Transcript Chapter Two 2.3

Section 2.3
Introduction to
Problem Solving
Copyright © 2013 Pearson Education, Inc.
Steps for Solving a Problem
Page 111
Step 1: Read the problem carefully and be sure that you
understand it. (You may need to read the problem more
than once.) Assign a variable to what you are being
asked to find. If necessary, write other quantities in terms
of this variable.
Step 2: Write an equation that relates the quantities described in
the problem. You may need to sketch a diagram or refer
to known formulas.
Step 3: Solve the equation. Use the solution to determine the
solution(s) to the original problem. Include any necessary
units.
Step 4: Check your solution in the original problem. Does it seem
reasonable?
Example
Page 112
Translate the sentence into an equation using the
variable x. Then solve the resulting equation.
a. Six times a number plus 7 is equal to 25.
b. The sum of one-third of a number and 9 is 18.
c. Twenty is 8 less than twice a number.
Solution
1
b.
x  9  18
a. 6x + 7 = 25
c. 20 = 2x − 8
3
6x = 18
6 x 18

6
6
x3
1
x9
3
x  27
28 = 2x
28 2 x

2
2
14 = x
Algebraic Translations of English Phrases
Consecutive Integers
Page 111
English Phrase
Algebraic
Expression
Example
Two consecutive integers
x, x + 1
5, 6
Three consecutive integers
x, x + 1, x + 2
8, 9, 10
Two consecutive even integers
x, x + 2
6, 8
Two consecutive odd integers
x, x + 2
-5, -3
Three consecutive even integers
x, x + 2, x + 4
2, 4, 6
Three consecutive odd integers
x, x + 2, x + 4
3, 5, 7
Numbers Example
Page 113
The sum of three consecutive integers is 126. Find the
three numbers.
Solution
Step 1: Assign a variable to an unknown quantity.
n: smallest of the three integers
n + 1: next integer
n + 2: largest integer
Step 2: Write an equation that relates these unknown
quantities.
n + (n + 1) + (n + 2) = 126
Numbers Example (cont)
Page 113
Step 3: Solve the equation in Step 2.
n + (n + 1) + (n + 2) = 126
(n + n + n) + (1 + 2) = 126
3n + 3 = 126
3n = 123
n = 41
So the numbers are 41, 42, and 43.
Step 4: Check your answer. The sum of these integers is
41 + 42 + 43 = 126. The answer checks.
Numbers Example 2
Page 113
The sum of three odd integers is 129. Find the three
numbers.
Solution
Step 1: Assign a variable to an unknown quantity.
n: smallest of the three integers
n + 2: next integer
n + 4: largest integer
Step 2: Write an equation that relates these unknown
quantities.
n + (n + 2) + (n + 4) = 129
Numbers Example 2 (cont)
Page 113
Step 3: Solve the equation in Step 2.
n + (n + 2) + (n + 4) = 129
(n + n + n) + (2 + 4) = 129
3n + 6 = 129
3n = 123
n = 41
So the numbers are 41, 43, and 45.
Step 4: Check your answer. The sum of these integers is
41 + 43 + 45 = 129. The answer checks.
Other types
Four subtracted from six times a number is 68. Find the number?
6 x  4  68
6 x  72
x  12
Add 4
Divide 6
A taxi charges $2.00 to turn on the meter plus $0.25 for each
eighth of a mile. If you have $10.00, how many eighths of a
mile can you go? How many miles is that?
2  .25 x  10
Sub 2
.25 x  8
Divide .25
x  32
32 eighths
4 miles
Page 114
Note: To write x% as a decimal number, move the
decimal point in the number x two places to the left
and then remove the % symbol.
Slide 10
Percent Example
Page 114
Convert each percentage to fraction and decimal
notation.
a. 47%
b. 9.8% c. 0.9%
Solution
Fraction Notation
Decimal Notation
47
a.
47%  0.47
47% 
b.
c.
100
9.8
98
49  2
49
9.8% 



100 1000 500  2 500
9.8%  0.098
0.9
9
0.9% 

100 1000
0.9%  0.009
Percent Example
Page 114
Convert each real number to a percentage.
2
a. 0.761 b.
c. 6.3
5
Solution
a. Move the decimal point two places to the right and
then insert the % symbol to obtain 0.761 = 76.1%
2
2
b.  0.40, so  40%.
5
5
c. Move the decimal point two places to the right and then
insert the % symbol to obtain 6.3 = 630%. Note that
percentages can be greater than 100%.
Percent Example
Page 115
The price of an oil change for an automobile increased
from $15 to $24. Calculate the percent increase.
Solution
new value - old value
100
old value
24 - 15

 100  60%
15
A = p times B
A  pB
Base = 15
amount = 24-15=9
(24  15)  p 15
9  p 15
9
15 p

15
15
p
9
 .6  60%
15
Using the Percent Formula
Amount = Percent times Base or a=pb
a.
What is 9% of 50?
a  9%  50
b.
amount  4.5
9  .6  b
9 is 60% of what?
9  60%  b
c.
a  .09  50
18 is what percent of 50?
18  p  50
9 .6b

.6
.6
base  15
18  p  50
18 50 p

50
50
percent  .36  36%
Using the Percent Formula
Amount = Percent times Base or a=pb
d.
A television regularly sells for
$940. The sale price is $611.
Find the percent decrease in the
television’s price?
Base = 940
amount = 940-611=329
Decrease = 940-611=329
329  p  940
329 940 p

940
940
p
329
 .35  35 %
940
Percent Example
Page 115
Number 72 on page 122, Tuition Increase:
Tuition is currently $125 per credit. There are plans to raise tuition
by 8% for next year. What will the new tuition be per credit?
New
tuition.
Solution
A = p times B
Base = $125
percent = 108%
Base = $125
percent = 8%
A  pB
A  108% 125
A  pB
A  8% 125
A  1.08125
A  $135
Tuition
increase.
A  0.08125
A  $10
New t uitionis $125plus $10  $135
Percent Example
Page 115
Number 69 on page 122, Voter Turnout:
In the 1980 election for president there were 86.5 million voters,
whereas in 2008 there were 132.6 million voters. Fine the percent
change in the number of voters.
Solution
A = p times B
Base = 86.5
Amount = 132.6 – 86.5 = 46.1
A  pB
46.1  p  86.5
46.1  p  86.5
46.1
p
 .533  53.3%
86.5
Percent Example
Page 115
A car salesman sells a total of 85 cars in the first and
second quarter of the year. In the second quarter, he had
an increase of 240% over the previous quarter. How
many cars did the salesman sell in the first quarter?
Solution
Step 1: Assign a variable.
x: the amount sold in the first quarter.
Step 2: Write an equation.
x + 2.4x = 85 (Note: A=pB which is A=240%*B)
Percent Example (cont)
Page 115
Step 3: Solve the equation in Step 2.
x + 2.4x = 85
3.4x = 85
x = 25
In the first quarter the salesman sold 25 cars.
Step 4: Check your answer.
An increase of 240% of 25 is 2.4 × 25 = 60.
Thus the amount of cars sold in the second
quarter would be 25 + 60 = 85.
Percent Example
After a 40% price reduction, an exercise machine sold for $564. What
was the exercise machine’s price before this reduction?
x  .4 x  564
.6 x  564
x  940
or
a  p b
564  .6 x
Divide .6
A Formula for Motion
Motion Formula
d=r·t
d = distance
r = rate
t = time
Distance equals rate times time.
Motion Example
Page 117
A truck driver travels for 4 hours and 30 minutes at a
constant speed and travels 252 miles. Find the speed of
the truck in miles per hour.
Solution
Step 1: Let r represent the truck’s rate, or speed, in
miles.
Step 2: The rate is to be given in miles per hour, so
change the 4 hours and 30 minutes to 4.5 or 9/2 hours.
d = rt
distance = 252
time = 4 hours 30 minutes
9
252   r
2
Motion Example (cont)
Page 117
Step 3: Solve the equation.
9
252   r
2
2
2 9
 252    r
9
9 2
56  r
Step 4:
The speed of the truck
is 56 miles per hour.
d = rt
9
 56   252 miles
2
The answer checks.
Motion Example
Page 117
Number 90 on page 123, Finding Speeds:
Two cars pass on a straight highway while traveling in opposite
directions. One car is traveling 6 miles per hour faster than the
other car. After 1.5 hours the two cars are 171 miles apart.
Find the speed of each car.
Rate
Time
Distance
Car one
r
1.5
1.5r
Car two
r+6
1.5
1.5(r+6)
171
1.5r  1.5(r  6)  171
Motion Example, cont
Page 117
Number 90 on page 123, Finding Speeds:
Two cars pass on a straight highway while traveling in opposite directions. One
car is traveling 6 miles per hour faster than the other car. After 1.5 hours the
two cars are 171 miles apart. Find the speed of each car.
Solve the equation and answer the question.
1.5r + 1.5(r+6) = 171
1.5r + 1.5r + 9 = 171
Distributive property
3r + 9 = 171
Combine like terms.
3r = 162
Subtract 9
r = 54
Divide both sides by 3
The first car will be 54mph, second car will be 60mph.
Motion Problem
Motion Problem:
Two cars leave from the same place, traveling in opposite directions. The rate of
the faster car is 55 miles per hour. The rate of the slower car is 50 miles per hour.
In how many hours will the cars be 420 miles apart?
Rate
Time
Distance
Faster
55 mph
X
55x
Slower
50 mph
X
50x
171
55x  50x  420
105x  420
105x
420

105
105
x  4 hours
Mixture Example
Page 118
A chemist mixes 100 mL of a 28% solution of alcohol with
another sample of 40% alcohol solution to obtain a
sample containing 36% alcohol. How much of the 40%
alcohol was used?
Concentration
Solution
Pure alcohol
Solution
Amount
(milliliters)
Step 1: Assign a variable.
x: milliliters of 40%
x + 100: milliliters of 36%
.
28%=0.28
100
40%=0.40
x
x+100
36%=0.36
28
.4x
0.36x+36
Mixture Example (cont)
A chemist mixes 100 mL of a 28% solution of alcohol with another sample of 40% alcohol solution to
obtain a sample containing 36% alcohol. How much of the 40% alcohol was used?
Concentration
0.28
0.40
0.36
Solution Amount
(milliliters)
100
x
x + 100
Pure alcohol
28
0.4x
0.36x + 36
Step 2: Write an equation.
0.28(100) + 0.4x = 0.36(x + 100)
Column
three.
Mixture Example (cont)
Step 3: Solve the equation in Step 2.
0.28(100) + 0.4x = 0.36(x + 100)
28(100) + 40x = 36(x + 100)
2800 + 40x = 36x + 3600
2800 + 4x = 3600
4x = 800
x = 200
200 mL of 40% alcohol solution was added to the 100 mL
of the 28% solution.
Mixture Example (cont)
Step 4: Check your answer.
If 200 mL of 40% solution are added to the 100 mL of
28% solution, there will be 300 mL of solution.
200(0.4) + 100(0.28) = 80 + 28 = 108 of pure alcohol.
108
The concentration is 300  0.36, or 36%.
Solving a Mixture Problem
Mixture Problem: How many ounces of a 50% alcohol solution must
mixed with a 20% alcohol solution to make 240 ounces of 40% alcohol
solution?
.
Quantity Percent Solution
50 percent sol
50%
.5x
20 percent sol
x
240-x
20%
.2(240-x)
New sol
240
40%
96
.5x  .2240 x   96
Solving a Mixture Problem
Mixture Problem continued: How many ounces of a 50% alcohol solution must
mixed with a 20% alcohol solution to make 240 ounces of 40% alcohol solution?
.
Quantity
Percent
50 percent sol
x
50%
.5X
20 percent sol
240-x
20%
.2(240-x)
240
40%
96
New sol
.5x  .2240 x   96
.5 x  48  .2 x  96
.5 x  .2 x  48  96
.3 x  48  96
.3 x  48
x  160 ounces
Solution
Distributive property
Combine like terms.
Subtract 48
Divide both sides by .3
Answer: 160 ounces of 50% and 80 ounces of 20%
Problems Involving Simple Interest
Suppose you invested $25,000, part at 9% simple interest and the
remainder at 12%. If the total yearly interest from these investments was
$2550, find the amount invested at each rate.
Principal
One
Two
x
25000-x
total
$25000
Rate
Time
Interest
9%
1
.09x
12%
1
.12(25000-x)
x = the amount in the first account.
25000 – x = amount in the second account.
.09x +.12(25000 – x) = 2550
$2550
Problems Involving Simple Interest
(continued)
Suppose you invested $25,000, part at 9% simple interest and the remainder at 12%. If the
total yearly interest from these investments was $2550, find the amount invested at each
rate.
Principal
Rate
Time Interest
4.
One
X
9%
1
.09X
Two
25000-X
12%
1
.12(25000-X)
total
$25000
$2550
Solve the equation and answer the question
.09x + .12(25000 – x) = 2550
.09x + .12(25000)-.12x = 2550
distributive property
-.03x + 3000 = 2550
combine like terms .09x-.12x
-3000
-.03x = -450
-3000
subtract 3000
divide by -.03
x = 15000
Answer is: $15,000 at 9% and $10,000 at 12%
Problems Involving Simple Interest
Teaching Example 12 on page 119, College Loans:
A student borrows two amounts of money. The interest rate on the first
amount is 4% and the interest rate on the second amount is 6%. If he
borrowed $5000 more at 4% than at 6%, then the total interest for one
year is $950. How much does the student owe at each rate?
One
Two
Principal
Rate
x
x-5000
4%
1
.04x
6%
1
.06(x-5000)
Time
total
Interest
$950
x = the amount in the first loan.
x-5000 = amount in the second loan.
Calculate interest
Problems Involving Simple Interest
Teaching Example 12 on page 119, College
Loans (continued):
A student borrows two amounts of money. The interest
rate on the first amount is 4% and the interest rate on
the second amount is 6%. If he borrowed $5000 more
at 4% than at 6%, then the total interest for one year is
$950. How much does the student owe at each rate?
One
Two
P
R
T Interest
x
x-5000
4%
1
.04x
6%
1
.06(x-5000)
total
$950
.04x +.06(x-5000) = 950
.04x +.06(x-5000) = 950
.04x +.06x-300 = 950
.10x -300 = 950
.10x = 1250
x = 12500
Distribute.
Combine like terms.
Add 300 from both sides.
Divide both sides by .1.
Invest $12500 in the 4% account and $7500 in the 6% account.
Solving a Mixture Problem
Mixture Problem:
How many pounds of cashews selling for $8.96 per pound
must be mixed with 12 pounds of chocolates selling for $4.48 per pound to create
a mixture that sells for $7.28 per pound?
.
Quantity
Cashews
Chocolates
Mix
Dollars or
price
Total price
x
$8.96
8.96x
12
$4.48
4.48(12)
x+12
$7.28
7.28(x+12)
8.96x  4.4812  7.28x  12
Solving a Mixture Problem (cont)
Mixture Problem (cont): How many pounds of cashews selling for $8.96 per pound must be mixed with 12
pounds of chocolates selling for $4.48 per pound to create a mixture that sells for $7.28 per pound?
Quantity
Dollars or
price
Total price
Cashews
X
$8.96
8.96X
Chocolates
12
$4.48
4.48(12)
X+12
$7.28
7.28(x+12)
Mix
8.96x  4.4812  7.28x  12
8.96 x  53.76  7.28 x  87.36
1.68 x  33 .60
x  20
Answer: 20 pounds
DONE
Objectives
•
Steps for Solving a Problem
•
Percent Problems
•
Distance Problems
•
Other Types of Problems