1.    Ratios and Proportion Write each fraction in simplest form.
ALGEBRA 1 LESSON 4-1
(For help, go to Skills Handbook pages 724 and 727.)
Write each fraction in simplest form.
Simplify each product. 49 84 24 42
135
180 35 25 40 14 99
144 96 88 21 81
108 56   
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2.       Ratios and Proportion 1. 2. 3. 4. 5. 6. Solutions = = = =
ALGEBRA 1 LESSON 4-1 1. 2. 3. 4. 5. 6.
Solutions
49 7 • 7 7
84 7 • 12 12 = =
24 6 • 4 4
42 6 • 7 7 = =
• 3 3
• 4 4 = =
• 7 5 • 8 5 • 7 • 5 • 8 8
• 5 7 • 2 5 • 5 • 7 • 2 2  =  = =
= 4
• • • 11 • 8 •
• 12 8 • • 12 • 8 •   = = = =
• • 3 • 3 • 4 3 • 7 •
• 3 • 3 • • 8 3 • 7 •  =  4 = = = 4
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3. Ratios and Proportion
ALGEBRA 1 LESSON 4-1
Another brand of apple juice costs $1.56 for 48 oz. Find the unit rate.
cost $1.56
ounces 48 oz
= $.0325/oz
The unit rate is 3.25¢/oz.
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4. Use appropriate conversion factors.
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?
40 mi
1 h
5280 ft
1 mi
60 min
1 min
60 s •
Use appropriate conversion factors.
40 mi
1 h
5280 ft
1 mi
60 min
1 min
60 s •
Divide the common units.
= 58.6 ft/s
Simplify.
The kangaroo’s speed is about 58.7 ft/s.
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5. Ratios and Proportion Solve = . • 12 =
ALGEBRA 1 LESSON 4-1
Solve = . y 3 3 4 y 3
• 12 = 4
Multiply each side by the least common multiple of 3 and 4, which is 12.
• 12
4y = 9
Simplify. 4y 4 = 9
Divide each side by 4.
y = 2.25
Simplify.
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6. Use cross products to solve the proportion = – .
Ratios and Proportion
ALGEBRA 1 LESSON 4-1 w
4.5 6 5
Use cross products to solve the proportion = – . w
4.5
= – 6 5
w(5) = (4.5)(–6)
Write cross products.
5w = –27
Simplify. 5w 5 =
–27
Divide each side by 5.
w = –5.4
Simplify.
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7. Define: Let t = time needed to ride 295 km.
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?
Define: Let t = time needed to ride 295 km.
Relate: Tour de France
average speed
Write:
equals =
295-km trip
3630
92.5
295 t
kilometers
hours
3630
92.5 =
295 t
3630t = 92.5(295) Write cross products.
t = Divide each side by 3630.
92.5(295)
3630
t Simplify. Round to the nearest tenth.
Traveling at his average speed, it would take Lance approximately 7.5 hours to cycle 295 km.
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8. Use the Distributive Property.
Ratios and Proportion
ALGEBRA 1 LESSON 4-1
z + 3 4
z – 4 6
Solve the proportion = .
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.
Divide each side by 2. = 2z 2
–34
z = –17
Simplify.
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9. Ratios and Proportion Solve.
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.
10.75¢/oz.
about 95.3 ft/s c 6 12 15 21 12 7 y =
4.8 = 4
3 + x 7 4 8 1 2
2 + x
x – 4 25 35 = =
–17
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10. Proportions and Similar Figures
ALGEBRA 1 LESSON 4-2
(For help, go to Skills Handbook and Lesson 4-1.)
Simplify
Solve each proportion. 36 42 81
108 26 52 x 12 7 30 y 12 8 45 w 15 12 27 = = = 9 a 81 10 25 75 z 30 n 9
n + 1 24 = = =
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11. Proportions and Similar Figures
ALGEBRA 1 LESSON 4-2
Solutions 2. 3. 4.
6. 8. 36 42
6 • 6 6
6 • 7 7 y 12 8 45 9 a 81 10 n 9
n + 1 24 = = = = =
45y = 12(8)
81a = 9(10)
24n = 9(n + 1) 81
108
27 • 3 3
27 • 4 4 = =
45y = 96
81a = 90
24n = 9n + 9 26 52
26 • 1 1
26 • 2 2
y = 96 45
a = 90 81 = =
15n = 9 9 15 2 15 1 9
n = x 12 7 30 =
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)
x = 84 30
27w = 180
75z = 750
w =
180 27
z =
750 75
x = 2 4 5 2 3
w = 6
z = 10
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12. Proportions and Similar Figures
ALGEBRA 1 LESSON 4-2
In the figure below, ABC ~ DEF. Find AB.
Write: =
Relate: = EF BC DE AB
Define: Let x = AB. 6 9 8 x
Write a proportion comparing the lengths of the corresponding sides.
Substitute 6 for EF, 9 for BC, 8 for DE, and x for AB.
6x = 9(8)
Write cross products. =
Divide each side by 6. 6x 6 72
x = 12
Simplify.
AB is 12 mm.
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13. 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? =
102 17 x 6
Write a proportion.
17x = 102 • 6
Write cross products.
17x = 612
Simplify.
Divide each side by 17. =
17x 17
612
x = 36
Simplify.
The flagpole is 36 ft tall.
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14. 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? =
map
actual 1 10
2.25 x
Write a proportion.
1 • x = 10 • 2.25
Write cross products.
x = 22.5
Simplify.
The actual distance from Valkaria to Gifford is approximately 22.5 mi.
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15. 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?
About 19.7 cm
12 ft
70 mi
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16. Proportions and Percent Equations
ALGEBRA 1 LESSON 4-3
(For help, go to skills handbook pages 727 and 728.)
Find each product.
• •
Write each fraction as a decimal and as a percent. 23 60 20 46 17
135 5 34 • • 7 10 23
100 2 5 13 20 35 40 7 16 4 25
170
200
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17. Proportions and Percent Equations
ALGEBRA 1 LESSON 4-3
Solutions
• 9 = 5.4
• 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%
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18. Proportions and Percent Equations
ALGEBRA 1 LESSON 4-3
Solutions (continued) 13 20
= 13 ÷ 20 =0.65; 0.65(100%) = 65% 35 40
= 35 ÷ 40 = 0.875; 0.875(100%) = 87.5% 7 16
= 7 ÷ 16 = ; (100%) = 43.75% 4 25
= 4 ÷ 25 = 0.16; 0.16(100%) = 16%
170
200
= 170 ÷ 200 = 0.85; 0.85(100%) = 85% 8. 9.
10.
11.
12.
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19. Proportions and Percent Equations
ALGEBRA 1 LESSON 4-3
What percent of 90 is 27?
percent = n
100 27 90
part
whole
90n = 2700
Find the cross products.
n = 30
Divide each side by 90.
30% of 90 is 27.
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20. Proportions and Percent Equations
ALGEBRA 1 LESSON 4-3
Find 25% of 480. = 25
100 n
480
part
whole
12,000 = 100n
Find the cross products.
120 = n
Divide each side by 100.
25% of 480 is 120.
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21. 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?
361,736,000 t
Relate: 70.8% of the total surface area is 361,736,000 km2.
Define: Let t the total surface area.
Write: = 
70.8
100
part
whole
70.8t = 361,736,000,000 
Find cross products.
t = 510,926,553.7
Divide each side by 70.8.
The total surface area of the earth is approximately 510,926,554 km2.
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22. Proportions and Percent Equations
ALGEBRA 1 LESSON 4-3
What percent of 140 is 84?
Relate: What percent of 140 is 84?
Define: Let p = the decimal form of the percent.
Write: p • 140 = 84
140p = 84
p = 0.6
Divide each side by 140.
p = 60%
Write the decimal as a percent.
60% of 140 is 84.
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23. Proportions and Percent Equations
ALGEBRA 1 LESSON 4-3
What percent of 60 is 144?
Relate: What percent of 60 is 114?
Define: Let n = the decimal form of the percent.
Write: n • 60 = 114
60n = 114
n = 1.90
Divide each side by 60.
n = 190%
Write the decimal as a percent.
190% of 60 is 114.
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24. Proportions and Percent Equations
ALGEBRA 1 LESSON 4-3
a. Estimate the number that is 19% of 323.
= 20%. So is a good approximation of 19%.
19% 1 5
325 and 5 are compatible numbers.
• 325 = 65 1 5
65 is approximately 19% of 323.
b. What is 73% of 125? Use fractions to estimate the answer.
= 75%. So is a good approximation of 73%.
73% 3 4
• 124 = 93 3 4
124 and 4 are compatible numbers.
93 is approximately 73% of 125.
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25. 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?
Relate: What is 18% of 8056?
Define: Let n = the unknown number.
Write: n = 0.18 • 8056
n = 0.18 • 8056
n =
Simplify.
About 1450 surveys were returned.
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26. 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
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