Presentation on theme: "2.7 Further Applications of Linear Equations. Objective 1 Use percent in solving problems involving rates. Slide 2.7-3."— Presentation transcript:
2.7 Further Applications of Linear Equations
Objective 1 Use percent in solving problems involving rates. Slide 2.7-3
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 Mixing different concentrations of a substance or different interest rates involves percents. To get the amount of pure substance or the interest, we multiply. In an equation, percent is always written as a decimal. Interest Problems (annual) principle × rate (%) = interest p × r = I Mixture Problems base × rate (%) = percentage b × r= p Slide 2.7-4 Use percent in solving problems involving rates.
How much pure acid is in 40 L of a 16% acid solution? Find the annual simple interest if $5000 is invested at 4%. Solution: Slide 2.7-5 Using Percents to Find Percentages CLASSROOM EXAMPLE 1
PROBLEM-SOLVING HINT Using a table helps organize the information in a problem and more easily set up an equation, which is usually the most difficult step. Slide 2.7-7 Use percent in solving problems involving rates. (cont’d) In a mixture problem, the concentration of the final mixture must be between the concentrations of the two solutions making up the mixture.
CLASSROOM EXAMPLE 2 Kg ofPercentageKg of Metal (as a decimal)Copper x0.40.4x 800.780(0.7)=56 x+800.50.5(x + 80) Solution: Let x = kg of 40% copper metal. 160 kg of the 40% copper metal is needed. 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? Slide 2.7-8 Solving a Mixture Problem
Solution: Let x = amount invested at 5%. $7000 was invested at 5% interest. 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 investments is $1710. Find the amount invested at 5%. CLASSROOM EXAMPLE 3 Amount InvestedRate ofInterest for in DollarsInterestOne Year x0.050.05x 2x + 30000.080.08(2x+3000) Slide 2.7-10 Solving a Simple Interest Problem
PROBLEM-SOLVING HINT To get the total value in problems that involve different denominations of money or items with different monetary values, we multiply. Money Denominations Problems number × value of one item = total value A table is also helpful for these problems. Slide 2.7-12 Solve problems involving denominations of money.
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? CLASSROOM EXAMPLE 4 Number ofDenominationTotal Coins(as a decimal)Value x0.250.25x x + 90.050.05(x + 9) Solution: Let x = amount of quarters. The man has 7 quarters and 16 nickels. Slide 2.7-13 Solving a Money Denominations Problem
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. distance = rate × time, Distance, Rate, and Time Relationship Slide 2.7-15 Solve problems involving distance, rate, and time.
A new world record in the men’s 100-m dash was set in 2008 by Usain Bolt of Jamaica, who ran it in 9.69 sec. What was his rate (to two decimal places)? (Source: World Almanac and Book of Facts.) Solution: Usain Bolt’s rate was 10.32 m per sec. Slide 2.7-16 Finding Distance, Rate, or Time CLASSROOM EXAMPLE 5
Two airplanes leave Boston at 12:00 noon and fly in opposite directions. If one flies at 410 mph and the other flies at 530 mph, 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. CLASSROOM EXAMPLE 6 Rate ×Time =Distance Faster plane410t410t Slower plane530t530t Slide 2.7-17 Solving a Motion Problem
PROBLEM-SOLVING HINT In motion problems, once you have filled in the first two pieces of information in each row of your table, we can automatically fill in the third piece of information, using the appropriate form of the distance formula. Then we set up the equation on the basis of our sketch and the information in the table. Slide 2.7-18 Solve problems involving distance, rate, and time. (cont’d)
Two buses left the downtown terminal, traveling in opposite directions. One had an average speed of 10 mph more than the other. Twelve minutes (1/5 hr)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. CLASSROOM EXAMPLE 7 Rate ×Time =Distance Faster busx + 101/5(1/5)(x + 10) Slower busx1/5(1/5)x Slide 2.7-19 Solving a Motion Problem