Chapter 2 Section 7 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Further Applications of Linear Equations Use percent in solving problems involving rates. Solve problems involving mixtures. Solve problems involving simple interest. Solve problems involving denominations of money. Solve problems involving distance, rate, and time. 5 5

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 1 Objective 1 Slide Use percent in solving problems involving rates.

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Use percent in solving problems involving rates. Recall that percent means “per hundred.” Thus, percents are ratios in which the second number is always 100. For example, 50% represents the ratio 50 to 100 and 27% represents the ratio 27 to 100. PROBLEM-SOLVING HINT Percents are often used in problems involving mixing different concentrations of a substance or different interest rates. In each case, to get the amount of pure substance or the interest, we multiply. Slide In an equation, percent is always written as a decimal. For example, 35% is written 0.35, not 35, and 7 % is written 0.07 not 7. Interest Problems (annual) principle × rate (%) = interest p × r = I Mixture Problems base × rate (%) = percentage b × r = p

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley What is the amount of pure acid in 40 L of a 16% acid solution? Find the annual interest if $5000 is invested at 4%. EXAMPLE 1 Using Percents to Find Percentages Solution: Slide

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide PROBLEM-SOLVING HINT In the examples that follow, we use tables to organize the information in the problems. A table enables us to more easily set up an equation, which is usually the most difficult step.

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 2 Objective 2 Solve problems involving mixtures. Slide

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 2 Kg ofPercentageKg of Metal(as a decimal)Copper x0.40.4x (0.7)=56 x (x + 80) Solution: Let x = kg of 40% copper metal. Solving a Mixture Problem 160 kg of the 40% copper metal is needed. Slide A certain metal is 40% copper. How many kilograms of this metal must be mixed with 80 kg of a metal that is 70% copper to get a metal that is 50% copper?

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 3 Objective 3 Solve problems involving simple interest. Slide

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 3 Solution: Let x = amount invested at 5%. Solving a Mixture Problem $7000 was invested at 5% interest. Slide With income earned by selling a patent, an engineer invests some money at 5% and $3000 more than twice as much at 8%. The total annual income from the investment is $1710. Find the amount invested at 5%. Amount InvestedRate ofInterest for in DollarsInterestOne Year x x 2x (2x+3000)

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 4 Objective 4 Solve problems involving denominations of money. Slide

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve problems involving denominations of money. Slide PROBLEM-SOLVING HINT Problems that involve different denominations of money or items with different monetary values are similar to mixture and interest problems. To get the total value, we multiply Money Denominations Problems number × value of one item = total value For example, 30 dimes have a monetary value of 30($0.10) = $3. Fifteen $5 bills have a value of 15($5) = $75. A table is helpful for these problems, too.

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 4 Solving a Money Denomination Problem Slide A man has $2.55 in quarters and nickels. He has 9 more nickels than quarters. How many nickels and how many quarters does he have? Number ofDenominationTotal Coins(as a decimal)Value x x x (x + 9) Solution: Let x = amount of quarters. The man has 7 quarters and 16 nickels.

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 5 Objective 5 Solve problems involving distance, rate, and time. Slide

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley If your car travels at an average rate of 50 mph for 2 hr, then it travels 50 × 2 = 100 mi. This is an example of the basic relationship between distance, rate, and time, given by the formula d=rt. By solving, in turn, for r and t in the formula, we obtain two other equivalent forms of the formula. The three forms are given here. Solve problems involving distance, rate, and time. distance = rate × time Slide

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 5 Finding Distance, Rate, or Time Slide A new world record in the men’s 100-m dash was set in 2005 by Asafa Powell of Jamaica, who ran it in 9.77 sec. What was his speed in meters per second? (Source: World Almanac and Book of Facts 2006.) Solution: Asafa Powell’s speed was m per sec.

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 6 Solving a Motion Problem Slide Two airplanes leave Boston at 12:00 noon and fly in opposite directions. If one flies at 410 mph and the other 120 mph faster, how long will it take them to be 3290 mi apart? Solution:Let t = time. It will take the planes 3.5 hr to be 3290 mi apart. Rate ×Time =Distance Faster plane410t410t Slower plane530t530t

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley In motion problems like the one in Example 6, once you have filled in the first two pieces of information in each row of your table, you should automatically fill in the third piece of information using the appropriate form of the formula relating distance, rate and time. Set up the equation on the basis of your sketch and the information in your table. Solving a Motion Problem. (cont’d) Slide

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 7 Solving a Motion Problem Slide Two buses left the downtown terminal, traveling in opposite directions. One had an average speed of 10 mph more than the other. Twelve min later, they were 12 mi apart. What were their speeds? Solution: Let x = rate of the slower bus. The slower bus was traveling at 25 mph and the faster bus at 35 mph. Rate ×Time =Distance Faster busx + 101/5(1/5)(x + 10) Slower busx1/5(1/5)x