Ratios and Proportion

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Ratios and Proportion. ALGEBRA 1 LESSON 4-1. (For help, go to Skills Handbook pages 724 and 727.). Write each fraction in simplest form. 1. 2. 3. Simplify each product. 4. 5. 6. . 49 84. 24 42. 135 180. 35 25. 40 14. 99 144. 96 88. 21 81. 108 56. . . . 4-1.

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Ratios and Proportion

ALGEBRA 1 LESSON 4-1

(For help, go to Skills Handbook pages 724 and 727.)

Write each fraction in simplest form.

1. 2. 3.

Simplify each product.

4. 5. 6.

49

84

24

42

135

180

35

25

40

14

99

144

96

88

21

81

108

56

4-1

Ratios and Proportion

ALGEBRA 1 LESSON 4-1

1.

2.

3.

4.

5.

6.

Solutions

49 7 • 7 7

84 7 • 12 12

=

=

246 • 4 4

42 6 • 7 7

=

=

13545 • 33

180 45 • 4 4

=

=

3540 5 • 7 5 • 8 5 • 7 • 5 • 8 8

25 14 5 • 5 7 • 2 5 • 5 • 7 • 2 2

=

=

=

= 4

99 96 9 • 11 8 • 12 9 • 11 • 8 • 12 9 3

144 88 12 • 12 8 • 11 12 • 12 • 8 • 11 12 4

=

=

=

=

21 108 3 • 7 3 • 3 • 3 • 4 3 • 7 • 4 4 1

81 56 3 • 3 • 3 • 3 7 • 8 3 • 7 • 8 8 2

4

=

=

=

=

4

4-1

cost \$1.56

ounces 48 oz

= \$.0325/oz

Ratios and Proportion

ALGEBRA 1 LESSON 4-1

Another brand of apple juice costs \$1.56 for 48 oz. Find the unit rate.

The unit rate is 3.25¢/oz.

4-1

40 mi

1 h

5280 ft

1 mi

1 h

60 min

1 min

60 s

Use appropriate conversion factors.

40 mi

1 h

5280 ft

1 mi

1 h

60 min

1 min

60 s

Divide the common units.

= 58.6 ft/s

Simplify.

Ratios and Proportion

ALGEBRA 1 LESSON 4-1

The fastest recorded speed for an eastern gray kangaroo is 40 mi per hour. What is the kangaroo’s speed in feet per second?

The kangaroo’s speed is about 58.7 ft/s.

4-1

y

3

3

4

• 12 =

• 12

4y = 9

Simplify.

4y

4

9

4

=

Divide each side by 4.

y = 2.25

Simplify.

Ratios and Proportion

ALGEBRA 1 LESSON 4-1

y

3

3

4

Solve = .

4-1

w

4.5

6

5

= –

w(5) = (4.5)(–6)

Write cross products.

5w = –27

Simplify.

5w

5

–27

5

=

Divide each side by 5.

w = –5.4

Simplify.

Ratios and Proportion

ALGEBRA 1 LESSON 4-1

w

4.5

6

5

Use cross products to solve the proportion = – .

4-1

Define: Let t = time needed to ride 295 km.

Relate: Tour de France

average speed

Write:

equals

=

295-km trip

average speed

kilometers

hours

3630

92.5

295

t

3630

92.5

295

t

=

3630t = 92.5(295) Write cross products.

92.5(295)

3630

t = Divide each side by 3630.

t 7.5 Simplify. Round to the nearest tenth.

Ratios and Proportion

ALGEBRA 1 LESSON 4-1

In 2000, Lance Armstrong completed the 3630-km Tour de France course in 92.5 hours. Traveling at his average speed, how long would it take Lance Armstrong to ride 295 km?

Traveling at his average speed, it would take Lance approximately 7.5 hours to cycle 295 km.

4-1

z + 3

4

z – 4

6

=

(z + 3)(6) = 4(z – 4)

Write cross products.

6z + 18 = 4z – 16

Use the Distributive Property.

2z + 18 = –16

Subtract 4z from each side.

2z = –34

Subtract 18 from each side.

2z

2

–34

2

Divide each side by 2.

=

z = –17

Simplify.

Ratios and Proportion

ALGEBRA 1 LESSON 4-1

z + 3

4

z – 4

6

Solve the proportion = .

4-1

Ratios and Proportion

ALGEBRA 1 LESSON 4-1

Solve.

1. Find the unit rate of a 12-oz bottle of orange juice that sells for \$1.29.

2. If you are driving 65 mi/h, how many feet per second are you driving?

Solve each proportion.

3. 4.

5. 6.

10.75¢/oz.

c

6

12

15

21

12

7

y

4.8

4

=

=

1

2

3 + x

7

4

8

2 + x

x – 4

25

35

–17

=

=

4-1

Proportions and Similar Figures

ALGEBRA 1 LESSON 4-2

(For help, go to Skills Handbook and Lesson 4-1.)

Simplify

1.2.3.

Solve each proportion.

4. 5. 6.

7. 8. 9.

36

42

81

108

26

52

x

12

7

30

y

12

8

45

w

15

12

27

=

=

=

n + 1

24

9

a

81

10

25

75

z

30

n

9

=

=

=

4-2

x

12

7

30

=

90

81

96

45

84

30

y =

x =

a =

180

27

750

75

4

5

z =

w =

x = 2

Proportions and Similar Figures

ALGEBRA 1 LESSON 4-2

1.5. 7. 9.

2.

3.

4.

6. 8.

Solutions

y

12

8

45

9

a

81

10

n

9

n + 1

24

36

42

6 • 66

6 • 7 7

=

=

=

=

=

45y = 12(8)

81a = 9(10)

24n = 9(n + 1)

81

108

27 • 33

27 • 4 4

=

=

45y = 96

81a = 90

24n = 9n + 9

26

52

26 • 11

26 • 2 2

15n = 9

=

=

9

15

2

15

1

9

n =

y = 2

a = 1

3

5

n =

30x = 12(7)

w

15

12

27

25

75

z

30

=

=

30x = 84

27w = 15(12)

75z = 25(30)

27w = 180

75z = 750

2

3

w = 6

z = 10

4-2

EF

BC

DE

AB

Relate: =

Write a proportion comparing the lengths of the corresponding sides.

Define: Let x = AB.

6

9

8

x

Write: =

Substitute 6 for EF, 9 for BC, 8 for DE, and x for AB.

6x = 9(8)

Write cross products.

6x

6

72

6

=

Divide each side by 6.

x = 12

Simplify.

Proportions and Similar Figures

ALGEBRA 1 LESSON 4-2

In the figure below, ABC ~ DEF. Find AB.

AB is 12 mm.

4-2

102

17

x

6

=

Write a proportion.

17x = 102 • 6

Write cross products.

17x = 612

Simplify.

17x

17

612

17

=

Divide each side by 17.

x = 36

Simplify.

Proportions and Similar Figures

ALGEBRA 1 LESSON 4-2

A flagpole casts a shadow 102 feet long. A 6 ft tall man casts a shadow 17 feet long. How tall is the flagpole?

The flagpole is 36 ft tall.

4-2

map

actual

map

actual

Write a proportion.

1

10

2.25

x

=

1 • x = 10 • 2.25

Write cross products.

x = 22.5

Simplify.

Proportions and Similar Figures

ALGEBRA 1 LESSON 4-2

The scale of a map is 1 inch : 10 miles. The map distance from Valkaria to Gifford is 2.25 inches. Approximately how far is the actual distance?

The actual distance from Valkaria to Gifford is approximately 22.5 mi.

4-2

Proportions and Similar Figures

ALGEBRA 1 LESSON 4-2

1. In the figure below, ABC ~ DEF. Find DF.

2. A boy who is 5.5 feet tall casts a shadow that is 8.25 feet long.

The tree next to him casts a shadow that is 18 feet long.

How tall is the tree?

3. The scale on a map is 1 in.: 20 mi. What is the actual distance

between two towns that are 3.5 inches apart on the map?

12 ft

70 mi

4-2

Proportions and Percent Equations

ALGEBRA 1 LESSON 4-3

(For help, go to skills handbook pages 727 and 728.)

Find each product.

1. 0.6 • 9 2. 3.8 • 6.8 3. 4.

Write each fraction as a decimal and as a percent.

5. 6. 7. 8.

9. 10. 11. 12.

23

60

20

46

17

135

5

34

7

10

23

100

2

5

13

20

35

40

7

16

4

25

170

200

4-3

Proportions and Percent Equations

ALGEBRA 1 LESSON 4-3

Solutions

1. 0.6 • 9 = 5.4

2. 3.8 • 6.8 = 25.84

3.

4.

5.

6.

7.

23

60

20

46

23 • 20

20 • 3 • 23 • 2

1

3 • 2

1

6

=

=

=

17

135

5

34

17 • 5

5 • 27 • 17 • 2

1

27 • 2

1

54

=

=

=

7

10

= 7 ÷ 10 = 0.7; 0.7(100%) = 70%

23

100

= 23 ÷ 10 = 0.23; 0.23(100%) = 23%

2

5

= 2 ÷ 5 = 0.4; 0.4(100%) = 40%

4-3

13

20

= 13 ÷ 20 =0.65; 0.65(100%) = 65%

8.

9.

10.

11.

12.

35

40

= 35 ÷ 40 = 0.875; 0.875(100%) = 87.5%

7

16

= 7 ÷ 16 = 0.4375; 0.4375(100%) = 43.75%

4

25

= 4 ÷ 25 = 0.16; 0.16(100%) = 16%

170

200

= 170 ÷ 200 = 0.85; 0.85(100%) = 85%

Proportions and Percent Equations

ALGEBRA 1 LESSON 4-3

Solutions (continued)

4-3

part

whole

n

100

27

90

percent =

90n = 2700

Find the cross products.

n = 30

Divide each side by 90.

Proportions and Percent Equations

ALGEBRA 1 LESSON 4-3

What percent of 90 is 27?

30% of 90 is 27.

4-3

part

whole

25

100

n

480

=

12,000 = 100n

Find the cross products.

120 = n

Divide each side by 100.

Proportions and Percent Equations

ALGEBRA 1 LESSON 4-3

Find 25% of 480.

25% of 480 is 120.

4-3

Relate: 70.8% of the total surface area is 361,736,000 km2.

Define: Let t the total surface area.

part

whole

70.8

100

361,736,000

t

Write: =

70.8t = 361,736,000,000

Find cross products.

t = 510,926,553.7

Divide each side by 70.8.

Proportions and Percent Equations

ALGEBRA 1 LESSON 4-3

Water covers about 361,736,000 km2, or about 70.8% of the earth’s surface area. Approximately what is the total surface area of the earth?

The total surface area of the earth is approximately 510,926,554 km2.

4-3

Relate: What percent of 140 is 84?

Define: Let p = the decimal form of the percent.

Write: p • 140 = 84

p = 0.6

Divide each side by 140.

p = 60%

Write the decimal as a percent.

Proportions and Percent Equations

ALGEBRA 1 LESSON 4-3

What percent of 140 is 84?

140p = 84

60% of 140 is 84.

4-3

Relate: What percent of 60 is 114?

Define: Let n = the decimal form of the percent.

Write: n • 60 = 114

n = 1.90

Divide each side by 60.

n = 190%

Write the decimal as a percent.

Proportions and Percent Equations

ALGEBRA 1 LESSON 4-3

What percent of 60 is 144?

60n = 114

190% of 60 is 114.

4-3

1

5

1

5

1

5

19%

= 20%. So is a good approximation of 19%.

323 325

325 and 5 are compatible numbers.

1

5

• 325 = 65

65 is approximately 19% of 323.

3

4

3

4

3

4

73%

= 75%. So is a good approximation of 73%.

125 124

124 and 4 are compatible numbers.

3

4

• 124 = 93

Proportions and Percent Equations

ALGEBRA 1 LESSON 4-3

a. Estimate the number that is 19% of 323.

b. What is 73% of 125? Use fractions to estimate the answer.

93 is approximately 73% of 125.

4-3

Relate: What is 18% of 8056?

Define: Let n = the unknown number.

Write: n = 0.18 • 8056

n = 1450.08

Simplify.

Proportions and Percent Equations

ALGEBRA 1 LESSON 4-3

A candidate for mayor sent out surveys to 8056 people in his city. After two weeks, about 18% of the surveys were returned. How many surveys were returned?

n = 0.18 • 8056

About 1450 surveys were returned.

4-3

Proportions and Percent Equations

ALGEBRA 1 LESSON 4-3

1. What is 35% of 160?

2. What percent of 450 is 36?

3. 32 is 80% of what number?

4. What is 0.03% of 260,000?

5. What percent of 50 is 75?

6. Estimate 62% of 83?

56

8%

40

78

150%

51

4-3