This document provides information on various financial mathematics concepts: 1) Simple interest, compound interest, linear depreciation, and reducing-balance depreciation formulas are defined. 2) Examples of simple interest, compound interest, and percentage gain/loss word problems are worked out in 3 steps or less. 3) Additional word problems involving topics like installment payments, savings deposits, cost price and selling price are solved concisely.

1. FINANCIAL MATHEMATICS

2. Simple
Interest
I =
𝑃×T×𝑅
100
Amount = Principal
Princip
al
Rate
Time

3. Compound
Interest
A = P 1 +
r
100
t
Amount
Principal
Rate
Time
If a principal, P is invested for 𝑛 years at
𝑟% per annum.

4. Linear Depreciation
A = P 1 − i × n
Depreciated
Value
Present value Rate of
depreciation
Number of
years
If the present value, P of an asset depreciates
for 𝑛 years at 𝑟% rate of depreciation.

5. Reducing-balance
depreciation
A = P 1 −
r
100
n
Depreciated
Present value
Rate of
depreciation
Time in
years
This is where a depreciation is also a depreciation on
previously depreciated value

6. SOLUTION1. A man bought a
television set on a hire
purchase for N25, 000,
out of which he paid
N10,000, if he is allowed
to pay the balance in
eight equal installments,
find the value of each
installment.
WAEC 2000 Q14
Sold price= N25, 000
Amount the man payed= N10,000
Amount left to be payed by the man
= N25, 000−N10,000
=N15,000
8 installments=N15,000
Hence each installment =
N15,000
8
= N1875

7. JAMB 2000 Q.6
1. A man wishes to keep his money in a
savings deposit at 25% compound
interest so that after three years he can
buy a car for N150,000. how much does
he need to deposit?
A. N112,000.50
B. N96,000.00
C. N85,714.28
D. N76,800.00
Ans; D
SOLUTION
From the question, P
= N150,000 , r = 25% and t = 3
Amount, A = P 1 + r t
150000 = P 1 +
25
100
3
150000 = P 1 +
1
4
3
150000 = P
5
4
3
150000 = P ×
125
64
150000 × 64 = P × 125
P =
9600000
125
P = N76800.00

8. WAEC 2000 Q49
1. If the simple interest on N2, 000, after 9
months is N60, at what rate per annum
is the interest charged?
A. 2
1
4
%
B. 4%
C. 5%
D. 6%
Ans; B
SOLUTIONS
Principal, P = N2,000, Interest, I = N60 and
Time, T = 9months =
9
12
yrs =
3
4
yrs
I =
PRT
100
We make R the subject
I × 100 = PRT
I × 100
PT
=
PRT
PT
R =
I × 100
PT
We subtitute the values of P, I and T to get R
R =
60 × 100
2000 ×
3
4
=
6000
1500
= 4%

9. JAMB 1999 Q6
A trader bought 100 oranges at 5 for
N1.20, 20 oranges got spoilt and the
remaining were sold at 4 for N1.50.
find the percentage gain or loss.
A. 30% gain
B. 25% gain
C. 30% loss
D. 25% loss
Ans: B
SOLUTION
Number of oranges bought = 100
cost of 5 oranges = N1.20
Cost of the 100 oranges =
100 × 1.2
5
= N24.00
Number of oranges spoilt = 20
number of oranges remained = 100 − 20 = 80
Selling price of 4 oranges = N1.50
Selling price of 80 oranges =
1.50 × 80
4
= N30.00
since the sllind price is more than the cost price,
there is gain.
So, gain percent =
S. P − C. P
C. P
× 100%
=
30 − 24
24
× 100%
= 25%

10. 1. Find the simple interest rate per
annum at which N1,000 accumulates to
N1,240 in 3 years.
A. 6%
B. 8%
C. 10%
D. 12%
Ans: B
SOLUTION
𝐹𝑟𝑜𝑚 𝑡ℎ𝑒 𝑞𝑢𝑒𝑠𝑡𝑖𝑜𝑛,
𝐴𝑚𝑜𝑢𝑛𝑡, 𝐴 = 𝑁1240; 𝑃𝑟𝑖𝑛𝑐𝑖𝑝𝑎𝑙, 𝑃
= 𝑁1000; 𝑇𝑖𝑚𝑒, 𝑇 = 3𝑦𝑟𝑠; 𝐼𝑛𝑡𝑒𝑟𝑒𝑠𝑡, 𝐼
=? , 𝑅𝑎𝑡𝑒, 𝑅 =?
𝐴 = 𝑃 + 𝐼
⇒ 𝑁1240 = 𝑁1000 + 𝐼
⇒ 𝐼 = 𝑁1240 − 𝑁1000𝐼 = 𝑁240
𝐼 =
𝑃𝑅𝑇
100
⇒ 240 =
1000 × 3 × 𝑅
100
⇒ 240 = 30 × 𝑅
⇒ 𝑅 =
240
30
= 8%

11. 2. A cinema has 200 seats. Ticket
prices are $5 for an adult and
$2.50 for a child. One evening
80% of the seats in the cinema
are occupied. Twenty of the
people present are children.
Calculate the total money taken
from the sale of tickets.
A. $50
B. $700
C. $750
D. $1600
E. $2000
Ans; C
SOLUTION
Totral number of seats in the cinema = 200
Ticket prices
Adult = $5 and Child = $2.50
One evening
Number of seats occupied = 80% of 200
=
80
100
× 200 = 160
Out of 160 peoples 20 are children
⇒ Number of adults = 140
Total money taken from tickets sold
= $5 140 + $2.50 20
= $700 + $50
= $750

12. 80% of the profit in a cake stall will
be to Joan. The ingredients for the
cake cost $100, and the total value
of the cake sold was $275.
Use the information above to
answer question 3 and 4
3. How much profit did the cake
stall make?
A. $50
B. $75
C. $125
D. $175
E. $375
Answer;D
SOLUTION
𝐶𝑜𝑠𝑡 𝑜𝑓 𝑖𝑛𝑔𝑟𝑒𝑑𝑖𝑒𝑛𝑡 = $100
𝑆𝑜𝑙𝑑 𝑣𝑎𝑙𝑢𝑒 = $275
⇒ 𝑃𝑟𝑜𝑓𝑖𝑡 𝑚𝑎𝑑𝑒 = $275 − $100 = $175

13. 4. What amount did Joan get ?
A. $15
B. $25
C. $140
D. $145
E. $165
Answer; C
SOLUTION
𝐶𝑜𝑠𝑡 𝑜𝑓 𝑖𝑛𝑔𝑟𝑒𝑑𝑖𝑒𝑛𝑡 = $100
𝑆𝑜𝑙𝑑 𝑣𝑎𝑙𝑢𝑒 = $275
⇒ 𝑃𝑟𝑜𝑓𝑖𝑡 𝑚𝑎𝑑𝑒 = $275 − $100 = $175
𝐴𝑚𝑝𝑜𝑢𝑛𝑡 𝑔𝑖𝑣𝑒𝑛 𝑡𝑜 𝐽𝑜𝑎𝑛 = 80% 𝑜𝑓$ 175
⇒
80
100
× $175 = $140

14. 5. A man car was bought for
GH¢15000.00. It was later sold at a
profit of 12.5%. What was the selling
price?
A. $15
B. $25
C. $140
D. $145
E. $165
Answer; C
SOLUTION
𝐶𝑜𝑠𝑡 𝑜𝑓 𝑖𝑛𝑔𝑟𝑒𝑑𝑖𝑒𝑛𝑡 = $100
𝑆𝑜𝑙𝑑 𝑣𝑎𝑙𝑢𝑒 = $275
⇒ 𝑃𝑟𝑜𝑓𝑖𝑡 𝑚𝑎𝑑𝑒 = $275 − $100 = $175
𝐴𝑚𝑝𝑜𝑢𝑛𝑡 𝑔𝑖𝑣𝑒𝑛 𝑡𝑜 𝐽𝑜𝑎𝑛 = 80% 𝑜𝑓$ 175
⇒
80
100
× $175 = $140