Math 134 Tutorial 7, 2011: Financial Maths
For each question, identify which of the formulae (a) to (g) applies. what you are asked to find, and what
information you have been given. Final answers can be worked out at home or during free periods.
Formulae used in Financial Maths:
(a) Amount of Simple Interest
I = P in
(b) Future Value using Simple Interest: S = P (1 + in)
(c) Future Value using Compound Interest: S = P (1 + i)n
(d) Effective Rate of Interest: ief f = (1 + mi )m − 1 where m is the no. of compounding periods in 1 year.
n
(1 + i) − 1
(e) Future Value of an Annuity: S = Rsnei = R
i
"
#
−n
1 − (1 + i)
(f) Present Value of an Annuity: A = Ranei = R
i
#
"
−n
1 − (1 + i)
(1 + i)
(g) Present Value of an Annuity paid in Advance: A = Ranei (1 + i) = R
i
(h) Total Interest = Difference between the Sum of Payments and the Present or Future Value.
1. Shona’s granny gives her a loan of R33 000 to be repaid after 6 years at 8% annual simple interest.
How much will Shona have to pay 6 years from now to settle the loan?
2. How much would Shona have to pay if interest was compounded (a) annually? (b) daily?
3. What is the annual simple interest rate that will make R400 grow to R450 in (a) 2 years? (b) 10
months?
4. Calculate the effective rate of interest for
(a) 16.7% p.a. compounded monthly
(b) 16.1% p.a. compounded daily
(c) 17% p.a. compounded semi-annually
(d) 16.9% p.a compounded quarterly
(e) 17.2% p.a. compounded annually.
5. A fixed deposit is known to have an effective rate of 8.5%. If interest is compounded monthly, find the
nominal rate of interest.
6. Your client just turned 35 and will retire at the end of the month she turns 55 years old. She starts
saving R1 000 at the end of each month, earning 7.2% p.a. compounded monthly. How much will she
have in her account when she retires?
7. Find the present value of R5 500 due in 7 years from now at an interest rate of 9% p.a. compounded
semi-annually.
8. Londiwe amortizes a home loan of R800 000 by obtaining a 20-year mortgage at the rate of 9.5% p.a.
compounded monthly. Find
(a) the monthly payment
(b) the total interest paid
(c) the principal remaining after 5 years
9. What nominal interest rate, compounded quarterly, would be needed for R1400 to grow to R2000 in 4
years?
1
10. What nominal interest rate, compounded daily, would be needed for R1400 to grow to R2000 in 4
years?
11. Yaseen bought a computer for R5 200 and agreed to pay it off by making monthly payments of R200.
If the shop charges interest at a rate of 12% p.a. compounded monthly, how many months would it
take to pay off the debt?
12. How many days will it take for an investment of R450 to grow to R600 at 18% p.a. compounded
daily?
13. How long will it take an investment to double if interest is 15% p.a. compounded monthly?
14. Andile is the beneficiary of a trust fund set up at his birth 21 years ago. Originally R10 000 was
invested in the trust fund. What will Andile receive from the trust fund if the money has earned
interest at 7% p.a. compounded (a) annually (b) monthly (c) daily?
15. Determine the finance charge (i.e. the interest paid) on a 36-month R120 000 car loan with monthly
payments if interest is charged at 11.5% p.a. compounded monthly.
16. A computer was bought for R12 000 on 23 March 2005. What was the annual rate of depreciation (on
reducing or diminishing balance) if the computer is worth R4 000 on 23 March 2010?
17. Suppose you want to have R1 million saved when you retire at age 60. How much should you save at
the end of each month if interest is 8% p.a. compounded monthly and you start saving when you are
(a) 40 years old (i.e. for 20 years)?
(b) 35 years old?
(c) 25 years old?
18. Suppose you have the choice of taking out a R500 000 loan at 10% p.a. compounded monthly for either
30 years or 15 years. How much savings is there in the finance charge if you choose the 15-year loan?
19. Simon can get married when he has saved R50 000. How much should he save at the end of each month
at 9% p.a. compounded monthly if he want to get married
(a) 4 years from now?
(b) 2 years from now?
20. Simon calculates that he can only save R250 per month for his wedding. How long will he have to save
(at 9% p.a. compunded monthly) before he has accumulated R50 000?
21. A loan of R1 000 000 is being amortized over 25 years at an interest rate of 12% p.a. compounded
monthly. Find
(a) the monthly payment
(b) the outstanding balance at the beginning of the 101st month.
(c) the interest paid in the 101st month
(d) the total interest paid in 25 years.
ANSWERS
1. b. Find S if P = 33 000, n = 6, i = 0.08; R48 840
2. (a) c. Find S if n = 6, i = 0.08; R52 367 (b) c. Find S if n = 2190, i =
0.08
365 ;
R53 328
3. (a) b. Find i if S = 450, P = 400, n = 2; 6.25% (b) b. Find i if S = 450, P = 400, n =
2
10
12 ;
15%
4. d or c (a) 18.04% (b) 17.46% (c) 17.72% (d) 18.00% (e) 17.2%
5. d. Find i if ief f = 0.085, m = 12; 8.19%
6. e. Find S if n = 240, i =
0.072
12 ,
R = 1000; R533 762.34
7. c. Find P if S = 5 500, n = 7×2 = 14, i =
8. (a) f. Find R if A = 800 000, i =
0.095
12 ,
0.09
2 ;
R2 969. 85
n = 20 × 12 = 240; R7457.05
(b) h. 240(7457.05) - 800 000 = R989 692
(c) f. Find A if i =
0.095
12 ,
n = 15 × 12 = 180, R = 7457.05; R714 123.13
9. c. Find i if S = 2000, P = 1400, n = 16 and i is replaced with 4i ; 9.02%
10. c. Find i if S = 2000, P = 1400, n = 4(365) = 1460 and i is replaced with
11. f. Find n if A = 5200, R = 200, i =
12. c. Find n if S = 600, P = 450, i =
0.12
12
0.18
365 ;
i
365 ;
8.92%
= 0.01; 30.26 months
583. 50 days
13. c. Find n if S = 2P, P = P, (or S = 2, P = 1, or S = 200, P - 100 etc.) i =
years 8 months
0.15
12 ;
55. 798 months ≈ 4
14. (a) c. Find S if P = 10 000, n = 21, i = 0.07; R41 405.62
0.07
12 ; R43 307.00
= 0.07
365 ; R43 486.22
(b) c. Find S if P = 10 000, n = 21(12) = 252 , i =
(c) c. Find S if P = 10 000, n = 21(365) = 252 , i
15. f and h. Find R if n = 36, A = 120 000, i =
0.115
12 ;
R = 3957.12. Interest = R22 456.32
16. c. Find i if P = 12 000, S = 4 000, n = 5; decrease of 19.73% per year.
17. (a) e. Find R if S = 1 000 000, i =
(b) e. Find R if S = 1 000 000, i =
(c) e. Find R if S = 1 000 000, i =
0.08
12 ,
0.08
12 ,
0.08
12 ,
18. f and h. Find R if A = 500 000, i =
Find R if A = 500000, i =
0.1
12 ,
0.1
12 ,
n = 240; R1 697.73
n = 300; R1 051.50
n = 420; R435.94
n = 360; R = 4 387.86. Interest = R1 079 629.60
n = 180; R = 5 373.03. Interest = R467 144.61
The savings from paying off in 15 years is 1 079 629.60 - 467 144.61 = R612 484.99
19. (a) e. Find R if S = 50 000, i =
0.09
12 ,
n = 48; R869.25
(b) e. Find R if S = 50 000, i =
0.09
12 ,
n = 24; R1909.24
20. e. Find n if R = 250, i =
0.09
12 ,
S = 50 000;.10 years 3 months
21. (a) f. Find R if A = 1 000 000, i =
(b) f. Find A if R = 10532.24, i =
0.12
12
0.12
12
= 0.01, n = 25 × 12 = 300; R10532.24
= 0.01, n = 200; R909 262.62
(c) a. Find I if P = 909262.62, i = 0.01, n = 1; R9092.63
(d) h. n = 300, R = 10532.24; R2 159 672
3
Math 134 Tutorial 7, 2011: Financial Maths Solutions
Tutors Version with questions and answers
1. Shona’s granny gives her a loan of R33 000 to be repaid after 6 years at 8% annual simple interest.
How much will Shona have to pay 6 years from now to settle the loan?
S = P (1 + in) = 33000 (1 + (0.08) (6)) = 48840
2. How much would Shona have to pay if interest was compounded (a) annually? (b) daily?
n
6
(a) S = P (1 + i) = 33000 (1 + 0.08) = 52367
6(365)
n
= 53328.
(b) S = P (1 + i) = 33000 1 + 0.08
365
3. What is the annual simple interest rate that will make R400 grow to R450 in (a) 2 years? (b) 10
months?
(a) S = P (1 + in) where S = 450, P = 400, n = 2
450 = 400 (1 + 2i) ∴ i = 12 450
400 − 1 = 0.0625 = 6.25%
(b) S = P (1 + in) where S = 450, P = 400, n = 10
12
12 450
450 = 400 1 + 10
i
∴
i
=
−
1
=
0.15
=
15%
12
10 400
4. Calculate the effective rate of interest for
(a) 16.7% p.a. compounded monthly
12
n
ief f = (1 + i) − 1 = 1 + 0.167
− 1 = 0.18039 = 18.04%
12
(b) 16.1% p.a. compounded daily
n
ief f = (1 + i) − 1 = 1 +
0.161 365
365
− 1 = 0.174 64 = 17.46%
(c) 17% p.a. compounded semi-annually
2
n
ief f = (1 + i) − 1 = 1 + 0.17
− 1 = 0.177 23 = 17.72%
2
(d) 16.9% p.a compounded quarterly
4
n
ief f = (1 + i) − 1 = 1 + 0.169
− 1 = 0.180 02 = 18.00%
4
(e) 17.2% p.a. compounded annually.
17.2% (no calculation needed - the effective rate is the percentage by which money actually grows
in a year)
5. A fixed deposit is known to have an effective rate of 8.5%. If interest is compounded monthly, find the
nominal rate of interest.
n
ief f = (1 + i) − 1
x 12
0.085 = 1 + 12
−1
0.08185792.... = 8.19%
6. Your client just turned 35 and plans to retire on her 55th birthday. At the end of each month she
saves R1 000, earning 7.2% p.a. compounded monthly. How much will she have in her account when
she retires?
R = 1000, i =
0.072
12 ,
S = Rsnei = R
n = 20 × 12 = 240
n
1+
(1 + i) − 1
= 1000
i
0.072 240
12
0.072
12
−1
!
= 533 762.34
7. Find the present value of R5 500 due in 7 years from now at an interest rate of 9% p.a. compounded
semi-annually.
S = 5500, n = 7 (2) = 14 semi-years, i =
n
14
S = P (1 + i) ∴ 5500 = P (1 + 0.045)
0.09
2
= 0.045
∴ P = 5500(1 + 0.045)−14 =R2 969. 85
4
8. Londiwe amortizes a home loan of R800 000 by obtaining a 20-year mortgage at the rate of 9.5% p.a.
compounded monthly. Find
(a) the monthly payment
(b) the total interest paid
(c) the principal remaining after 5 years
(a) A = 800 000, i = 0.095
12 , n = 20 × 12 = 240
−240 1 − 1 + 0.095
12
800 000 = R
0.095
12
R = 7 457.05
(b) Total amount paid = 240 (7457.05) = 1 789 692
Loan amount = 800 000.
The interest is the difference between what is paid and what the loan amount was.
1 789 692 − 800000 = 989 692
(c) Find the present value of the remaining 15 years of payments after 5 years of payments have passed.
−(15×12) 1 − 1 + 0.095
12
= 714 123.13
A = 7457.05
0.095
12
9. What nominal interest rate, compounded quarterly, would be needed for R1400 to grow to R2000 in 4
years?
P = 1400, S = 2000, n = 4 (4) = 16
16
n
Divide by 1400 first.
S = P (1 + i) ∴ 2000 = 1400 1 + 4i
1
16
− 1 = 0.09017 = 9.02%
∴ i = 4 2000
1400
10. What nominal interest rate, compounded daily, would be needed for R1400 to grow to R2000 in 4
years?
P = 1400, S = 2000, n = 4 (365) = 1460
n
i 1460
S = P (1 + i) ∴ 2000 = 1400 1 + 365
Divide by 1400 first.
1
1460
∴ i = 365 2000
− 1 = 0.08918 = 8.92%
1400
11. Yaseen bought a computer for R5 200 and agreed to pay it off by making monthly payments of R200.
If the shop charges interest at a rate of 12% p.a. compounded monthly, how many months would it
take to pay off the debt?
A = 5200, R = 200, i =
0.12
12
= 0.01
−n
(1 − 1.01 )
0.01
Before taking logs, isolate 1.01−n
5200 = 200
0.26 = 1 − 1.01−n
log 0.74 = log 1.01−n
log 0.74
−n =
log 1.01
n = 30.26 months
In reality, there will either be 31 payments (with the last payment smaller), or the 30th payment will
be increased to reduce the balance of the debt to 0.
5
12. How many days will it take for an investment of R450 to grow to R600 at 18% p.a. compounded
daily?
P = 450, S = 600, i =
0.18
365
n
S = P (1 + i) ∴ 600 = 450 1 + 0.18
365
600
log 450
= 583. 50 days
n=
log 1 + 0.18
365
n
13. How long will it take an investment to double if interest is 15% p.a. compounded monthly?
P = 100 (or any value, or just use P ), S = 200 (or S = 2P ), i =
n
n
n
S = P (1 + i) ∴ 200 = 100 1 + 0.15
∴ 2 = 1 + 0.15
12
12
n=
0.15
12
log 2
= 55. 798 months ≈ 4 years 8 months
log 1 + 0.15
12
14. Andile is the beneficiary of a trust fund set up at his birth 21 years ago. Originally R10 000 was
invested in the trust fund. What will Andile receive from the trust fund if the money has earned
interest at 7% p.a. compounded (a) annually (b) monthly (c) daily?
n
21
(a) S = P (1 + i) ∴ S = 10000 (1 + 0.07) = 41405.62
12(21)
n
= 43307.00
(b) S = P (1 + i) ∴ S = 10000 1 + 0.07
12
365(21)
n
(c) S = P (1 + i) ∴ S = 10000 1 + 0.07
= 43486.22
365
15. Determine the finance charge (i.e. the interest paid) on a 36-month R120 000 car loan with monthly
payments if interest is charged at 11.5% p.a. compounded monthly.
First find the payments:
A = 120000, n = 36, i = 0.115
12
0.115 −36
1 − 1 + 12
120000 = R
0.115
12
R = 3957.12
Total amount paid = (36) (3957.12) = 142 456.32
Subtract the amount loaned: 120 000
Finance charge = 22 456.32
(Without rounding the answer is 22 456.35, which is slightly more accurate.)
16. A computer was bought for R12 000 on 23 March 2005. What was the annual rate of depreciation (on
reducing or diminishing balance) if the computer is worth R4 000 on 23 March 2010?
P = 12000, S = 4000, n = 5
n
5
S = P (1 − i) ∴ 4000 = 12000 (1 − i) ∴
4
12
15
=1−i∴i=1−
4
12
15
= 0.197 26 = 19.73%
17. Suppose you want to have R1 million saved when you retire at age 60. How much should you save at
the end of each month if interest is 8% p.a. compounded monthly and you start saving when you are
(a) 40 years old (i.e. for 20 years)?
S = 1 000 000, i = 0.08
12 , n = 20 × 12 = 240 payments
!
240
1 + 0.08
−
1
12
1 000 000 = R
0.08
12
1000000 0.08
12
0.08 240
1 + 12
−
R= 1
= 1 697.73
6
(b) 35 years old (i.e. for 25 years)?
S = 1 000 000, i = 0.08
12 , n = 25 × 12 = 300 payments
!
300
1 + 0.08
−1
12
1 000 000 = R
0.08
12
R = 1 051.50
(c) 25 years old (i.e. for 35 years)?
S = 1 000 000, i = 0.08
12 , n = 35 × 12 = 420 payments
!
420
1 + 0.08
−1
12
1 000 000 = R
0.08
12
R = 435.94
18. Suppose you have the choice of taking out a R500 000 loan at 10% p.a. compounded monthly for either
30 years or 15 years. How much savings is there in the finance charge if you choose the 15-year loan?
Find the payments for each case, then work out the interest paid in each case and subtract one from
the other.
For 30 years: A = 500 000, i =
−360 1 − 1 + 0.1
12
500000 = R
0.1
0.1
12 ,
n = 360
12
R = 4 387.86
Total amount paid = (360) (4387.86) = 1 579 629.60
Interest paid = Total amount paid - loan amount = 1 079 629.60
For 15 years: A = 500 000, i =
−180 1 − 1 + 0.1
12
500000 = R
0.1
0.1
12 ,
n = 180
12
R = 5 373.03
Total amount paid = 180(5373.03) = 967 145.40
Interest paid = Total amount paid - loan amount = 467 144.61
The savings from paying off in 15 years is 612 484.99
19. Miguel can get married when he has saved R50 000. How much should he save at the end of each
month at 9% p.a. compounded monthly if he want to get married
(a) 4 years from now?
S = 50 000, i = 0.09
12 , n = 4 × 12 = 48 payments
!
48
−
1
1 + 0.09
12
50 000 = R
0.09
12
R = 869.25
(b) 2 years from now?
S = 50 000, i = 0.09
12 , n = 2 × 12 = 24 payments
!
24
−1
1 + 0.09
12
50 000 = R
0.09
12
R = 1909.24
7
20. Miguel calculates that he can only save R250 per month for his wedding. How long will he have to
save (at 9% p.a. compunded monthly) before he has accumulated R50 000?
S = 50 000, i =
50 000 = 250
0.09
12 ,
R = 250
!
n
1 + 0.09
−
1
12
0.09
12
n
−1
1 + 0.09
50000
12
=
0.09
250
12
n
50000 0.09 −1
= 1 + 0.09
12
12
250
n
50000 0.09 + 1 = 1 + 0.09
12
12
250
Now take logs (or write in log form)
50000 0.09 log
+ 1 = n log 1 + 0.09
12
12
250
50000 0.09 log
+
1
12
250
n=
= 122.6 months = 10 years, 3 months
0.09
log 1 + 12
21. A loan of R1 000 000 is being amortized over 25 years at an interest rate of 12% p.a. compounded
monthly. Find
(a) the monthly payment
A = 1 000 000, i = 0.12
12 = 0.01, n= 25 × 12 = 300 payments
−300
1 − (1 + 0.01)
1 000 000 = R
0.01
R = 10532.24
(b) the outstanding balance at the beginning of the 101st month.
This is the present value of the 200 payments that still need to be paid (after 100 payments have
been made).
−200
1 − (1 + 0.01)
A = 10532.24
0.01
= 909 262.62
(c) the interest paid in the 101st month
(909262.62) (0.01) = 9092.63
(d) the total interest paid in 25 years.
This is the difference between the total amount paid (300 payments of R10 532.24) and the loan
amount.
(300) (10532.24) − 1000000 = 2 159 672
8