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MATHEMATICS STUDY GUIDE
Practice Materials for the Developmental Mathematics Parts of the
Mathematics Placement Test at California University of Pennsylvania
Dear Prospective Student:
When you come to New Student Registration at California University of
Pennsylvania, you will be required to take a mathematics placement test.
The attached practice materials are intended to help you review for the Basic
Mathematics and Introductory Algebra tests. If your score on the first test (Part A)
is below 11 of 17, you are required to work with the tutors in the Math Lab on Part
I of this study guide and take Part A test again. The Math Lab is located in 115
Noss Hall – phone (724) 938-5893 for an appointment. If your score on the second
part (Part B) is below 12 of 17, you will be scheduled for Introductory Algebra
(DMA 092) if you are in a math-intensive major. You may study Part II of this
study guide and retake Part B of the math placement test just once. Please contact
Claire Pizer at (724) 938-5779 or [email protected] to set up an appointment to
retake the test.
It is important for you to know that the credit for Introductory Algebra will
not count toward graduation. However, the grade earned in the course does count
toward your grade point average.
Also, included are study topics for Parts C and D covering College Algebra
and Trigonometry if you want to test for Trigonometry, Pre-Calculus or Calculus.
We hope you will spend some time reviewing the attached materials because
they will be helpful to you in preparing for the placement test. (Calculators may
not be used during the test.)
Best wishes to you. We expect you will have enjoyable and productive
experiences here at California. If you have any questions, please call Dr. Paul
Williams at (724) 938-5894.
Sincerely,
The Mathematics Curriculum Committee
Test Date:
Test Time:
Location: Noss Hall 215
2
Dear Student:
In preparation for the developmental mathematics parts of the math
placement test (Part A and/or Part B), construct at least a dozen practice tests. Do
so by;
• marking 17 of the Practice Problems with a pencil,
• taking the phone off the hook,
• timing yourself for 15 minutes,
• taking the test,
• checking your answers, and
• referring to the worked out problem type of each question you missed.
Construct the next practice test by erasing the marks you placed by the 17
practice problems and by marking another 17 questions, whether or not you
marked them on the last test.
A score of 11 is passing on Part A and 12 is passing on Part B, but you
should try to get scores of 13 on the practice tests you take on your own. Call the
Math Lab in 115 Noss Hall at (724) 938-5893 for free help when you are having
difficulty with particular problems.
Sincerely,
Dr. Paul Williams
Math Lab Director
3
PART I – Basic Mathematics
Type of Question
Example
Solution
1. Add whole numbers
Find 126 + 5,123 + 57.
126
5123
+ 57
5306
2. Subtract whole numbers
Compute 753 – 68.
753
- 68
685
3. Multiply whole numbers
Jake worked 8 hours during
each of 6 days last week. He
received time and one-half
for weekends. How many
hours did he work last week?
64
4. Divide whole numbers
8
x6
48
Hours
608
54 32,832
Divide: 54 32,832
- 324
432
- 432
0
5. Add fractions or mixed
numbers
Find:
5 7 5
+ +
6 8 12
LCD = 24
=
=
=
5 7 5
+ +
6 8 12
5x4
7x3
6x4
8x3
20 21 10
+
+
24 24 24
51
24
2
24 51
- 48
3
=
=
2 3/24
2 1/8
5x2
12 x 2
4
Type of Question
6. Subtract fractions or mixed
numbers
Example
Sue missed 5/16 of a shift.
For how much of the work
shift should she be paid?
Solution
1–
5
16
16 5
16 16
11
=
shift
16
=
7. Multiply fractions or mixed
numbers
Find: 4
8. Divide fractions or mixed
numbers
Divide: 16 ÷
9. Add decimals
Find 0.27 + 6 + 0.3
10. Subtract decimals
Find 8.2 - .67
2
x 6.
3
7
8
2
x6
3
= 14 x 6 2
31
1
28
=
1
= 28
7
16 ÷
8
16 8
=
x
1
7
128
=
7
2
= 18
7
4
0.27
6.00
+ 0.30
6.57
7 11
8.20
- .67
7.53
11. Multiply decimals
Find 4.63 x .015
4.63 2 places
x.015 3 places
2315
463
.06945 5 places in
answer
5
12. Divide decimals
Find 16.03 ÷ .014
1145.
.014 16.030
Move
decimal
point
three
places to
the right.
13. Using decimals
If 6 cans of soda costs $1.20,
how much will 4 cans cost?
.20
6 1.20
14
20
14
63
56
70
70
0
per can
.20 x 4 = .80 for 4 cans
14. Renaming percents
Name 24% as a fraction and
a decimal.
24 out of 100 =
24 or reduce to 6
100
25
or 0.24
15. Finding percents of a
number
Find 62% of 120.
62% =
62
= 0.62
100
120 x 0.62 =
120
x 0.62
240
720
74.40
62% of 120 is 74.4
16. Using percents
A television priced at $360 is
on sale for 20% off.
Find the sale price.
Find 20% of 360
360 x .20
72.00
Regular price = 360.00
- 72.00
$288.00
6
PART I – Basic Math
Practice Problems
Problem Type
1. Find: 423 + 5,469 + 32
(1)
2. Subtract:
5,023
- 3,697
(2)
3. Multiply:
786
x 709
(3)
4. Divide: 63 44,667
5. Find:
7 1
1
- +
15 5
25
6. Calculate:
5
13
–
24 16
7. Multiply: 4
8. Compute:
1
7
x 5
8
13
7
5
÷
6
12
(4)
(5 & 6)
(6)
(7)
(8)
9. Find: 32 + 0.58 + 0.064
(9)
10. Find: 108.6 – 65.93
(10)
11. Multiply: 0.675 x 3.82
(11)
12. Find: 10.224 ÷ 3.6
(12)
13. Sarah can ride her bicycle 1.35 miles in 45 minutes. At
this rate, how far can Sarah ride in one minute?
(13)
14. Write 48% as a decimal
(14)
15. Find: 54% of 208
16. Everything in Marty’s hardware store is on sale for a 25%
discount. What would a hammer that was originally
priced at $16.90 cost?
(15)
17. Joe ran 5 miles, walked 2 miles, ran 3 miles, and drove 75
miles. How far did Joe run?
(1)
(16)
7
18. The Civic Club added 8,227 new members last year. They
now have 30,419 members. How many members did the
Civic Club have before last year?
(2)
19. Compute:
694 x 780
(3)
20. Calculate: 9,318 ÷ 49
(4)
21. Compute: 42
1
5
+ 35
6
8
(5)
22. Subtract: 34
1
7
- 18
12
8
(6)
2
1
pounds of grapes and sold
of them.
3
3
How many pounds of these grapes does John have left?
(7)
1
3
sections of a fence in 8 hours. If
7
2
she continues to paint at this rate, how many more sections
can she paint in one more hour?
(8)
23. John bought 7
24. Joan can paint 5
25. Peter ran 0.16 of a mile, walked 0.5 of a mile and then
jogged back the same distance. How far did Peter travel?
(9)
26. Joyce lost 16.8 kg on her new diet. If Joyce originally
weighed 160 kg, how much does she weigh now?
(10)
27. Mrs. Reynolds bought 6.5 yards of material costing $1.98
per yard. How much did she pay for the material?
(11)
28. Compute: 66 - .0033
(12)
29. Mr. King charges $36.95 for a new automobile tire. How
much would four tires cost?
(13)
30. What fraction represents 65%?
(14)
31. What is 6% of 72?
(15)
32. How much sales tax would you pay on a refrigerator that
costs $478.00 at a rate of 6% tax?
(16)
8
Answers to Practice Problems
1.
5,924
17. 8 miles
2.
1,326
18. 22,192 members
3.
557,274
19. 541,320
4.
709
20. 190 R8
5.
23
75
21. 77
19
24
6.
11
48
22. 15
5
24
7.
24
8.
1
9.
32.644
3
4
3
7
23. 2
24.
4
pounds
9
76
sections
119
25. 1.32 miles
10. 42.67
26. 143.2 kg
11. 2.5785
27. $12.87
12. 2.84
28. 65.9967
13. 0.03 miles
29. $147.80
14. 0.48
30.
65
13
or
100
20
15. 112.32
31. 4.32
16. $12.68
32. $28.68
9
PART II – Introductory Algebra
Type
1.
Problem
Adding integers with like signs
-14 + (-19)
Method: To add integers with the same sign –
(1) add the absolute value of the addends,
(2) attach to the result the common sign of the addends.
(1) |−14| = 14
19
|−19| = 19
+ 14
33
(2) Since both addends are negative, the result is negative.
-14 + (-19) = -33
2.
Add integers with unlike signs
23 + (-35)
Method: To add integers with different signs:
(1) subtract these absolute values,
(2) give the result the same sign as the addend with the greater absolute value
23
+
(-35)
positive
negative
|23|
(1)
= 23
35
|−35| = 35
- 23
12
(2) Since |−35| > |23|, the result is the same sign as -35; negative
23 + (-35) = -12
3.
Finding the sum of more than two
integers
Method:
4.
-6 + (-8) + 13 + (-4)
-6 + (-8) + 13 + (-4)
=
-14 + 13 + (-4)
=
-1
+ (-4)
=
-5
Subtracting Integers
-17 – (-20)
Method: To subtract two integers,
(1) change the subtraction sign to addition,
(2) find the opposite of the second number, and
(3) add:
-17 – (-20)
= -17 + 20
= 3
10
Type
5.
Problem
Subtraction occurring several times
Method:
18 – 24 – 7
18 – 24 – 7
= 18 + (-24) – 7
=
-6 – 7
= -6 + (-7)
=
6.
-13
Multiplying or dividing integers with
unlike signs.
(1) 6 (-13)
(2) 120 ÷ (-8)
Method: If the signs are different, the result is negative.
(1) 6 (-13) = -78
(2) 120 ÷ (-8) = -15
7.
Multiplying or dividing integers with
like signs.
(1) -8 (-7)
(2) -162 ÷ (-9)
Method: If the signs are the same, the result is positive.
(1) -8 (-7) = 56
(2) -162 ÷ (-9) = 18
8.
Multiplying more than two factors.
(1) -5 (6) (-7) (-2)
Method: (1) -5 (6) (-7) (-2)
= -30 (-7) (-2)
=
=
9.
210 (-2)
-420
Solving two-step equations.
Method: Solve: 7x – 8 = -29
7x – 8 + 8 = -29 + 8
7x = -21
7x − 21
=
7
7
x = -3
Solve: 7x – 8 = -29
Check: 7 (-3) – 8 = -29
-21 - 8 = -29
-29 = -29
Therefore; -3 is the solution.
11
10.
Solving equations of the form
ax + b = cx + d
Method:
11.
Solving equations containing
parentheses
Method:
12.
Solve: 6a – (3a – 4) = 5 (a + 2)
6a – 3a + 4 = 5a + 10
3a + 4= 5a + 10
3a – 5a + 4= 5a -5a + 10
-2a + 4= 10
-2a + 4 – 4 = 10 – 4
-2a = 6
-2a = 6
-2 -2
a = -3
Adding polynomials
Method:
13.
Solve: 4x - 5 = 6x + 11
4x – 6x – 5 = 6x – 6x +11
-2x – 5 = 11
-2x – 5 + 5 = 11 + 5
-2x = 16
-2x = 16
-2
-2
x = -8
Subtracting polynomials
=
=
=
=
Solve: 6a – (3a - 4) = 5 (a + 2)
Check; the solution is -3
(12x2 – 17x + 4) + (9x2 + 13x – 5)
(3x2 + 2x – 2) – (5x2 – 8x – 6)
(3x2 + 2x – 2) – (5x2 – 8x – 6)
(3x2 + 2x – 2) + - (5x2 – 8x – 6)
3x2 + 2x – 2 – 5x2 + 8x + 6
3x2 – 5x + 2x + 8x - 2 + 6
-2x2 + 10x + 4
Multiplying monomials
Method:
Check; the solution is -8
(12x2 – 17x + 4) + (9x2 + 13x – 5)
= 12x2 – 17x + 4 + 9x2 + 13x – 5
= 12x2 + 9x2 – 17x + 13x + 4 – 5
= 21x2 – 4x – 1
Method:
14.
Solve: 4x – 5 = 6x +11
(-3x6y2z) (-5x2y3z)
= (-3) (-5) (x6  x2) (y2  y3) (z  z)
= 15x6+2y2+3z1+1
= 15x8y5z2
(-3x6y2z) (-5x2y3z)
12
15.
Method:
16.
(-2x4y2z)3
= (-2)3 (x4)3 (y2)3 (z1)3
= -8x43 y23 z13
= -8 x12 y6 z3
Multiplying a monomial by a
polynomial.
Method:
17.
(-2x4y2z)3
Simplifying powers of monomials
4x3 (3x2 – 2x + 6)
4x3 (3x2 – 2x + 6)
= 4x3 (3x2) + 4x3 (-2x) + (4x3) (6)
= 12x5 – 8x4 + 24x3
(2x – 3) (x2 – 3x – 5)
Multiplying two polynomials
x2 – 3x – 5
Method:
2x – 3
-3x2 + 9x + 15
2x3 – 6x2 – 10x
2x3 – 9x2 – x + 15
18.
Multiplying two binomials
Method:
(5a – 2b) (3a + 4b)
To find the product of two binomials use the “FOIL” method:
(5a – 2b) (3a + 4b)
product of the
first terms
(5a) (3a)
+
product of the
outer terms
(5a) (4b)
+
product of the
inner terms
(-2b) (3a)
+
= 15a2 + 20ab – 6ab – 8b2
= 15a2 + 14ab – 8b2
19.
Finding the square of the binomial
Method:
(1)
(2)
(3a – 5)2
(3a + 5)2
(1)
(3a – 5)2
=
=
=
(3a – 5) (3a – 5)
9a2 – 15a – 15a + 25
9a2 – 30a + 25
(2)
(3a + 5)2
=
=
=
(3a + 5) (3a + 5)
9a2 + 15a + 15a +25
9a2 + 30a + 25
product of the
last terms
(-2b) (4b)
13
20.
Dividing a polynomial by a monomial
Method:
= 12b7
3b3
+
36b5
3b3
12b7 + 36b5 – 3b3
3b3
-3b3
3b3
+
= 4b4 + 12b2 - 1
21.
Factoring difference of 2 squares
Method:
Factor: 4x2 – 9
Find the square roots of both terms. Since 9 is negative, one parenthesis will have a
positive (+) in the middle and the other a negative (-).
√4𝑥 2 = 2x
(2x + 3) (2x – 3)
√9 = 3
The result can be checked by multiplying (2x + 3) (2x – 3)
4x2 – 6x + 6x – 9 = 4x2 – 9
22.
Factor: 6ax2 – 3a2 x3 – 9a2 x2
Common factoring
Method:
Find the greatest common factor of all the terms in the polynomial. Divide each term
by the greatest common factor.
3ax2 divides into all 3 terms
6ax2 - 3a2 x3
3ax2
3ax2
2 - ax - 3a
3ax2 (2 – ax – 3a)
-
9a2 x2 =
3ax2
The result can be checked by multiplying.
23.
Common factoring
Method:
Factor: 3x (a + b) – 2 (a + b)
The parenthesis in the original problem groups together a + b so that they act like a
monomial. Divide the terms by the common binomial.
3x (a + b) - 2 (a + b)
a + b
a + b
(a + b) (3x – 2)
24.
Factoring by grouping
Method:
Factor: 2x2 – 2xy + 3x – 3y
1. Group the first 2 terms and the last 2 terms together
(2x2 – 2xy) + (3x – y)
2. Common factor each group of terms
2x (x – y) + 3 (x – y)
3. Common factor the common binomial
(x – y) (2x + 3). (Refer to problem 23.)
14
25.
Factoring a perfect square trinomial
Method:
26.
27.
A perfect square trinomial is a trinomial where
(1) first and last terms are squares
(2) middle term – 2 times square root of first term times square root of last term.
To factor – find the square roots of the first and last terms. The sign inside the
parenthesis is the same as the sign in front of the middle term.
√9𝑥 2 = 3x
√4 = 2
2
(3x + 2)
Factoring a trinomial
Method:
Factor: 6x2 – x – 12
This trinomial can be factored using trial and error.
(1) find two factors of the first term
6x2 = 3x  2x
(2) find two factors of the last term
-12 = -6  2
(3x – 6) (2x + 2)
Check your guess through multiplication
(3x – 6) (2x + 2)
6x2 + 6x – 12x – 12 =
6x2 – 6x – 12 Incorrect.
If your first guess is unsuccessful, try again.
Second try: 6x2 = 3x  2x
(3x
) (2x )
-12 = 4  -3
(3x + 4) (2x – 3)
Check: (3x + 4) (2x – 3) =
6x2 – 9x + 8x – 12 =
6x2 – x – 12 Correct.
Reducing a rational algebraic
expression
Method:
Factor: 9x2 + 12x + 4
Reduce:
2x2 – 7x + 3
3x2 – 7x – 6
This fraction is reduced by factoring the numerator and denominator.
2x2 – 7x + 3 = (2x – 1) (x – 3)
3x2 – 7x – 6
(x – 3) (3x + 2)
Common binomial factors are divided out
(2x – 1) (x – 3) = 2x – 1
(x – 3) (3x +2)
3x + 2
28.
Factoring a common factor with
factoring a trinomial
Method:
Factor: 45x2 + 6x – 24
In any type of factoring, common factoring should be done first. 3(15x2 + 2x – 8).
Then, use an appropriate factoring method to factor the result. 3(5x + 4) (3x – 2).
Trial and error should be used.
15
PART 2 – Introductory Algebra
Practice Problem
Problem Type
1.
14 – 26
#4
2.
-25 + (-59)
#1
3.
13 + (-22) + 9 + (-6)
#3
4.
13 – (-23)
#4
5.
#6
6.
135 ÷ (-9)
#7
7.
-42 ÷ (-3)
-6 – (-19) – (-9)
#5
8.
-5 (23)
#6
9.
#7
10.
-84 ÷ (-7)
(-7) (8) (-2) (-3)
#8
11.
-12 – (-19)
#4
12.
Solve: -8x + 3 = -29
#9
13.
Solve: -5a – 3 = 2a + 18
#10
14.
Solve: 5 – (9 – 6x) = 2 (x – 1)
#11
15.
(8a4 b3) (-7ab5)
#14
16.
(-8y2 + y – 15) – (6y2 + 2y + 7)
#13
17.
15b4 – 5b2 + 10b
5b
#20
18.
(4a3b4c)2
#15
19.
-2x (4x2 + 7x – 9)
#16
20.
(4x – 3)2
#19
21.
(3x3 – 2x2 – 4x) + (8x2 – 8x +7)
#11
16
Practice Problem
Problem Type
22.
(4x + 3)2
#19
23.
(3x2 + 2x – 5) (4x – 3)
#17
24.
(2x – 3) (4x +7)
#18
25.
Factor: 5x2y – 10x3y2 – 15x2y5
#22
26.
Factor: 15x2 – 35x + 6x – 14
#24
27.
Factor: 6x2 – 25x – 9
#26
28.
Factor: 49x2 – 28x + 4
#25
29.
Reduce:
x2 – 25
2x + 7x – 15
#27
2
30.
Factor: 6x2 – 2x – 9x + 3
#24
31.
Factor: 3y(x – 2) + 5(x – 2)
#23
32.
Factor: 10x2 – 55x + 25
#28
33.
Factor: 64y2 – 25
#21
34.
Factor: 25x2 + 40x + 16
#25
35.
Factor: 9 – y2
#21
36.
Factor: 27x2 – 18x + 3
#28
37.
Reduce:
38.
Factor: 14b2c3 – 12b4c5d + 6b2c2d2
#22
39.
Factor: 21x2 + 23xy + 6y2
#26
40.
Factor: 5(y – x) – 3x (y – x)
#23
8x2 – 26x + 15
12x2 – 5x – 3
#27
17
ANSWER KEY
1.
-12
21.
3x3 + 6x2 – 12x + 7
2.
-84
22.
16x2 + 24x + 9
3.
-6
23.
12x3 – x2 – 26x + 15
4.
36
24.
8x2 + 2x – 21
5.
-15
25.
5x2y (1 – 2xy – 3y4)
6.
14
26.
(5x + 2) (3x – 7)
7.
22
27.
(3x + 1) (2x – 9)
8.
-115
28.
(7x – 2)2
9.
12
29.
x–5
2x - 3
10.
-336
30.
(2x – 3) (3x – 1)
11.
7
31.
(x – 2) (3y + 5)
12.
4=x
32.
5 (2x – 1) (x – 5)
13.
-3 = a
33.
(8y + 5) (8y – 5)
14.
½=x
34.
(5x + 4)2
15.
-56 a5b8
35.
(3 – y) (3 + y)
16.
-14y2 – y – 22
36.
3 (3x – 1)2
17.
3b3 – b + 2
37.
18.
16a6b8c2
38.
2b2c2 (7c – 6b2c3d + 3d2)
19.
-8x3 – 14x2 + 18x
39.
(3x + 2y) (7x + 3y)
20.
16x2 – 24x + 9
40.
(y – x) (5 – 3x)
2x – 5
3x + 1
18
Study Guide for California University of Pennsylvania Math
Placement C and D Tests
For Math Placement Test C, take any college algebra book and do the chapter review problems
for chapters that include the following;
•
•
•
•
•
•
Special products
Factoring trinomials
Simplification of radicals
Positive, negative, and fractional exponents
Complex numbers
Solution of systems of linear equations
For Math Placement Test D, take any college trigonometry book and do the chapter review
problems for chapters that include the following;
•
•
•
•
•
•
The trig functions
Trig functions of an acute angle
Trig identities
Trig functions of any special angle w/o a calculator
Properties of logarithms
Inverse trig functions