac{19}{12}\) = \(\frac{38}{12}\) square Feet .

Total Area of two beds = \(\frac{28}{12}\) + \(\frac{38}{12}\) = \(\frac{66}{12}\) = \(\frac{11}{2}\) square feet.

Therefore,Total Area of two beds = \(\frac{11}{2}\) square feet .

b.

Length of George’s bed = 2 feet

Width of George’s bed = 1\(\frac{2}{12}\) = \(\frac{14}{12}\) Feet .

Perimeter of George’s bed = 2 ( l + w ) = 2 ( 2 + \(\frac{14}{12}\) ) = 2 ( \(\frac{24}{12}\) + \(\frac{14}{12}\) ) = 2 ( \(\frac{38}{24}\) ) = \(\frac{38}{12}\) = \(\frac{19}{6}\) feet .

Therefore, Perimeter of George’s bed = \(\frac{19}{6}\) feet

Question 3.

Perimeter: 12\(\frac{2}{10}\) inches

Area: __ square inches

Answer:

Perimeter of the Rectangle = 12\(\frac{2}{10}\) = \(\frac{122}{10}\) inches

Length of the Rectangle = x inches .

Width of the Rectangle = 2\(\frac{1}{10}\) = \(\frac{21}{10}\) inches

Perimeter of the Rectangle = 2 ( l + b) = 2 ( x + \(\frac{21}{10}\) )

=> \(\frac{122}{10}\) = 2x + \(\frac{42}{10}\)

=>\(\frac{122}{10}\) – \(\frac{42}{10}\) = 2x

=> \(\frac{80}{10}\) = 2x

=> 8 = 2x

=> x = 4 inches.

Therefore, length of the Rectangle = 4 inches .

Question 4.

Area: 9\(\frac{3}{8}\) square feet

Width: ___ feet

Answer :

Area of the Rectangle = 9\(\frac{3}{8}\) = \(\frac{75}{8}\) square feet

Length of Rectangle = x feet.

Width of the Rectangle = 3 feet .

Area of the Rectangle = length × Width = x × 3

=> \(\frac{75}{8}\) = x × 3

=> \(\frac{75}{8 × 3 }\) = x

= > x = \(\frac{75}{24}\) = \(\frac{25}{8}\) feet

Therefore, Length of Rectangle = x = \(\frac{25}{8}\) feet .

**Practice**

Question 5.

\(\frac{5}{6}\) – \(\frac{1}{6}\) = ___

Answer:

\(\frac{5}{6}\) – \(\frac{1}{6}\) = \(\frac{4}{6}\) = \(\frac{2}{3}\)

Question 6.

\(\frac{8}{8}\) – \(\frac{3}{8}\) = __

Answer:

\(\frac{8}{8}\) – \(\frac{3}{8}\) = \(\frac{5}{8}\)

Question 7.

\(\frac{9}{10}\) – \(\frac{5}{10}\) = __

Answer:

\(\frac{9}{10}\) – \(\frac{5}{10}\) = \(\frac{4}{10}\) = \(\frac{2}{5}\)

Question 8.

\(\frac{11}{12}\) – \(\frac{5}{12}\) = __

Answer:

\(\frac{11}{12}\) – \(\frac{5}{12}\) = \(\frac{6}{12}\) = \(\frac{1}{2}\)

### Everyday Math Grade 4 Home Link 8.9 Answer Key

**Using Doghouse Dimensions**

Dan and Diane’s Doghouse Dynasty builds doghouses to order. They can change the length and width for doghouses, but they always build them to have the same height. Solve the number stories about doghouses built to certain widths and lengths based on the information given in the table. Use drawings or equations to show how you solved each problem.

Question 1.

Mrs. Swift ordered 3 medium-size doghouses. What will their combined width be? __ feet

Answer:

Length of the Medium size dog house = 4 feet

Width of the Medium size dog house = 1\(\frac{3}{4}\) = \(\frac{7}{4}\) .

Width of 3 medium size dog house = 3 × \(\frac{7}{4}\) = \(\frac{21}{4}\) = 5\(\frac{1}{4}\) .

Therefore, The Combined width = 5\(\frac{1}{4}\) feet .

Question 2.

Kisa’s Kennel has a space that is 18 feet wide in which they want to place doghouses side by side. If they order 5 small and 4 medium doghouses, will they all fit in the space? ___

Answer :

Width of the space = 18 feet .

Width of small dog house = 1\(\frac{1}{2}\) = \(\frac{3}{2}\) feet .

Width of 5 small dog houses = 5 × \(\frac{3}{2}\) = \(\frac{15}{2}\) feet = 7.5 feet

Width of the Medium size dog house = 1\(\frac{3}{4}\) = \(\frac{7}{4}\) .

Width of the 4 Medium size dog houses = 4 × \(\frac{7}{4}\) = 7 .

Total width of 5 small and 4 medium doghouses = 7.5 + 7 = 14.5 feet

14.5 < 18 feet so, Yes, they will fit in the 18 feet space.

**Practice
**

Question 3.

2 ∗ \(\frac{3}{6}\) = ___

Answer :

2 ∗ \(\frac{3}{6}\) = 1 ∗ \(\frac{3}{3}\) = 1

Question 4.

5 ∗ \(\frac{7}{10}\) = __

Answer:

5 ∗ \(\frac{7}{10}\) = 1 ∗ \(\frac{7}{2}\) = \(\frac{7}{2}\) = 3\(\frac{1}{2}\)

Question 5.

9 ∗ \(\frac{6}{100}\) = ___

Answer:

9 ∗ \(\frac{6}{100}\) = \(\frac{54}{100}\)

Question 6.

7 ∗ \(\frac{8}{12}\) = __

Answer:

7 ∗ \(\frac{8}{12}\) = 7 ∗ \(\frac{2}{3}\) = \(\frac{14}{3}\)

### Everyday Math Grade 4 Home Link 8.10 Answer Key

**Liquid Measurement and Fractions**

Complete the “What’s My Rule?” tables and state the rules.

Question 1.

Rule: __

Answer:

Rule: Out(pints) is 8 times in(gallons) .

Explanation :

=> in = 3\(\frac{1}{2}\) = \(\frac{7}{2}\)

out = 8 × \(\frac{7}{2}\) = 4 × \(\frac{7}{1}\) = 28 .

=> out = 48 .

in = \(\frac{out}{8}\) = \(\frac{48}{8}\) = 6 .

=> in = 7\(\frac{1}{4}\) = \(\frac{29}{4}\)

out = 8 × \(\frac{29}{4}\) = 2 × \(\frac{29}{1}\) = 58 .

=> out = 80 .

in = \(\frac{out}{8}\) = \(\frac{80}{8}\) = 10 .

Question 2.

Rule: __

Answer:

Rule: 4 times in (quarts) is the out ( cups) .

Explanation :

=> in = 4\(\frac{1}{2}\) = \(\frac{9}{2}\)

out = 4 × \(\frac{9}{2}\) = 2 × \(\frac{9}{1}\) = 18 .

=> out = 32

in = \(\frac{out}{4}\) = \(\frac{32}{4}\) = 8 .

=> in = 4\(\frac{1}{2}\) = \(\frac{9}{2}\)

out = 4 × \(\frac{9}{2}\) = 2 × \(\frac{9}{1}\) = 18 .

=> in = 9\(\frac{3}{4}\) = \(\frac{39}{4}\)

out = 4 × \(\frac{39}{4}\) = 1 × \(\frac{39}{1}\) = 39 .

=> in = 12\(\frac{1}{4}\) = \(\frac{49}{4}\)

out = 4 × \(\frac{49}{4}\) = 1 × \(\frac{49}{1}\) = 49 .

Use this recipe for a Creamsicle Smoothie to solve the problems below. \(\frac{3}{4}\) cup orange juice 4 fluid ounces cold water 1 cup vanilla ice cream

Combine all ingredients.

Question 3.

a. Will this recipe fit in a glass that holds 24 fluid ounces? _____

Explain your thinking.

Answer:

Yes, the Sample will fill .

Explanation :

The total amount of all combined ingredients is 18 fluid ounces, so, the smoothie will fit in the 24-fluid ounce glass .

b. About how many more cup(s) of smoothie could fit in the glass? ___ cup(s)

Answer:

glass Quantity = 24 ounces .

Ingredients Quantity = 18 ounces .

More Quantity can fill = 24 – 18 ounces = 6 ounces .

6 ounces = \(\frac{3}{4}\) cup

c. Frank wants to triple the recipe. How much of each ingredient will he need?

__ orange juice

__ cold water

___ vanilla ice cream

Answer:

\(\frac{3}{4}\) cup orange juice , 4 fluid ounces cold water 1 cup vanilla ice cream

triple the given quantities by multiplying into 3 , we get.

\(\frac{9}{4}\) cup orange juice , 12 fluid ounces cold water 3 cup vanilla ice cream

d. After tripling the recipe, how much smoothie will Frank have? __ fluid ounces

Answer:

The total amount of all combined ingredients is 18 fluid ounces

After tripling the Quantity is

The total amount of all combined ingredients after tripling= 3 ×18 fluid ounces = 54 ounces.

**Practice**

Question 4.

3,560 ÷ 3 → __

Answer:

3,560 ÷ 3 → 1186 + 2 Remainder .

Explanation :

Question 5.

9,295 ÷ 5 → __

Answer:

9,295 ÷ 5 → 1,859

Explanation :

Question 6.

Answer:

= 1174 + 2 Remainder .

Question 7.

Answer:

= 519

Explanation :

### Everyday Math Grade 4 Home Link 8.11 Answer Key

**Planning a Cookout**

The Whispering Lakes Neighborhood Association is having a hamburger cookout. Each family can choose whether to order the hamburgers or bring their own. Use the information in the table to solve the number stories. Use drawings, tables, or equations to show what you did.

Question 1.

a. What is the combined weight of 1 of each size hamburger? __ pounds

Answer:

Weight of Small size Hamburger = \(\frac{1}{8}\)

Weight of Medium size Hamburger = \(\frac{1}{4}\)

Weight of Large size Hamburger = \(\frac{1}{2}\)

Weight of Jumbo size Hamburger = \(\frac{3}{4}\)

Weight of king of the burgers of one = 1\(\frac{1}{2}\) = \(\frac{3}{2}\) .

Total Weight of all burgers = \(\frac{1}{8}\) + \(\frac{1}{4}\) + \(\frac{1}{2}\) + \(\frac{3}{4}\) + \(\frac{3}{2}\) = \(\frac{1}{8}\) + \(\frac{4}{4}\) + \(\frac{4}{2}\) = \(\frac{1}{8}\) + \(\frac{8}{8}\) + \(\frac{16}{8}\) = \(\frac{25}{8}\) = 3\(\frac{1}{8}\) .

Therefore, Total Weight of all hamburgers = 3\(\frac{1}{8}\) pounds .

b. How many ounces is that? __ ounces

Answer:

50 ounces .

Explanation :

\(\frac{25}{8}\) pounds = ? ounces .

1 pound = 16 ounces .

Multiply the pounds value into 16 to get in ounces .

=> \(\frac{25}{8}\) × 16 = \(\frac{25}{1}\) × 2 = 50 ounces .

c. Mrs. Ward found 80- ounce packages of hamburger on sale. If she needs to make 2 of each size hamburger, how many packages of meat will she need to buy?

__ packages

Answer:

2 packages .

Explanation :

1 of each size hamburgers weights = 50 ounces .

2 of each size hamburgers weights = 2 × 50 ounces = 100 ounces .

total weight of 1 packages = 80 ounces.

80< 100 ounces.

Therefore, we require 2 packages of meat .

Question 2.

The Finch family ordered 2 small hamburgers, 1 medium hamburger, and 1 jumbo hamburger. How many pounds of hamburger meat does the neighborhood association need to buy for this family?

__ pounds

Answer:

Weight of Small size Hamburger = \(\frac{1}{8}\)

Weight of Medium size Hamburger = \(\frac{1}{4}\)

Weight of Jumbo size Hamburger = \(\frac{3}{4}\)

Total pounds of hamburger ordered = \(\frac{1}{8}\) + \(\frac{1}{8}\) + \(\frac{1}{4}\) + \(\frac{3}{4}\) = \(\frac{2}{8}\) + \(\frac{4}{4}\) = \(\frac{1}{4}\) + \(\frac{4}{4}\) = \(\frac{5}{4}\) = 1 \(\frac{1}{4}\) pounds .

Therefore, pounds of hamburger meat does the neighborhood association need to buy for this family = 1 \(\frac{1}{4}\) pounds

**Practice**

Question 3.

5,107 ∗ 3 = __

Answer :

5,107 ∗ 3 = 15,321

Explanation :

Question 4.

4,794 ∗ 6 = __

Answer:

4,794 ∗ 6 = 28,764

Explanation :

Question 5.

74 ∗ 29 = __

Answer:

74 ∗ 29 = 2,146

Explanation :

Step I: Arrange the numbers vertically.

Step II: Multiply 74 by 9 ones

74 × 9 = 666

Step III: Multiply 74 by 2 tens

74 × 20 = 1480

Step IV: Add

666 + 1480 = 2146

Question 6.

93 ∗ 48 = __

Answer:

93 ∗ 48 =4,464

Explanation :

Step I: Arrange the numbers vertically.

Step II: Multiply 93 by 8 ones

93 × 8 = 744

Step III: Multiply 93 by 4 tens

93 × 40 = 3720

Step IV: Add

744 + 3720 = 4464

### Everyday Math Grade 4 Home Link 8.12 Answer Key

**Number-Tile Computations**

Cut out the 0–9 number tiles at the bottom of the page. Use them to help you solve the problems. Each of the 20 tiles can only be used once.

Question 1.

Use odd-numbered tiles 1, 3, 5, 7, and 9 to make the largest sum.

Answer:

add 973 and 51 gives largest sum 1024 .

Question 2.

Use even-numbered tiles 0, 2, 4, 6, and 8 to make the smallest difference.

Answer:

Subtract 204 – 86 we get difference as 118

Question 3.

Use number tiles 0, 4, 6, and 8 to make the largest product.

Answer:

Multiply 84 and 60 we get largest product = 5,040.

Question 4.

Use number tiles 1, 2, 5, and 7 to make the smallest whole-number quotient. The answer may have a remainder.

Answer:

125 divided by 7 = 17 + 6 Remainder .

Question 5.

Answer the following questions using only the unused tiles and any operation. Write number sentences to show your work.

a. What is the largest answer you can find?

Answer :

3 and 9 numbers are unused

To get largest number we need to multiply those number 9 × 3 = 27

b. What is the smallest answer you can find?

Answer :

3 and 9 numbers are unused

To get smallest number we need to divide those number 9 ÷ 3 = 3

**Practice**

Question 6.

4\(\frac{3}{5}\) + 3\(\frac{4}{5}\) = ____

Answer:

4\(\frac{3}{5}\) + 3\(\frac{4}{5}\) = \(\frac{23}{5}\) + \(\frac{19}{5}\) = \(\frac{42}{5}\) = 8\(\frac{2}{5}\)

Question 7.

1\(\frac{5}{8}\) + 3\(\frac{5}{8}\) = __

Answer:

1\(\frac{5}{8}\) + 3\(\frac{5}{8}\) = \(\frac{13}{8}\) + \(\frac{29}{8}\) = \(\frac{42}{8}\) = 5\(\frac{2}{8}\)

Question 8.

2\(\frac{9}{12}\) + 4\(\frac{5}{12}\) = __

Answer:

2\(\frac{9}{12}\) + 4\(\frac{5}{12}\) = \(\frac{33}{12}\) + \(\frac{53}{12}\) = \(\frac{86}{12}\) = \(\frac{43}{6}\) = 7\(\frac{1}{6}\)

Question 9.

5\(\frac{89}{100}\) + 5\(\frac{92}{100}\) = __

Answer:

5\(\frac{89}{100}\) + 5\(\frac{92}{100}\) = \(\frac{589}{100}\) + \(\frac{592}{100}\) = \(\frac{1181}{100}\) = 11\(\frac{81}{100}\)

### Everyday Math Grade 4 Home Link 8.13 Answer Key

**Many Names for Numbers**

Write five names in each box below. Use as many different kinds of numbers (such as whole numbers, fractions, decimals) and different operations (+, -, ∗, ÷) as you can.

Question 1.

Answer:

Question 2.

Answer:

Make up your own name-collection boxes.

Question 3.

Answer:

Question 4.

Answer:

**Practice**

Question 5.

5\(\frac{1}{4}\) – 1\(\frac{3}{4}\) = __

Answer:

5\(\frac{1}{4}\) – 1\(\frac{3}{4}\) = \(\frac{21}{4}\) – \(\frac{7}{4}\) = \(\frac{14}{4}\) = \(\frac{7}{2}\) = 3\(\frac{1}{2}\)

Question 6.

4\(\frac{3}{10}\) – 2\(\frac{7}{10}\) = __

Answer:

4\(\frac{3}{10}\) – 2\(\frac{7}{10}\) = \(\frac{43}{10}\) – \(\frac{27}{10}\) = \(\frac{16}{10}\) = \(\frac{8}{5}\) = 1\(\frac{3}{5}\)

Question 7.

6\(\frac{7}{12}\) – 3\(\frac{11}{12}\) = __

Answer:

6\(\frac{7}{12}\) – 3\(\frac{11}{12}\) = \(\frac{79}{12}\) – \(\frac{47}{12}\) = \(\frac{32}{12}\) = \(\frac{8}{3}\) = 2\(\frac{2}{3}\) .

Question 8.

8\(\frac{1}{6}\) – 4\(\frac{5}{6}\) = __

Answer:

8\(\frac{1}{6}\) – 4\(\frac{5}{6}\) = \(\frac{49}{6}\) – \(\frac{29}{6}\) = \(\frac{20}{6}\) = \(\frac{10}{3}\) = 3\(\frac{1}{3}\) .