Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
2.6 Ratio, Proportion, and Percent Write ratios. A ratio is a comparison of two quantities using a quotient. Ratio The ratio of the number a to the number b (b ≠ 0) is written atob, a : b, or a . b The last way of writing a ratio is most common in algebra. Slide 2.6-4 CLASSROOM EXAMPLE 1 Writing Word Phrases as Ratios Write a ratio for each word phrase. 3 days to 2 weeks Solution: 2weeks 7days 14days 3days 3days 14days weeks 3 14 12 hr to 4 days 4days 24hours 96hours 12hours 1 hours 96hours 8 4days Slide 2.6-5 CLASSROOM EXAMPLE 2 Finding Price per Unit A supermarket charges the following prices for pancake syrup. Which size is the best buy? What is the unit cost for that size? Solution: $3.89 $0.108 36 $2.79 $0.116 24 $1.89 $0.158 12 The 36 oz. size is the best buy. The unit price is $0.108 per oz. Slide 2.6-6 Solve proportions. A ratio is used to compare two numbers or amounts. A proportion says that two ratios are equal, so it is a special type of equation. For example, 3 15 4 20 is a proportion which says that the ratios 3 4 and 15 are equal. 20 In the proportion a c b, d 0 , b d a, b, c, and d are the terms of the proportion. The terms a and d are called the extremes, and the terms b and c are called the means. We read the proportions a c as “a is to b as c is to d.” b d Slide 2.6-8 Solve proportions. (cont’d) Beginning with this proportion and multiplying each side by the common denominator, bd, gives a c b d a c bd bd b d ad bc. We can also find the products ad and bc by multiplying diagonally. bc a c b d ad For this reason, ad and bc are called cross products. Slide 2.6-9 Solve proportions. (cont’d) Cross Products a c the cross products ad and bc are equal—that is, the product then , b d of the extremes equals the product of the means. If Also, ifad If bc,then a c where b, d 0 . b d a b then ad = cb, or ad = bc. This means that the two proportions , c d are equivalent, and the proportion a ccan also be written as b d a b c, d 0 . c d Sometimes one form is more convenient to work with than the other. Slide 2.6-10 CLASSROOM EXAMPLE 3 Deciding Whether Proportions Are True Decide whether the proportion is true or false. 21 62 15 45 13 91 17 119 15 62 930 Solution: False 21 45 945 17 91 1547 Solution: True 13 119 1547 Slide 2.6-11 CLASSROOM EXAMPLE 4 Solve the proportion Finding an Unknown in a Proportion x 35 . 6 42 Solution: x 42 6 35 42 x 210 42 42 x5 The solution set is {5}. The cross-product method cannot be used directly if there is more than one term on either side of the equals symbol. Slide 2.6-12 CLASSROOM EXAMPLE 5 Solve Solving an Equation by Using Cross Products x6 2 . 2 5 Solution: x 6 5 2 2 5 x 30 30 4 30 5 x 26 5 5 26 x 5 26 The solution set is . 5 When you set cross products equal to each other, you are really multiplying each ratio in the proportion by a common denominator. Slide 2.6-13 Objective 3 Solve applied problems by using proportions. Slide 2.6-14 CLASSROOM EXAMPLE 6 Applying Proportions Twelve gallons of diesel fuel costs $37.68. How much would 16.5 gal of the same fuel cost? Solution: Let x = the price of 16.5 gal of fuel. $37.68 x 12 gal 16.5 gal 12 x 621.72 12 12 x 51.81 16.5 gal of diesel fuel costs $51.81. Slide 2.6-15 Objective 4 Find percents and percentages. Slide 2.6-16 Write ratios. A percent is a ratio where the second number is always 100. Since the word percent means “per 100,” one percent means “one per one hundred.” 1% 0.01, or 1 1% 100 Slide 2.6-17 CLASSROOM EXAMPLE 7 Converting Between Decimals and Percents Convert. 310% to a decimal Solution: 3.1 8% to a decimal .08 0.685 to a percent 68.5% Slide 2.6-18 CLASSROOM EXAMPLE 8 Solving Percent Equations Solve each problem. What is 6% of 80? Solution: x .06 80 x 4.8 16% of what number is 12? .16 x 1200 1200 x 16 x 75 What percent of 75 is 90? 9000 x 75 9000 x 75 x 1.2 or 120% Slide 2.6-19 CLASSROOM EXAMPLE 9 Solving Applied Percent Problems Mark scored 34 points on a test, which was 85% of the possible points. How many possible points were on the test? Solution: Let x = the number of possible points on the test. 34 85 x 100 3400 85 x 85 85 x 40 There were 40 possible points on the test. Slide 2.6-20