chapter 2 section 7 copyright © 2008 pearson education, inc. publishing as pearson addison-wesley

Chapter Chapter 22Section Section 77

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley


Further Applications of Linear Equations

11

44

33

22

2.72.72.72.7Use 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.

55


Objective 11

Slide 2.7 - 3

Use percent in solving problems involving rates.


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 HINTPROBLEM-SOLVING HINTPercents 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 2.7 - 4

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


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 1Using Percents to Find Percentages

Solution:

Solution:

Slide 2.7 - 5

40 0.16 6.4L L

$5000 0.04 $200

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2.7 - 6

PROBLEM-SOLVING HINTPROBLEM-SOLVING HINTIn 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.


Objective 22

Solve problems involving mixtures.

Slide 2.7 - 7


EXAMPLE 2

Kg of Percentage Kg of

Metal (as a decimal) Copper

x 0.4 0.4x

80 0.7 80(0.7)=56

x+80 0.5 0.5(x + 80)

Solution: Let x = kg of 40% copper metal.

Solving a Mixture Problem

160 kg of the 40% copper metal is needed.

Slide 2.7 - 8

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?

4 15 55 60x xx x 1 16

1 1

0x

160x

0.4 5610 10 10 0.5 80x x 5604 560 5 400 560x x

0.4 56 0.5 80x x

Then, 0.4 56 .5 80 .x x


Objective 33

Solve problems involving simple interest.

Slide 2.7 - 9


EXAMPLE 3

Solution: Let x = amount invested at 5%.

Solving a Mixture Problem

$7000 was invested at 5% interest.Slide 2.7 - 10

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 Invested Rate of Interest for

in Dollars Interest One Year

x 0.05 0.05x

2x + 3000 0.08 0.08(2x+3000)

21 1470

21 21

00x

7000x

100 1000.08 2 3000 0.05 0 171010x x 2400016 24000 5 1710 2400000x x

Then, 0.08 2 3 0.05 1710.x x


Objective 44

Solve problems involving denominations of money.

Slide 2.7 - 11


Solve problems involving denominations of money.

Slide 2.7 - 12

PROBLEM-SOLVING HINTPROBLEM-SOLVING HINTProblems 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.


EXAMPLE 4 Solving a Money Denomination Problem

Slide 2.7 - 13

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 of Denomination Total

Coins (as a decimal) Value

x 0.25 0.25x

x + 9 0.05 0.05(x + 9)

Solution:Let x = amount of quarters.

0.25 0.05( 9) 2.5100 100 0 510x x 25 5 4 2 5 45 545 5x x

3

30 30

0 210x

7x The man has 7 quarters and 16 nickels.

7 9 16

Then, 0.25 0.05 9 2.55.x x


Objective 55

Solve problems involving distance, rate, and time.

Slide 2.7 - 14


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 2.7 - 15

d rtd

rt

d

tr


EXAMPLE 5Finding Distance, Rate, or Time

Slide 2.7 - 16

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:

dr

t

Asafa Powell’s speed was 10.24 m per sec.

100

9.77r 10.24r


EXAMPLE 6 Solving a Motion Problem

Slide 2.7 - 17

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.

530 410 3290t t

9

940 940

40 3290t

3.5t

It will take the planes 3.5 hr to be 3290 mi apart.

Rate × Time = Distance

Faster plane 410 t 410t

Slower plane 530 t 530t


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 2.7 - 18


EXAMPLE 7 Solving a Motion Problem

Slide 2.7 - 19

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.

1 110 12

5 5x x

1 12 12

52 2

5x x

210

5

5

5

2 2x

The slower bus was traveling at 25 mph and the faster bus at 35 mph.

Rate × Time = Distance

Faster bus x + 10 1/5 (1/5)(x + 10)

Slower bus x 1/5 (1/5)x

25x 25 10 35

1

5hr

chapter 2 section 7 copyright © 2008 pearson education, inc. publishing as pearson addison-wesley

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