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Solution and Answer Guide For Discovering Mathematics A Quantitative Reasoning Approach 2nd Edition by Richard Aufmann All Chapters A+

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Solution and Answer Guide: Aufmann, Discovering Mathematics: A Quantitative Approach, 2e, 9780357760031;
Chapter 1: Introduction To Problem Solving



Solution and Answer Guide
AUFMANN, DISCOVERING MATHEMATICS: A QUANTITATIVE APPROACH, 2e, 9780357760031;
CHAPTER 1: INTRODUCTION TO PROBLEM SOLVING


TABLE OF CONTENTS
Chapter 1: Introduction to Problem Solving.............................................................................. 1
Think About It 1.1 ........................................... 1
Section 1.1 Exercise Solutions ...................................................................................................... 1
Think About It 1.2 ........................................... 4
Section 1.2 Exercise Solutions ...................................................................................................... 4
Think About It 1.3 ........................................... 6
Section 1.3 Exercise Solutions ...................................................................................................... 6
Think About It 1.4 .......................................... 10
Section 1.4 Exercise Solutions .................................................................................................... 10
Chapter 1 Review Exercises ....................................................................................................... 27



CHAPTER 1: INTRODUCTION TO PROBLEM SOLVING

THINK ABOUT IT 1.1
1. Inductive
2. Specific
3. One example is 5, which is an odd number.


SECTION 1.1 EXERCISE SOLUTIONS
1. 64. The numbers are the squares of consecutive integers. 8 2 = 64.
2. 35. Subtract 1 less than the integer subtracted from the previous integer.
13
3. . Add 2 to the numerator and denominator.
15
6
4. . Add 1 to the numerator and denominator.
7
5. –13. Use the pattern of adding 5, then subtracting 10 to obtain the next pair of numbers.
6. 21. Subtract 8 from 5, add 12 to –3, subtract 16 from 9, add 20 to –7, etc., increasing the amount by 4
each time while alternating between adding and subtracting.
7. Each image has a smaller square inside a larger square. The smaller square moves to a new position in
a counterclockwise direction. The next figure is shown.




© 2023 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly 1
accessible website, in whole or in part.

, Solution and Answer Guide: Aufmann, Discovering Mathematics: A Quantitative Approach, 2e, 9780357760031;
Chapter 1: Introduction To Problem Solving

8. Each image has a smaller square inside a larger square. The smaller square moves to a new position in
a counterclockwise direction. The next figure is shown.




9. Each image has a smaller square and a circle inside a larger square. The smaller square moves to a new
position in a corner in a counterclockwise direction. The circle moves to a new position in a
counterclockwise direction. The next figure is shown.




10. Each image has a smaller square and a circle inside a larger square. The smaller square moves to a new
position in a corner in a counterclockwise direction. The circle moves to a new position in a clockwise
direction. The next figure is shown.




11. Each figure is a circle with a polygon, alternating positions inside and outside. The first figure is a
triangle (3 sides) with a circle inside. The second is a circle with a square (4 sides) inside. The next
figure is shown.




12. Each figure is a polygon within a polygon, alternating blue and yellow
interiors. The first figure is a 7-sided figure inside an 8-sided figure. The second figure is a 6-sided
figure inside a 7-sided figure. The next figure is shown.




13. The amount is decreasing by $1000 per month. Thus, the amount in year 6 will be $5000.
14. a. Greater since the temperature is increasing.
b. No. Fall and winter would come and the temperature would decrease.
15. More than. The increase in average annual salary increases from $7763 to $8927 to $10,266. Thus the
increase from 20 years to 25 years will be more than $10,266, giving an average annual salary for a
teacher having 25 years of experience of greater than $78,705 + $10,266 = $88,971, which is greater
than $88,000.
16. Fewer. From the bar graph, it appears that as the price of cell phones increases, the number of cell
phones sold decreases. Assuming the same trend, if the price of the cell phone is $700, the company
will sell fewer than 333,000 cell phones.
17. For every 10 seconds, the distance is increasing by 300 feet. Therefore, after 70 seconds, the athlete
will run a distance of 2100 feet.
18. For every 2 hours, the distance is decreasing by 100 miles. Therefore, after 6 hours, the trip will have
200 miles remaining.
19. From the table, as the depth increases, the temperature decreases. Since the last entry of the table is 40
m and 11°C, then for 45 m, the temperature should be less than 11°C.
20. Each year the car depreciates by a smaller amount: $5336, $5191, $2930, and $2878. The value of the
car at year 5 is $11,743. After year 6, the value of the car will be less than $11,000.



© 2023 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly 2
accessible website, in whole or in part.

, Solution and Answer Guide: Aufmann, Discovering Mathematics: A Quantitative Approach, 2e, 9780357760031;
Chapter 1: Introduction To Problem Solving

21. From the table, the distance (in feet) is equal to the time (in seconds) squared. So at 7 seconds the
distance is 49 feet, and at 8 seconds the distance is 64 feet.
22. The pattern from 0 seconds to 7 seconds, repeats again starting at 8 seconds. So at 14 seconds, the
distance will be the same as at 6 seconds, which is 0 inches.
1 1 1
23. Answers will vary. For instance,   .
4 2 8
24. Answers will vary. For instance, 5 – (–3) = 8.
25. A diamond shape can have four equal sides and is not a square.
26. Answers will vary. For instance, 20.
27. Whales are mammals and do not have legs.
28. A penguin is a bird that does not fly.
29. a. Deductive
b. Inductive
30. a. Inductive
b. Deductive
31. Place 4 coins in the left balance pan, 4 coins in the right balance pan, and set 4 coins aside. There are three
possibilities. The right side goes down, the left side goes down, or the scale balances (the heavier coin is set
aside). Take the 4 coins that include the heavier coin and place 2 on the left balance pan and 2 on the right
balance pan.
From the 2 coins on the heavier pan, place one coin on each balance pan to determine the heavier coin.
32. Label the coins 1, 2, 3, … 8. Set coins 5, 6, 7, and 8 aside.
Weighing 1: Place coins 1 and 2 on one balance and coins 3 and 4 on the other.
If the scale is balanced, the heavier or lighter coin is 5, 6, 7, or 8. Coins 1, 2, 3, or 4 are of equal
weight.
If the scale is unbalanced, the heavier or lighter coin is 1, 2, 3, or 4. Coins 5, 6, 7, or 8 are of equal
weight.
Weighing 2: From the four coins that contain the heavier or lighter coin, choose two coins, placing
one coin on each balance, and set the other two coins aside. Suppose that coins 5, 6, 7, or 8 are not of
equal weight. Choose coins 5 and 6 and place one on each side of the balance.
If the scale is balanced, then coins 5 and 6 are of equal weight and the coin that is heavier or
lighter is either coin 7 or 8.
If the scale is unbalanced, then coins 7 and 8 are of equal weight and the coin that is heavier or
lighter is either coin 5 or 6.
Weighing 3: Suppose the coin that is heavier or lighter is either coin 7 or 8. Place coin 7 on one
balance and coin 1 (or any one of coins 2 to 6, since they are of equal weight), on the other balance.
If the scale is balanced, then coin 8 is the coin that is heavier or lighter.
If the scale is unbalanced, then coin 7 is the coin that is heavier or lighter.
33. Label each stack of coins as 1, 2, 3, … 10. From each stack select the same amount of coins as the
labels, so select 1 coin from stack 1, 2 coins from stack 2, and so on to 10 coins from stack 10. Weigh
the selected coins. Since the counterfeit coins each weigh 0.1 g more, the multiple of 0.1 g over the
weight expected will determine which stack contains the counterfeit coins.
34. Label each stack of coins as 0, 1, 2, 3, … 10. Note that there are 11 stacks. From each stack select the
same amount of coins as the labels, so select 0 coins from stack 0, 1 coin from stack 1, 2 coins from
stack 2, and so on to 10 coins from stack 10. Weigh the selected coins. Since the counterfeit coins each
weigh 0.1 g more, the multiple of 0.1 g over the weight expected will determine which stack contains
the counterfeit coins. Note that if the weight is not more than expected, the counterfeit coins are in
stack 0.
35.
Util Auto Tech Oil



© 2023 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly 3
accessible website, in whole or in part.

, Solution and Answer Guide: Aufmann, Discovering Mathematics: A Quantitative Approach, 2e, 9780357760031;
Chapter 1: Introduction To Problem Solving

A X1 X1  X3
T X1 X1 X3 
M  X2 X3 X3
J X2  X2 X2
Maria: the utility stock; Jose: the automotive stock; Anita: the technology stock; Tony: the oil stock
36.
Soup Entree Salad Dessert
C X2  X3 X2
S X2 X3  X2
O X1 X2 X2 
G  X2 X2 X2
Changs: entrée; Steinbergs: salad; Ontkeans: dessert; Gonzaleses: soup
37.
Coin Stamp Comic Baseball
A X3  X3 X4
C X2 X4 X1 
P  X3 X3 X2
S X2 X3  X3
Atlanta: stamps; Chicago: baseball cards; Philadelphia: coins; San Diego: comic books
38. a. Yes. Change the color of Iowa to yellow and Oklahoma to blue.
b. No. One possible explanation: Oklahoma, Arkansas, and Louisiana must each have a different
color than the color of Texas and they cannot all be the same color. Thus, the map cannot be
colored using only two colors.


THINK ABOUT IT 1.2
1. a. Estimate
b. Estimate
c. Exact
d. Estimate
e. Exact
f. Exact
2. 5.32 × 10–6 = 0.00000532
The answer is b, greater than 0 but less than 1.
3. True
1
4. False; 10 3  0.001 
1000
5. Yes


SECTION 1.2 EXERCISE SOLUTIONS
1. 1487 miles ≈ 1500 miles
1500 m iles
 500 m iles per day
3 days
2. 11 children ≈ 10 children; 3 slices ≈ 4 slices
4 slices
10 children   40 slices
1child


© 2023 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly 4
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