How To Do Word Problems
Rahim Faradineh
East Los Angeles College
Monterey Park, CA
2012
Contents
Why Word Problems? And What Can We Do About It? . . . . . . . . . . . .
2
1 Translations
1.1 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Homework & Quizzes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
3
5
2 Basic Percent
2.1 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Homework & Quizzes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
8
10
3 Proportions
13
3.1 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
3.2 Homework & Quizzes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
4 Parts
19
4.1 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
4.2 Homework & Quizzes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
5 Geometric Perimeters
25
5.1 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
5.2 Homework & Quizzes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
6 Consecutive Integers
31
6.1 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
6.2 Homework & Quizzes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
7 Motion: Linear Equation
37
7.1 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
7.2 Homework & Quizzes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
8 Money
44
8.1 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
8.2 Homework & Quizzes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
1
Why Word Problems? And What Can We Do About It?
Problem solving is the cornerstone of mathematics and other subjects in all science related fields. Unless
students can solve problems, the facts, concepts, and procedures they know are of little use. Students can
learn mathematical procedures, but without real-world applications, these skills are rendered meaningless
and are forgotten readily. However, research has shown that students have difficulty solving word problems.
At least three reasons have been proposed for why students have little success solving word problems:
1. limited experience with word problems,
2. lack of motivation to solve word problems,
3. and irrelevance of word problems to students lives.
These factors should be addressed in an effort to improve student performance on word problems, a fundamental component of mathematics education. Personalizing word problems, and replacing selected information
with students personal information can address the latter two, motivation and relevance, which may in turn
lead to the first, greater experience with word problems.
2
Chapter
1
Translations
1.1
Solved Problems
Problem: 1.2. The difference between a number and 6 is −4.
Solution:. We first start by using the Let statement to represent the unknown.
Let x be the number,
Now we should identify the keywords difference and is.
So the phrase can be translated to
x − 6 = −4
Recommendation:. Look up keywords in this problem and how they are used.
Problem: 1.9. If 3 is subtracted from 4 times a number, the result is the sum of 5 times the number and
10.
Solution:. We first start by using the Let statement to represent the unknown.
Let x be the number,
Now we should identify the keywords subtracted from, times, result is, and sum of.
So the phrase can be translated to
4x − 3 = 5x + 10
Recommendation:. Look up keywords in this problem and how they are used.
3
Problem: 1.14. If 3 is added to twice a number, the result is the difference of 4 times the number and 5.
Solution:. We first start by using the Let statement to represent the unknown.
Let x be the number,
Now we should identify the keywords added to, twice, result is, difference and times.
So the phrase can be translated to
2x + 3 = 4x − 5
Recommendation:. Look up keywords in this problem and how they are used.
Problem: 1.18. The product of 6 and a number is added to 3 is equal to the number squared increased by
8.
Solution:. We first start by using the Let statement to represent the unknown.
Let x be the number,
Now we should identify the keywords product, added to, is equal to, squared and increased by.
So the phrase can be translated to
3 + 6x = x2 + 5
Recommendation:. Look up keywords in this problem and how they are used.
4
1.2
Homework & Quizzes
1. There are two lecture quizzes in this section, do both quizzes.
2. Solve all the problems in exercise set for this section, including the ones that were solved in this section.
3. You will be asked to attach one of the lecture quizzes in this section to your solutions for the problems
in the exercise set and turn them in as your homework.
5
How to Do Word Problems
Lecture Quiz 1.1
Name:
No Work ⇔ No Points
Use Pencil Only ⇔ Be Neat & Organized
1. (2 points) Translate: Some number is reduced by 5.
1.
2. (2 points) Translate: Twice some number is increased by −10.
2.
3. (2 points) Translate: 5 less than half of some number.
3.
4. (2 points) Translate: Square of some number is reduced by 5, the result is
equal to 25.
4.
5. (2 points) Translate −3x + 4 = −17 into words.
5.
Page 1 of 1
Lecture Quiz 1.1 Prepared by R. Faradineh
Total Points = 10
How to Do Word Problems
Lecture Quiz 1.2
Name:
No Work ⇔ No Points
Use Pencil Only ⇔ Be Neat & Organized
1. (2 points) Translate: 5 times some number is increased by −5.
1.
2. (2 points) Translate: half some number is increased by
1
.
4
2.
3. (2 points) Translate: 5 less than half of some number is equal to the number
increased by 5.
3.
4. (2 points) Translate: Square of some number is reduced by 5, the result is
equal to 25 more than twice the number.
4.
5. (2 points) Translate 5x − 3 = 17 into words.
5.
Page 1 of 1
Lecture Quiz 1.2 Prepared by R. Faradineh
Total Points = 10
Chapter
2
Basic Percent
2.1
Solved Problems
Problem: 2.2. What is 20% of 150?
Solution:. We can use direct translation to solve this problem.
We let x to represent what, and identify the keywords is and of. We can also rewrite percentage in
fraction or decimal notation. Here is the translation and solution.
x
20
• 150
100
= 0.20 • 150
=
=
30
simplify
simplify
Now we can answer the question that 30 is 20% of 150.
Recommendation:. Look up methods on how to solve linear equation.
Problem: 2.6. 16 is what percent of 120?
Solution:. We can use direct translation to solve this problem.
p
We let 100 to represent what percent, and identify the keywords is and of. Here is the translation
and solution.
16
=
16
=
p
• 120
100
0.01p • 120
convert to decimals
16 = 1.20 • p
16
= p
1.20
13.3 = p
13.3
simplify
isolate
≈ p
8
Now we can answer the question that 16 is 13.3% of 120.
Recommendation:. Look up methods on solving linear equations and working with repeating decimals.
Problem: 2.16. 5 is 200% of what number?
Solution:. We can use direct translation to solve this problem.
We let x to represent what number, and identify the keywords is and of. We can also rewrite percentage in fraction or decimal notation. Here is the translation and solution.
5
=
5 =
5
=
2
2.5 =
200
•x
100
2•x
Simplify
Isolate
x
x
Now we can answer the question that 5 is 200% of 2.5.
Recommendation:. Look up how to simplify percent, and solve linear equation.
Problem: 2.18. 84 is what percent of 21?
Solution:. We can use direct translation to solve this problem.
p
We let 100 to represent what percent, and identify the keywords is and of. Here is the translation
and solution.
84
=
84
=
p
• 21
100
0.01p • 21 Convert to decimals
84 = 0.21 • p
84
= p
0.21
400 = p
Simplify
Isolate
Now we can answer the question that 84 is 400% of 21.
Recommendation:. Look up methods on solving linear equations and working with repeating decimals.
9
2.2
Homework & Quizzes
1. There are two lecture quizzes in this section, do both quizzes.
2. Solve all the problems in exercise set for this section, including the ones that were solved in this section.
3. You will be asked to attach one of the lecture quizzes in this section to your solutions for the problems
in the exercise set and turn them in as your homework.
10
How to Do Word Problems
Lecture Quiz 2.1
Name:
No Work ⇔ No Points
Use Pencil Only ⇔ Be Neat & Organized
1. (3 points) Use translation to solve: What is 2% of $40?
1.
2. (3 points) Use translation to solve: 54 is 24% of what number?
2.
3. (3 points) Use translation to solve: What percent of 150 is 39?
3.
4. Maria plans to buy a HDTV at regular price of $800, however it is on sale at
25% off, and the city sales tax rate is 6.5% on the selling price.
(a) (2 points) What is the discounted price?
(a)
(b) (2 points) What is the amount of sales tax?
(b)
(c) (2 points) How much in total Maria has to pay to purchase this HDTV?
(c)
Page 1 of 1
Lecture Quiz 2.1 Prepared by R. Faradineh
Total Points = 15
How to Do Word Problems
Lecture Quiz 2.2
Name:
No Work ⇔ No Points
Use Pencil Only ⇔ Be Neat & Organized
1. (3 points) Use translation to solve: What is 3.5% of $700?
1.
2. (3 points) Use translation to solve: 125 is 0.5% of what number?
2.
3. (3 points) Use translation to solve: What percent of 4200 is 5250?
3.
4. Jose plans to buy a dishwasher at regular price of $680, however it is on sale
at 20% off, and the city sales tax rate is 6.25% on the selling price.
(a) (2 points) What is the discounted price?
(a)
(b) (2 points) What is the amount of sales tax?
(b)
(c) (2 points) How much in total Maria has to pay to purchase this dishwasher?
(c)
Page 1 of 1
Lecture Quiz 2.2 Prepared by R. Faradineh
Total Points = 15
Chapter
3
Proportions
3.1
Solved Problems
Problem: 3.2. Katelyn drives her car 235 miles in 5 hours. At this rate, how far will she travel in 7 hours?
Solution:. We need to set up two ratios of how far to how long. She drove 235 miles in 5 hours and in ratio
we have
235 miles
(3.1)
5 hours
We let x to represent how far she can drive in 7 hours. Here is the second ratio.
x
7
miles
hours
(3.2)
Now we can use the keyword at this rate in the problem which allows us to equate these two ratios and
then solve for the unknown.
235
x
=
5
7
235 • 7 = 5 • x
235 • 7
5•x
=
5
5
329 = x
cross-multiply
isolate
So Katelyn can travel 329 miles in 7 hours.
Recommendation:. Look up how to simplify fractions and do cross-multiplication.
Problem: 3.9. In a 10 game season, Willis McGahee rushed for 1250 yards. On the average how many
yards did he rush in 5 games?
13
Solution:. We need to set up two ratios of how many yards to how many games. He rushed for 1250 yards
in 10 games and in ratio we have
1250 yards
(3.3)
10 games
We let x to represent how many yards that he can rush in 5 games. Here is the second ratio.
x
5
yards
games
(3.4)
Now we can use the keyword on the average in the problem which allows us to equate these two ratios
and then solve for the unknown.
1250
x
=
10
5
1250 • 5 = 10 • x
1250 • 5
10 • x
=
10
10
625 = x
cross-multiply
isolate
So he can rush 625 yards in 5 games.
Recommendation:. Look up methods on solving linear equations and working with repeating decimals.
Problem: 3.13. Seven cups of oatmeal contains 378 grams of carbohydrates. How many grams of carbohydrate would there be in 4 cups of oatmeal?
Solution:. We need to set up two ratios of how many oatmeals to how many games of carbohydrates. The
problem indicates that 7 oatmeals has 378 grams of carbohydrates and in ratio we have
7
378
oatmeals
grams of carbs.
(3.5)
We let x to represent how many grams of carbohydrates in 4 oatmeals. Here is the second ratio.
4
x
oatmeals
grams of carbs.
(3.6)
Now we can use the keyword on the average in the problem which allows us to equate these two ratios
and then solve for the unknown.
7
378
7•x
7•x
7
x
=
=
=
=
4
x
378 • 4
378 • 4
7
216
cross-multiply
isolate
So there are 216 grams of carbohydrates in 4 oatmeals.
Recommendation:. Look up methods on solving linear equations and working with repeating decimals.
14
Problem: 3.20. On a road map, 4 inches represents 50 miles. How many inches would represent 125 miles?
Solution:. We need to set up two ratios of how many inches to how many miles. The problem indicates
that 4 inches represents 50 miles and in ratio we have
4 inches
50 miles
(3.7)
We let x to represent how many inches for 125 miles. Here is the second ratio.
x inches
(3.8)
125 miles
Now we can use the keyword on the average in the problem which allows us to equate these two ratios
and then solve for the unknown.
x
4
=
50
125
4 • 125 = 50 • x
4 • 125
50 • x
=
50
50
10 = x
cross-multiply
isolate
So 10 inches would represent 125 miles.
Recommendation:. Look up methods on solving linear equations and working with repeating decimals.
15
3.2
Homework & Quizzes
1. There are two lecture quizzes in this section, do both quizzes.
2. Solve all the problems in exercise set for this section, including the ones that were solved in this section.
3. You will be asked to attach one of the lecture quizzes in this section to your solutions for the problems
in the exercise set and turn them in as your homework.
16
How to Do Word Problems
Lecture Quiz 3.1
Name:
No Work ⇔ No Points
Use Pencil Only ⇔ Be Neat & Organized
1. (3 points) Use cross-multiplication to solve:
1
x
=
12
5
1.
2. (4 points) Use cross-multiplication to solve:
2
x−2
=
x
3
2.
3. (4 points) It took John 3 hours to paint 5 walls of the same size. At this
rate, how long does it take John to paint 24 walls of the same size?
3.
4. (4 points) Kobe had 75 points in 4 games. At this rate, how many points
would he have in a season of 82 games.
4.
Page 1 of 1
Lecture Quiz 3.1 Prepared by R. Faradineh
Total Points = 15
How to Do Word Problems
Lecture Quiz 3.2
Name:
No Work ⇔ No Points
Use Pencil Only ⇔ Be Neat & Organized
1. (3 points) Use cross-multiplication to solve:
3
7.5
=
x
8
1.
2. (4 points) Use cross-multiplication to solve:
3
2x + 3
=
x+5
2
2.
3. (4 points) Jose made $275 to paint 4 rooms of the same size. At this rate,
how much would Jose make to paint 10 rooms of the same size?
3.
4. (4 points) Kobe had 20 assists in 3 games. At this rate, how many assists
would he have in a season of 82 games.
4.
Page 1 of 1
Lecture Quiz 3.2 Prepared by R. Faradineh
Total Points = 15
Chapter
4
Parts
4.1
Solved Problems
Problem: 4.3. A History textbook costs $7 less than a Sociology textbook. If the total cost of both
textbooks is $73, what is the cost of the history textbook?
Solution:. We let x to represent the cost of the sociology textbook, and since the cost of the history
textbook is $7 less than we can conclude that the history textbook would cost x − 7 dollars.
Now we can use the keyword total cost in the problem which implies to add the costs of both textbooks
and then solve for the unknown.
x+x−7
2x − 7
2x − 7 + 7
2x
2x
2
x
=
73
=
73
combine like terms
=
73 + 7
use addition property of equation
=
80
80
=
2
= 40
simplify
isolate
So sociology textbook costs $40 and the history textbook costs $33.
Recommendation:. Look up how to use properties of equation.
Problem: 4.5. A box of candy contains 28 pieces. If the number of pieces of milk chocolate is 4 less than
3 times the number of pieces of dark chocolate, how many pieces of each kind are there?
Solution:. We let x to represent the number of pieces of the dark chocolate, and since the number of
pieces of milk chocolate is 4 less than 3 times the number of pieces of dark chocolate we can conclude that
the number of pieces of milk chocolate would be 3x − 4.
19
Now we can use the keyword box contains in the problem which implies the total number of pieces of two
kinds of chocolate and then solve for the unknown.
x + 3x − 4
=
28
=
28
combine like terms
4x − 4 + 4
=
28 + 4
use addition property of equation
4x − 4
4x
4x
4
x
=
32
32
=
4
= 8
simplify
isolate
So the box contains $8 pieces of dark chocolate and 3(8) − 4 = 20 pieces of milk chocolate.
Recommendation:. Review on collecting like terms and solving linear equations.
Problem: 4.15. A 61-foot board will be cut into three pieces. The second piece will be 6 feet shorter than
twice the first piece, and the third piece will be 7 feet longer than the first piece. How long is the second
piece of the board be?
Solution:. We let x be the measure of the first piece, and since the second piece is 6 feet shorter than
twice the first piece we can conclude that the length of the second piece would be 2x − 6 feet.
The third piece is 7 feet longer than the first piece, so we can conclude that the length of the third piece
would be x + 7 feet.
we also know that the board itself before cutting into three pieces was 61 feet which suggests that we can
add add add all three pieces, equate that to the length of the original board, and then solve for the unknown.
x + 2x − 6 + x + 7
=
4x
4x
4
x
=
61
4x + 1 = 61
4x + 1 − 1 = 61 − 17
60
60
=
4
= 15
combine like terms
use subtraction property of equation
simplify
isolate
So the first piece is 15 feet, the second piece is 2(15) − 6 = 30 − 6 = 24 feet.
Recommendation:. Look up properties of equation, and review solving linear equations.
Problem: 4.20. In an episode of Powerpuff Girls, Buttercup defeated twice as many villains as Blossom.
Bubbles defeated 6 less villains than Blossom. If the Powerpuff Girls defeated a total of 74 villains, how
many villains did Buttercup defeat?
Solution:. We let x represent the number of villains that the Blossom defeated, and since the Buttercup
defeated twice as many as Blossom, then we can say that the buttercup defeated $2x villains. The Bubbles
defeated 6 less villains than Blossom, therefore we now can say that the Bubbles defeated x − 6 villains.
20
Now we can use the keyword total in the problem which implies to add the number of villains defeated by
each group, and then solve for the unknown.
x + 2x + x − 6
=
74
=
74
combine like terms
4x − 6 + 6
=
74 + 6
use addition property of equation
=
80
80
4
20
simplify
4x − 6
4x
4x
4
x
=
=
isolate
So the Blossom girls defeated 20 villains, and Buttercup girls defeated twice as many villains as Blossom
girls, so they defeated 40 villains.
Recommendation:. Look up how to use properties of equation.
21
4.2
Homework & Quizzes
1. There are two lecture quizzes in this section, do both quizzes.
2. Solve all the problems in exercise set for this section, including the ones that were solved in this section.
3. You will be asked to attach one of the lecture quizzes in this section to your solutions for the problems
in the exercise set and turn them in as your homework.
22
How to Do Word Problems
Lecture Quiz 4.1
Name:
No Work ⇔ No Points
Use Pencil Only ⇔ Be Neat & Organized
1. (3 points) John is 5 years older than Mary. The sum of their ages is 39. How
old is John?
1.
2. (4 points) The number of women in a class was 7 fewer than the number of
men. Find the number of women in the class if there are 55 students in the
class.
2.
3. (4 points) Sholeh has 40 coins in dimes and nickels. The number of nickels
is 5 less than twice the number of dimes. How many of each coin does she
have?
3.
4. (4 points) Two sides of a triangle are equal and the third side is 2cm less
than each one of the equal sides. Find the measure of all three sides of this
triangle if its perimeter is 19cm.
4.
Page 1 of 1
Lecture Quiz 4.1 Prepared by R. Faradineh
Total Points = 15
How to Do Word Problems
Lecture Quiz 4.2
Name:
No Work ⇔ No Points
Use Pencil Only ⇔ Be Neat & Organized
1. (3 points) John is 8 years younger than Mary. The sum of their ages is 56.
How old is John?
1.
2. (4 points) The number of women in a class was 4 more than the number of
men. Find the number of women in the class if there are 64 students in the
class.
2.
3. (4 points) Nasreen has 39 coins in dimes and nickels. The number of nickels
is 7 more than three times the number of dimes. How many of each coin
does she have?
3.
4. (4 points) Two sides of a triangle are equal and the third side is 2cm less
than twice the measure of each one of the equal sides. Find the measure of
all three sides of this triangle if its perimeter is 22cm.
4.
Page 1 of 1
Lecture Quiz 4.2 Prepared by R. Faradineh
Total Points = 15
Chapter
5
Geometric Perimeters
5.1
Solved Problems
Problem: 5.3. The length of a rectangle is 11 feet less than twice its width. Its perimeter is 26 feet. What
is the length and the width of the rectangle?
Solution:. We let x be the measure for the width of the rectangle, and since the length of this rectangle
is 11 less than twice the width we can conclude that the measure of the length to be 2x − 11 feet.
Now we can use the formula for the perimeter of the rectangle and then solve for the unknown.
P
=
26
2L + 2W
=
26
perimeter formula for the rectangle
2(2x − 11) + 2x
=
26
make substitution for length & width
=
26
distribute to remove (. . . )
6x − 22
=
26
collect like terms
=
26 + 22
addition property for equation
4x − 22 + 2x
6x − 22 + 22
6x
6x
6
x
=
48
48
=
6
= 8
simplify
isolate
So the width of the rectangle is 8 feet and the length of the rectangle is 2(8) − 11 = 5 feet.
Recommendation:. Review the definition of rectangle and look up formulas for perimeter and area of any
rectangle.
Problem: 5.6. In a triangle, the length of the second side is twice the length the first side. The length of
the third side is 8 less then 3 times the length of the first side. If the perimeter of this triangle is 34 inches,
how long is each side?
25
Solution:. We let x to represent the length of first side, and since the length of the second side is 2x
and the length of the third side is than 3x − 8.
Now we can use the information perimeter of this triangle is 34 in the problem which implies the sum
of all three sides of the triangle is equal to 34 and then solve for the unknown.
x + 2x + 3x − 8
6x − 8
6x − 8 + 8
6x
6x
6
x
=
34
=
34
combine like terms
=
34 + 8
use addition property of equation
=
42
42
=
6
= 7
simplify
isolate
So the first side is $7 inches, the second side is 2(7) = 14 inches, and the third side is 3(7) − 8 = 13 inches.
Recommendation:. Review on collecting like terms, solving linear equations and all properties of a triangle.
Problem: 5.11. The largest known map is a the map of Asia. It is rectangular in shape and its length is
25 times its width. The perimeter is 936 feet. What is the length of the map?
Solution:. We let x be the measure of the width of this map, and since the length of this map is 25 times
the the width we can conclude that the length would be 25x.
Now we can use the formula for the perimeter of the rectangle and then solve for the unknown.
P
=
936
2L + 2W
=
936
perimeter formula for the rectangle
2(25x) + 2x =
936
make substitution for length & width
25x + 2x =
936
simplify
52x =
52x
=
52
x =
936
936
52
18
collect like terms
isolate
So the the length of this rectangular map is 25(18) = 450 feet.
Recommendation:. Review the definition of rectangle and look up formulas for perimeter and area of any
rectangle.
Problem: 5.18. In the entranceway of the Sunrise Cinema, there is a large rectangular postern advertising
the upcoming movie ”Terminator III”. If the poster’s length is 3 feet less than twice its width, and its
perimeter is 24 feet, what is the length of the poster?
Solution:. We let x be the measure of the width of the poster, and since the length of this poster is 3
feet less than twice its the width we can conclude that the length of the poster would be 2x − 3 feet.
26
Now we can use the formula for the perimeter of the rectangle and then solve for the unknown.
P
=
24
2L + 2W
=
24
perimeter formula for the rectangle
2(2x − 3) + 2x =
24
make substitution for length & width
4x − 6 + 2x =
6x − 6
6x − 6 + 6
24
distribute to remove (. . . )
=
24
collect like terms
=
24 + 6
6x = 30
6x
30
=
6
6
x = 5
use addition property of equation
simplify
isolate
So the the length of this rectangular poster is 2(5) − 3 = 7 feet.
Recommendation:. Review the definition of rectangle and look up formulas for perimeter and area of any
rectangle.
27
5.2
Homework & Quizzes
1. There are two lecture quizzes in this section, do both quizzes.
2. Solve all the problems in exercise set for this section, including the ones that were solved in this section.
3. You will be asked to attach one of the lecture quizzes in this section to your solutions for the problems
in the exercise set and turn them in as your homework.
28
How to Do Word Problems
Lecture Quiz 5.1
Name:
No Work ⇔ No Points
Use Pencil Only ⇔ Be Neat & Organized
1. (5 points) The length of a rectangular playground is 1 foot longer than its
width. Find its length f the the perimeter of the playground is 78 feet.
1.
2. (5 points) The length of a rectangular garden is 5 meters more than three
times the measure of its width. Find its dimensions if the perimeter of the
garden is 114 meters.
2.
3. (5 points) The perimeter of an isosceles triangle is 44 inches. The base of
this triangle is 4 inches less than twice the measure of each of the equal sides.
Find all three sides of this triangle.
3.
Page 1 of 1
Lecture Quiz 5.1 Prepared by R. Faradineh
Total Points = 15
How to Do Word Problems
Lecture Quiz 5.2
Name:
No Work ⇔ No Points
Use Pencil Only ⇔ Be Neat & Organized
1. (5 points) The length of a rectangular pool is 3 foot longer than twice its
width. Find its length f the the perimeter of the pool is 54 feet.
1.
2. (5 points) The length of a rectangular garden is 12 meters shorter than five
times the measure of its width. Find its dimensions if the perimeter of the
garden is 96 meters.
2.
3. (5 points) The perimeter of an isosceles triangle is 28 inches. The base of
this triangle is 12 inches less than the sum of the other two sides. Find all
three sides of this triangle.
3.
Page 1 of 1
Lecture Quiz 5.2 Prepared by R. Faradineh
Total Points = 15
Chapter
6
Consecutive Integers
6.1
Solved Problems
Problem: 6.2. The sum of two consecutive integers is 193. Find the integers.
Solution:. Let x be the first integer and now we can set up a table for this problem:
Integers
First
Second
Consecutive Integers
x
x+1
Now we can use the keyword sum in the problem which implies to add these two consecutive integers and
then solve for the unknown.
First + Second
=
193
x+x+1
=
193
use translation and make the substitutions
2x + 1
=
193
combine like terms
2x + 1 − 1
=
193 − 1
use subtraction property of equation
2x
2x
2
x
=
192
192
=
2
= 96
simplify
isolate
So the two consecutive integers are 96 and 96 + 1 = 97.
Recommendation:. Look up how to use properties of equation and main keywords for translation.
Problem: 6.6. The sum of two consecutive even integers is 90. Find the integers.
Solution:. Let x be the first even integer and now we can set up a table for this problem:
31
Integers
First
Second
Consecutive Even Integers
x
x+2
Now we can use the keyword sum in the problem which implies to add these two consecutive integers and
then solve for the unknown.
First + Second
=
90
x+x+2
=
90
use translation and make the substitutions
2x + 2
=
90
combine like terms
2x + 2 − 2
=
90 − 2
use subtraction property of equation
2x = 88
2x
88
=
2
2
x = 44
simplify
isolate
So the two consecutive even integers are 44 and 44 + 2 = 46.
Recommendation:. Look up how to use properties of equation and main keywords for translation.
Problem: 6.10. The sum of three consecutive odd integers is 153. Find the integers.
Solution:. Let x be the first odd integer and now we can set up a table for this problem:
Integers
First
Second
Third
Consecutive Odd Integers
x
x+2
x+4
Now we can use the keyword sum in the problem which implies to add these two consecutive integers and
then solve for the unknown.
First + Second + Third
=
153
x+x+2+x+4
=
153
use translation and make the substitutions
3x + 6
=
153
combine like terms
3x + 6 − 6
=
153 − 6
use subtraction property of equation
3x = 147
147
3x
=
3
3
x = 49
simplify
isolate
So the three consecutive odd integers are 47, 47 + 2 = 49, and 47 + 4 = 51.
Recommendation:. Look up how to use properties of equation and main keywords for translation.
Problem: 6.15. Find the smaller of two consecutive odd integers if the larger is 20 less than three times
the smaller.
32
Solution:. Let x be the first odd integer and now we can set up a table for this problem:
Integers
First
Second
Consecutive Odd Integers
x
x+2
Now we can use the keyword is, less than, and times in the problem and then solve for the unknown. It
is worth noted that the second odd integer is the larger integer of the two integers.
Larger
=
x+2
=
x + 2 − 3x
=
−2x + 2
−2x + 2 − 2
−2x
−2x
−2
x
=
3 • Smaller − 20
3 • x − 20
use translation and make the substitutions
3x − 20 − 3x
use subtraction property of equation
−20
simplify
= −20 − 2
use subtraction property of equation
= −22
−22
=
−2
= 11
simplify
isolate
So the smaller of these two consecutive odd integer is 11.
Recommendation:. Look up how to use properties of equation and main keywords for translation.
Problem: 6.19. Find the second of three consecutive integers if the sum of twice the first and 4 times the
second is equal to 20 more than twice the third.
Solution:. Let x be the first integer and now we can set up a table for this problem:
Integers
First
Second
Third
Consecutive Integers
x
x+1
x+2
Now we can use the keyword sum, twice, times, is equal to , and more than in the problem and then
solve for the unknown.
2 • First + 4 • Second
=
2 • Third + 20
use translation and make the substitutions
=
2 • (x + 2) + 20
2x + 4 + 20
use distribution to remove (. . . )
=
2x + 24
simplify
6x + 4 − 4
=
2x + 24 − 4
use subtraction property of equation
4x
4x
4
x
=
2x + 4(x + 1)
=
2x + 4x + 4
6x + 4
6x − 2x
=
2x + 20 − 2x
use subtraction property of equation
20
20
=
4
= 5
simplify
isolate
So the second of these three consecutive integer is 5 + 1 = 6.
Recommendation:. Look up how to use properties of equation and main keywords for translation.
33
6.2
Homework & Quizzes
1. There are two lecture quizzes in this section, do both quizzes.
2. Solve all the problems in exercise set for this section, including the ones that were solved in this section.
3. You will be asked to attach one of the lecture quizzes in this section to your solutions for the problems
in the exercise set and turn them in as your homework.
34
How to Do Word Problems
Lecture Quiz 6.1
Name:
No Work ⇔ No Points
Use Pencil Only ⇔ Be Neat & Organized
1. (3 points) Find three consecutive integers such that their sum is 180.
1.
2. (4 points) The sum of three consecutive even integers is 210. Find the largest
one.
2.
3. (4 points) Find two consecutive odd integers such that the sum of five times
the smaller one and three times the larger one is 94.
3.
4. (4 points) Find two consecutive even integers such that the difference between five times the larger one and three times the smaller one is 38.
4.
Page 1 of 1
Lecture Quiz 6.1 Prepared by R. Faradineh
Total Points = 15
How to Do Word Problems
Lecture Quiz 6.2
Name:
No Work ⇔ No Points
Use Pencil Only ⇔ Be Neat & Organized
1. (3 points) Find three consecutive integers such that the sum of the first one,
twice the second one and three times the third one is 182.
1.
2. (4 points) The sum of first and third of three consecutive even integers is
80. Find the middle one.
2.
3. (4 points) Find two consecutive odd integers such that the sum of six times
the smaller one and the larger one is 107.
3.
4. (4 points) Find five consecutive even integers such that their sum is 0.
4.
Page 1 of 1
Lecture Quiz 6.2 Prepared by R. Faradineh
Total Points = 15
Chapter
7
Motion: Linear Equation
7.1
Solved Problems
Problem: 7.2. Two planes leave Miami International Airport at the same time flying in opposite directions.
One plane is traveling at the speed of 550 mph and the other plane is traveling at a speed of 600 mph. In
how many hours will they be 5750 miles apart?
Solution:. Let t be the time that it will take for both planes to be 5750 miles apart and now we can set up
a table for this problem:
Objects
Plane 1
Plane 2
Rate(Speed)
550 mph
600 mph
Time
t hours
t hours
Distance
550t miles
600t miles
Let d1 = 550t be the distance travelled by plane 1 and d2 = 600t be the distance travelled by plane 2.
Since these two places are going in opposite directions as indicated in the problem which implies that
the total distance between them is 5750, so we can set up the equation and then solve for the unknown.
d1 + d2
=
5750
550t + 600t =
5750
make the substitutions
1150t = 5750
5750
1150t
=
1150
1150
t = 5
combine like terms
isolate
So after 5 hours, these two planes will be 5750 miles apart.
Recommendation:. Look up motion formula and linear equation.
Problem: 7.4. On University Drive in Lauderhill, two buses leave the bus stop at the same time traveling
in the same direction. One bus is traveling at 40 mph and the other bus is traveling at 45 mph. How long
will it be before they are 30 miles apart?
37
Solution:. Let t be the time that it will take for both buses to be 30 miles apart, so now we can set up a
table for this problem:
Objects
Bus 1
Bus 2
Rate(Speed)
40 mph
45 mph
Time
t hours
t hours
Distance
40t miles
45t miles
Let d1 = 40t be the distance travelled by bus 1 and d2 = 45t be the distance travelled by bus 2.
The second bus with 45 mph speed would travel a longer distance than the other bus in the same time
since it is going faster Since these two places are going in same direction as indicated in the problem which
implies that the difference of distances between them is 30, so we can set up the equation and then solve for
the unknown.
d2 − d1
=
30
45t − 40t =
30
make the substitutions
5t = 30
5t
30
=
5
5
t = 6
combine like terms
isolate
So after 6 hours, these two buses will be 30 miles apart.
Recommendation:. Look up motion formula and linear equation.
Problem: 7.11. Two joggers are 10 miles apart. At the same time, they begin running towards each other.
One jogger is running at 4 mph and the other is running at 6 mph. How long after they begin will they
meet?
Solution:. Let t be the time that it will take for both joggers to meet and now we can set up a table for
this problem:
Objects
Jogger 1
Jogger 2
Rate(Speed)
4 mph
6 mph
Time
t hours
t hours
Distance
4t miles
6t miles
Let d1 = 4t be the distance travelled by jogger 1 and d2 = 6t be the distance travelled by jogger 2.
Since these two joggers are running towards each other as indicated in the problem which implies that
they are going in opposite direction hence when they meet, the total distance travelled between them is 10,
so we can set up the equation and then solve for the unknown.
d1 + d2
=
10
4t + 6t =
10
10t = 10
10t
10
=
10
10
t = 1
make the substitutions
combine like terms
isolate
So after 1 hour, these two joggers will meet.
38
Recommendation:. Look up motion formula and linear equation.
Problem: 7.15. Victor and Ingrid leave their house at the same time riding their bicycles in opposites
directions. Victor rides 5 mph faster that Ingrid. After two hours, they are 58 miles apart. At what speed
was Victor riding?
Solution:. Let x be the speed that Ingrid was riding, therefore Victor’s speed must be x + 5 since he was
riding 5 mph faster than Ingrid and now we can set up a table for this problem:
Objects
Ingrid
Victor
Rate(Speed)
x mph
(x + 5) mph
Time
2 hours
2 hours
Distance
2x miles
2(x + 5) miles
Let d1 = 2x be the distance travelled by Ingrid and d2 = 2(x + 5) be the distance travelled by Victor.
Since they are going in opposite directions as indicated in the problem which implies that the total distance
between them is 58, so we can set up the equation and then solve for the unknown.
d1 + d2
=
58
2x + 2(x + 5)
=
58
make the substitutions
2x + 2x + 10
=
58
distribute to to remove ( . . . )
4x + 10
=
58
combine like terms
4x + 10 − 10
=
58 − 10
4x = 48
48
4x
=
4
4
x = 12
use subtraction property of equation
simplify
isolate
So Ingrid is riding at 12 mph, therefore Victor was riding at 12 + 5 = 17 mph.
Recommendation:. Look up motion formula and linear equation.
Problem: 7.20. Two cross country skiers begin at the same time, traveling in opposite directions. One
skier is traveling at a rate of 64 mph slower than the other skier. In 8 hours, hey are 96 miles apart. At
what speed was the faster skier traveling?
Solution:. Let x be the speed of the faster skier, therefore x − 6 is the speed for the slower skier since it is
traveling 6 mph slower as indicated in the problem and now we can set up a table for this problem:
Objects
Faster skier
Slower skier
Rate(Speed)
x mph
(x − 6) mph
Time
8 hours
8 hours
Distance
8x miles
8(x − 6) miles
Let d1 = 8x be the distance travelled by faster skier and d2 = 8(x−6) be the distance travelled by slower skier.
39
Since these two skiers are going in opposite directions as indicated in the problem which implies that
the total distance between them is 96, so we can set up the equation and then solve for the unknown.
d1 + d2
=
96
8x + 8(x − 6)
=
96
make the substitutions
=
96
distribute to to remove ( . . . )
=
96
combine like terms
8x + 8x − 48
16x − 48
16x − 48 + 48
16x
16x
16
x
= 96 + 48
= 144
144
=
16
= 9
use addition property of equation
simplify
isolate
So the faster skier is traveling at 9 mph.
Recommendation:. Look up motion formula and linear equation.
40
7.2
Homework & Quizzes
1. There are two lecture quizzes in this section, do both quizzes.
2. Solve all the problems in exercise set for this section, including the ones that were solved in this section.
3. You will be asked to attach one of the lecture quizzes in this section to your solutions for the problems
in the exercise set and turn them in as your homework.
41
How to Do Word Problems
Lecture Quiz 7.1
Name:
No Work ⇔ No Points
Use Pencil Only ⇔ Be Neat & Organized
1. (3 points) John and Maria were 220 miles apart. They left at the same time
and started driving towards each other. John was driving 10 mph faster than
Maria and they met after 2 hours. Find how fast was John driving?
1.
2. (4 points) Two buses left the bus stop at the same time, going in opposite
directions. One bus was driving 30 mph slower than twice the speed of the
other bus. Find their speed if they were 360 miles apart after 3 hours.
2.
3. (3 points) Two joggers started jogging in the same direction. One was jogging at 2.5 mph while the other one was jogging at 4 mph. How long does it
take before they are 6 miles apart?
3.
Page 1 of 1
Lecture Quiz 7.1 Prepared by R. Faradineh
Total Points = 10
How to Do Word Problems
Lecture Quiz 7.2
Name:
No Work ⇔ No Points
Use Pencil Only ⇔ Be Neat & Organized
1. (3 points) John and Maria were 920 miles apart. They left at the same time
and started driving towards each other. John was driving 15 mph slower
than Maria and they met after 8 hours. Find how fast were they driving?
1.
2. (4 points) Two buses left the bus stop at the same time, going in opposite
directions. One bus was driving 5 mph faster than two times the speed of
the other bus. Find their speed if they were 1100 miles apart after 10 hours.
2.
3. (3 points) Two joggers started jogging in the same direction. One was jogging at 6.5 mph while the other one was jogging at 4 mph. How long does it
take before they are 10 miles apart?
3.
Page 1 of 1
Lecture Quiz 7.2 Prepared by R. Faradineh
Total Points = 10
Chapter
8
Money
8.1
Solved Problems
Problem: 8.2. A collection of coins consists of dimes and nickels. The number of dimes is two more than
twice number of nickels. The value of collection is $2.70. How many dimes are there in the collection?
Solution:. Let x be the number of nickels, and since the number of dimes is two more than twice the number
of nickels, therefore 2x + 2 would be the number of dimes and now we can set up a table for this problem:
Denomination
Dimes
Nickels
Value
.10
.05
Number
2x + 2
x
Denomination Worth
.10(2x + 2)
.05x
Let W1 = .10(2x + 2) be the denomination worth for dimes and W2 = .05x be the denomination worth for
nickels.
Since we are given the value of collection to be $2.70 as indicated in the problem which implies that
the sum of both denominations worth is $2.70, so we can set up the equation and then solve for the unknown.
=
2.70
.10(2x + 2) + 0.05x =
W1 + W2
2.70
make the substitutions
10(2x + 2) + 5x =
270
multiply by 100 to remove the decimal points
20x + 20 + 5x =
270
distribute to remove parenthesis
25x + 20
=
270
combine like terms
25x + 20 − 20
=
270 − 20
use subtraction property of equation
25x = 250
25x
250
=
25
25
x = 10
simplify
isolate
So there are 10 nickels in the collection, but the number of dimes in the collection is 2(10) + 2 = 20 + 2 = 22.
44
Recommendation:. Look up denomination values, removing decimal points and linear equation.
Problem: 8.6. A collection of 57 coins is made of quarters and dimes. If the collection totals $9.00, how
many quarters are there?
Solution:. Let x be the number of quarters, and since the total number of coins is 57, therefore 57 − x
would be the number of dimes and now we can set up a table for this problem:
Denomination
Dimes
Quarters
Value
.10
.25
Number
57 − x
x
Denomination Worth
.10(57 − x)
.25x
Let W1 = .10(57 − x) be the denomination worth for dimes and W2 = .25x be the denomination worth for
quarters.
Since we are given the value of collection to be $9.00 as indicated in the problem which implies that
the sum of both denominations worth is $9.00, so we can set up the equation and then solve for the unknown.
W1 + W2
=
9.00
.10(57 − x) + 0.25x =
9.00
make the substitutions
900
multiply by 100 to remove the decimal points
570 − 10x + 25x =
900
distribute to remove parenthesis
combine like terms
10(57 − x) + 25x =
15x + 570
=
900
15x + 570 − 570
=
900 − 570
15x = 330
330
15x
=
15
15
x = 22
use subtraction property of equation
simplify
isolate
So there are 22 quarters in the collection.
Recommendation:. Look up denomination values, removing decimal points and linear equation.
Problem: 8.8. Mario’s Pizzeria made a total of $220 during lunch, consisting of one-dollar bills, and fivedollar bills. If the number of fives was 4 less than three times the number of ones, how many one-dollar bills
did Mario’s Pizzeria have after lunch?
Solution:. Let x be the number of one-dollar bills, and since the number of fives was 4 less than three times
the number of ones, therefore 3x − 4 would be the number of five-dollar bills and now we can set up a table
for this problem:
Denomination
Ones
Fives
Value
1
5
Number
x
3x − 4
45
Denomination Worth
1•x
5(3x − 4)
Let W1 = 1 • x be the denomination worth for ones and W2 = 5(3x − 4) be the denomination worth for fives.
Since we are given the total to be $220 as indicated in the problem which implies that the sum of both
denominations worth is $220, so we can set up the equation and then solve for the unknown.
W1 + W2
=
220
1 • x + 5(3x − 4)
=
220
=
220
x + 15x − 20
16x − 20
16x − 20 + 20
make the substitutions
distribute to remove parenthesis
=
220
combine like terms
=
220 + 20
use addition property of equation
16x = 240
16x
240
=
16
16
x = 15
simplify
isolate
So there are 15 one-dollar bills in the collection.
Recommendation:. Look up denomination values, removing decimal points and linear equation.
Problem: 8.12. At the end of each week, Jos´e would take the ones and the fives out of his wallet to save
up for spending money on his vacation. After a year, he had a total of 220 bills that were both $600. How
many of each type of bill did Jos´e have?
Solution:. Let x be the number of one-dollar bills, and since he had a total of 220 bills, therefore 220 − x
would be the number of five-dollar bills and now we can set up a table for this problem:
Denomination
Ones
Fives
Value
1
5
Number
x
220 − x
Denomination Worth
1•x
5(220 − x)
Let W1 = 1 • x be the denomination worth for ones and W2 = 5(220 − x) be the denomination worth for fives.
Since we are given the total to be $600 as indicated in the problem which implies that the sum of both
denominations worth is $600, so we can set up the equation and then solve for the unknown.
W1 + W2
=
600
1 • x + 5(220 − x)
=
600
=
600
x + 1100 − 5x
make the substitutions
distribute to remove parenthesis
−4x + 1100 = 600
combine like terms
−4x + 1100 − 1100 = 600 − 1100 use subtraction property of equation
−4x
−4x
−4
x
= −500
−500
=
−4
= 125
simplify
isolate
So Jos´e has 125 one-dollar bills and 220 − 125 = 95 five-dollar bills.
46
Recommendation:. Look up denomination values, removing decimal points and linear equation.
Problem: 8.16. Suppose you have $6.05 in nickels and quarters. If you have twice as many quarters as
nickels, how many nickels do you have?
Solution:. Let x be the number of nickels, and since we have twice as many quarters as nickels, therefore
2x would be the number of quarters and now we can set up a table for this problem:
Denomination
Nickels
Quarters
Value
.05
.25
Number
x
2x
Denomination Worth
.05x
.25(2x)
Let W1 = .05x be the denomination worth for nickels and W2 = .25(2x) be the denomination worth for
quarters.
Since we are given the value of collection to be $6.05 as indicated in the problem which implies that
the sum of both denominations worth is $6.05, so we can set up the equation and then solve for the unknown.
W1 + W2
=
6.05
.05x + 0.25(2x)
=
6.05
5x + 25(2x)
=
605
multiply by 100 to remove the decimal points
5x + 50x =
605
multiply
55x = 605
605
55x
=
55
55
x = 11
make the substitutions
combine like terms
isolate
So you must have 11 nickels.
Recommendation:. Look up denomination values, removing decimal points and linear equation.
47
8.2
Homework & Quizzes
1. There are two lecture quizzes in this section, do both quizzes.
2. Solve all the problems in exercise set for this section, including the ones that were solved in this section.
3. You will be asked to attach one of the lecture quizzes in this section to your solutions for the problems
in the exercise set and turn them in as your homework.
48
How to Do Word Problems
Lecture Quiz 8.1
Name:
No Work ⇔ No Points
Use Pencil Only ⇔ Be Neat & Organized
1. (3 points) Nasreen has 17 coins, consists of dimes and quarters with a total
value of $3.05. How many of each coin does she have?
1.
2. (4 points) Gina cashed her paycheck of $1490 and received in twenty-dollar
bills and fifty-dollar bills. The number of fifty-dollar bills were one more
than twice the number of twenty-dollar bills. Find how many fifty-dollar
bills she received.
2.
3. (3 points) Sholeh has twice as many nickels as dimes, and three times as
many quarters as nickels. If she has $6.80, find how many quarters she has.
3.
Page 1 of 1
Lecture Quiz 8.1 Prepared by R. Faradineh
Total Points = 10
How to Do Word Problems
Lecture Quiz 8.2
Name:
No Work ⇔ No Points
Use Pencil Only ⇔ Be Neat & Organized
1. (3 points) Nasreen has 3 more dimes than twice the number of quarters.
How many of each coin does she have if the total value is $3.45?
1.
2. (4 points) Gina cashed her paycheck of $420 and received in twenty-dollar
bills and five-dollar bills. Find how many five-dollar bills she received if she
has a total of 30 bills.
2.
3. (3 points) Sholeh has twice as many nickels as dimes, and six more quarters
as nickels. If she has a total of $6.40, find how many quarters she has.
3.
Page 1 of 1
Lecture Quiz 8.2 Prepared by R. Faradineh
Total Points = 10