Practice Exam 2
Math 1010-001
July 19, 2005
Name:
This practice exam is a modified copy of a rough draft of the real exam. Do not expect
the real exam to be exactly like this. It is likely that there will be fewer problems but
that a few problems may become harder. It is also possible that I may decide that a
particular type of problem is too difficult or not as important as something elso so it may
be dropped from the exam. Solutions to this practice exam will be availible after the class
on Wednesday. The reveiw for the test will be on Wednesday as well. Since this is so
similar to the real thing let me know if something is unclear.
Read all of the following information before starting the real exam:
• Show all work, clearly and in order, if you want to get full credit. I reserve the
right to take off points if I cannot see how you arrived at your answer (even if your
final answer is correct).
• This exam is closed book and no calculators are allowed.
• Justify your answers algebraically whenever possible to ensure full credit. You
may also use pictures whenever appropriate. Remember that you are trying to
communicate to me that you understand the material. I am not a mind reader.
• Circle or otherwise indicate your final answers.
• Please keep your written answers brief; be clear and to the point. I will take points
off for rambling and for incorrect or irrelevant statements.
• This test has 12 problems and is worth 100 points, plus some extra credit at the
end. It is your responsibility to make sure that you have all of the pages!
• Good luck!
1
1.
(10 points) Your friend goes to the auto parts store to get a distributor cap and
rotor. She tells you that together they were $30 and the rotor is $5 less than half the
price of the cap. Instead of asking your friend how much the each item was, you decide
to figure it out yourself, so you look cool.
a. (5 pts) Set up a system of linear equations that describe this situation. (don’t
solve them)
solution:
c + r = 30
r = c/2 − 5 →
−c/2 + r = −5 →
So we have
c + r = 30
− 21 c + r = −5
b. (5 pts) Solve the system of linear equations using your method of choice.
solution: We will use elimination to solve this problem. To do this we write the system
in matrix form.
1
1 30 −−−−→ 1 1 30 −−−−−→ 1 1 30 −−−→
R2/3
R1 + R2
2 × R2
0 3 20
−1 2 −10
−1/2 1 −5
−−−−−−−→ 1 0 30 − 20/3
1 1 30
−R2 + R1
0 1
20/3
0 1 20/3
So c = 30 − 20/3 = 90/3 − 20/3 = 70/3 = 23 31 and r = 20/3 = 6 23 . So the disributor cap
cost $23.33 and the rotor cost $6.67.
2.
(14 points) Rewrite the following expressions using just one rational exponent or if
possible write it as an integer.
4
a. (4 pts) 8 3
4
4
solution: 8 3 = (23 ) 3 = 24 = 16
5
b. (4 pts) (32 ) 2
solution:
5
2
5
(32 ) 2 = 2 1 × 2 = 35
c. (3 pts)
solution:
6
5
(z 25 ) 4
6
5
6
5
30
3
(z 25 ) 4 = z 25 × 4 = z 100 = z 10
2
d. (3 pts)
solution:
1
u3 u3
3
u4
1
2
u3 u3
u
3
4
=
u1
u
3
4
3
1
= u1− 4 = u 4
3.
(4 points) Write 2, 030, 400, 000, 000 in scientific notation.
solution: 2.0304E12
4.
(8 points) Let the polynomial p be defined by
p(x) = (x − 4)(2x + 3)
a. (4 pts) Write p in standard form (p(x) = ax2 + bx + c).
solution: We will expand the poly nomial into standard form using f.o.i.l.
x2 + 3x − 8x − 12 = x2 + 5x − 12
b. (4 pts)
solution:
What is p(4)?
p(4) = (4 − 4)(2(4) + 3) = 0
5. (10 points) On video games it is not uncommon for characters to jump their own
height and more. For a 6 foot tall human how fast would they have to leave the ground
to jump 9 feet (150% of their height) into the air. Remember that
h(t) = h0 + v0 t − 16t2
v(t) = v0 − 32t
solution: We substitute all of the information we have into the equations given and see if
we can make sense of it.
h(t) = 0 + v0 t − 16t2
v(t) = v0 − 32t
This didn’t get us very far so we try to think of other information given in the problem.
We are looking for the maximum height. From class we discussed the fact that an object
is momentarily motionless at the apex of fligt. So we can use this to find the time when
the person reaches the maximum height. Also, we know the maximum height is 9 ft.
9 = v0 t − 16t2
0 = v0 − 32t
To solve the problem we try substitution. We solve for t in the second equation and
substitute this into the first equation.
0 = v0 − 32t → 32t = v0 → t = v0 /32
Now we substitute
v0
v0
− 16( )2 →
32
32
32 × 9 = v02 − 16 × v02 /32 →
1
32 × 9 = v02 − v02 →
2
1 2
32 × 9 = v0 →
2
64 × 9 = v02 →
√ √
√
v0 = ± 64 × 9 = ± 64 9 = ±24
Only the positive answer makes sense here so v0 = 24f t/sec is the answer.
9 = v0 ×
(Be sure to study the homework problem 28-31 from assignment 6).
6.
(10 points) Simplify
1
7
− 2
x − 3 x + x − 12
solution: First we find a common denominator. To do this we factor both denominators
and find the least common multiple. The donominators factor to (x−3) and (x−3)(x+4).
Since the latter contains both the LCM is (x − 3)(x + 4).
7
x+4−7
x−3
1
1 × (x + 4)
−
→
→
→
(x − 3)(x + 4) (x − 3)(x + 4)
(x − 3)(x + 4)
(x − 3)(x + 4)
x+4
7.
(10 points) Simplify
1
1
×
+ x − 12 x + 4
solution: Don’t over think multiplication. don’t find an LCM. Just factor and multiply.
x2
1
1
1
×
=
(x − 3)(x + 4) x + 4
(x + 4)2 (x − 3)
8.
(10 points) A circular swimming pool has a depth d and a raduis that is 10 times
its depth. A square pool has the same depth as the circular pool its width 2 less than
10 times its depth. What is the ratio of the square pool’s volume to the circular pools
volume?
solution: Just try to write down equations for the problem using the information in the
problem and the volume equations.
For the circular pool its volume is Vc = πr2 d. The problem says that r = 10d. Use
substitution to get Vc = π(10d)2 d = 100πd3
For the square pool the volume is Vs = w2 d. The problem says that width of the pool is
w = 10d − 2. Again we use substitution and we find Vs = (10d − 2)2 d.
The problem asked for the ratio of the square pool to the circular pool. This is
(10d − 2)2 d
(10d − 2)2
=
100πd3
100πd2
9.
(10 points) Solve the following system of equations

z
= −1
 x
−x −y −2z = −2

x +y
=2
solution: To solve this linear system we could use substitution or elimination. I will solve
it using elimination.




1
0
1 −1 −−−−−→
1
0
1 −1 −−−−−→
 −1 −1 −2 −2  R2 + R3  −1 −1 −2 −2  R1 + R2
0
0 −2 0
1
1
0
2




1 0
1 −1 −−−−−→ 1 0
1 −1 −−−−−−−−−−−−−−−−−→
 0 −1 −1 −3  R3/(−2)  0 −1 −1 −3  R3 + R2 & − R3 + R1
0 0 −2 0
0 0
1
0




1 0 0 −1 −−→ 1 0 0 −1
 0 −1 0 −3  −R2  0 1 0 3 
0 0 1 0
0 0 1 0
We can see that x = −1, y = 3 and z = 0.
10.
(10 points) You have just landed on a small planet where gravity is
on earth. Because of this the projectile motion equations become.
1
16
what it is
h(t) = h0 + v0 t − 1t2
v(t) = v0 − 2t
To celebrate your landing you throw a rock straight up it the air with an initial velocity
of 30 f t/sec. How high does the rock go?
solution: Just as with the other story problems we start by writing down equation that
go with this type of problem with the appropriate values inserted in it.
h(t) = 0 + 30t − t2
v(t) = 30 − 2t
Again we use the fact that at the highest point v(t) = 0.
0 = 20 − 2t → 2t = 30 → t = 15
Now we know when it happens. Also, height is a function of time so if we give it the time
it will give us back the height.
h(15) = 30 × 15 − 152 = 2 × 152 − 152 = 152 = 225
11.
(4 points) Solve the folllowing equation.
√
3x − 4 = 7
solution: We will use the golden rule of algebra repeatedly to solve for x.
√
2
→
→
−
→
−
3x − 4 = 7 −
(·)2 3x − 4 = 49 +4 3x = 53 ÷3 x = 53/3 = 17
3
12.
(4 points) Factor p(x) = x2 + x − 20
solution: Since the last sign is negative the factors will have opposite signs. Since the
middle sign is positive the larger of the two is positive.
We proceed as we have in class. First we factor 20 and then subtract its factors to see if
we can get 1.
(1)
(2)
(3)
1 × 20 = 20
2 × 10 = 20
4 × 5 = 20
20 − 1 = 19
10 − 2 = 8
5−4=1
The last attempt was a success. So we have (x − 4)(x + 5) as the factors.
Bonus Question (2pt)
Add every number from 1 to 100 together. (For example here is the sum of all the numbers
from 1 to 5. 1+2+3+4+5=15 )
solution: Group the numbers into pairs that add up to 100. Then count how many pairs
you have and multiply this number by 100. There is one number left unpaired. The result
is 5050.
The bonus questions on the tests are almost never worth the time it takes to solve them.
Only attempt them if you are confident in the rest of the test. Also, there is always a
trick that makes the calculations simple.