Math 1090-004
Midterm 3
April 24, 2015
Please show all of your work as partial credit will be given where
appropriate, and there may be no credit given for problems where there is no
work shown. You may use a scientific calculator and a 4 × 6 inch note card.
Scratch paper will be provided but NOT collected, so please transfer all
finished work onto the proper page in the test. This exam totals 70 points.
Name:
UID:
Formulas
i If a1 , a2 · · · is an arithmetic sequence where at = at−1 + d, then a1 + a2 + · · · + an =
n
(a1 + an ).
2
n
−1
ii If a1 , a2 · · · is a geometric sequence where at = dat−1 , then a1 + a2 + · · · + an = a1 dd−1
.
iii Simple interest: S = P (1 + rt).
nt
iv Compound interest: R = P 1 + nr .
n
v AP Y = 1 + nr − 1 or AP Y = er − 1.
R((1+rc )N −1)
, where rc = nr , N = nt.
vi Future value of an ordinary annuity S =
rc
i
h
N +1
vii Future value of an annuity due: S = R (1+rc )rc −1 − R, where rc = nr , N = nt.
R(1−(1+rc )−N )
,
rc
where rc = nr , N = nt.
R(1+rc )(1−(1+rc )−N )
,
rc
where rc = nr , N = nt.
viii Present value of an ordinary annuity: S =
ix Present value of an annuity due: S =
x Present value of a deferred ordinary annuity: P =
−(N −k)
xi Amortization formula: SN −k = R 1−(1+rrcc)
.
R(1−(1+rc )−N )
.
rc (1+rc )m
1. (10 pts) Solve the equation
2x+1
x−5
=4−
3
x−5
Solution.
3
2x + 1
=4−
x−5
x−5
2x + 1 = 4(x − 5) − 3
2x + 1 = 4x − 20 − 3
2x = 24
x = 12
2. (10 pts) Given a geometric sequence with a1 =
10 terms of the sequence.
3
2
and a6 =
3
,
64
find the sum of the first
It suffices to write down the final expression that you would plug into your calculator.
Solution. The sum of the first n terms of a geometric sequence that starts with a1 and
such that each term is obtained from the previous one multiplying by d is given by
S = a1
In our case, a1 =
so we have
3
2
dn − 1
d−1
and we want to find the sum of the first 10 terms of the sequence,
S=
3 d10 − 1
2 d−1
and it suffices to find d. For this we use the fact that a6 =
3
.
64
Recall that the n-th term
of a geometric sequence is given by an = a1 dn−1 . In our case, a1 =
have
a6 = a1 d 5
3
3
= d5
64
2
1
= d5
32
1
d=
2
We thus conclude that
3
S=
2
1 10
−
2
1
−1
2
Page 2
1
3
2
so for n = 6 we
3. (10 pts) Solve and graph the solutions of the following system of inequalities
x + 7y ≤ −15
5x − y ≥ −3
x − 2y ≤ 12
You need to: graph the lines, find the coordinates of the vertices of the resulting region, and
shade it.
Solution. Each of the three inequalities determines a half plane and the solution of the
system is the triangular region depicted below. In order to find the coordinates of the
vertices of the triangle, we need to find the intersection of each pair of lines.
Lines 1 and 2. We solve the system
x + 7y = −15
5x − y = −3
Multiplying the second equation by 7 and adding it to the first one we eliminate the
variable y and we obtain
36x = −15 − 21 = −36
x = −1
y = 5x + 3 = −2
The point of intersection of lines 1 and 2 is thus given by (−1, −2) .
Lines 1 and 3. We solve the system
x + 7y = −15
x − 2y = 12
Subtracting the second equation from the first one we eliminate the variable x and we
obtain
9y = −15 − 12 = −27
y = −3
x = 2y + 12 = 6
The point of intersection of lines 1 and 3 is thus given by (6, −3) .
Lines 2 and 3. We solve the system
5x − y = −3
x − 2y = 12
Page 3
Multiplying the first equation by 2 and subtracting the second one we eliminate the
variable y and we obtain
9x = −6 − 12 = −18
x = −2
y = 5x + 3 = −7
The point of intersection of lines 2 and 3 is thus given by (−2, −7) .
Page 4
4. (a) (5 pts) How much should be invested quarterly (at the end of each quarter) at 12%
compounded quarterly to pay off a debt of $30, 000 in 6 years?
Start by writing down the formula that you will be using.
Solution. We are being asked about the future value of an annuity: namely, we
need to determine how much we should invest into an account (at the end of every
quarter) so that its future value is of $30, 000. The formula for the future value of
an ordinaty annuity is
R (1 + rc )N − 1
r
S=
, rc = , N = nt
rc
n
where S = $30, 000 is the future value of the annuity and where R is the amount
of money that should be regularly invested. Solving for R and plugging our data
we obtain
30, 000 · 0.12
Src
=
= $871.42
R=
4·64
N
0.12
(1 + rc ) − 1
1+
−1
4
(b) (5 pts) If we invest $700 into the account in (a) at the end of each quarter, how
many years will it take to pay off the debt?
Start by writing down the formula that you will be using. It suffices to write down the
final expression that you would plug into your calculator.
Solution. In this case we are being told how much we invest at the end of every
quarter and we are being asked to determine how long it will take to pay off the
debt. We need to apply the same formula as before
S=
R ((1 + rc )nt − 1)
rc
but in this case we need to solve for t (namely, the number of years that we need
to pay off the debt if we invest $700 at the end of every quarter into the account).
R (1 + rc )N − 1
S=
rc
Src = R (1 + rc )nt − 1
Src
+ 1 = (1 + rc )nt
R Src
log
+ 1 = nt log(1 + rc )
R
c
log Sr
+
1
R
t=
n log(1 + rc )
30,000· 0.12
4
log
+
1
700
t=
4 log 1 + 0.12
4
Page 5
5. (a) (3 pts) What is the annual percentage yield (APY)? Write a sentence explaining
this, not a formula.
Solution. The APY of an account is the simple interest rate that yields the same
return
(b) (7 pts) Which is the better investment deal?
(i)
(ii)
(iii)
(iv)
An
An
An
An
account
account
account
account
earning
earning
earning
earning
10% compounded annually.
9% interest compounded quarterly.
8.6% interest compounded monthly.
8% interest compounded continuously.
Solution. We simply compute the APY for each account: the best investment deal
will be the one with the highest APY. Recall that the APY is given by
AP Y = (1 + rc )n − 1,
AP Y = er − 1
depending on whether the compounding is discrete or continuous.
(i) An account earning 10% compounded annually: AP Y = 0.1
(ii) An account earning 9% interest compounded quarterly:
AP Y =
0.09
1+
4
4
− 1 = 0.0931
(iii) An account earning 8.6% interest compounded monthly:
12
0.086
AP Y = 1 +
− 1 = 0.0895
12
(iv) An account earning 8% interest compounded continuously:
AP Y = e0.08 − 1 = 0.0833
We conclude that the best investment option is the account in (a).
Page 6
6. (10 pts) George and Mary have a friendly contest between them to see who can earn the
most money in the next 5 years. They agree not to invest more than $10, 500 each in an
account earning 10% interest compounded monthly, but neither of them has that much
money up front. George decides to invest $5, 000 at the beginning, and then add $85 at
the end of each month to his account. Mary invests $170 at the end of every month for
the entire 5 years. Answer the questions below to see who has more money at the end
of the 5 years.
(a) (1 pt) How much money did George and Mary invest?
Solution. George invested
$5, 000 + $85 · 12 · 5 = $10, 100
and Mary invested
$170 · 12 · 5 = $10, 200
(b) (5 pts) How much money does George have at the end of the 5 years?
Start by writing down the formula(s) that you will be using.
Solution. The $5, 000 that he invests at the beginning will earn interest during
the full 5 years, and we may apply the compound interest formula to these
12·5
0.1
r nt
= 5000 1 +
= 8226.54
S =P 1+
n
12
After that he adds $85 to the account at the end of each month for 5 years, and for
this we may apply the formula for the future value of an ordinary annuity
nt
12
−1
1 + 0.1
−1
1 + nr
12
=
85
= $6582.15
S=R
r
0.1
n
12
In total, George’s account will be worth $8226.54 + $6582.15 = $14, 808.69
(c) (4 pts) How much money does Mary have at the end of the 5 years?
Start by writing down the formula(s) that you will be using.
Solution. Mary simply invests $170 at the end of every month during the 5 years,
so we may compute the future value of her annuity as
nt
12·5
1 + nr
−1
1 + 0.1
−1
12
S=R
= 170
= $13164.3
r
0.1
n
12
Page 7
7. For the following 3 questions, you don’t need to use your calculator: it suffices to write
down the final expression that you would type in. Make sure to always indicate which
formula(s) you are using.
(a) (3 pts) How much will an inheritance of $100, 000 provide at the end of each month
for 10 years if money is worth 7.2% compounded monthly?
Solution. We are being asked to compute the present value of an ordinary annuity,
so we apply the formula
R 1 − (1 + rc )−N
r
, rc = , N = nt
S=
rc
n
and solve for R:
R=
100, 000 · 0.072
Src
=
12
0.072 −12·10
1 − (1 + rc )−N
1 − 1 + 12
(b) (4 pts) How much will an inheritance of $100, 000 provide at the end of each month
for 10 years if money is worth 7.2% compounded monthly and we don’t start withdrawing until the 5th month of the 3rd year?
Solution. In this case, we are being asked to compute the present value of a
deferred ordinary annuity, so we apply he formula
R 1 − (1 + rc )−N
S=
rc (1 + rc )m
and solve for R. In this formula, m indicates the number of deferred periods
(namely, the number of months that the $100, 000 sit in the account before we start
withdrawing), which is 2 years and 5 months, so m = 2 · 12 + 5 = 29. Similarly,
N indicates the number of withdrawals we will make. Note that we will be making
8 withdrawals in the third year, and 12 withdrawals each of the next 7 years, so
N = 12 · 7 + 8 = 84 + 8 = 92. Therefore, at the end of every month we will be able
to withdraw
2·12+5
100, 000 · 0.072
· 1 + 0.072
Src (1 + rc )m
12
12
=
R=
−(12·7+8)
1 − (1 + rc )−N
1 − 1 + 0.072
12
Page 8
(c) (3 pts) How much will an inheritance of $100, 000 provide at the beginning of each
month for 10 years if money is worth 7.2% compounded monthly and we don’t start
withdrawing until the 5th month of the 3rd year?1
Solution. In this case, we are being asked to compute the present value of a
deferred annuity due, so we apply he formula
R(1 + rc ) 1 − (1 + rc )−N
S=
rc (1 + rc )m
and solve for R. In this formula, m indicates the number of deferred periods
(namely, the number of months that the $100, 000 sit in the account before we start
withdrawing), which in this case is 2 years and 4 months, so m = 2 · 12 + 4 = 28.
Similarly, N indicates the number of withdrawals we will make. Note that we will
be making 8 withdrawals in the third year, and 12 withdrawals each of the next 7
years, so N = 12 · 7 + 8 = 84 + 8 = 92. Therefore, at the end of every month we
will be able to withdraw
0.072 2·12+4
100, 000 · 0.072
·
1
+
Src (1 + rc )m
12
12
=
R=
0.072
0.072 −(12·7+8)
(1 + rc ) (1 − (1 + rc )−N )
1 + 12
1 − 1 + 12
1
I highlighted in red the differences with respect to part (b).
Page 9