Transcript Ch. 2.7 power point

Chapter 2
Section 7
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2.7
1
2
3
4
5
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.
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Objective 1
Use percent in solving problems
involving rates.
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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.
Interest Problems (annual)
Mixture Problems
base × rate (%) = percentage principle × rate (%) = interest
p
×
r
=
I
b ×
r
=
p
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.
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EXAMPLE 1
Using Percents to Find
Percentages
What is the amount of pure acid in 40 L of a 16% acid
solution?
Solution: 40 L  0.16  6.4 L
Find the annual interest if $5000 is invested at 4%.
Solution: $5000  0.04  $200
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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.
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Objective 2
Solve problems involving
mixtures.
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EXAMPLE 2
Solving a Mixture Problem
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?
Solution:
Let x = kg of 40% copper metal.
Then, 0.4 x  56  .5  x  80 .
Kg of
Percentage
Kg of
Metal
(as a decimal)
Copper
0.4x  56  0.5  x  80
10 0.4x  10 56  10 0.5 x  80
x
0.4
0.4x
4 x  560  560  5 x  400  560
80
0.7
80(0.7)=56
x+80
0.5
0.5(x + 80)
4 x  5 x  5 x  160  5 x
1x 160
160 kg of the 40% copper

1
1
metal is needed.
x  160
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Objective 3
Solve problems involving
simple interest.
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EXAMPLE 3
Solving a Mixture Problem
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%.
Solution:
Let x = amount invested at 5%.
Then, 0.08  2 x  3  0.05 x 1710.
100 0.082x  3000  100 0.05x  1001710
Amount Invested
Rate of
in Dollars
Interest
x
0.05
2x + 3000
0.08
16 x  24000  5 x  24000  171000  24000
Interest for
21x 147000

One Year
21
21
0.05x
x  7000
0.08(2x+3000)
$7000 was invested at 5% interest.
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Objective 4
Solve problems involving
denominations of money.
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Solve problems involving denominations
of money.
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.
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EXAMPLE 4
Solving a Money
Denomination Problem
A man has $2.55 in quarters and nickels. He has 9 more
nickels than quarters. How many nickels and how many
Number of
Denomination
Total
quarters does he have?
Coins
(as a decimal)
Value
0.25
0.25x
0.05
0.05(x + 9)
Solution:
x
Let x = amount of quarters.
x+9
Then, 0.25 x  0.05  x  9   2.55.
100 0.25x  100 0.05(x  9)  100 2.55
25 x  5 x  45  45  255  45
30 x 210

30 30
x7
7  9  16
The man has 7 quarters and 16 nickels.
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Objective 5
Solve problems involving
distance, rate, and time.
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Solve problems involving distance, rate, and time.
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,
distance = rate × 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.
d  rt
d
r
t
d
t
r
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EXAMPLE 5
Finding Distance, Rate, or Time
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:
d
r
t
100
r
9.77
r  10.24
Asafa Powell’s speed was 10.24 m per sec.
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EXAMPLE 6
Solving a Motion Problem
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?
Rate
× Time
= Distance
Solution:
Let t = time.
Faster plane
410
t
410t
Slower plane
530
t
530t
530t  410t  3290
940t 3290

940 940
t  3.5
It will take the planes 3.5 hr
to be 3290 mi apart.
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Solving a Motion Problem. (cont’d)
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.
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EXAMPLE 7
Solving a Motion Problem
Two buses left the downtown terminal, traveling in
opposite directions. One had an average speed of
10 mph more than the other. Twelve min  15 hr  later, they
were 12 mi apart. What were their speeds?
Solution: Let x = rate of
the slower bus.
1
1
x   x  10   12
5
5
1
1
x  x  2  2  12  2
5
5
52
5
x

10
 
 
25
 2
×
Rate
Time
= Distance
Faster bus
x + 10
1/5
(1/5)(x + 10)
Slower bus
x
1/5
(1/5)x
x  25
25  10  35
The slower bus was traveling at 25
mph and the faster bus at 35 mph.
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