# review problems

```mathematics of finance
Faculty of commerce – English Section
Second year
Review Problems
Derive the formula Sn (for) = R[(1+i)
n+p
– (1+i) ]/ i
p
Sn (for)
1
…
.
….
n-2
n-1
n
1
2
Payment interval of n periods
3
P-1
P
P interest periods after last payment
Sn (for) = R sn+ p┐- Rsp┐
= R (sn+ p┐- sp┐)
= R [((1+i)n+p-1)/i - ((1+i)p-1)/i]
= R [((1+i)n+p - (1+i)p)/i]
= R ((1+i)n – 1 /i) (1+i)p
= R sn┐(1+i)p
Derive the formula An (def) = R an┐(1+i)-m
….
An
1
2
….
m
1
Deferment interval of m periods
2
3
n-1
n
Payment interval of n periods
An (def) = R am+n┐- Ram┐
= R (am+n┐- am)
= R [(1-(1+i)-(m+n)/ i) – (1-(1+i)-m/ i)]
= R [( (1+i)-m- (1+i)-(m+n)/ i)]
= R [1 - (1+i)-n / i] (1+i)-m
= R an┐(1+i)-m
A man aged 25 is considering 2 types of life insurance policies. The first is an ordinary life with an annual
premium of \$200. The second is a 20-year endowment with an annual premium of \$510. He decides to take the
ordinary life and to deposit the difference in premiums each year into an investment paying 51/2% compounded
annually. He dies just before making his 17th deposit but after receiving interest for the first 16 deposits. How
much was in the account at that time?
Sn (due) = R (1+i)n-1 (1+i)
i
= 310 [(1.055)16 – 1 / .055] (1.055) = \$8058.88
(a) You need a loan. Which is the best deal?
Bank A: 7.15% compounded monthly, Bank B: 7% comp. annually, or Bank C: 6.85% comp. quarterly
(b) Convert 9% compounded monthly to an equivalent nominal rate converted semi-annually
Bank A: 7.15% comp. monthly  i = 7.15%/12  r = (1+7.15%/12)12-1  r = 7.39%
Bank B: 7% comp. annually  r = 7%
Bank C: 6.85% comp. quarterly  i = 6.85%/4  r = (1+6.85%/4)4-1  r = 7.03%
(c) (1+.09/12)12 = (1+J/2)2  (1.0075)6 = (1+J/2)  J = 9.17%
1
Faculty of commerce – English Section
mathematics of finance
Second year
Review Problems
On February 18, 2002, a person deposited \$4500 in an account earning 3.5% compounded quarterly. If simple
interest is allowed for part of an interest conversion period, find the amount that will be in the account on
August 28, 2010. Also, state the total amount of interest earned
18/2/2002
\$4500
18/8/2002
Compound
28/8/2010
Simple
Year
2010
2002
Month
8
2
Day
28
18
8 Years
6 months
10 days
n = (8 &times; 4)+6/3 = 32+2=34
Simple interest
S = P(1+i)n= 4500 (1.00875)34 = \$6,051.36
I = Prt = 6,051.36 &times; 3.5/100 &times; 10/360 = \$5.88
Final amount = 6,051.36 + 5.88 = \$6057.24
OR: Final amount = P(1+i)n + (1+rt) = 4500 (1.00875)34 (1+3.5%&times;10/360) = \$6057.24
A person deposits \$250 monthly in a saving account earning 6% compounded monthly. The first deposit was
made on April 1, 1989 and the last deposit was made on February 1, 2004
a. Find the amount in the account just after the last deposit on February 1, 2004.
b. If no more deposits are made after February 1, 2004 but the money is allowed to continue earning
interest, find the amount in the account on October 1, 2010
Year
2004
1989
Month
2
4
14 Years
10 months
Day
1
1
0 days
n = 14 &times; 12 + 10 + 1 = 179
Sn = R (1+i)n-1
i
= 250 [(1.005)179 – 1 / .005] = \$72,094.21
Year
2010
2004
Month
10
2
Day
1
1
6 Years
8 months
0 days
P = 6 &times; 12 + 8 = 80
Sn (for) = R (1+i)n-1 (1+i)p
i
179
= 250 [(1.005) – 1 / .005] (1.005)80 = \$107,444.78
2
mathematics of finance
Faculty of commerce – English Section
Second year
Review Problems
A person buys a TV that costs \$1200. The payment plan requires \$150 paid down and the balance financed at
18% compounded monthly for 2 years. Find the size of the payments made at the end of each month. Also state
the total interest paid on this loan
1200 150
1050
R
R
R
R
2Y
An = R an┐
= R (1-(1+i)-n/ i)
1050= R &times;(1-(1+.015)-24/.015)
R = \$52.42
Total interest = (52.42 &times; 24) – 1050 = \$208.08
Determine the amount of money you need to deposit in an account earning 7% compounded monthly on the
day your child is born so that on the child's 18th birthday, he/she can begin withdrawing \$500 per month for 4
years to help cover college expenses. Also state the total amount of interest earned on this account.
18 Y
R R
P.v. comp.
R
R
R
22 Y
4Y
An(due) = R an┐(1+i)
= R (1-(1+i)-n/ i) (1+i)
= 500 &times; (1-(1+0.07/12)-48/0.07/12) (1+0.07/12)
= \$21,001.9
P = S(1+i)-n
= 21,001.9 (1+0.07/12) -216 = \$5,979.12
Interest = (500 &times; 48) - 5,979.12 = \$18,020.88
Your uncle deposited \$10,000 in an account for you that earns 8% compounded monthly. How much can you
withdraw monthly for 5 years if the first withdrawal made immediately?
An(due) = R an┐(1+i)
10,000= R (1-(1+.08/12)-60/ .08/12i) (1+.08/12)
R = \$201.42
When fred turned 30 he deposited \$45,000 into his savings account which earns 7% interest compounded
annually. How much can be withdrawn each year for 15 years to empty his account, if his first withdrawal is
Y.30
45,000
Y.65
R
Y.80
…..
R
An (def) = R an┐(1+i)-m
R = 45,000 / [1-(1+.07)-15/.07)] (1.07)-34
= 45,000/ 0.913
= \$49,299.45
Phil was loaned \$50,000 at 4% compounded monthly. He makes regular monthly payments of \$900
3
Faculty of commerce – English Section
mathematics of finance
Second year
Review Problems
(a) How many full payments must he make?
(b) What is the size of the smaller concluding payment?
An = R an┐
50,000 = 900 an┐  an┐= 55.556  1-(1+.04/12)-n/.04/12 = 55.556  (1+.04/12)-n = 0.814813333
-n log (1+.04/12) = log 0.814813333  n = 61.54
This means 61 full payment + smaller concluding payment
50,000 (1+.04/12) 62 =
61,457.51
Sn = R (1+i)n-1 (1+i)
i
= 900 [(1+.04/12) 61 – 1 / .04/12] (1+.04/12) = \$60,970.54
Size of concluding payment = 61,457.51 - 60,970.54 = \$486.97
Lex decides to give the Kents a break. He loans them \$50,000 and asks them to pay him back just \$53,880 at
the end of 3 years. What interest rate converted quarterly did lex charge the kents?
S = P(1+i)n  53,880 = 50,000 (1+i) 12  (1+i) 12 = 1.0776  i = 0.62  J4 = 0.62 &times; 4 = 2.50%
You just deposited \$50,000 into an account which earns 5% compounded annually. What is the most you can
withdraw each year so that the account is never depleted, if your first withdrawal is made 20 years from now?
50,000
n
20
R
R
R
19
S = P (1+i) = 50,000 (1.05) = 126,347.51
1 Period = 1 Year, so, we can use simple interest formula
I = Prt = 126,347.51 &times; .05 &times; 1 = \$6,317.38
Mr. Bigshot wants to set up an annual scholarship of \$5,000 in his honor at NCSU. The university can earn 7%
interest compounded monthly. If the first scholarship is awarded at the end of 1 year, how much does Mr.
Bigshot need to donate?
First, we convert 7% compounded monthly to % compounded annually
r = (1+i)m-1 = (1+.07/12) 12-1= 7.229%
OR (1+.07/12) 12 = (1+j)  j = 7.229%
We need to deposit enough money so that the amount of interest earned each year is \$5,000. But since 1 year is
1 conversion period, simple and compound interest are the same. So,
I = Prt  5,000 = P &times; .07229 &times; 1  P = \$69,165.86
Simon's Construction Company (SCC) borrows \$80,000 at 10% simple interest from the first bank of America.
The bank requires the SCC to make quarterly interest-only payments and pay the full \$80,000 at the end of 5
4
Faculty of commerce – English Section
mathematics of finance
Second year
Review Problems
years. In order to meet the 5 year obligation of \$80,000, SCC makes equal deposits at the end of each quarter
into a sinking fund which earns 7% compounded quarterly?
(a) Find SCC's total quarterly obligation?
(b) Find the equivalent amortization rate compounded quarterly for this loan?
The quarterly interest payments must be: I = Prt = 80,000 &times; 10% &times; 3/12 = \$2,000
Sn = R (1+i)n-1
i
80,000 = R [(1+.07/4) 20 – 1 / .07/4]  R = \$3,375.30
SCC's total quarterly obligation is : 2,000 + 3,375.3 = \$5,375.3
An = R 1-(1+i)n
i
80,000 = 5,375.3 sn┐ a20┐= 14.883................................... The equivalent amortization rate = 11.984%
Converting compound interest rates. You don't need to draw a time diagram for this problem.
(a) 5.75% compounded quarterly is equivalent to what rate compounded annually?
(b) 10% compounded daily (use Banker's Rule) is equivalent to what rate compounded semiannually?
(a) i = .0575/4 = 0.014375
r = (1.014375)4 – 1 = 5.88%
OR (1.014375) 4 = (1+J)  J2 = 5.88%
(b) i = 0.1/360
r = (1+0.1/360)360 – 1 = 10.52%
J = (1+10.52/100)1/2 – 1 = 5.128%  J2 = 10.25%
OR (1+0.1/360) 360 = (1+ J/2)2  (1+0.1/360) 180 = (1+ J/2)  J2 = 10.25%
In June, 1986, A person owes \$8,000 due in 3 years, \$15,000 due in 8 years and \$12500 due in 15 years. The
debtor want replace these, with 2 payments and twenty annuities, The first of all \$9000 due in 4 years, the
second \$7000 due in 13 years, and he has to pay annuities will be start in June 1, 1992 and at the end of each 6
months, If money is worth 9% converted annually. What is the ordinary annuity? [Putting the focal date on Jen
3Y
.
1, 1996]
1/6/1986
4Y
.
8000
r = (1+i)2 – 1
8Y
.
10Y.
13Y.
15Y.
15000
9000
r = 9%
6Y
.
RRR
R
12500
R
7000
Focal date
i = 4.403%
….R
.
8000(1.04403)14+15000(1.04403)4 + 12500(1.04403)-10 = 9000(1.04403)12 + 7000(1.04403)-6+[ R a9┐+ R s11┐]
[ R a9┐+ R s11┐] = 20,071.62  R [(1-(1.04403)-9/.04403) + ((1.04403)11-1/.04403)]
R = \$952.48
On June 1, 1995, a woman owes \$8000, which she is unable to repay. Her creditor charges her 12%
compounded quarterly from that date. If the debtor pays \$3000 on June 1, 1997, and she have to pay annuity at
1/6/1995
5
1/6/1997
1/6/2001
Faculty of commerce – English Section
mathematics of finance
Second year
Review Problems
the end of each 3 months for 4 years. What is the ordinary annuity at the end of each 3 months? (Putting the
focal date on June 1, 2001)
8000
3000
R
………
………
R
8000(1.03)24 = 3000(1.03)16 + R s16┐
R
R s16┐ = 11448.23
R =\$567.96
A company can put \$9500 in an investment that is expected to result in cash inflows of \$3000 in 3 months and
\$5000 in 8 months and \$2000 in 1 year. Find the internal rate of return.
3m.
9500
3000
8m.
1 Y.
5000
2000
At 10%: P.V. = 3000/(1+0.1&times;3/12) + 5000/(1+0.1&times;8/12) + 2000/(1+0.1&times;1) = 9432.51
At 8%: P.V. = 3000/(1+0.08&times;3/12) + 5000/(1+0.08&times;8/12) + 2000/(1+0.08&times;1) = 9539.86
Rate
8%
2%
d
r
10%
Present value of inflows
9539.86
9500
39.86
107.35
9432.51
d/2 = 39.86/107.35
d = 2&times; 39.86/107.35 =.74
r = 8+.74 = 8.74%
An investment of \$8000 is made for 15 years and 2 months. During the first 7 years the interest rate is 9%
converted semiannually. Then the rate drops to 7% converted semiannually for the remainder of the time. What
is the final amount?
7 Y.
9% semi
15 Y.
15 Y.+2m.
7% semi
8000
S1 = P (1+i)n = 8000 (1.045)14 = \$14,815.56  Using 9% semiannually
S2 = P (1+i)n = 14,815.56 (1.035)16 = \$25,689.97  Using 7% semiannually
I = Prt = 25,689.97 &times; .07 &times; 2/12 = \$299.72
Final amount = 25,689.97 + 299.72 = \$ 25,989.69
Calculate the present value of LE 1000 due in 3 years
(a) at a simple interest rate of 10% P.a.  P = S/(1+rt) = 1000/(1+.1&times;3) = \$769.23
(b) at a simple discount rate of 9% P.a.  P = S(1-dt) = 1000(1-.09&times;3) = \$730
Example 3: A couple gets an \$80,000, 30-year, 12% loan. The monthly payment is \$822.9. How much of the
first two payments goes to interest and how much to principal?
6
Faculty of commerce – English Section
mathematics of finance
Second year
Review Problems
As to first payment :
I = Prt
= 80,000 &times; .12 &times; 1/12
= \$800
Payment to principal = \$822.9 – 800 = \$22.9
As to second payment :
New P = 80,000 – 22.9 = \$ 79, 977.1
I = Prt
= 79, 977.1&times; .12 &times; 1/12
= \$799.77
Payment to principal = \$822.9 – 799.77 = \$23.13
Payments of LE 10 are due for the next 6 years. What is the present value of these payments at 1% per month?
What is the amount value if the LE 10 payments are due at :
(a) The end of each month
(b) The beginning of each month
 Ordinary annuity
Sn= Rn+ n / 2 &times; R &times; r &times; t (n-1)
1
1
Sn= 10 &times; 72 + 72 / 2 &times; 10 &times; .12 &times; 1/12 (72-1)
= 720 + 36 &times; 1.2 &times; 1/12 &times; 71
= 720 + 255.6 = 975.6
 annuity due
Sn= Rn+ n / 2 &times; R &times; r &times; t (n+1)
1
1
Sn= 10 &times; 72 + 72 / 2 &times; 10 &times; .12 &times; 1/12 (72+1)
= 720 + 36 &times; .1 &times; 73
= 720 + 262.8 = 982.8
d= r/(1+rt)=.12/(1+.12&times;1)= 10.71%
An= Rn- n / 2 &times; R &times; d (t1+ tn)
1
 Present value
Present value of ordinary =10&times;72–72/2&times;10&times;.1071(1/12+72/12)
An= Rn- n / 2 &times; R &times; d (t1+ tn)
 Present value
Present value due = 10&times;72–72/2 &times;10&times;.1071 (0 + 71/12)
1
Find the amount of LE 100 accumulated for 20 years at the following rate of interest:-
7
Faculty of commerce – English Section
mathematics of finance
Second year
Review Problems
(a) Effective rate of interest corresponding to a nominal rate of 4% P.a. convertible half yearly?
(b) 6% p.a. convertible three times p.a. for 12 years, thereafter 3.5% p.a. convertible every second year
(c) the rate of interest p.a. at which a sum of money will treble itself in 21 years
(a) r = (1+ .04/2)2 -1 = 4.04%
S = P (1+i) n = 100(1.0404)20 = \$220.80
(b) S1 = P (1+i) n = 100(1+.06/3)36 = \$203.99
S2 = P (1+i) n = 203.99(1+.035/360&times;24&times;60&times;60)31104000 = \$211.22
(c) S = P (1+i) n
3 P = P (1+i)21
(1+i) 21 = 3  i = 5.37%
S = P (1+i) n
S = 100 (1+.0537)20 = \$284.67
8
mathematics of finance
Faculty of commerce – English Section
Second year
Review Problems
An automobile is purchased for \$12,000. The down payment, including the trade-in value of the old car, was
worth \$3,000. The interest is 10% on an add-on basis. The loan is to be repaid in 3 years. Find the nominal
interest rate the buyer is paying.
Amount owed = 12,000 – 3,000 = \$9,000
Add-on = 9,000 &times; 3 &times; 10%
= 2,700
Total amount owed
= 11,700
Monthly payments = 11,700 / 36 = \$325
An = R an┐
9,000 = 325 a36┐  a36┐= 27.6923
j
15%
3%
d
Present value of inflows
28.8473
i
27.6923
18%
28.8473
1.155
1.1866
d / 3 = 1.155/1.1866
d = 2.92
j = 15+2.92 = 17.92 %
The parents of four children, a 13 year-old, a 12-year-old, and 10-year-old twins, die suddenly. Their wills call
for a trust fund of \$200,000 to be held, with equal amounts of the money to be given to each of the children at
age 18. If the fund earns 10% converted quarterly, find the amount each child will get
Y.10
8 Y.
Y.12
Y.13
5 Y.
200000
6 Y.
X
X
2X
X (1.025)-20 + X (1.025)-24 + 2X (1.025)-32 = 200,000
X [(1.025)-20 + (1.025)-24 + 2 (1.025)-32] = 200,000
X = \$96,586.28
Sixty days after borrowing, a person pays back exactly LE 200. How much was borrowed if the LE 200
payment includes the principal and simple interest at 9% P.a.
S = P (1+rt)
200 = P (1+.09&times;60/360)
P =200/1.015 = \$197.04
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Faculty of commerce – English Section
mathematics of finance
Second year
Review Problems
Deposits of LE 400 are made at the beginning of each half year for 5 years into an account paying 16%
compounded half yearly. How much is in the account:
a. at the end of 5 years
b. Just before the sixth deposit
(a)
Sn (due) = R (1+i)n-1 (1+i)
i
= 400 [(1+.08) 10 – 1 / .08] (1+.08) = \$6,258.19
(b)
Sn (due) = R (1+i)n-1 (1+i)
i
= 400 [(1+.08) 5 – 1 / .08] (1+.08) = \$2,534.37
Eight hundred pounds is due at the end of 2 years and LE700 at the end of 8 years at rate 12% compounded
monthly, find an equivalent single amount at the end of:
(a)
2 Years
(b) 6
Years
(c) 10
Years
2 Y.
(a)
8 Y.
800
700
Focal
date
X = 800 + 700(1.01)-72 = \$1141.95
What amount should be invested at outset to give an accumulated amount of L.E. 5,000 at the end of 15 years, if
the interest rate is 9% p.a. for the first 8 years, 7% p.a. for the next 4 years, and 5% p.a. for the last 3 years?
-n
8Y.
4Y.
3Y.
9%
7%
5%
5000
-3
P1 = S (1+i) = 5000 (1+.05) = \$4319.19
P2 = S (1+i)-n = 4319.19 (1+.07)-4 = \$3295.09
P3 = S (1+i)-n = 3295.09 (1+.09)-8 = \$1653.69
A person borrows \$9000 on 3 years and \$11000 on 5 years at 19% interest. The debt will be repaid with three
payments, X, Y, and Z, that Y equal 3X and Z equal 2Y, A one on 7 years, the second on 9 years, and the other
on 12 years. Put the focal date on 7years. And find the size of the payments (using time diagram)
Y = 3X
Z = 2Y=6X
3 Y.
5 Y.
7 Y.
X
9 Y.
Y
12 Y.
Z
9000
11000
9000(1+0.19&times;4)+11000(1+0.19&times;2) = X+3X/(1+0.19&times;2)+6X/(1+0.19&times;5)
15840 + 15180 = X + 3/1.38X + 6/1.95X
31020 = 6.25 X  X = 4963.2  Y = 3 &times; 4963.2 = 14889.6  Z = 6 &times; 4963.2 = 29779.2
10
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