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Module 7: Notes and Solutions to Practice Problems for Unit 2.
This material was prepared by Mr. John Hemmerling and with revisions by Mr. Louis Aquila.
Since word problems can be challenging for many students, I will try to convince you that you are well
equipped to handle them. A word problem describes a problem in a language, such as English, and then
asks the reader to answer one, or more, questions. What you have to do is to translate the words into
equation(s), solve it (them) and answer the question(s).
To give you plenty of practice, I will set up every problem by going through steps 1 & 2 of the 4-step
solution process that is described in Module 7. You’ll usually have an equation to solve by the end of step
2, so you should all be able to complete the problems from there. There are a several problems that have
two equations to solve so I’ll throw in a step 3 as needed to help you get past the rough parts. The solutions
below are not complete. You have to solve the equations.
[Unit 1 is very basic and is thoroughly covered in the course guide so we will start with Unit 2.]
Important information in each problem below is denoted by *...*.
1. *One number is* *three times another number*, and *their sum is 68*. What are the two numbers?
x = 3y
and
x + y = 68
There are two equations: 1) x = 3y and 2) x + y = 68
Substitute 1) into 2):
3y + y = 68
Solve and answer the question.
2. *One number is* *two less than five times another* and *their sum is 76*. What are the numbers?
x = 5y – 2
and
x + y = 76
There are two equations: 1) x = 5y – 2 and
2) x + y = 76
Substitute 1) into 2): 5y – 2 + y = 76
Solve and answer the question.
3. *One number is* *four times another number* and *their difference is 69*. What are the two numbers?
x = 4y
and
x – y = 69
There are two equations: 1) x = 4y
Substitute 1) into 2):
2) x – y = 69
and
4y -y = 69
Solve and answer the question.
4.*Divide the number 40 into two parts* so that *three times the smaller part is equal* to *twice the larger
part*.
L + S = 40
3S
=
2L
Dividing 40 into two parts is tricky. The word divide means break 40 into two parts. It’s similar to having
$40 to spend and buying a shirt for $15 and pants for $25. This $40 was divided into a $15 part and a $25
part.
There are two equations: 1) L + S = 40
and
2) 3S = 2L
We want to substitute so we need one variable alone in one equation. Solve equation 1) for L: L + S – S =
40 – S so L = 40 – S.
Substitute 1) into 2): 3S = 2(40 – S)
Solve and answer the question.
5. Find *four consecutive integers* whose *sum is 270*.
Four consecutive integers are: x, x + 1, x + 2, x + 3
Their sum is: x + (x + 1) + (x + 2) + (x + 3) = 270
Solve and answer the question.
6. Find * five consecutive even integers* whose *sum is -240*.
Five consecutive even integers are: x, x + 2, x + 4, x + 6, x + 8.
Their sum is: x + (x + 2) + (x + 4) + (x + 6) + (x + 8) = -240
Solve and answer the question.
7. Find *six consecutive odd integers* whose *sum is 252*.
Six consecutive odd integers are: x, x + 2, x + 4, x + 6, x + 8, x + 10.
Their sum is: x + (x + 2) + (x + 4) + (x + 6) + (x + 8) + (x + 10) = 252
Solve and answer the question.
8. The *sum of two numbers is 71*. The *larger number is* *five less than three times the smaller*.
L + S = 71
L = 3S - 5
What are the two numbers?
There are two equations: 1) L + S = 71
and
2) L = 3S – 5
Substitute 2) into 1): 3S – 5 + S = 71
Solve and answer the question.
9. The* sum of three numbers is 89*. The *second number is three times the first* and the *third is four
more than the first*. What are the three numbers?
*sum of three numbers is 89* means x + y + z = 89
*second number is three times the first* means y = 3x
*third is four more than the first* means z = x + 4
There are three equations. 1) x + y + z = 89, 2) y = 3x and 3) z = x + 4
Substitute 2) and 3) into 1): x + 3x + x + 4 = 89
Solve and answer the question.
10. The *sum of three consecutive even integers** is 258*. What are the numbers?
Three consecutive even integers are: x, x + 2, x + 4
Their sum is: x + (x + 2) + (x + 4) = 258
Solve and answer the question.
11. Find five consecutive odd integers whose sum are 455.
Five consecutive odd integers are: x, x + 2, x + 4, x + 6, x + 8.
Their sum is: x + (x + 2) + (x + 4) + (x + 6) + (x + 8) = 455
Solve and answer the question.
12. *One number* *exceeds another by 8* and *their sum is 62*. What are the two numbers?
x= y+8
and
x + y = 62
There are two equations: 1) x = y + 8 and 2) x + y = 62
Substitute 1) into 2): y + 8 + y = 62
Solve and answer the question.
13. *Divide 99 books into two groups* so that* four-fifth of one group is equal* in number *to two thirds
of the other group*.
*Divide 99 books into two groups* means x + y = 99
4
*four-fifth of one group is equal* means   x 
5
2
*to two thirds of the other group*. means   y
3
2
4
There are two equations: 1) x + y = 99 and 2)   x    y
5
3
We want substitute so we need to solve one of the equations for x or y to get a variable alone.
2)
 5  4   5  2 
5
x

y

or

x

     
 y
 4  5   4  3 
6
5
 y  y  99
6
Substitute 2) into 1): 
(Hint: You can multiply everything by 6 to remove the denominator.
Solve and answer the question.
14. *A sum of $85 is divided among A, B and C* so that *B has $10 more than A* and* C has 3 times as
much as A*. How much cash does each have?
*A sum of $85 is divided among A, B and C* means A + B + C = 85.
*B has $10 more than A* means B = A + 10
*C has 3 times as much as A* means C = 3A
There are three equations: 1) A + B + C = 85, 2) B = A + 10 & 3) C = 3A
Substitute 2) and 3) into 1): A + A + 10 + 3A = 85
15. *Twice a number*, *increased by 55 is 89*. Find the number.
2x
+ 55
= 89
There is one equation: 2x + 55 = 89
Solve it and get the answer .
16. A student in a College Algebra class has test scores 82,79, 85 and 92.(Each test is worth 100points)
What score must she get on the final exam in order to have an average score of 85 in the course?
You calculate an average by adding the scores and dividing by the number of scores. We are given 4 of 5
scores and we want the average to be 85.
85 
82  79  85  92  x
5
Solve the equation and answer the question.
17. The *sum of two numbers is 38*. If the *larger is divided by the smaller*, the *quotient is 4 and the
remainder is 3*. Find these two numbers.
The first part is simple: *sum of two numbers is 38* means L + S = 38.
The second part is tricky: *larger is divided by the smaller* means
L
and *quotient is 4 and the
S
remainder is 3* means…... ?
Let’s look at an example. Consider the following fraction:
20
7
This quotient is similar to what has been described above. Write this fraction as a mixed number.
20
3
6
7
7
Most people would read this mixed number as three and six-sevenths so let’s use that “and” word and slip it
into the mixed number as addition.
20
6
 3  Notice that we have a larger number divided by a smaller number. The quotient is 3 and the
7
7
remainder is 6 and that the smaller number is the denominator in both fractions in the equation.
Now replace the numbers we used in the example with the information we were given in this problem.
L
3
 4  Multiply everything by S and we have a much simpler equation: L = 4S + 3.
S
S
There are two equations: 1) L + S = 38 and
2) L = 4S + 3
Substitute 2) into 1): 4S + 3+ S = 38
Solve the equation and answer the question.
18. The *sum of three consecutive integers** is 378*. Find these consecutive integers.
Three consecutive integers are: x, x + 1, x + 2
Their sum is: x + (x + 1) + (x + 2) = 378
Solve and answer the question.
19. The *sum of four consecutive even integers** is 364). Find these four integers.
Four consecutive even integers are: x, x + 2, x + 4, x + 6.
Their sum is: x + (x + 2) + (x + 4) + (x + 6) = 364
Solve and answer the question.
20. *Four times a number*, *decreased by 37 is 75*. Find the number.
4x
-37
= 75
There is one equation: 4x – 37 = 75
Solve it.