Algebra Topics COMPASS Review – revised Summer 2013
You will be allowed to use a calculator on the COMPASS test. Acceptable calculators are basic calculators, scientific
calculators, and approved graphing calculators. For more information, see the JCCC Testing Services website at
http://www.jccc.edu/testing/ or call 913-469-4439.
Work out these problems and select the correct answer.
x 9
1. What are the solutions to the equation
?

4 2x
b.  4.5
c.  18
a.  2 3
d.  3 2
2. The solution of the equation 5x  2 x  11 falls between what two consecutive integers?
a. 0 and 1
b. 1 and 2
c. 2 and 3
d. 3 and 4
3. If
e.  6
e. 4 and 5
1
is subtracted from 6 times the reciprocal of a number x , the result is 12. Which of the following equations
3
determines the correct value of x ?
1
1 1
a.  6 x  12
b. 6     12
3
 x 3
4. Evaluate 3x2  5x  6 when x  4
a. 34
b. 50
4
7
and x 
5
10
17
b. 
10
 1  1
c.     12
 6x  3
d.
1  1 
    12
3  6x 
c. 38
d. 74
e. 6 x 
1
 12
3
e. 22
5. Evaluate 3y  x when y 
a. 
31
10
c. 
3
2
d. 
19
5
e. 
19
10
6. What is the value of the expression 2 y  xy  8 when x  0 and y  5 ?
d. 30
a. 23
b. 18
c. 8
e. 2
7. If x  5 and y  4 what is the value of 3 y  4 x 2 ?
a. 79
b. 28
c. 88
e. 52
1
8. If x  3  9 , then x  ?
3
1
1
a. 6
b. 12
3
3
9. What is the value of the expression
a.
5
32
b.
5
11
5 x 2  kx  3
 5 x  3 then k  ?
x 1
a. 16
b. 8
c. 6
2
3
d. 112
d. 5
2
3
e. 5
1
3
x 2  xy
when x  5 and y  3 ?
x  2 xy  y 2
2
5
7
c. 10
d. 
c. 2
d. 15
e.
5
2
10. If
e. 1
11. If 2x + 3, 4x – 2, and 3x +11 represent 3 test scores, which of the following algebraic expressions represents the
average (arithmetic mean) of the three test scores?
a. 3x + 4
b. 80
d. 92
c. 24 x  66
e. 9 x  12
x 2  x  42
?
x 2  36
x7
c.
x6
12. For all x  6 , which of the following is equivalent to
a.
x7
x6
x6
x6
b.
d.
x7
x6
e.
x7
x6
13. Given the following graph, how long is AB ?
d. 3 2
a. 6
e. 3
b. 3 5
B
c. 6
A
14. What is the distance between the x and y intercepts of the line y  4 x  8 ?
b. 8
a. 2 17
c. 10
d. 2 5

15. Find the distance between  4,0  and 2, 5
a.
16. If
10
29
b.
41
e. 4 5

61
c.
d. 9
e. 3
d. 0
e. 6
d. x  2
e. x  4
d. 57 x  64
e. 9 x2  48x  64
 5 then x 2  ?
x  10
1
a.
b. 2
c. 14
2
17. For all x ,  7 x3  2 x 2  x  4    5x3  4 x 2  3x  4  =
2
C
a. 2 x  x 2  3x 
d. 2 x 2  x 4  6 x 2  2 
b. 2  x3  x 2  2 x  4 
e. 2  x3  3x 2  x  4 
c. 2 x  x 2  3x  1
18. Which of the following is a factor of 3x2  10 x  8 ?
a. 3x  2
b. 3x  4
c. 3x  2
19. Which of the following is equivalent to  3x  8 ?
2
a.
20. Add
a.
9 x2  64
50 5 5

=?
3
6
50  5 5
6
b.
10 2  5 5
6
21. What is the solution of the equation
a. x  1
c. 9 x2  48x  64
b. 6 x2  16
b. x 
7
4
c.
15 7
6
d.
5 55
18
e.
5 2 5 5
18
1 5
1
  x ?
2 4
4
c. x  4
2
d. x  1
e. x  
3
4
22. A car salesman received a weekly salary of W dollars plus a 6% commission on his total sales S. Which expression
best describes his weekly pay?
a. W + S
b. (W + S) (0.6)
c. 0.06W + S
d. W + 0.06S
e. W + 6S
23. Which of following is equivalent to  y  5  2 y 2  2 y  3 ?
a. 2 y3  8 y 2  7 y  15
b. 2 y 2  3 y  2
c. 2 y3  10 y  3
d. 2 y 2  y  8
c. x  3
d. x  
e. 2 y3  2 y 2  15
24. Solve for x: 2 x  11  17
a. x  3
b. x  3
5
2
e. x  8
25. John can complete a job alone in 2 hours, and it takes Tim 3 hours to do the same job alone. How long will it take
them if they work together?
b. 1
a. 1 hour
1
hours
d. 2
c. 5 hours
5
1
hours
2
26. What is the value of 4 x 3x  5 x3  2 x 27 x when x  3 ?
b. 75
a. 75 3
c. 18  15 3
2
1
 2
x  x 6 x 9
1
2
a.
 x  3 x  2  x  3 b. x  x  12
5
hour
6
d. 90  15 3
27. The graph of the equation y  2 x  1 will lie in which quadrants of the coordinate plane?
a. I and II
b. I and IV
c. I, II, and III
d. I, II, and IV
28. Subtract:
e.
e. 147 3
e. I, II, III, and IV
2
x2  6 x  9
x2  4

x2  x  2 x2  2x  3
 x  3 x  2 
a.
b.  x  3 x  2 
2
 x  1
c.
 x  4
 x  3 x  2  x  3
d.
1
 x  x  6 x2  9
2
e.
1
x3
29. Multiply:
30. Solve this equation:
a. x  1
 x  3
 x  1 x  3
2
c.
d. 5
e. 1
2  3x  1  3 x  3  4  2 x  1  2
b. x  
13
5
c. x  1
2
x4 ?
3
2
c. y   x  4
3
d. x 
3
5
e. x  0
d. y 
3
x2
2
e. y  4 x 
31. Which of the following lines is perpendicular to y 
3
a. y   x  1
2
b. y 
2
x5
3
2
3
32. How many ounces of a 50% alcohol solution must be mixed with 20 ounces of a 20% solution, to make a 40%
solution?
a. 40 ounces
b. 20 ounces
c. 50 ounces
d. 25 ounces
e. 30 ounces
33. Multiply and write your answer in a + bi form:  2  3i  4  5i  .
a. 8  13i
b. 23  2i
c. 10  12i
3
d. 8  15i
e. 8  15i
34. What are the solutions of the equation: x2  x  12 ?
a. x  4, x  3
b. x  4, x  3
c. x  6, x  2
d. x  6, x  2
e. x  0, x  1
35. What are the solutions of the equation: 2 x2  9 x  10  0 ?
19
5
b. x  1, 5
a. x  10,
c. x   ,1
2
2
5
d. x   , 2
2
5
e. x  , 2
2
36. What is the sum of the solutions to x2  7 x  12  0 ?
a. 12
b. 12
c. 8
d. 7
e. 7
d  0,5 
e.
e. 10  6 2
37. The graph of 5x  3 y  15 has an x -intercept of:
a.
 3,0 
38. Multiply:
b.
 2  8 5 
a. 2
39. Solve for x:
18
c.
 3,5

b. 12  10 2
c. 10  12
d. 2  4 2
b. x  z  y
c. x   yz
d. x 
c. 8x  4
d. 5x  3
e. 5x2  3x
c. 3abc
d. 9ab2 c
e. 3a 2b3c 3c
d. 5
e. 20
10 x 2  6 x
?
2x
b. 5x2  6 x
40. For all x  0 ,
a. 5x  6
b. 9a 2b3c
42. If x2  10 x  25 , which of the following is a value for x 2  x ?
a. 5
b. 20
c. 30
a.
yz
yz
e. x 
yz
yz
27a 4b6 c3
a. 3ab2c a
43. Simplify:
 3,0 
1 1 1
 
x y z
a. x  yz
41. Simplify:
 0,3
 5a

1 3 2
b
5
ab
10
a 3b 7
for all a  0, b  0
b.
1
25a 3b7
c.
10b5
a3
d.
25
a 3b 7
e. 25a 7b5
44. A car rental agency charges $15 per day plus $0.30 per mile. Which of the following is an expression for the total
charges in dollars of renting a car for one day and driving m miles
a. 15  0.30m
b. 15m  0.30
c. 15.30m
d. 15  3m
e. 0.30m
45. At what  x, y  coordinates do the following lines intersect?
2x  y  1
3x  2 y  12
a. x  1, y  3
b. x  2, y  3
c. x  5, y  9
4
d. x  2, y  9
e. x  1, y  1
46. Which of the following equations represents the line graphed?
a. 2 x  3 y  6 d. 2 x  3 y  6
b. 3x  2 y  6 e. 2 x  3 y  6
c. 3x  2 y  6
x2  2x  3
x2
47. Simplify:
x 1
x2  4
a.
x
2
 3  x  4 
b.
 x  3  x 2  4 
c.
1
 x  2
d.
2
 x  3 x  2
e.
 x  2 x  3
48. John and Susan are reading final exam research papers. John can read at a constant rate of x pages per day. Susan
reads 8 pages a day more than John. If it takes John 16 full days to read all the final exam research papers and it takes
Susan 12 full days, how many pages a day is John reading?
a. 20 pages
b. 4 pages
c. 16 pages
d. 32 pages
e. 24 pages
49. Ten times the reciprocal of a number is 3 less than the original number. Which equation can be used to solve for the
original number?
a. 10 
1
 3 x
x
b. 10 
1
 x3
x
c. 10  x  3 
1
x
50. If r  3 and l  5 , which of the following is equivalent to  rl   r 2 ?
a. 21
b. 24
c. 135
Answers to the Algebra Topics COMPASS Review
1. d
14. a
27.
2. d
15. e
28.
3. b
16. c
29.
4. d
17. c
30.
5. a
18. c
31.
6. b
19. e
32.
7. c
20. b
33.
8. d
21. a
34.
9. e
22. d
35.
10. b
23. a
36.
11. a
24. a
37.
12. e
25. b
38.
13. b
26. d
d
c
a
c
a
a
b
a
d
e
a
d
d. 10  x  3  x
1
e. 10   3  x
x
d. 17
e. 14
39.
40.
41.
42.
43.
44.
45.
46.
47.
48.
49.
50.
5
e
d
e
c
b
a
b
d
e
e
b
b
Solutions to the Algebra Topics COMPASS Review
x 9
1. What are the solutions to the equation
?

4 2x
Cross multiply to obtain:
2 x 2  36
2 x 2 36

2
2
2
x 18
x   18
x 92
x 3 2
2. Solve the linear equation.
5 x  2 x  11
5 x  2 x  11
3 x  11
11
2
x 3
3
3
So the solution falls in between 3 and 4.
1
1 1
3. “ is subtracted from 6 times the reciprocal of a number x ” is written as 6   
3
 x 3
“the result is 12” is written as =12
1 1
Putting this together gives 6     12
 x 3
4. Given x  4 , substitute 4 in place of x .
3(4) 2  5(4)  6 
3(16)  5(4)  6  Use order of operations to simplify
48  20  6  74
4
7
and x  , substitute these values in place of x and y :
5
10
 4   7 
3     
 5   10 
12 7
(Find common denominator)
 
5 10
24 7 31
 
10 10 10
5. Given y 
6. Given x  0 and y  5 , substitute these values in place of x and y :
2  5    0  5   8 
10  0  8  18
(Use order of operations to simplify)
6
7. Given x  5 and y  4 , substitute these values in place of x and y :
3  4   4  5  
2
12  4  25  
(Use order of operations to simplify)
12  100  88
8. Solve the linear equation:
1
x3 9
3
1
x 93
3
3
1 
3
x  8  3  Rewriting 9 as 8 
3
3 
3
2
x5
3
9. Given x  5 and y  3 substitute these values in place of x and y :
 5
 5
2
2
  5 3
 2  5 3   3
2

25  15
10 5
 
25  30  9 4 2
10. When we cross multiply, we obtain 5x2  kx  3   5x  3 x  1 .
Multiplying the right side and combining like terms gives: 5x2  kx  3  5x2  8x  3
Therefore k  8 .
11. The average or (arithmetic mean) is found by adding up the test scores and dividing by 3 since there are three tests.
 2 x  3   4 x  2    3x  11  9 x  12  3 3x  4   3x  4
3
3
3
12. Factor both the numerator and the denominator.
 x  7  x  6   x  7 
x 2  x  42

2
x  36
 x  6  x  6   x  6 
13. One way to find the length uses the distance formula. The distance formula is d 
 x2  x1 
2
  y2  y1 
2
Point A has coordinates  4,1 and Point B has coordinates  2, 4  . Substituting these values into the formula gives:
 2  4   4  1
2

2

 6
14. The distance formula is d 
2
  3  36  9  45  9  5  3 5
2
 x2  x1 
2
  y2  y1 
2
To find the x  intercept , substitute y  0 into y  4 x  8 .
0  4x  8  8  4x  2  x . The x  intercept is (2,0)
To find the y  intercept , substitute x  0 into y  4 x  8 .
y  4(0)  8  y  0  8  y  8 . The y -intercept is (0,-8)
Substituting these values into the formula gives:
 0  2
2
  8  0  
2
 2
2
  8  4  64  68  4 17  2 17
2
7
15. Substitute the values given into the distance formula.
 2  4
2


5 0

2

 2 
2

 5
2
 45  9 3
16. Cross multiplying gives 5 x2  10  10  x2  10  2
Squaring both sides gives x2  10  4  x2  14
17. To simplify, you need to distribute the minus sign first and then combine like terms.
 7 x 3  2 x 2  x  4    5 x 3  4 x 2  3x  4  =
7 x3  2 x 2  x  4  5 x3  4 x 2  3 x  4 =
2 x3  6 x 2  2 x =
2 x  x 2  3x  1 by factoring.
18. To factor 3x2  10 x  8 ,
Method 1
Start with 3x and x at the beginning of the parentheses:  3x
 x  .
nd
Try different pairs of factors of -8 for the 2 number in the parentheses. Then decide which will FOIL out correctly.
The pairs of factors are:
-1 and 8
1 and -8
-2 and 4
2 and -4
Of those choices, “-2 and 4” give the correct factorization:.  3x  2  x  4 
Therefore, 3x – 2 is a factor.
Method 2
Multiply the a and the c: 3 times -8 = -24.
Listing the pairs of factors of -24 results in:
-1 and 24,
1 and -24,
-2 and 12,
2 and -12,
-3 and 8,
3 and -8,
-4 and 6,
-4 and 6
Choose the pair that adds to +10: -2 and 12.
Rewrite the quadratic expression by splitting up the b term into the pair you found above:
3x2  10 x  8
3x2  2 x  12 x  8 (Now factor by grouping)
x(3x  2)  4  3x  2 
Factor out  3x  2  to yield:  3x  2  x  4 
Therefore,
 3x  2 is a factor
19. To square a binomial, multiply it by itself.
 3x  8 
2
  3x  8 3x  8  9 x2  24 x  24 x  64  9 x 2  48 x  64
20. Simplify each of the radicals, find a common denominator, and combine.
50 5 5
25  2 5 5
5 2 5 5 10 2  5 5






3
6
3
6
3
6
6
8
21. The solution is found by multiplying every term by the Least Common Denominator and then solving.
LCD = 4
1 5
1
 x
2 4
4
1
1 5
4   x  
4
2 4
1 5
1
4   4   4 x  4 
2 4
4
2  5  4x  1
3  4 x  1
4  4 x
1  x
22. The salesman will receive W dollars, plus whatever he makes on commission.
His commission is (0.06)(S).
Adding these together results in: W + 0.06S.
23. To find the one that is equivalent to ( y  5)(2 y 2  2 y  3) :
Multiply (distribute) y to each of the terms in 2 y 2  2 y  3 , and
then multiply (distribute) 5 to each of the terms in 2 y 2  2 y  3 .
2 y3  2 y 2  3 y  10 y 2  10 y  15 
2 y3  8 y 2  7 y  15
24. To solve the inequality, isolate x .
2 x  11  17
2 x
 17-11
 2x  6
(Note that dividing by a negative value reverses the inequality sign)
2 x 6

2 2
x  3
9
1
of the job done in 1 hour.
2
1
Since Tim can complete the job in 3 hours, he would get of the job done in 1 hour.
3
1
Let x stand for the time it takes them to complete the job together. Then
of the job could get finished in 1 hour.
x
1 1 1
The equation would then be  
2 3 x
Multiply by the least common denominator 6x on both sides of the equation.
 1 1
1
6x     6x  
 2 3
 x
1
1
1
6x    6x    6x  
2
 3
 x
3x  2 x  6
5x  6
6
1
x   1 hours to complete the job together.
5
5
25. Since John can complete the job in 2 hours, he would get
26. Given x  3 , substitute 3 in place of x
4  3 3  3  5
 3
3
 2  3 27  3  
12 9  5 27  6 81 
12  3  5 9  3  6  9 
36  5  3 3  54 
90  15 3
2
27. y  2 x  1 has a y-intercept of 1, and a slope of  , so it is graphed as indicated below.
1
10
5
0
-10
-5
0
5
10
-5
-10
Quadrant I is the upper right, Quadrant II is the upper left, Quadrant III is the lower left, and Quadrant IV is the lower
right. Therefore, the graph lies in Quadrants I, II, and IV.
10
28. To subtract
2
1
, you must have common denominators.
 2
x  x6 x 9
2
Factor all denominators to find the least common denominator
2
1
Note that the LCD is  x  3 x  2 x  3 .

( x  3)( x  2) ( x  3)( x  3)
Rewrite each fraction with the new LCD by multiplying each fraction by term(s) that are in the LCD but not in the
original denominator.
2
1
 x 3
 x2
 



 x  3  ( x  3)( x  2)  x  2  ( x  3)( x  3)
(2 x  6)  ( x  2)
x4

( x  3)( x  2)( x  3) ( x  3)( x  2)( x  3)
( x  3)( x  3) ( x  2)( x  2)
.

( x  2)( x  1) ( x  3)( x  1)
Cancel like factors that appear in both the numerator and the denominator leaving you with:
( x  3) ( x  2)  x  3 x  2 


2
( x  1) ( x  1)
 x  1
29. Factor all the numerators and denominators to obtain
30. To solve, isolate x .
2  3 x  1  3  x  3  4  2 x  1  2
6 x  2  3x  9  8 x  4  2
3x  11  8 x  6
5  5x
1 x
2
31. The slope of the given line is . A line perpendicular to the given line will have a slope that is the negative
3
3
3
3
reciprocal,  . The only line with a slope of  is y   x  1 .
2
2
2
32. Let x  the amount of 50% solution. Let x  20 = the amount of the final 40% solution.
 0.50  x   0.20  20   0.40  x  20 
0.50 x  4  0.40 x  8
0.10 x  4
0.10 x
4

0.10 0.10
x  40 ounces of the 50% soluition
33. Multiply  2  3i  4  5i  (Note that i 2  1 )
8  10i  12i  15i 2  8  2i  15  1  8  2i  15  23  2i
11
34. Solve x2  x  12
x 2  x  12  0
 x  4  x  3  0
x  4  0 or x  3  0
x  4, 3
35. Solve 2 x2  9 x  10  0 (Same steps as #18 for the factoring part)
 2 x  5 x  2   0
2 x  5  0 or x  2  0
5
x   , 2
2
36. Solve x2  7 x  12  0
 x  3 x  4   0
x  3  0 or x  4  0
x  3, 4
The sum of the solutions, 3 and 4, is 7.
37. At the x -intercept, the y value will be zero.
5 x  3  0   15
5 x  15
x3
The ordered pair is  3,0  , since the x value is 3, and the y value is 0.
38. To multiply (2  8)(5  18) , simplify the radicals first, then distribute.
(2  2 2)(5  3 2) = 10  6 2  10 2  6  2 = 10  4 2  12 = 2  4 2
39. To solve
1 1 1
  ,start by multiplying all terms by the LCD to eliminate fractions. The LCD is xyz .
x y z
1 1
1
xyz     xyz  
z
x y
yz  xz  xy
xz  xy   yz
x  z  y    yz
x
(Factor out x and then divide to isolate x )
 yz
yz
or x 
zy
yz
40. When dividing by a monomial, separate each term over the denominator and reduce.
10 x 2  6 x 10 x 2 6 x


 5x  3
2x
2x
2x
12
41. To simplify, take out anything that has a nice square root
32  3  a 2  a 2  b2  b2  b2  c 2  c  3  a  a  b  b  b  c 3c  3a 2 b3 c 3c
42. First, solve for x .
x2  10 x  25  x2  10 x  25  0   x  5 x  5  0  x  5
Since x  5 , we will substitute -5 for x
x2  x   5   5  25  5  30
2
(5a 1b3 )2
a 5b
a2
1
 2 5 6 
5 a bb
25a3b7
43. Use rules of exponents to simplify
(5a 1b 3 ) 2
52 a 2b6
=
a 5b
a 5b
44. Let m represent the number of miles. Since the car agency charges $0.30 per mile, part of the cost would be $0.30m.
They also charge the renter $15 per day, so you need to add to get the total charges. 15  0.30m
45. One method is the addition method: Multiply the top equation by 2 with the intent of eliminating the y when this
equation is added to second equation.
4x  2 y  2
2(2 x  y )  1(2)
(add the two equations)

3x  2 y  12
3x  2 y  12
7 x  14
x2
Since x  2 substitute this x-value into one of the two original equations, to find y .
2  2  y  1
4  y 1
y  3
The solution is: x  2, y  3 .
13
46. Method 1
The y -intercept is -2. Count up and to the right from that point, (0, -2) to another point, such as (3,0) to find the
2
2
slope. The slope is . Substitute these two values into the slope-intercept form, y  mx  b , to get y  x  2 .
3
3
Put this equation into standard form, since the choices are all in standard form ( Ax  By  C )
2

3( y )  3  x  2 
3


3y  2x  6
2 x  3 y  6
2x  3y  6
Method 2
Use the two intercepts (3,0), and (0,-2), to find the slope, using m 
y2  y1 2  0 2


x2  x1
03 3
Substitute this slope and one of the ordered pairs into the in the point-slope form, y  y1  m  x  x1  , to get
y0
2
 x  3
3
2
x2
3
Now change the form of the answer to standard form as in Method 1 above.
y
47. To simplify, factor all numerators and denominators. Then take the reciprocal of the second fraction, and change
division to multiplication.
x2  2 x  3
( x  1)( x  3)
( x  1)( x  3) ( x  2)( x  2)
x2
x2



 ( x  3)( x  2)
x 1
x 1
( x  2)
( x  1)
x2  4
( x  2)( x  2)
48. Let x be the number of pages John reads per day. Since it takes John 16 full days, his total number of pages would be
16x .
Susan reads 8 pages more a day then John so her rate would be x  8 . Since it takes Susan 12 full days, her total
number of pages would be 12  x  8 .
Since they are reading the same research papers, the total number of pages will be equal, so
16 x  12  x  8  16 x  12 x  96  4 x  96  x  24
John reads 24 pages a day.
1
= the reciprocal of the original number.
x
1
“Ten times the reciprocal of a number” will be written as 10  .
x
“3 less than the original number” will be written as x  3 .
1
The equation becomes 10   x  3 .
x
49. Let x = the original number. Then,
50. Substitute in for the given variables, r  3 and l  5 and then simplify
 (3)  5    3   15    9   15  9  24
2
14