Answer
After reading question 17, you’re likely to come up
with the equation in answer a. Since a is correct, it is
not the right choice. Now manipulate the equation to
see whether you can find an equivalent equation. If
you subtract 3 from each side, answer b will result.
From there, dividing both sides by 10, you come up
with c. All those are equivalent equations. Choice d can
be derived by using b and subtracting v from both
sides. Choice e is not an equivalent and is therefore the
correct answer.
Distance, Rate, and
Time Problems
One type of problem made simpler by algebra are
those involving distance, rate, and time. Your math
review would not be complete unless you had at least
one problem about trains leaving the station.
Sample Distance Problem
18. A train left the station near your home and went
at a speed of 50 miles per hour for 3 hours. How
far did it travel?
a. 50 miles
b. 100 miles
c. 150 miles
d. 200 miles
e. 250 miles
Answer
Use the three Success Steps to work through the
problem.
1. D = R × T
2. D = 50 × 3

3. 50 × 3 = 150
Practice
Try these:
19. How fast does a dirt bike go if it goes 60 miles
every 3 hours?
20. How long does it take to go 180 miles at 60 miles
per hour?
Answers
19. R = 20
20. T = 3
HOT TIP
Another way to look at the distance formula is
When you’re working out a problem, cross out the let-
ter that represents the value you need to find. What
remains will tell you the operation you need to perform
to get the answer: the horizontal line means divide and
the vertical line means multiply. For example, if you
need to find R, cross it out. You’re left with D and T.
The line between them tells you to divide, so that’s
how you’ll find R. This is a handy way to remember the
formula, especially on tests, but use the method that
makes the most sense to you.
Three Success Steps for Distance,
Rate, and Time Problems
1. First, write the formula. Don’t skip this step!
The formula for Distance, Rate, and Time is D
= R × T. Remember this by putting all the let-
ters in alphabetical order and putting in the
equal sign as soon as possible. Or think of the
word DIRT where the I stands for is, which is

always an equal sign.
2. Fill in the information.
3. Work the problem.
–CBEST MINI-COURSE–
118
D
R T

Math 8: Averages, Probability,
and Combinations
In this lesson, you’ll have a chance to do sample aver-
age questions as well as problems on probability and
on the number of possible combinations. They may be
a little more advanced than those you did in school,
but they will not be difficult for you if you master the
information in this section.
Averages
You probably remember how you solved average prob-
lems way back in elementary school. You added up the
numbers, divided by the number of numbers, and the
average popped out. Here’s this process in algebraic
form:
= Average
What makes CBEST average problems more dif-
ficult is that not all the numbers will be given for you
to add. You’ll have to find some of the numbers.
Sample Average Question
1. Sean loved to go out with his friends, but he
knew he’d be grounded if he didn’t get 80% for
the semester in his English class. His test scores

were as follows: 67%, 79%, 75%, 82%, and 78%.
He had two more tests left to go. One was tomor-
row, but his best friend Jason had invited him to
his birthday party tonight. If he studied very
hard and got 100% on his last test, what could he
get by with tomorrow and still have a chance at
the 80%?
a. 65%
b. 72%
c. 76.2%
d. 79%
e. 80.2%
Answer
Use the four Success Steps to solve the problem.
1. Draw the horizontal line:
ᎏᎏ
2. Write in the information:
ᎏ
7
ᎏ
= 80%
3. Multiply the number of numbers by the average to
obtain the sum of the numbers: 7 × 80 = 560
4. 560 has to be the final sum of the numbers. So far,
if you add up all the scores, Sean has a total of 381.
With the 100 he plans to get on the last test, his
total will be 481. Since he needs a sum of 560 for
the average of the seven tests to come out 80, he
needs 79 more points. The answer is d.
Sample Average Question

2. On an overseas trip, Jackie and her husband are
allowed five suitcases that average 110 pounds
each. They want to pack in all the peanut butter
and mango nectar they can carry to their family
in Italy. They weighed their first four suitcases
and the weights were as follows: 135, 75, 90, and
Four Success Steps
for Average Problems
1. In order to use the formula above, draw the
horizontal line that is under the sum and over
the number of numbers in the average
formula.
2. Write in all the information you know. Put the
number of numbers under the line and the
average beside the line. Unless you know the
whole sum, leave the top of the line blank.
3. Multiply the number of numbers by the aver-
age. This will give you the sum of the
numbers.
4. Using this sum, solve the problem.
Sum of the Numbers
ᎏᎏᎏ
Number of Numbers
–CBEST MINI-COURSE–
119
120. How much weight are they allowed to stuff
into their fifth bag?
Answer
1. Draw the horizontal line:
ᎏᎏ

2. Write in the information:
ᎏ
5
ᎏ
= 110
3. Multiply the number of numbers by the average to
obtain the sum of the numbers:
5 × 110 = 550.
4. Since the total weight they can carry is 550 lb. and
they already have 420 lb. (135 + 75 + 90 + 120), the
fifth suitcase can weigh as much as 550 − 420, or
130 lbs.
Frequency Charts
Some average problems on the CBEST use frequency
charts.
Sample Frequency Chart Question
3. The following list shows class scores for an easy
Science 101 quiz. What is the average of the
scores?
a. 28.5
b. 85
c. 91
d. 95
e. 100
Answer
Use the four Success Steps to solve the problem.
1. In this frequency chart, the test score is given on
the right, and the number of students who
received each grade is on the left: 10 students got
100, 15 got a 90, etc.

2. Multiply the number of students by the score,
because to find the average, each student’s grade
has to be added individually:
10 × 100 = 1,000
15 × 90 = 1,350
3 × 80 = 240
2 × 70 = 140
3. Then add the multiplied scores:
1,000 + 1,350 + 240 + 140 = 2,730
4. Then divide the total number of students,
30 (10 + 15 + 3 + 2), into 2,730 to get the average:
ᎏ
2,
3
7
0
30
ᎏ
= 91.
10
15
3
2
10
15
3
2
10
15
3

2
100
90
80
70
# of
Students Score
Four Success Steps for
Frequency Chart Questions
1. Read the question and look at the chart. Make
sure you understand what the different
columns represent.
2. If a question asks you to find the average,
multiply the numbers in the first column by the
numbers in the second column.
3. Add the figures you got by multiplying.
4. Divide the total sum by the sum of the left col-
umn. This will give you the average.
–CBEST MINI-COURSE–
120
Other Average Problems
There are other kinds of averages besides the mean,
which is usually what is meant when the word average
is used:
■
Median is the middle number in a range.
■
Mode is the number that occurs most frequently.
■
Range is the difference between the highest and

lowest number.
Sample Median Question
4. What is the median of 6, 8, 3, 9, 4, 3, and 12?
a. 2
b. 6
c. 9
d. 10
e. 12
Answer
To get a median, put the numbers in order—3, 3, 4, 6,
8, 9, 12—and choose the middle number: 6. If there
are an even number of numbers, average the middle
two (you probably won’t have to do that on the
CBEST).
Mode
The mode is the number used most frequently in a
series of numbers. In the above example, the mode is
3 because 3 appears twice and all other numbers are
used only once. Look again at the frequency table from
the frequency chart sample question. Can you find the
mode? More students (15) earned a score of 90 than
any other score. Therefore the mode is 15.
Range
To obtain the range, subtract the smallest number from
the largest number. The range in the median sample
question is 12 − 3 or 9. The range in the frequency
chart sample question is 100 − 70, or 30.
Probability
Suppose you put one entry into a drawing that had 700
entrants. What would be your chances of winning? 1 in

700 of course. Suppose you put in two entries. Your
chances would then be 2 in 700, or reduced, 1 in 350.
Probabilities are fairly simple if you remember the few
tricks that are explained in this section.
Sample Probability Question
5. If a nickel were flipped thirteen times, what is the
probability that heads would come up the thir-
teenth time?
a. 1:3
b. 1:2
c. 1:9
d. 1:27
e. 1:8
Answer
Use the four Success Steps to solve the problem.
1. Form a fraction.
2. Each time a coin is flipped, there are 2 possibili-
ties—heads or tails—so 2 goes on the bottom of
the fraction. The thirteenth time, there are still
going to be only two possibilities.
3. The number of chances given is 1. There is only
one head on a coin. Therefore, the fraction is
ᎏ
1
2
ᎏ
.
Four Success Steps for
Probability Questions
1. Make a fraction.

2. Place the total number of different possibilities
on the bottom.
3. Place the number of the chances given on the
top.
4. If the answers are in a:b form, place the
numerator of the fraction first, and the denom-
inator second.
–CBEST MINI-COURSE–
121
“Thirteen times” is extra information and does
not have a bearing on this case.
4. The answer is b, 1:2. The numerator goes to the
left of the colon and the denominator to the
right.
Sample Probability Question
6. A spinner is divided into 6 parts. The parts are
numbered 1–6. When a player spins the spinner,
what are the chances the player will spin a num-
ber less than 3?
Answer
Once again, use the four Success Steps.
1. Form a fraction.
2. Total number of possibilities = 6. Therefore, 6 goes
on the bottom.
3. Two goes on top, since there are 2 numbers less
than 3: 1 and 2.
4. The answer is
ᎏ
2
6

ᎏ
or reduced
ᎏ
1
3
ᎏ
= 1:3.
Combinations
Combination problems require the solver to make as
many groups as possible given certain criteria. There
are many different types of combination problems, so
these questions need to be read carefully before
attempting to solve them. One of the easiest ways to
make combination problems into CBEST points is to
make a chart and list in a pattern all the possibilities.
The following sample question is a typical CBEST
combination problem.
Sample Combination Question
7. Shirley had three pairs of slacks and four
blouses. How many different combinations of
one pair of slacks and one blouse could she
make?
a. 3
b. 4
c. 7
d. 12
e. 15
Answer
To see this problem more clearly, you may want to
make a chart:

Each pair of slacks can be matched to 4 different
blouses, making 4 different outfits for each pair of the
3 pairs of slacks, 3 × 4, making a total of 12 possible
combinations.
Sample Combination Question
9. Five tennis players each played each other once.
How many games were played?
a. 25
b. 20
c. 15
d. 10
e. 5
–CBEST MINI-COURSE–
122
Answer
This combination problem is a little trickier in that
there are not separate groups of items as there were for
the slacks and blouses. This question involves the same
players playing each other. But solving it is not diffi-
cult. First, take the total number of players and sub-
tract one: 5 − 1 = 4. Add the numbers from 4 down: 1
+ 2 + 3 + 4 = 10. To learn how this works, take a look
at the following chart:
Letter the five players from A to E:
■
A plays B, C, D, and E (4 games)
■
B has already played A, so needs to play C, D, E (3
games)
■

C has already played A and B, so needs to play D,
E (2 games)
■
D has already played A, B, and C, so needs to play
E (1 game)
■
E has played everyone
Adding up the number of games played (1 + 2 + 3 + 4)
gives a total of 10, choice d.
This same question might be asked on the
CBEST using the number of games 5 chess players
played or the number of handshakes that occur when
5 people shake hands with each other once.
Other Combination Problems
Although the above combination problems are the
most common, other kinds of problems are possible.
The best way to solve other combination problems is
to make a chart. When you notice a pattern, stop and
multiply. For example, if you’re asked to make all the
possible combinations of three letters using the letters
A through D, start with A:
AAA ABA ACA ADA
AAB ABB ACB ADB
AAC ABC ACC ADC
AAD ABD ACD ADD
There seem to be 16 possibilities that begin with
A, so probably there are 16 that begin with B and 16
that begin with C and D, so multiplying 16 × 4 will give
you the total possible combinations: 64.


Math 9: The Word Problem
Game
The directions for the word problem game are simple:
While carefully observing a word problem, find all the
math words and numbers in the problem. Eliminate
the nonessential words and facts in order to find your
answer.
Operations in Word Problems
To prepare for the game, make five columns on a sheet
of paper. Write one of these words on the top of each
column: Add, Subtract, Multiply, Divide, Equals.
Now try to think of five words that tell you to add, five
that tell you to subtract, and so on. If you can think of
five for each column, you win the first round. If you
can’t think of five, you can cheat by looking at the list
below.
How did you do?
0 = keep studying
1–3 for each = good
4–6 for each = excellent
7+ for each = Why are you reading this book?
■
Add: sum, plus, more than, larger than, greater
than, and, increased by, added to, in all,
altogether, total, combined with, together, length-
ened by
–CBEST MINI-COURSE–
123