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CHAPTER
3
Problem Solving
3.1
3.2
3.3
3.4
3.5
Ratios and Proportions
Percents
Problems with Two or More Unknowns
Rates
Investment and Mixture
Copyright © 2011 Pearson Education, Inc.
3.1
1.
2.
3.
4.
Ratios and Proportions
Solve problems involving ratios.
Solve for a missing number in a proportion.
Solve proportion problems.
Use proportions to solve for missing lengths in figures
that are similar.
Copyright © 2011 Pearson Education, Inc.
Ratio: A comparison of two quantities
using a quotient.
Ratios are usually expressed in fraction
form or with a colon. When a ratio is
written in English, the word to separates
the numerator and denominator
quantities.
12
The ratio of 12 to 17 translates to .
17
Numerator Denominator
Unit ratio: A ratio with a denominator of 1.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 3
Example 1
A bin at a hardware store contains 120 washers and
85 bolts. Write the ratio of washers to bolts in
simplest form.
Solution
120
24

The ratio of washers to bolts =
85
17
Copyright © 2011 Pearson Education, Inc.
Slide 3- 4
Example 2
A bin at a hardware store contains 120 washers and
85 bolts. Write the ratio of washers to bolts in
simplest form. Express the ratio as a unit ratio.
Interpret the answer.
Solution
120 24 1.41


85
1
17
The unit ratio indicates that there are 1.41 washers
for every bolt.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 5
a c
Proportion: An equation in the form  , where
b d
b  0 and d  0.
Proportions and Their Cross Products
a c
If
 where b  0 and d  0, then ad = bc.
b d
Copyright © 2011 Pearson Education, Inc.
Slide 3- 6
Solving a Proportion
To solve a proportion using cross products:
1. Calculate the cross products.
2. Set the cross products equal to each other.
3. Use the multiplication principle of equality to isolate
the variable.
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Slide 3- 7
Example 3a
Solve. 3  x
5
Solution
8
3  8  24
5  x  5x
3 x

5 8
24  5x
24 5 x

5
5
Calculate the cross
products.
Set the cross
products equal to one
another.
Divide both sides by 5
to isolate x.
24
x
5
Copyright © 2011 Pearson Education, Inc.
Slide 3- 8
Example 3b
1
5
Solve. 5  6
15 w
Solution
5 15  w 
26
5
15  6  90
w
Calculate the cross
products.
5
6

15 w
Set the cross
products equal to one
26
another.
w  90
5
5 26
90 5
Multiply both sides by
 w 
5/26 to isolate w.
26 5
1 26
450
8
4
w
 17
 17
26
26
13
1
5
Copyright © 2011 Pearson Education, Inc.
Slide 3- 9
Example 4
Pedro drives his semi truck 2200 miles in 4 days. At
this rate, how many miles will he drive in 15 days?
Understand We are to find the amount of miles that
Pedro will drive in 15 days if he drives
2200 miles in 4 days.
Translating 2200 miles every 4 days to a
ratio, we have 2200 miles .
4 days
Plan At this rate indicates 2200 miles stays the same
4 days
for 15 days, so we can use a proportion.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 10
continued
Execute
2200 miles
x

4 days
15 days
15  2200  33, 000
4  x  4x
2200 x

4
15
33, 000  4 x
33, 000 4 x

4
4
Set the cross
products equal to one
another.
Divide both sides by 4
to isolate x.
8250  x
Copyright © 2011 Pearson Education, Inc.
Slide 3- 11
continued
Answer Pedro will drive 8250 miles in 15 days.
Check Verify the reasonableness of the answer. Pedro
drives about 550 miles per day so
550 15  8250 is correct.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 12
Solving Proportion Application Problems
To solve proportion problems:
1. Set up a proportion in which the numerators and
denominators of the ratios correspond in a logical
manner.
2. Solve using cross products.
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Slide 3- 13
Congruent angles: Angles that have the same
measure. The symbol for congruent is  .
Similar figures: Figures with congruent angles
and proportional side lengths.
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Slide 3- 14
Example 5
The two figures are similar. Find the missing lengths.
4m
c
4m
8m
3m
a
b
7m
Understand Since the figures are similar, the lengths
of the corresponding sides are proportional.
Ratio of larger
to
smaller
4 8 7 4
  
3 a b c
Copyright © 2011 Pearson Education, Inc.
Slide 3- 15
continued
Plan Write a proportion to find each missing side
length.
Execute It is not necessary to write a proportion for c.
The corresponding sides must be equal so we
can conclude that c = 3.
To find a:
To
find
b:
4 7
4 8

3 a
24  4a
24 4a

4
4
6a

3 b
21  4b
21 4b

4
4
5.25  b
Copyright © 2011 Pearson Education, Inc.
Slide 3- 16
continued
Answer The missing lengths are a = 6 m, b = 5.25 m,
and c = 3 m.
Check Verify the reasonableness of the answers by
quickly doing estimates of the missing side
lengths. These do check out.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 17
Solve.
x 16

20 15
a) 12
b) 18 ¾
c) 21.3
d) 21 ¾
3.1
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Slide 1- 18
Solve.
x 16

20 15
a) 12
b) 18 ¾
c) 21.3
d) 21 ¾
3.1
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Slide 1- 19
On a game preserve 112 deer were caught,
tagged, and released. Later, 82 were caught
in which 32 had tags. Estimate the number
of deer in the preserve.
a) 23
b) 44
c) 214
d) 287
3.1
Copyright © 2011 Pearson Education, Inc.
Slide 1- 20
On a game preserve 112 deer were caught,
tagged, and released. Later, 82 were caught
in which 32 had tags. Estimate the number
of deer in the preserve.
a) 23
b) 44
c) 214
d) 287
3.1
Copyright © 2011 Pearson Education, Inc.
Slide 1- 21
3.2
1.
2.
3.
4.
Percents
Write a percent as a fraction or decimal number.
Write a fraction or decimal number as a percent.
Translate and solve percent sentences.
Solve problems involving percents.
Copyright © 2011 Pearson Education, Inc.
Percent: A ratio representing some part out of
100.
The symbol for percent is %. For example, 20
percent, which means 20 out of 100, is written
20%.
20% = 20 out of 100
20

100
1
 or 0.2
5
Copyright © 2011 Pearson Education, Inc.
Slide 3- 23
Rewriting a Percent
To write a percent as a fraction or decimal:
1. Write the percent as a ratio with 100 in the
denominator.
2. Simplify to the desired form.
Note: When simplifying, remember that dividing a
decimal number by 100 moves the decimal point two
places to the left.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 24
Example 1
Write each percent as a decimal number and a fraction.
a. 14.4%
Solution Write 14.4 over 100, then simplify.
14.4% = 14.4 Write as a ratio with 100 in the denominator.
100
= 0.144
144

1000
18

125
Write the decimal form.
Write the fraction form.
Simplify to lowest terms.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 25
Example 1
1
b. 22 %
8
1
Solution Write 22 over 100, then simplify.
8
1
22
1
22 %  8
Write as a ratio with 100 in the
8
denominator.
100
1
 22  100
Rewrite the division.
8
177 1

Write an equivalent multiplication.
8 100
177

or 0.22125
Multiply.
800
Copyright © 2011 Pearson Education, Inc.
Slide 3- 26
Writing a Fraction or Decimal as a Percent
To write a fraction or decimal number as a percent:
1. Multiply by 100%.
2. Simplify.
Note: Multiplying a decimal number by 100 moves the
decimal point two places to the right.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 27
Example 2
Write as a percent.
a. 3
b. 0.348
8
Solution
a. 3 3
8

2
8
b. 0.348 = 0.348•100%
25
100
%
1
= 34.8%
75
 %
2
1
 37 % or 37.5%
2
Copyright © 2011 Pearson Education, Inc.
Slide 3- 28
A percent of a whole is a part of the whole.
A percent of
a whole
is a part.
Unknown part:
42% of
68
is what amount?
Unknown whole:
76% of
what number
is 63.84?
Unknown percent:
What percent of
72
is 63?
Copyright © 2011 Pearson Education, Inc.
Slide 3- 29
Translating Simple Percent Sentences
Method 1. Translate the sentence word for word.
1. Select a variable for the unknown.
2. Translate the word is to an equal sign.
3. If of is preceded by the percent, translate it to
multiplication. If of is preceded by a whole number,
translate it to division.
Method 2. Translate to a proportion by writing the
following form:
Part
Percent =
Whole
,
where the percent is expressed as a fraction with a
denominator of 100.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 30
Example 3
36% of 72 is what number?
Solution
Notice that the part is unknown.
36%
of
72
is
what number?
Percent
of
the whole
is
the part
Copyright © 2011 Pearson Education, Inc.
Slide 3- 31
continued
Method 1: Word-for-word translation
36% of 72 is what number?
0.36  72 = n
25.92 = n
Copyright © 2011 Pearson Education, Inc.
Slide 3- 32
continued
Method 2. Proportion:
Part
Percent =
Whole
36
n

100
72
100n  2592
100n
2592

100
100
 Part
 Whole
Equate the cross products.
Divide both sides by 100 to isolate n.
n  25.92
Copyright © 2011 Pearson Education, Inc.
Slide 3- 33
Example 4
42% of what number is 31.92?
Solution
Note the three pieces:
42%
of
what number
is
31.92?
Percent
of
the whole
is
the part
Copyright © 2011 Pearson Education, Inc.
Slide 3- 34
continued
Method 1. Word-for-word translation:
42% of what number is 31.92?
0.42 •
n
= 31.92
0.42n = 31.92
0.42n 31.92

0.42
0.42
n = 76
Copyright © 2011 Pearson Education, Inc.
Slide 3- 35
continued
Method 2. Proportion:
Part
Percent =
Whole
42
31.92  Part

100
n
 Whole
42n  3192
42n
3192

42
42
Equate the cross products.
Divide both sides by 42 to isolate n.
n  76
Copyright © 2011 Pearson Education, Inc.
Slide 3- 36
Example 5
What percent of 63 is 56?
Solution
Note the pieces:
What percent
of
63
is
56?
Percent
of
the whole
is
the part
Copyright © 2011 Pearson Education, Inc.
Slide 3- 37
continued
Method 1. Word-for-word translation:
What percent of 63 is 56?
p
• 63 = 56
63p = 56
63 p 56

63 63
p = 0.889
Answer 0.889  100% = 88.9%
Copyright © 2011 Pearson Education, Inc.
Slide 3- 38
continued
Method 2. Proportion:
Part
Percent =
Whole
p
56

100
63
63n  5600
63 p
5600

63
63
 Part
 Whole
Equate the cross products.
Divide both sides by 63 to isolate p.
p  88.889
Answer 88.9%
Copyright © 2011 Pearson Education, Inc.
Slide 3- 39
Solving Percent Application Problems
To solve problems involving percent,
1. Determine whether the percent, the whole, or the part
is unknown.
2. Write the problem as a simple percent sentence (if
needed).
3. Translate to an equation (word for word or
proportion).
4. Solve for the unknown.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 40
Example 6
Mia gets 97 questions correct on a test of 120
questions. What percent correct did she receive on
the test?
Understand The unknown is the percent of the
questions she got correct. There
were 120 questions, which is the
whole. She got 97 correct, which is
the part.
Plan Write a simple percent sentence, then solve.
We will let P represent the unknown
percent, and we will translate to a
proportion.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 41
continued
Execute What percent of 120 is 97?
P

100
120P 
120 P

120
P
 Part
97
 Whole
120
9700
9700
120
80.8
Answer Mia got about 80.8% on the test.
Check 0.808 • 120 = 96.96, therefore 80.8% is correct.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 42
Example 7
Jamal purchases a new surfboard that costs $570.
The sales tax is 7%, what is the total cost of the
surfboard?
Understand Determine the amount of sales tax
and add to the price of the
surfboard.
Plan Write a simple percent sentence, then solve.
We will let P represent the unknown
percent, and translate word for word.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 43
continued
Execute The sales tax is 7% of $570.
p  0.07  570
p  39.9
Sales tax is $39.90
Answer The total cost is $570 + $39.90 = $609.90.
Check The check is left to the student.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 44
Example 8
A pediatrician plots the growth of a child as growing
from 27.5 inches to 31.25 inches. What is the
percent of increase?
Understand We are to determine the percent of
increase in growth.
Plan To determine the percent of increase in
growth, we need the amount of the
increase, which is found by subtracting
the initial amount from the final amount.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 45
continued
Execute Amount of increase = 31.25 – 27.5 = 3.75
What percent of 27.5 is 3.75?
P
3.75  Part

100
27.5  Whole
27.5P  375
27.5 P
375

27.5
27.5
P  13.6
Answer The percent of increase in growth is 13.6%.
Check 0.136 • 27.5 = 3.74. Add the increase: 27.5 +
3.74 = 31.24. The answer is reasonable.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 46
What is 7/8 as a percent?
a) 78%
b) 87%
c) 87.5%
d) 718%
3.2
Copyright © 2011 Pearson Education, Inc.
Slide 1- 47
What is 7/8 as a percent?
a) 78%
b) 87%
c) 87.5%
d) 718%
3.2
Copyright © 2011 Pearson Education, Inc.
Slide 1- 48
2.4% of 130 is what number?
a) 31.2
b) 3.12
c) 1.83
d) 0.18
3.2
Copyright © 2011 Pearson Education, Inc.
Slide 1- 49
2.4% of 130 is what number?
a) 31.2
b) 3.12
c) 1.83
d) 0.18
3.2
Copyright © 2011 Pearson Education, Inc.
Slide 1- 50
What percent of 48 is 51.84?
a) 93%
b) 98%
c) 108%
d) 109%
3.2
Copyright © 2011 Pearson Education, Inc.
Slide 1- 51
What percent of 48 is 51.84?
a) 93%
b) 98%
c) 108%
d) 109%
3.2
Copyright © 2011 Pearson Education, Inc.
Slide 1- 52
3.3
Problems with Two or More
Unknowns
1. Solve problems involving two unknowns.
2. Use a table in solving problems with two unknowns.
Copyright © 2011 Pearson Education, Inc.
Solving Problems with Two or More Unknowns
To solve problems with two or more unknowns:
1. Determine which unknown will be represented by a
variable.
Tip: Let the unknown that is acted on be represented
by the variable.
2. Use one of the relationships to describe the other
unknown(s) in terms of the variable.
3. Use the other relationship to write an equation.
4. Solve the equation.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 54
Example 1
At a book sale, Franklin buys one paperback and
one hardback book. The hardback book is $3.00
more than the paperback. If his total purchase is
$7.80 before tax, how much does each book cost?
Understand There are two unknowns in the problem.
We must find the price of the paperback
book as well as the hardback book.
Relationship 1: The cost of the hardback book is $3
more than the paperback book.
Relationship 2: The total cost for both books is $7.80.
Plan Translate each relationship into an equation and
then solve.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 55
continued
Execute Use the first relationship to determine the
variable.
Relationship 1: The cost of the hardback book is $3
more than the paperback book.
Cost of hardback = $3 + cost of paperback
Cost of hardback = $3 + n
Now use the second relationship to write an equation.
Relationship 2: The total cost for both books is $7.80.
Cost of paperback + Cost of hardback = $7.80
n
+
3+n
= 7.80
2n + 3 = 7.80
2n = 4.80
n = 2.40
Copyright © 2011 Pearson Education, Inc.
Slide 3- 56
continued
Answer Since n represents the cost of the paperback
book, the paperback costs $2.40. The hardback
book costs $3.00 more than the paperback, so
$2.40 +$ 3.00 = $5.40.
Check Verify the total cost is $7.80, which it is
because $5.40 + $2.40 = $7.80.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 57
Example 3
At rectangular patio is to be built so that the length
is 12 feet more than the width. The perimeter of the
patio is to be 72 feet. Find the dimensions of the
patio.
Understand Draw a picture and list the relationships.
Relationship 1: The length is 12 feet more than the
width.
Relationship 2: The perimeter must be 72 feet.
Plan Translate to an equation using the key words and
then solve the equation.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 58
continued
Execute Relationship 1:
The length is 12 feet more than the width.
Length = Width + 12
Length = w + 12
Relationship 2: The perimeter is 72 feet. The formula
w + 12
for perimeter is P = 2l + 2w
w
w
w + 12
Perimeter = 72
2l + 2w = 72
2(w + 12) + 2w = 72
2w + 24 + 2w = 72
4w + 24 = 72
4w = 48
w = 12
Substitute w + 12 for l.
Distribute.
Combine like terms.
Subtract 24 from both sides.
Divide both sides by 4 to isolate w.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 59
continued
Answer The width is 12 feet. To determine the length
use Relationship 1. If the length = w + 12, then the
length = 12 + 12 = 24 feet.
Check Verify that both conditions in the problem are
satisfied.
The length is 12 feet more than the width.
The perimeter must be 72 feet.
This is true: 2(24) + 2(12) = 72 feet.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 60
Complementary angles: Two angles are
complementary if the sum of their measures is 90.
A
D
ABD and DBC are
complementary because
32 + 58 = 90
32
58
B
C
Supplementary angles: Two angles are supplementary
if the sum of their measures is 180.
ABD and DBC are
supplementary because
D
20 + 160 = 180
160
20


A
B
C
Copyright © 2011 Pearson Education, Inc.
Slide 3- 61
Example 4
Two angles are complementary. If the measure of
one angle is 32 degrees less than the other angle,
find the measure of the two angles.
Understand We must find the measure of the two
angles.
Relationship 1: One angle is 32 degrees less than the
other.
Relationship 2: Angle 1 + Angle 2 = 90.
Plan Translate the relationships to an equation and
then solve.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 62
continued
Execute Relationship 1:
One angle is 32 degrees less than the other.
a – 32
Relationship 2: The two angles are complementary.
Angle 1 + Angle 2 = 90
a + a – 32 = 90
2a – 32 = 90
2a = 122
a = 61
Combine like terms.
Add 32 to both sides.
Divide both side by 2 to isolate a.
Answer One angle is 61 degrees the other angle is
61 – 32 = 29 degrees.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 63
continued
Check Verify that both conditions in the problem are
satisfied.
One angle is 32 degrees less than the other. This is
true: 29 is 32 less than 61.
The two angles are complementary:
29 + 61 = 90
Copyright © 2011 Pearson Education, Inc.
Slide 3- 64
Another group of problems in which the
relationships are not obvious involve consecutive
integers. The word consecutive means that the
numbers are in sequence.
Patterns of consecutive integers…
With 1 as the
First Integer
With n as the
First Integer
First integer:
1
n
Next integer:
1+1=2
n+1
Next integer:
1+2=3
n+2
Copyright © 2011 Pearson Education, Inc.
Slide 3- 65
Consecutive Odd Integers
With 1 as the
First Odd
Integer
With n as the First
Odd Integer
First odd integer:
1
n
Next odd integer:
1+2=3
n+2
Next odd integer:
1+4=5
n+4
Consecutive Even Integers
With 2 as the
First Even
Integer
With n as the First
Even Integer
First even integer:
2
n
Next even integer:
2+2=4
n+2
Next even integer:
2+4=6
n+4
Copyright © 2011 Pearson Education, Inc.
Slide 3- 66
Example 5
The sum of three consecutive even integers is 384.
What are the integers?
Understand The key word sum means to “add”. If we
let n represent the smallest of the even
integers, then the pattern for three
consecutive even integers is:
Smallest unknown even integer: n
Next even integer: n + 2
Third even integer: n + 4
Plan Translate to an equation, then solve.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 67
continued
Execute Translate: The sum of three consecutive even
integers is 384.
Smallest even
integer
n
+
Next even
integer
n+2
+
Third even
integer
= 384
= 384
n+4
3n + 6 = 384
–6 –6
3n = 378
3n 378

3
3
n = 126
Copyright © 2011 Pearson Education, Inc.
Slide 3- 68
continued
Answer The smallest of the unknown integers,
represented by n, is 126.
Next even integer: n + 2 = 126 + 2 = 128
Third even integer: n + 4 = 126 + 4 = 130
Check The numbers 126, 128, and 130 are
three consecutive integers and
126 + 128 + 130 = 384.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 69
Example 6
A day spa sells two sizes of bottles of lotion. A
small bottle sells for $10.25 and a large bottle sells
for $16.75. One weekend the spa sells twice as many
large bottles as small bottles. If the total revenue that
weekend for these bottles of lotion was $393.75,
how many bottles of each size were sold?
Understand We know the total revenue is $393.75.
Revenue from the + Revenue from the
= $393.75
small bottles
Price per bottle
large bottles
•
Number of bottles
sold
Copyright © 2011 Pearson Education, Inc.
=
Revenue for that
size bottle
Slide 3- 70
continued
With so much information, it is helpful to use a table
to list it all.
Categories
Price per Bottle
Number of Bottles
Revenue
Small
$10.25
n
10.25n
Large
$16.75
2n
16.75(2n)
Givens
n represents the
number of small
bottles, then
translated “twice
as many large
bottles as small”
Price per • Number of = Revenue
bottle
bottles
from each
size of
bottle
Copyright © 2011 Pearson Education, Inc.
Slide 3- 71
continued
Plan Translate the information to an equation, and then
solve.
Execute Now we can use our initial relationship:
Revenue from the + Revenue from the = 393.75
small bottles
10.25n
large bottles
+
16.75  2n   393.75
10.25n  33.5n  393.75
43.75n  393.75
43.75n 393.75

43.75
43.75
n9
Copyright © 2011 Pearson Education, Inc.
Slide 3- 72
continued
Answer The number of small bottles sold, n, is 9. The
number of large bottles sold, 2n, is 18.
Check First, verify that twice the number of small
bottles sold, 9, is 18, which is true. Second,
verify that the total revenue from the sale of
lotion is $393.75.
10.25(9) + 16.75(18)= 92.25 + 301.50 = 393.75.
It checks.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 73
Example 7
A bakery sells muffins and bagels. The muffins sell
for $6 per dozen and the bagels sell for $8 per
dozen. The number of dozen of each kind was lost.
The bakery knows that there were 32 dozen bagels
and muffins sold and that the total revenue was
$222. How many dozen of each were sold?
Understand We know the total revenue is $222.
$6 times dozens
+ $8 times dozens
= $222
of muffins
of
bagels
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Slide 3- 74
continued
With so much information, it is helpful to use a table
to list it all.
Categories
Price per Dozen
Number
Revenue
Muffins
$6.00
32  n
6(32  n)
Bagels
$8.00
n
8n
Total number of dozen sold was 32.
Dozens of muffins + dozens of bagels = 32
Let n represent the number of dozen bagels.
muffins = 32 – n
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Slide 3- 75
continued
Plan Translate the information to an equation, and then
solve.
Execute Now we can use our initial relationship:
Revenue from the +
Revenue from
= 222
muffins
6(32  n)
bagels
+
8n
 222
192  6n  8n  222
192  2n  222
2n 30

2n 2
n  15
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Slide 3- 76
continued
Answer The number of dozens of bagels sold, n is 15.
The number of dozen of muffins sold is
17 dozen (found by 32 – 15)
Check First, verify that the sum is 32 dozen. and the
revenue is correct.
15 + 17 = 32
17(6) + 15(8) = 102 + 120 = 222
It checks.
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Slide 3- 77
One number is 4 more than the other.
The sum of the numbers is 58. What is
the larger of the two numbers?
a) 24
b) 27
c) 31
d) 34
3.3
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Slide 1- 78
One number is 4 more than the other.
The sum of the numbers is 58. What is
the larger of the two numbers?
a) 24
b) 27
c) 31
d) 34
3.3
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Slide 1- 79
Adrian is 3 times as old as his younger
brother George. In five years, Adrian
will only be twice as old as George.
How old is Adrian now?
a) 5 years old
b) 10 years old
c) 15 years old
d) 20 years old
3.3
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Slide 1- 80
Adrian is 3 times as old as his younger
brother George. In five years, Adrian
will only be twice as old as George.
How old is Adrian now?
a) 5 years old
b) 10 years old
c) 15 years old
d) 20 years old
3.3
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Slide 1- 81
3.4
Rates
1. Solve problems involving two objects traveling in
opposite directions.
2. Solve problems involving two objects traveling in the
same direction.
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Example 1
Two trains are traveling on parallel tracks toward each
other from a distance of 533 miles. If the freight train
is traveling at 27 miles per hour and the passenger train
is moving at 55 miles per hour, how long will it take
for them to pass each other?
Understand Draw a picture of the situation:
27 mph
55 mph
533 miles
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Slide 3- 83
continued
Distance traveled
by passenger train
+
Distance traveled
by freight train
= 533
Use a table to find expressions for the individual
distances.
Categories
Rate
Time
Distance
Passenger
55 mph
t
55t
Freight
27 mph
t
27t
Both trains start at the same
time and meet at the same
moment in time, so they
traveled the same amount of
time, t.
Multiplying the rate value
by the time value gives the
distance value.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 84
continued
Plan Use the information to write an equation, then
solve.
Execute
55t + 27t = 533
82t = 533
82
82
t = 6.5
Answer The trains will meet in 6.5 hours.
Check Verify that in 6.5 hours, the trains will travel a
combined distance of 533 miles.
55(6.5) + 27(6.5) ? 533
357.5 + 175.5 = 533
Copyright © 2011 Pearson Education, Inc.
Slide 3- 85
Two Objects Traveling in Opposite Directions
To solve for time when two objects are moving in
opposite directions:
1. Use a table with columns for categories, rate,
time, and distance. Use the fact that
rate  time = distance.
2. Write an equation that is the sum of the individual
distances equal to the total distance of separation.
3. Solve the equation.
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Slide 3- 86
Example 2
Michael and Renee are traveling north in separate cars
on the same highway. Michael is traveling at 55 mph
and Renee is traveling at 65 mph. Michael passes Exit
54 at 2:30 p.m. Renee passes the same exit at 2:45 p.m.
At what time will Renee catch up with Michael?
Understand Calculate the amount of time it will take
Renee to catch up to Michael. Add that time to 2:45.
15 minutes = ¼ of an hour or 0.25
Copyright © 2011 Pearson Education, Inc.
Slide 3- 87
continued
Categories
Rate
Michael
55 mph
Renee
65 mph
Time
t + 0.25
t
Distance
55(t + 0.25)
65t
The distance Michael and Renee will have traveled
when they catch up to each other is the same amount,
so their distances are equal.
Plan Set the expressions of their individual distances
equal, and solve for t.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 88
continued
Execute
65t = 55(t + 0.25)
65t = 55t + 13.75
–55t –55t
10t = 13.75
10
10
t = 1.375
Answer It will take Renee 1.375 hours, which is 1
hour and 22.5 minutes to catch up to Michael.
2:45 + 1 hour 22.5 minutes = 4:07 p.m.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 89
continued
Check Verify that Renee and Michael are equal
distances from the exit after traveling their
respective times.
Renee: d = 65(1.375)
= 89.375 miles
Michael: d = 55(1.625)
= 89.375 miles
Copyright © 2011 Pearson Education, Inc.
(1.375 + 0.25)
Slide 3- 90
Two Objects Traveling in the Same Direction
To solve problems involving two objects traveling in
the same direction in which the objective is to
determine the time for one object to catch up to the
other:
1. Use a table to organize the rates and times. Let t
represent the time for the object to catch up, and
add the time difference to t to represent the other
object’s time.
2. Set the expressions for the individual distances
equal.
3. Solve the equation.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 91
Two trains are traveling toward each other from a distance
of 208 miles. One train is traveling at 18 miles per hour
and the other at 46 miles per hour. How long will it take
for them to pass each other?
a) 3.25 hours
b) 4.5 hours
c) 6.75 hours
d) 11.6 hours
3.4
Copyright © 2011 Pearson Education, Inc.
Slide 1- 92
Two trains are traveling toward each other from a distance
of 208 miles. One train is traveling at 18 miles per hour
and the other at 46 miles per hour. How long will it take
for them to pass each other?
a) 3.25 hours
b) 4.5 hours
c) 6.75 hours
d) 11.6 hours
3.4
Copyright © 2011 Pearson Education, Inc.
Slide 1- 93
Two trains are delivering to the same site. One leaves at
8:00 a.m. and the other leaves at 8:15 a.m. If the first train
is traveling at 55 mph and the second at 60 mph, at what
time will the second catch up to the first?
a) 9:45 a.m.
b) 11:00 a.m.
c) 10:15 a.m.
d) 11:30 a.m.
3.4
Copyright © 2011 Pearson Education, Inc.
Slide 1- 94
Two trains are delivering to the same site. One leaves at
8:00 a.m. and the other leaves at 8:15 a.m. If the first train
is traveling at 55 mph and the second at 60 mph, at what
time will the second catch up to the first?
a) 9:45 a.m.
b) 11:00 a.m.
c) 10:15 a.m.
d) 11:30 a.m.
3.4
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Slide 1- 95
3.5
Investment and Mixture
1. Use a table to solve problems involving two
investments.
2. Use a table to solve problems involving mixtures.
Copyright © 2011 Pearson Education, Inc.
Example 1
Geraldine invests a total of $4000 in two different
accounts. The first account earns 6% while the second
account earns 4%. If the total interest earned is $210
after one year, what principle was invested in each
account?
Understand Since Geraldine invests a total of $4000,
we can say
Principal invested
Principal invested
+ in second account
in first account
Copyright © 2011 Pearson Education, Inc.
= 4000
Slide 3- 97
continued
We can isolate one of the unknown amounts by writing
a related subtraction statement.
Principal invested
in first account
Accounts
Principal invested
= 4000 
in second account
= 4000 – P
Rate
Principal
Interest
First
0.06
4000 – P
0.06(4000 – P)
Second
0.04
P
0.04P
Copyright © 2011 Pearson Education, Inc.
Slide 3- 98
continued
Plan Write an equation, then solve for p.
Execute
0.06(4000 – p) + 0.04p = 210
240 – 0.06p + 0.04p = 210
240 – 0.02p = 210
–240
–240
– 0.02p = – 30
– 0.02
–0.02
p = 1500
Answer The principal invested in the second account
is $1500.
In the first account is $4000  $1500 = $2500.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 99
continued
Check Verify that investing $2500 at 6% and $1500 at
4% result in a total of $210 in interest.
2500(0.06) + 1500(0.04) ? 210
50 + 60 = 210
It checks.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 100
Solving Problems Involving Investment
To solve problems involving two interest rates, in
which the total interest is given:
1. Use a table to organize the interest rates and
principals. Multiply the individual rates and
principals to get expressions for the individual
interests.
2. Write an equation that is the sum of the
expressions for interest set equal to the given total
interest.
3. Solve the equation.
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Slide 3- 101
Chemicals are often mixed to achieve a solution that
has a particular concentration. Concentration refers to
the portion of a solution that is pure.
Concentration  Whole solution =
Volume
volume
particular chemical
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Slide 3- 102
Example 2
Mitchell has a bottle containing 80 milliliters of 40%
HCl solution and a bottle of 15% HCl solution. He
wants a 25% HCl solution. How much of the 15%
solution must be added to the 40% solution so that a
25% concentration is created?
Understand Since more than one solution is involved,
a table is helpful in organizing the information.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 103
continued
Solution
40%
Concentration
of HCl
0.40
Volume of
Solution
80
15%
0.15
x
25%
0.25
80 + x
Percent written in
decimal form.
The 40% and 15%
solutions are
combined to get
the 25% solution,
so we add.
Copyright © 2011 Pearson Education, Inc.
Volume of
HCl
0.4(80)
0.15x
0.25(80 + x)
Multiply straight
across to generate
the expressions
for the volume of
HCl in each
solution.
Slide 3- 104
continued
Plan Write an equation, then solve for x.
Execute
0.4(80) + 0.15x = 0.25(80 + x)
32 + 0.15x = 20 + 0.25x
20
–20
12 + 0.15x = 0 + 0.25x
 0.15x
0.15x
12 = 0.1x
0.1 0.1
120 = x
Answer 120 milliliters of the 15% solution must be
added to 80 milliliters of 40% solution to create a
25% solution.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 105
continued
Check Verify that the volume of HCl in the two
original solutions combined is equal to the volume
of HCl in the combined solution.
0.40(80) + 0.15(120) ? 0.25(80 + 120)
32 + 18 = 0.25(200)
50 = 50
It checks.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 106
Solving Mixture Problems
To solve problems involving mixing solutions:
1. Use a table to organize the concentrations and
volumes. Multiply the individual concentrations
and solution volumes to get expressions for the
volume of the particular chemical.
2. Write an equation that the sum of the
volumes of the chemical in each solution set is
equal to the volume in the combined solution.
3. Solve the equation.
Copyright © 2011 Pearson Education, Inc.
Slide 3- 107
Lisa invests a total of $6000 in two different accounts.
The first account earns 8% while the second account earns
3%. If the total interest earned is $390 after one year, what
amount is invested at 8%?
a) $1800
b) $2100
c) $4200
d) $4800
3.5
Copyright © 2011 Pearson Education, Inc.
Slide 1- 108
Lisa invests a total of $6000 in two different accounts.
The first account earns 8% while the second account earns
3%. If the total interest earned is $390 after one year, what
amount is invested at 8%?
a) $1800
b) $2100
c) $4200
d) $4800
3.5
Copyright © 2011 Pearson Education, Inc.
Slide 1- 109
Martin has a bottle containing 120 milliliters of 30% HCl
solution and a bottle of 15% HCl solution. He wants a
25% HCl solution. How much of the 15% solution must
be added to the 30% solution so that a 25% concentration
is created?
a) 30 milliliters
b) 45 milliliters
c) 60 milliliters
d) 75 milliliters
3.5
Copyright © 2011 Pearson Education, Inc.
Slide 1- 110
Martin has a bottle containing 120 milliliters of 30% HCl
solution and a bottle of 15% HCl solution. He wants a
25% HCl solution. How much of the 15% solution must
be added to the 30% solution so that a 25% concentration
is created?
a) 30 milliliters
b) 45 milliliters
c) 60 milliliters
d) 75 milliliters
3.5
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Slide 1- 111