Download Math 115 Practice Test rubrics - Los Angeles Southwest College

Los Angeles Southwest College
Mathematics Department
Math 115 – Common Final Exam
Practice TEST (rubrics)
This Final Exam consists of 36 problems with a total of 150 points. Show any necessary work neatly
and clearly in the space provided. Please circle or box your final answer.
Part A: No partial credit will be given to the problems 1 through 10
2
1
Simplify: 26  (37)  13
2
Decide whether the statement is true or false:  | 29 |  | 18 | True
For the following word phrase write an expression using x as the variable and simplify.
3
(2 pts)
"Thirty two subtracted from the difference between twice a number and twelve "
(2 pts)
(2 pts)
2 x  44
4
Evaluate: 4 x 2 y  5 xy 2 , if x  3 and y  2
5
6
Combine like terms: 5m 2  8m  3m 2  12m  15
Graph 2  x  3 on a number line.
7
Find the slope of the line through (2, 6) and ( 4, 2) and write answer in the simplest form
132
2m 2  4m  15
(2 pts)
(2 pts)
(2 pts)
(2 pts)
4
8
Multiply: 2 y 2 (4 y 3  3 y 2  7)
8 y 5  6 y 4  14 y 2
(2 pts)
9
Factor: 9 y 2  64
(3 y  8)(3 y  8)
(2 pts)
10
Find the LCM of 18 x 4 y 3 and 24 x5 y 2
72 x 5 y 3 or 23  32 x 5 y 3
(2 pts)
Part B: Partial credit may be given to the problems 11 through 36. Answers without supporting
work will be given no credit. Please circle or box your final answer.
11
Simplify:
(4)  32  2  7
(3)  6  13
(3 pts)
a) 32  9
1pt
b)  4  9  36 


(3)  6  18
36  14 50

 10
18  13
5
2  7  14
c)
12
13
1pt
1pt
Solve the equation: 3( x  4)  5( x  2)  14
(4 pts)
3x 12  5x 10  14
2x  22  14
2 x  8
1pt
1pt
1pt
2 x 8

2 2
1pt
x  4
Solve the following word problem by drawing a picture, defining the variable(s),
setting up the equation(s), and then solve. Don’t forget to include units.
The length of a rectangle is three more than twice the width.
If the perimeter is 78 inches, find the width and the length.
2l  2w  P


l  2w  3

2(2w  3)  2 w  78

6 w  6  78

6 w  72


w  12 inches 
l  2(12)  3  27 inches
3 pts
2 pt
1pt
2
(6 pts)
14
Subtract: (6 x3  4 x 2  11x  8)  (5 x3  2 x 2  8 x  17)
6 x3  4 x 2  11x  8  5 x3  2 x 2  8 x  17
1pt
6 x  5 x  4 x  2 x  11x  8 x  8  17
1pt
x3  6 x 2  3x  9
1pt
3
15
3
2
2
Solve the formula A  p  prt for t.
(3 pts)
A  p  prt
A  p prt

pr
pr
1pt
1pt
A p
 t or
pr
16
(3 pts)
A 1
 t
pr r
1pt
Write the equation of the line passing through the point ( 3, 4) and having slope  5 . Give
(4 pts)
the final answer in the slope - intercept form.
y  y1  m( x  x1 )
y  mx  b


 or

4  (5)(3)  b 
y  4  (5)[ x  (3)]
y  4  (5)( x  3) 

y  4  5 x  15 
y  5x  11
17
or
4  15  b 

b  11 
2 pts
1pt
1pt
Write the equation of the line passing through the pair of points (2, 7) and ( 4,5) . Give the
final answer in the slope - intercept form.
m
y2  y1
x2  x1
m
5  (7)
4  2
m
12
6
m  2
y  y1  m( x  x1 )
y  mx  b


 or

7  (2)2  b 
y  (7)  (2)( x  2) 
y  7  (2)( x  2) 
7  4  b 
or


y  7  2 x  4 
b  3

y  2 x  3
3
1pt
2 pts
1pt
1pt
(5 pts)
18
Graph the linear inequality 3 x  2 y  6
x0
y  3
2 pt
(8 pts)
y0
x2
2 pt
choose the broken line
graph the line
2 pt
The test point (0, 0), 0  6 is a true, shade a right region
19
(a)Solve the linear inequality, (b) graph the solution, and (c) write the solution set as interval
notation: 7( x  1)  2( x  4)  0
7 x  7  2x  8  0
5x 15  0
1pt
1pt
5 x  15
1pt
1pt
x3
(,3]
20
1pt
1pt
1pt
2 x  5 y  16
3 x  4 y  17
(4 pts)
Solve the system of equations: 
(2 x  5 y  16)  (3)

(3x  4 y  17)  2
6 x  15 y  48

6 x  8 y  34
7
14
7 y  14
y
7
7
2 x  5(2)  16 

2 x  10  16 

2x  6 x  3 
(5 pts)
1pt
1pt
y  2
1pt
1pt
4
21
Simplify and write the answer using only positive exponents:
for
for
for
for
canceling 27 with 63 by 9
dividing x5 by x4
dividing y7 by y6
moving x9 to denominator as x9
27 x 5 y 7
63x 4 y 6
(4 pts)
1pt
1pt
1pt
1pt
3 y13
7 x9
22
Multiply: (2 y  3)(2 y 3  4 y 2  2 y  3)
4 y 4  8 y 3  4 y 2  6 y  6 y 3  12 y 2  6 y  9
2 pts
4 y  8 y  6 y  4 y  12 y  6 y  6 y  9
2 pts
4 y 4  14 y 3  16 y 2  12 y  9
1pt
4
23
24
(5 pts)
3
3
2
2
Perform the indicated operation: (2 x  3 y )2
(3 pts)
(2 x  3 y )(2 x  3 y )
1pt
4 x 2  6 xy  6 xy  9 y 2
1pt
4 x 2  12 xy  9 y 2
1pt
6 x 2  13x  3
Use the long division to perform the division:
x 3
2 pts 2 pts
5
6x
x  3 6 x  13 x  3
2
 6 x 2  18 x
5x  3
 5 x  15
12
2 pts
5
(6 pts)
25
The product of the second and third of three consecutive integers is 2 more than 10 times the
first integer. Find the integers.
( x  1)( x  2)  10 x  2
3 pts
x 2  3x  2  10 x  2
1pt
x2  7 x  0
1pt
x( x  7)  0
x0
x7
1pt
1pt
{0,1, 2} or {7,8,9)
26
Solve by factoring: 3 y 2  10 y  8
(5 pts)



2
3 y  12 y  2 y  8  0 

3 y ( y  4)  2( y  4)  0
( y  4)(3 y  2)  0
3 y 2  10 y  8  0
y40
3y  2  0
27
(7 pts)
2 pts
1pt
1pt
y4 

2
y 
3
1pt
x 2  2 x  24
x 2  x  12
x 2  2 x  24  ( x  6)( x  4)
1.5 pts
x  x  12  ( x  3)( x  4)
1.5 pts
( x  6)( x  4) ( x  6)

( x  3)( x  4) ( x  3)
1pt
(4 pts)
Simplify into lowest terms:
2
6
28
x 2  5 x  6 x 2  x  20
Multiply and simplify: 2

x  3x  28 x 2  2 x  15
x 2  5 x  6  ( x  2)( x  3)
(7 pts)
1.5 pts
x 2  x  20  ( x  5)( x  4)
1.5 pts
x  3x  28  ( x  7)( x  4)
1.5 pts
x 2  2 x  15  ( x  5)( x  3)
1.5 pts
( x  2)( x  3) ( x  5)( x  4) ( x  2)


( x  7)( x  4) ( x  5)( x  3) ( x  7)
1pt
2
29
Subtract. Write answer in lowest terms:
(6 pts)
3
9
 2
x2 x x2
( x  1)3
9
LCD  ( x  1)( x  2) 1pt

( x  2)( x  1) ( x  2)( x  1)
3x  3
9

( x  2)( x  1) ( x  2)( x  1)
3x  3  9
3x  6
1pt
( x  2)( x  1)
( x  2)( x  1)
3( x  2)
( x  2)( x  1)
30
3
( x  1)
Solve and check for any extraneous solution:
1pt
x
2
18


LCD  ( x  5)( x  4)
x  5 x  4 ( x  5)( x  4)
 x

2
18
 x  5  x  4  ( x  5)( x  4)  ( x  5)( x  4)


x( x  4)  2( x  5)  18
x 2  4 x  2 x  10  18


2
x  2x  8  0


( x  4)( x  2)  0
x4
N /S
(8 pts)
x
2
18

 2
x  5 x  4 x  x  20
3 pts
1pt
1pt
1pt
1pt
x  2
1pt
7
31
32
Nancy Johnson invested $18,000. Part of it was invested at 3% annual simple
interest, and the rest was invested at 4%. Her interest income for the first year was
$630. How much did she invest at each rate?
 x  y  18000
3 pts

0.03 x  0.04 y  630
 x  y  18000
( x  y  18000)(3) 3x  3 y  54000



3x  4 y  63000 3x  4 y  63000
3 x  4 y  63000
y  9000
1pt
x  9000  18000
1pt
Multiply and simplify:
x  9000
2 pts
(4 pts)
3x3 y  27 xy 3
3x3 y  27 xy3
33
1pt
81x4 y 4
1pt
92 ( x 2 ) 2 ( y 2 ) 2  9 x 2 y 2
2 pts
Solve: x 2  21  3
x  18 or
2
(4 pts)
x  18  0
2
1pt
x 2   18 or ( x  18)( x  18)  0
1pt
x  3 2
34
2 pts
Solve and check each potential solution:

x7

2
(7 pts)
x7  4
 (4)2
x  7  16
97  4
(4 pts)
2 pts
x9
1pt
94
1pt
8
35
Use the quadratic formula to solve: x 2  4 x  2  0
b  b2  4ac
x
2a
a  1 b  4 c  2
(4)  (4) 2  4  1  2
x
2 1
2(2  2)
x
x  2 2
2
36
(5 pts)
x
2 pts
4 8
2
2 pts
1pt
(a) Find the x -intercept = (3, 0) ,(b) and the y –intercept = (0, 4) of the equation
4 x  3 y  12 .
x0
(c) Graph the equation using the intercepts.
y  4
y0
2 pts
x3
2 pts
2 pts
9
(6 pts)