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Module 7: Notes and Solutions to Practice Problems for Unit 2.
Since word problems are usually pretty traumatic for students, I will try to convince you all that you are
well equipped to handle them. A word problem describes an equation(s), then asks a question. All you
have to do is find the equation(s), solve it (them) and answer the question.
Instead of providing solutions for a few of the problems, I’m going to 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. These solutions are not complete. You have to solve the equations.
Unit 1 is very basic and is covered very well in the course guide so we’ll start with Unit 2. This is the nocolor version so important information is denoted by *...* in each problem.
1. *One number is* *three times another number*, and *their sum is 68*. What are the two numbers?
x
=
3y
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
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
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 kind of like 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. 1) 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 isthree 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 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
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.
 5  4   2  5 
5
2)    x     y _& _ x    y
 4  5   3  4 
6
5
 y  y  99
6
Substitute 2) into 1): 
(Hint: You can multiply everything by 6 to get rid of the fraction.)
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…... What the heck does that mean?
Let’s 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.