### Practice Exercises

```Practice Exercises
MAT 091/093/095 – Algebra
Application Problems – Part I (Solutions)
1.
Consecutive odd integers. Find four consecutive odd integers whose sum is 352.
STEP 1: Read the problem and think about what a correct answer might look like. What is its approximate
value? Can it be negative? Can it be a decimal or fraction?
All of the answers must be positive integers (no decimals or fractions), and they will all be about ¼ of
the total.
STEP 2: If possible, draw a diagram to illustrate the problem. (Not useful in this case)
STEP 3: Identify all of the unknowns. Pick one of the unknowns as a variable. Write down what the variable
means. Pay attention to the units (e.g., feet, mph, pounds, etc.).
There are four unknowns – the four consecutive odd integers. Choose the smallest integer as the
variable.
Let x = the smallest of the four consecutive odd integers.
STEP 4: If there are other unknowns, write algebraic expressions for the other unknowns in terms of the
variable.
Each of the other integers is two more than the previous integer, therefore:
x + 2 = Second consecutive odd integer.
x + 4 = Third consecutive odd integer.
x + 6 = Fourth consecutive odd integer.
STEP 5: Write an equation to model the problem statement.
We haven’t yet used the fact that the sum of the numbers is 352. So write an expression for the sum of
the four numbers and set it equal to 352.
x  x  2  x  4  x  6  352
STEP 6: Solve the equation.
4 x  12  352
4 x  340
x  85
Combine like terms.
Subtract 12 from both sides.
Divide both sides by 4.
STEP 7: Answer the original question in an English sentence.
The variable x represents the smallest number; therefore:
“The four numbers are 85, 87, 89 and 91.”
STEP 8: Check your answer against the original word problem (not the equation).
Check that 85, 87, 89 and 91 are four consecutive odd integers.
Check that the four integers add up to 352.
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Iss 2.0
Practice Exercises
MAT 091/093/095 - Algebra
2.
Page 2
Application Problems – Part I
(Solutions)
Dimensions of a lot. Mr. Larsen put a fence around three sides of a rectangular piece of property he owns.
The fourth side borders a canal and does not require fencing. If the length is 5 feet less than twice the width
and he used 315 feet of fencing, what are the dimensions of the property? (See diagram.)
STEP 1: Read the problem and think about what a correct answer might look like. What is its approximate
value? Can it be negative? Can it be a decimal or fraction?
All of the asnwers be positive numbers less than 315. Lengths can be fractions or decimals.
STEP 2: If possible, draw a diagram to illustrate the problem. (Diagram is provided).
STEP 3: Identify all of the unknowns. Pick one of the unknowns as a variable. Write down what the variable
means. Pay attention to the units (e.g., feet, mph, pounds, etc.).
There are two unknowns – the length and width of the property. Choose width as the variable, because
the length is described in terms of the width.
Let x = the width of the property (in feet).
STEP 4: If there are other unknowns, write algebraic expressions for the other unknowns in terms of the
variable.
Express the length in terms of the variable x.
2x – 5 = Length of property.
STEP 5: Write an equation to model the problem statement.
It helps to write our expressions for length and width directly on
the diagram. We haven’t yet used the fact that Mr. Larsen used
315 feet of fencing. From the diagram, we get the following
equation:
2x – 5
x
x  2 x  5  x  315
Property
x
CANAL
STEP 6: Solve the equation.
4 x  5  315
4 x  320
x  80
Combine like terms.
Divide both sides by 4.
STEP 7: Answer the original question in an English sentence.
The variable x represents the width of the property. To find the length, substitute x=80 into the
expression for the length. We get:
“The dimensions of the property are 80 ft. (width) x 155 ft. (length).”
STEP 8: Check your answer against the original word problem (not the equation).
Check that the length (155 ft.) is five feet less than twice the width (80 ft.).
Check that the three sides add up to 315 ft.
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Iss. 2.0
Practice Exercises
MAT 091/093/095 - Algebra
3.
Application Problems – Part I
(Solutions)
Page 3
Pieces of a wire. A 57-foot wire is cut into three pieces. The second piece is 7 feet longer than the first
piece and the third piece is twice as long as the second piece. How long is each piece?
STEP 1: Read the problem and think about what a correct answer might look like. What is its approximate
value? Can it be negative? Can it be a decimal or fraction?
The answers all must be positive numbers less than 57. Lengths can be decimals or fractions.
STEP 2: If possible, draw a diagram to illustrate the problem.
x
x+7
2(x + 7)
57 feet
STEP 3: Identify all of the unknowns. Pick one of the unknowns as a variable. Write down what the variable
means. Pay attention to the units (e.g., feet, mph, pounds, etc.)
There are three unknowns – the three lengths of wire. Choose the first piece as the variable, because
the other pieces can be described in terms of that piece.
Let x = the length of the first piece of wire (in feet).
STEP 4: If there are other unknowns, write algebraic expressions for the other unknowns in terms of the
variable.
Express the lengths of the other two pieces in terms of the first piece:
x+7
= Length of second piece.
2(x + 7) = Length of the third piece.
STEP 5: Write an equation to model the problem statement.
It helps to write our expressions for the three lengths directly on the diagram. We haven’t yet used the
fact that the total length of the wire is 57 feet. From the diagram, we get the following equation:
x  x  7  2( x  7)  57
STEP 6: Solve the equation.
x  x  7  2 x  14  57
4 x  21  57
4 x  36
x9
Use Distributive Property to remove parentheses.
Combine like terms.
Subtract 21 from both sides.
Divide both sides by 4.
STEP 7: Answer the original question in an English sentence.
The variable x represents the length of the first piece of wire. To find the lengths of the other two
pieces, substitute x=9 into the expressions for the other two lengths. We get:
“The lengths of the three pieces of wire are 9 feet, 16 feet and 32 feet.”
STEP 8: Check your answer against the original word problem (not the equation).
Check that the second piece is 7 feet longer than the first piece.
Check that the third piece is twice as long as the second piece.
Check that the three pieces add up to 57 feet.
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Iss. 2.0
Practice Exercises
MAT 091/093/095 - Algebra
4.
Application Problems – Part I
(Solutions)
Page 4
Local election. In a local referendum, a property tax increase passed by 137 votes. If 993 votes were cast,
how many voters voted for and against the tax increase?
STEP 1: Read the problem and think about what a correct answer might look like. What is its approximate
value? Can it be negative? Can it be a decimal or fraction?
The answers all must be positive integers (no decimals or fractions) because you can’t have a negative
number of voters or fractions of a voter.
STEP 2: If possible, draw a diagram to illustrate the problem. (Not useful in this case)
STEP 3: Identify all of the unknowns. Pick one of the unknowns as a variable. Write down what the variable
means. Pay attention to the units (e.g., feet, mph, pounds, etc.).
There are two unknowns – the number of “for” votes and the number of “against” votes. You can
choose either one as the variable.
Let x = the number of “for” votes.
STEP 4: If there are other unknowns, write algebraic expressions for the other unknowns in terms of the
variable.
The number of “against” votes was 137 less than the number of “for” votes. Therefore:
x – 137 = Number of “against” votes.
STEP 5: Write an equation to model the problem statement.
We haven’t yet used the fact that the total number of votes was 993. So write an expression for the sum
of the “for” and “against” votes and set it equal to 993.
x  x  137  993
STEP 6: Solve the equation.
2 x  137  993
2 x  1130
x  565
Combine like terms.
Divide both sides by 2.
STEP 7: Answer the original question in an English sentence.
The variable x represents the number of “for” votes. To find the number of “against” votes, substitute
x=565 into the expression for the number of “against” votes:
“565 people voted for the tax increase and 428 people voted against it.”
STEP 8: Check your answer against the original word problem (not the equation).
Check that the margin of victory was 137 votes.
Check that the total number of votes is 993.
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Iss. 2.0
Practice Exercises
MAT 091/093/095 - Algebra
5.
Application Problems – Part I
(Solutions)
Page 5
Pickup truck. A pickup truck weighs 3,965 lbs. without any cargo. You plan to load bags of topsoil in the
cargo bed. Each bag weighs 40 lbs. If the truck loaded with topsoil weighs 4,485 lbs., how many bags is it
carrying?
STEP 1: Read the problem and think about what a correct answer might look like. What is its approximate
value? Can it be negative? Can it be a decimal or fraction?
The answer must be an integer (because we are counting the number of bags) and it cannot be negative.
STEP 2: If possible, draw a diagram to illustrate the problem. (Not useful in this case)
STEP 3: Identify all of the unknowns. Pick one of the unknowns as a variable. Write down what the variable
means. Pay attention to the units (e.g., feet, mph, pounds, etc.).
There is one unknown – the number of bags loaded on the truck.
Let n = the number of bags of topsoil.
STEP 4: If there are other unknowns, write algebraic expressions for the other unknowns in terms of the
variable. (No other unknowns).
STEP 5: Write an equation to model the problem statement.
We haven’t yet used the information that the truck loaded with topsoil weighs 4,485 lbs. Thus, we need
to come up with an expression for the weight of the truck with the topsoil. To write the expression,
write out how you would calculate the weight of the truck:
Weight of truck = 3,965 + 40  No. of bags of topsoil
Replacing “number of bags of topsoil” with the variable n, we get the expression 3965  40n and the
equation:
3965  40n  4485
STEP 6: Solve the equation.
3965  40n  4485
40n  520
n  13
Original equation
Subtract 3965 from each side.
Divide both sides by 40.
STEP 7: Answer the original question in an English sentence.
The variable n represents the number of bags of topsoil. Therefore, we can answer the question:
“13 bags of topsoil were loaded onto the truck.”
STEP 8: Check your answer against the original word problem (not the equation).
Calculate the weight of the truck with 13 bags of topsoil and verify that it equals 4485.
3965  40(13)  4485
3965  520  4485
4485  4485
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Iss. 2.0
Practice Exercises
MAT 091/093/095 - Algebra
6.
Application Problems – Part I
(Solutions)
Page 6
Car rental. A rental car company charges \$39 per day plus 24 cents per mile for an intermediate car. If the
rental car bill for one day is \$67.80, how many miles were driven?
STEP 1: Read the problem and think about what a correct answer might look like. What is its approximate
value? Can it be negative? Can it be a decimal or fraction?
The answer cannot be negative, but it may be a fraction or decimal.
STEP 2: If possible, draw a diagram to illustrate the problem. (Not useful in this case)
STEP 3: Identify all of the unknowns. Pick one of the unknowns as a variable. Write down what the variable
means. Pay attention to the units (e.g., feet, mph, pounds, etc.).
There is one unknown – the distance driven.
Let x = the distance driven (measured in miles)
STEP 4: If there are other unknowns, write algebraic expressions for the other unknowns in terms of the
variable. (No other unknowns).
STEP 5: Write an equation to model the problem statement.
We haven’t yet used the information that the rental car bill was \$67.80. Thus, we need to come up with
an expression for the rental car bill. To write the expression, write out how you would calculate the
bill:
Rental car bill = 39.00 + 0.24  No. of miles
Notice that we converted 24 cents to \$0.24, so all the terms in the equation are expressed in the same
units (dollars). Replacing “number of miles” with the variable x, we get the expression 39  .24x
and the equation:
39  .24 x  67.80
STEP 6: Solve the equation.
39  .24 x  67.80
.24 x  28.80
x  120
Original equation
Subtract 39 from each side.
Divide both sides by .24.
STEP 7: Answer the original question in an English sentence.
The variable x represents the number of miles driven. Therefore, we can answer the question:
“120 miles were driven.”
STEP 8: Check your answer against the original word problem (not the equation).
Calculate the rental car bill for 120 miles and verify that it equals 4485.
39  .24(120)  67.80
39  28.80  67.80
67.80  67.80
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Iss. 2.0
Practice Exercises
MAT 091/093/095 - Algebra
7.
Application Problems – Part I
(Solutions)
Page 7
Distance. If a car is traveling at a constant speed of 65 mph, how long would it take to travel 208 miles?
STEP 1: Read the problem and think about what a correct answer might look like. What is its approximate
value? Can it be negative? Can it be a decimal or fraction?
The answer cannot be negative because we are measuring time, but it may be a fraction or decimal of
an hour.
STEP 2: If possible, draw a diagram to illustrate the problem. (Not useful in this case)
STEP 3: Identify all of the unknowns. Pick one of the unknowns as a variable. Write down what the variable
means. Pay attention to the units (e.g., feet, mph, pounds, etc.).
There is one unknown – the time (measured in hours, because the speed is quoted in miles per hour.)
Let t = the driving time (measured in hours)
STEP 4: If there are other unknowns, write algebraic expressions for the other unknowns in terms of the
variable. (No other unknowns).
STEP 5: Write an equation to model the problem statement.
We haven’t yet used the fact that 208 miles was the distance. We can use the formula d  rt
(distance = rate  time) to compute the distance given the rate of speed and the driving time. We know
the distance and rate; therefore, we can find the time using the equation:
208  65t
STEP 6: Solve the equation.
208  65t
t  3.2
Original equation
Divide both sides by .65.
STEP 7: Answer the original question in an English sentence.
The variable t represents the time it took to drive 208 miles. Therefore, we can answer the question in
English: “It took 3.2 hours to drive 208 miles.”
STEP 8: Check your answer against the original word problem (not the equation).
Use d  rt to calculate the distance traveled in 3.2 hours, and verify that it is indeed 208 miles.
d  rt  (65)(3.2)  208
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Iss. 2.0
Practice Exercises
MAT 091/093/095 - Algebra
8.
Application Problems – Part I
(Solutions)
Page 8
Sales tax. Hoping to avoid the Illinois sales tax, a Waukegan resident bought a new car at a dealership in
Kenosha, Wisconsin. He later received a bill from the Illinois Department of Revenue for \$1770 to cover
the 7.5% sales tax. What was the price of the car?
STEP 1: Read the problem and think about what a correct answer might look like. What is its approximate
value? Can it be negative? Can it be a decimal or fraction?
The answer must be positive and it can be a decimal (representing dollars and cents). The price of the
car will be much larger than the sales tax.
STEP 2: If possible, draw a diagram to illustrate the problem. (Not useful in this case)
STEP 3: Identify all of the unknowns. Pick one of the unknowns as a variable. Write down what the variable
means. Pay attention to the units (e.g., feet, mph, pounds, etc.).
There is only one unknown: the price of the car.
Let x = the price of the car (in dollars)
STEP 4: If there are other unknowns, write algebraic expressions for the other unknowns in terms of the
variable. (No other unknowns).
STEP 5: Write an equation to model the problem statement.
We haven’t yet used the fact that sales tax rate is 7.5% and the sales tax amount is \$1770. Sales tax is
determined by multiplying the sales tax rate by the price of the car (which is our variable x). Therefore:
.075 x  1770
Remember to convert 7.5% to a decimal before multiplying by x.
STEP 6: Solve the equation.
.075 x 1770

.075
.075
x  23600
Divide both sides by .075.
STEP 7: Answer the original question in an English sentence.
The variable x represents the price of the car. Therefore:
“The price of the car was \$23,600.”
STEP 8: Check your answer against the original word problem (not the equation).
Check that sales tax on a \$23,600 car is \$1,770, by multiplying \$23,600 by .075.
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Iss. 2.0
Practice Exercises
MAT 091/093/095 - Algebra
9.
Application Problems – Part I
(Solutions)
Page 9
Shoe sale. Maria paid \$82.50 for a pair of boots at the Shoe Emporium’s fall clearance sale, where
everything was marked 40% off. What was the regular price of the boots?
STEP 1: Read the problem and think about what a correct answer might look like. What is its approximate
value? Can it be negative? Can it be a decimal or fraction?
The answer must be positive and it can be a decimal (representing dollars and cents). The regular price
will be greater than the sale price.
STEP 2: If possible, draw a diagram to illustrate the problem. (Not useful in this case)
STEP 3: Identify all of the unknowns. Pick one of the unknowns as a variable. Write down what the variable
means. Pay attention to the units (e.g., feet, mph, pounds, etc.).
There is only one unknown: the regular price of the boots.
Let x = the regular price of the boots (in dollars)
STEP 4: If there are other unknowns, write algebraic expressions for the other unknowns in terms of the
variable. (No other unknowns).
STEP 5: Write an equation to model the problem statement.
We haven’t yet used the fact that the discount rate is 40% or the sale price is \$82.50. The sale price
(\$82.50) is computed by subtracting the discount (40% of the regular price) from the regular price. The
regular price is represented by the variable x. Therefore:
x  .40 x  82.50
Remember to convert 40% to a decimal and multiply 40% by x.
STEP 6: Solve the equation.
.60 x  82.50
x  137.50
Combine like terms.
Divide both sides by .60.
STEP 7: Answer the original question in an English sentence.
The variable x represents the regular price of the boots.. Therefore:
“The regular price of the boots is \$137.50.”
STEP 8: Check your answer against the original word problem (not the equation).
Check that a pair of boots regularly priced at \$137.50 would sell for \$87.50 if marked “40% off.”
137.50
 55.00
87.50
Regular price
Subtract discount (40% of 137.50).
Sale price
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