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
■
Subtract: difference, minus, decreased by, reduced
by, diminished by, less, take away, subtract, low-
ered by, dropped by, shortened by, lightened by,
less, less than, subtracted from, take from,
deducted from. Note: The words in bold are
backwardswords. (See below.)
■
Multiply: product, times, of, multiplied by, twice,
thrice, squared, cubed, doubled, tripled, rows of,
columns of
■
Divide: quotient of, ratio of, halved, per, split,
equal parts of, divided by, divided into, recipro-
cal. Note: The words in bold are backwardswords.
(See below.)
■

Equals: is, equal to, the same as, amounts to,
equivalent to, gives us, represents
Backwardswords
Backwardswords are words in a word problem that
tend to throw off test takers; they indicate the opposite
of which the numbers appear in the problem. Only
subtraction and division have backwardswords. Addi-
tion and multiplication come out the same no matter
which number is written first: 2 + 6 is the same as 6 +
2, but 2 – 6 is not the same as 6 – 2. Using the numbers
10 and 7, notice the following translations:
Subtraction:
10 minus 7 is the same as 10 – 7
10 take away 7 is 10 – 7
10 less 7 is the same as 10 – 7
But 10 less than 7 is the opposite, 7 – 10
10 subtracted from 7 is also 7 – 10
Division:
10 over 7 is written
ᎏ
1
7
0
ᎏ
The quotient of 10 and 7 is
ᎏ
1
7
0
ᎏ

But 10 divided into 7 is written
ᎏ
1
7
0
ᎏ
And the reciprocal of
ᎏ
1
7
0
ᎏ
is
ᎏ
1
7
0
ᎏ
Writing Word Problems
in Algebraic Form
Sample Word Conversion Questions
The following are simple problems to rewrite in alge-
braic form. Using N for a number, try writing out the
problems below. Remember to add parentheses as
needed to avoid order of operation problems.
1. Three added to a number represents 6.
2. Six subtracted from a number is 50.
Answers
Use the four Success Steps to find the answer to ques-
tion 1.

1. “Represents” is the verb. Put in an equal sign: =
2. 3, 6, and N are the numbers: 3 N =6
3. Added means +: 3 + N = 6
4. No parentheses are needed.
Follow the Success Steps for question 2.
Four Success Steps for
Converting Words to Algebra
In order to make an equation out of words use
these steps:
1. Find the verb. The verb is always the = sign.
2. Write in the numbers.
3. Write in the symbols for the other code words.
Be careful of backwardswords.
4. If necessary, add parentheses.
HOT TIP
When setting up division problems in algebra, avoid using
the division sign: ÷. Instead, use the division line:
ᎏ
3
4
ᎏ
.
–CBEST MINI-COURSE–
124
1. “Is” is the verb. Put in an equal sign: =
2. 6, N, and 50 are the numbers: 6 N = 50.
3. Subtracted from means –, but it is a back-
wardsword: N – 6 = 50.
4. No parentheses are needed.
Practice

Underline the backwardswords, then write the
equations.
3. A number subtracted from 19 is 7.
4. 3 less a number is 5.
5. 3 less than a number is 5.
6. 9 less a number is –8.
7. A number taken from 6 is –10.
8. 30 deducted from a number is 99.
9. The quotient of 4 and a number equals 2.
10. The reciprocal of 5 over a number is 10.
11. 6 divided into a number is 3.
Change the following sentences into algebraic
equations.
12. The sum of 60 and a number all multiplied by 2
amounts to 128.
13. Forty combined with twice a number is 46.
14. $9 fewer than a number costs $29.
15. 7 feet lengthened by a number of feet all divided
by 5 is equivalent to 4 feet.
16. 90 subtracted from the sum of a number and
one gives us 10.
17. Half a number plus 12 is the same as 36.
Answers
3. subtracted from, 19 – N = 7
4. 3 – N = 5
5. less than, N – 3 = 5
6. 9 – N = –8
7. taken from, 6 – N = –10
8. deducted from, N – 30 = 99
9.

ᎏ
N
4
ᎏ
= 2
10. reciprocal of,
ᎏ
N
5
ᎏ
= 10
11. divided into,
ᎏ
N
6
ᎏ
= 3
12. (60 + N)2 = 128 or 2(60 + N) = 128
13. 40 + 2N = 46
14. N – $9 = $29
15.
ᎏ
7+
5
N
ᎏ
= 4
16. (N +1) – 90 = 10
17.
ᎏ

N
2
ᎏ
+12 = 36 or
ᎏ
1
2
ᎏ
N + 12 = 36
Words or Numbers?
Try these two problems and determine which is easier
for you.
1. Three more than five times a number equals 23.
2. Jack had three more than five times the number
of golf balls than Ralph had. If Jack had 23 golf
balls, how many did Ralph have?
Answers: 1. 23 = 3 + 5N
2. 23 = 3 + 5N
Did you notice that the two problems were the
same, but the second one was more wordy? If question
1 was easier, you can work word problems more easily
by eliminating non-essential words. If question 2 was
easier, you can work out problems more easily by pic-
turing actual situations. If they were both equally easy,
–CBEST MINI-COURSE–
125
then you have mastered this section. Go on to the sec-
tion on two-variable problems, which is a little more
difficult.
Practice

If you found wordy word problems difficult, here are
some more to try:
18. Sally bought 6 less than twice the number of
boxes of CDs that Raphael (R) bought. If Sally
bought 4 boxes, how many did Raphael buy?
19. A 1-inch by 13-inch rectangle is cut off a piece of
linoleum that was made up of three squares;
each had N inches on a side. This left 62 square
inches left to the original piece of linoleum. How
long was each of the sides of the squares?
20. Six was added to the number of sugar cubes in a
jar. After that, the number was divided by 5. The
result was 6. How many sugar cubes were in the
jar?
Answers
18. Sally = 2R – 6. Substitute 4 for Sally: 4 = 2R – 6
19. (N
2
× 3) – (1 × 13) = 62. N is the side of a square
so the area of the square is N
2
. There were three
squares, so you have 3N
2
.1 × 13 was taken away
(–). Notice that the parentheses, while not strictly
necessary if you follow the order of operations,
will help you keep track of the numbers.
20.
ᎏ

6+
5
N
ᎏ
= 6
Problems with Two Variables
In solving problems with two variables, you have to
watch out for another backwards phrase: as many as.
Sample Two-Variable Questions
The following equations require the use of two vari-
ables. Choose the answers from the following:
a. 2x = y
b. 2y = x
c. 2 + x = y
d. 2 + y = x
e. none of the above
21. Twice the number of letters Joey has equals the
number of letters Tina has. Joey = x, Tina = y.
22. Tuli corrected twice as many homework assign-
ments as tests. Homework =x, tests = y.
Answers
21.a. “Equals”is the verb. Joey or x is on one side of
the verb, Tina or y is on the other. A straight ren-
dering will give you the answer a, or 2x = y,
because Tina has twice as many letters. To check,
plug in 6 for y. If Tina has 6 letters, Joey will have
6 ÷ 2, or 12. The answer makes sense.
22. b. “Corrected” is the verb. Which did Tuli correct
fewer of? Tests. You need to multiply 2 times the
tests to reach the homework assignments. Check:

Three Success Steps for Problems
with Two Variables
When turning “as many as” sentences into equa-
tions, consider the following steps.
1. Read the problem to decide which variable is
least.
2. Combine the number given with the least
variable.
3. Make the combined number equal to the
larger amount.
–CBEST MINI-COURSE–
126
If there are 6 tests, then there are 12 homework
assignments: 2 × 6 = 12. This answer makes sense.
Practice
Now that you are clued in, try the following using the
same answer choices as above.
a. 2x = y
b. 2y = x
c. 2 + x = y
d. 2 + y = x
e. none of the above
23. Sandra found two times as many conch shells as
mussel shells. Conch = x, mussel = y.
24. Sharon walked two more miles today than she
walked yesterday. Today = x, yesterday = y.
25. Martin won two more chess games than his
brother won. His brother = x, Martin = y.
Answers
23. b.

24. d.
25. c.

Math 10: The CA Approach
to Word Problems
Of course, it helps to know the formula or method
needed to solve a problem. But there are always those
problems on the test that you don’t recognize or can’t
remember how to do, and this may cause you a little
anxiety. Even experienced math teachers experience
that paralyzing feeling at times. But you shouldn’t
allow anxiety to conquer you. Nor should you jump
into a problem and start figuring madly without a
careful reading and analysis of the problem.
The CA SOLVE Approach
When approaching a word problem, you need the
skills of a detective. Follow the CA SOLVE method to
uncover the mystery behind a problem that is unfa-
miliar to you.
C Stands for Conquer
Conquer that queasy feeling—don’t let it conquer you.
To squelch it, try step A.
A Stands for Answer
Look at the answers and see if there are any similarities
among them. Notice the form in which the answers are
written. Are they all in cubic inches? Do they all con-
tain pi? Are they formulas?
S Stands for Subject Experience
Many problems are taken from real life situations or
are based on methods you already know. Ask: “Do I

have any experience with this subject or with this type
of problem? What might a problem about the subject
be asking me? Can I remember anything that might
relate to this problem?”
Eliminate experiences or methods of solving that
don’t seem to work. But be careful; sometimes sorting
through your experiences and methods memory takes
a long time.
O Stands for Organize the Facts
Here are some ways to Organize your data:
1. Look for clue words in the problem that tell you
to add, subtract, multiply, or divide.
2. Try out each answer to see which one works.
Look for answers to eliminate.
3. Think of formulas or methods that have worked
for you in solving problems like this in the past.
Write them down. There should be plenty of
room on your test booklet for this.
–CBEST MINI-COURSE–
127
L Stands for Live
Living the problem means pretending you’re actually
in the situation described in the word problem. To do
this effectively, make up details concerning the events
and the people in the problem as if you were part of
the picture. This process can be done as you are read-
ing the problem and should take only a few seconds.
V Stands for View
View the problem with different numbers while keep-
ing the relationships between the numbers the same.

Use the simplest numbers you can think of. If a prob-
lem asked how long it would take a rocket to go
1,300,000 miles at 650 MPH, change the numbers to
300 miles at 30 MPH. Solve the simple problem, and
then solve the problem with the larger numbers the
same way.
E Stands for Eliminate
Eliminate answers you know are wrong. You may also
spend a short time checking your answer if there is
time.
Sample Question
Solve this problem using the SOLVE steps described
above.
1. There are 651 children in a school. The ratio of
boys to girls is 4:3. How many boys are there in
the school?
a. 40
b. 325
c. 372
d. 400
e. 468
Answer
1. Subject Experience: You know that 4 and 3 are
only one apart and 4 is more. You can conclude
from this that boys are a little over half the school
population. Following up on that, you can cut 651
in half and eliminate any answers that are under
half. Furthermore, since there are three numbers
in the problem and two are paired in a ratio, you
can conclude that this is a ratio problem. Then

you can think about what methods you used for
ratio problems in the past.
2. Organize: The clue word total means to add. In
the context in which it is used, it must mean girls
plus boys equals 651. Also, since boys is written
before girls, the ratio should be written Boys:Girls.
3. Live: Picture a group of three girls and four boys.
Now picture more of these groups, so many that
the total would equal 651.
4. View: If there were only 4 boys and 3 girls in the
school, there would still be a ratio of 4 to 3. Think
of other numbers that have a ratio of 4:3, like 40
and 30. If there were 40 boys and 30 girls, there
would be 70 students in total, so the answer has to
be more than 40 boys. Move on to 400 boys and
300 girls—700 total students. Since the total in the
problem is 651, 700 is too large, but it is close, so
HOT TIP
Don’t try to keep a formula in your head as you solve the
problem. Although writing does take time and effort, jot-
ting down a formula is well worth it for three reasons: 1) A
formula on paper will clear your head to work with the
numbers; 2) You will have a visual image of the formula
you can refer to and plug numbers into; 3) The formula will
help you see exactly what operations you will need to per-
form to solve the problem.
–CBEST MINI-COURSE–
128