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Math 251: Practice Exam 2
Answers
Disclaimer
You should use this practice exam to assess your speed and to
improve your ability to correctly identify different problem types.
The questions on this practice exam are taken from exams given in
previous semesters, but they may not be representative of the
questions that will appear on this semester's exam. You should also
invest time re-reading the relevant parts of your textbook, reviewing
your notes, and practicing homework problems.
Math 251: Practice Exam 2
Answers
You will not receive full credit if you do not clearly show work as demonstrated
in class. Show all work in the space provided on this exam. Circle your answers.
1.
Consider the inequality,
2
5
( p + 3) > ( p − 4) .
3
6
a.
Solve for p.
⎡2
⎤
⎡5
⎤
6 ⎢ ( p + 3) ⎥ > 6 ⎢ ( p − 4) ⎥
⎣3
⎦
⎣6
⎦
4( p + 3) > 5( p − 4)
4 p + 12 > 5 p − 20
4 p + 32 > 5 p
32 > p
p < 32
b.
Graph the solution set on the number line provided.
0
c.
Write the solution in interval notation.
( −∞, 32 )
8 16 24 32
(8 points)
Page 2
2.
⎧2y = 3x
Solve the system of equations by graphing: ⎨
⎩ y + 3x = 9
2y = 3x
3 y= x
2
(8 points)
y + 3x = 9
y = −3x + 9
6
5
4
3
2
1
So the solution is the point (2, 3) .
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
-1
-2
-3
-4
-5
3.
Solve each problem.
a.
42% of what number is 294?
(42%)x = 294
(0.42)x = 294
294
x=
0.42
x = 700
b.
4 is what percent of 50?
4 = x(50)
4
=x
50
0.08 = x
8% = x
(4 points)
Page 3
4.
A coin collector has $1.70 in dimes and nickels. She has two more dimes than nickels.
(10 points)
How many nickels does she have? Let x = the number of nickels. (The table is optional,
but you must set up and solve an equation. Guess and check is not an acceptable method.)
Number of coins
Value per coin
Value
Dimes
x + 2
10¢
10(x + 2)
Nickels
x
5¢
5x
Total
170¢
10(x + 2) + 5x = 170
10x + 20 + 5x = 170
15x + 20 = 170
15x = 150
x = 10
She has 10 nickels.
Page 4
5.
A line l1 passes through the points: A(-2, 4) and B(4, -6).
a.
Write the equation of l1 in slope-intercept form.
m1 =
(22 points)
y2 − y1 (−6) − (4) −10
5
=
=
=−
x2 − x1 (4) − (−2)
6
3
y − y1 = m1 (x − x1 )
5
y − (4) = − (x − (−2))
3
5
y − 4 = − (x + 2)
3
12
5
10
y− = − x−
3
3
3
5
2
y=− x+
3
3
b.
Find the equation of the line l2 which is perpendicular to l1 and has y-intercept: (0, -1).
m2 = −
y=
1 3
=
m1 5
3
x −1
5
l1
6
5
4
l2
3
2
1
-5
-4
-3
-2
-1
0
-1
-2
-3
-4
c.
Graph both lines on the coordinate plane.
(Be sure to label each line.)
-5
1
2
3
4
5
6
Page 5
6.
⎧2x = 12 + 6y
⎪
Solve the system of equations by the elimination method: ⎨ 2
x
+
y
=
3
⎪⎩ 3
⎧2x = 12 + 6y
⎧ 2x − 6y = 12
⎪
⎯−3
⎯→ ⎨
⎨2
⎩−2x − 3y = −9
⎪⎩ 3 x + y = 3
−9y = 3
3
y=
−9
1
y=−
3
1⎞
⎛
So the solution is the point ⎜ 5, − ⎟
⎝
3⎠
7.
Solve the equation
z z +1
=
.
2
3
z z +1
=
2
3
⎛ z⎞
⎛ z + 1⎞
6⎜ ⎟ = 6⎜
⎝ 2⎠
⎝ 3 ⎟⎠
3z = 2(z + 1)
3z = 2z + 2
z=2
(8 points)
⎧2x = 12 + 6y
⎧ 2x − 6y = 12
⎪
⎯6⎯
→⎨
⎨2
⎩ 4x + 6y = 18
⎪⎩ 3 x + y = 3
6x = 30
30
x=
6
x=5
(3 points)
Page 6
8.
A motorboat traveling with the current went 88 km in 4 hours. Against the current,
the boat could go only 64 km in the same amount of time. Find the rate of the boat
in still water and the rate of the current. Let x = the rate of the boat in still water and
y = the rate of the current. (The table is optional, but you must set up and solve a
system of equations. Guess and check is not an acceptable method.)
(10 points)
time
rate
distance
Downstream
4
x + y
88
Upstream
4
x - y
64
⎧ 4(x + y) = 88 1/4 ⎧ x + y = 22
⎯1/4
⎯→ ⎨
⎨
⎩ 4(x − y) = 64
⎩ x − y = 16
2x = 38
x = 19
⎧ x + y = 22
⎧ 4(x + y) = 88 1/4 ⎧ x + y = 22
⎯1/4
⎯→ ⎨
⎯−1
⎯
→⎨
⎨
⎩ 4(x − y) = 64
⎩ x − y = 16
⎩−x + y = −16
2y = 6
y=3
So the speed of the boat in still water is 19 kph and the speed of the current is 3 kph.
Page 7
9.
⎧x < 3
Graph the solution set: ⎨
⎩ 3x + 2y ≥ 3
3x + 2y ≥ 3
2y ≥ −3x + 3
−3x + 3
y≥
2
3
3
y≥− x+
2
2
(8 points)
6
5
4
3
2
1
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
-1
-2
-3
-4
-5
10.
Use f (x) = 2x − 3 and g(x) = −x + 3 to complete each table below.
x
f(x)
x
g(x)
-2 -7
-2
5
0 -3
0
3
2 1
2
1
4 5
4
-1
(8 points)
Page 8
⎧2x − y = −13
11.
Solve the system of equations by substitution: ⎨
⎩ x + 3y = 18
x + 3y = 18
x = 18 − 3y
x = 18 − 3(7)
x = 18 − 21
x = −3
(8 points)
2x − y = −13
2(18 − 3y) − y = −13
36 − 6y − y = −13
36 − 7y = −13
−7y = −49
y=7
So the solution is the point ( −3, 7 ) .
12.
Solve the equation
49 x
=
56 8
⎛ 49 ⎞
⎛ x⎞
56 ⎜ ⎟ = 56 ⎜ ⎟
⎝ 56 ⎠
⎝ 8⎠
49 = 7x
49 7x
=
7
7
7=x
49 x
= .
56 8
(3 points)