Learning Objectives
Chapter Overview
Key skills you will develop by the end of Chapter 2.
Calculate simple interest and amount using \(I = \frac{PRT}{100}\)
Calculate compound interest using \(A = P\left(1+\frac{R}{100}\right)^n\)
Work out hire-purchase price, downpayment and instalments
Convert one currency to another using an exchange rate
Find insurance premiums, allowing for depreciation
Distribute inheritance (Islamic law) and share partnership profit
Mathematics: Financial Transaction
Complete chapter notes with all formulae, worked examples and full step-by-step exercises: PDF format
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Topic 2.1
Simple & Compound Interest
When you save or borrow money, extra money called interest is added.
Key concepts
What is the difference between simple and compound interest?
Simple interest is always calculated on the original principal, so the interest earned each year is the same. Compound interest is calculated on the principal and on the interest already earned, so the amount keeps growing faster.
Simple: \(I = \dfrac{P \times R \times T}{100}\), \(A = P + I\)
Compound: \(A = P\left(1+\dfrac{R}{100}\right)^{n}\), \(\text{C.I.} = A - P\)
If the time is in months, convert to years: \(T = \dfrac{\text{months}}{12}\). Compounded half-yearly → rate \(\dfrac{R}{2}\), periods \(2n\); monthly → rate \(\dfrac{R}{12}\), periods \(12n\).
Worked Example: find the simple interest and the time
(a) Jun Wei borrows \($1000\) at \(4\%\) per annum simple interest. Find the interest after 3 years.
(b) Huixian invests \($4000\) at \(2\%\) per annum; find the time to grow to \($4400\).
(b) Huixian invests \($4000\) at \(2\%\) per annum; find the time to grow to \($4400\).
(a) \(I = 1000 \times \dfrac{4}{100} = 40\) per year
Interest for 3 years \(= 40 \times 3 = $120\)
(b) Interest needed \(= 4400 - 4000 = 400\)
Interest per year \(= 4000 \times \dfrac{2}{100} = 80\)
Time \(= \dfrac{400}{80} = 5\) years
Interest for 3 years \(= 40 \times 3 = $120\)
(b) Interest needed \(= 4400 - 4000 = 400\)
Interest per year \(= 4000 \times \dfrac{2}{100} = 80\)
Time \(= \dfrac{400}{80} = 5\) years
Worked Example: compound interest for 7 years
Find the compound interest on \($5000\) for 7 years at \(3\%\) per annum, compounded annually.
\(A = P\left(1+\dfrac{R}{100}\right)^{n} = 5000\left(1+\dfrac{3}{100}\right)^{7}\)
\(A = $6149.37\) (to the nearest cent)
\(\text{C.I.} = A - P = 6149.37 - 5000 = $1149.37\)
\(A = $6149.37\) (to the nearest cent)
\(\text{C.I.} = A - P = 6149.37 - 5000 = $1149.37\)
The compound formula gives the amount \(A\) first — subtract \(P\) to get the interest.
Worked Example: find the interest rate (compound)
Mrs Lee places \($4000\) for 2 years, compounded yearly, and receives \($4243.60\). Find the rate.
\(4243.60 = 4000\left(1+\dfrac{R}{100}\right)^{2}\)
\(\left(1+\dfrac{R}{100}\right)^{2} = \dfrac{4243.60}{4000} = 1.0609\)
\(1+\dfrac{R}{100} = \sqrt{1.0609} = 1.03 \Rightarrow R = 3\%\) per annum
\(\left(1+\dfrac{R}{100}\right)^{2} = \dfrac{4243.60}{4000} = 1.0609\)
\(1+\dfrac{R}{100} = \sqrt{1.0609} = 1.03 \Rightarrow R = 3\%\) per annum
Topic 2.2
Hire Purchase
Buying goods by paying a deposit first, then regular instalments.
Key concepts
What are the key hire-purchase formulae?
The downpayment is the first lump sum; the balance is paid in instalments which include interest, so the total (the hire-purchase price) is more than the cash price.
\(\text{Interest} = \dfrac{\text{Balance} \times R \times T}{100}\)
\(\text{HP price} = \text{Downpayment} + (\text{Instalment} \times \text{No. of instalments})\)
\(\text{Extra paid} = \text{HP price} - \text{Cash price}\)
Worked Example: monthly instalment on a washing machine
Cash price \($450\); \(15\%\) downpayment; balance over 2 years at \(10\tfrac{2}{3}\%\) per annum. Find (i) the monthly instalment, (ii) the total paid.
Downpayment \(= \dfrac{15}{100}\times 450 = $67.50\)
Balance \(= 450 - 67.50 = $382.50\)
Interest (1 yr) \(= 382.50 \times \dfrac{10\frac{2}{3}}{100} = 40.80\); for 2 yr \(= 81.60\)
Total instalments \(= 382.50 + 81.60 = 464.10\)
(i) Monthly \(= \dfrac{464.10}{24} = $19.34\)
(ii) Total paid \(= 67.50 + 464.10 = $531.60\)
Balance \(= 450 - 67.50 = $382.50\)
Interest (1 yr) \(= 382.50 \times \dfrac{10\frac{2}{3}}{100} = 40.80\); for 2 yr \(= 81.60\)
Total instalments \(= 382.50 + 81.60 = 464.10\)
(i) Monthly \(= \dfrac{464.10}{24} = $19.34\)
(ii) Total paid \(= 67.50 + 464.10 = $531.60\)
Topic 2.3
Money Exchange
Converting one currency into another using an exchange rate.
Key concepts
How do I convert between two currencies?
First find the value of one unit, then multiply by the amount.
If A\$1 = S\$1.0318, then A\$543 = \(1.0318 \times 543\). If RM100 = S\$32.368, then RM1 = \(\dfrac{32.368}{100}\), so RM4850 = \(\dfrac{32.368}{100} \times 4850\).
Worked Example: convert into Singapore dollars
Given A\$1 = S\$1.0318 and RM100 = S\$32.368, convert (i) A\$543, (ii) RM4850 into S\$.
(i) A\$543 \(= 1.0318 \times 543 = \) S\$560.27
(ii) RM4850 \(= \dfrac{32.368}{100} \times 4850 = \) S\$1569.85
(ii) RM4850 \(= \dfrac{32.368}{100} \times 4850 = \) S\$1569.85
Worked Example: two-way exchange with spending
A family changes HK\$35 000 at HK\$100 = S\$16.235, spends S\$3500, and converts the rest back at HK\$100 = S\$16.242. Find the HK\$ received.
S\$ obtained \(= 35000 \times \dfrac{16.235}{100} = 5682.25\)
Remaining \(= 5682.25 - 3500 = 2182.25\)
HK\$ received \(= \dfrac{2182.25 \times 100}{16.242} = \) HK\$13 436
Remaining \(= 5682.25 - 3500 = 2182.25\)
HK\$ received \(= \dfrac{2182.25 \times 100}{16.242} = \) HK\$13 436
Topic 2.4
Insurance, Inheritance & Partnership
Premiums and depreciation, dividing an inheritance, and sharing business profit.
Insurance
How is an insurance premium calculated?
A premium is the amount paid for insurance cover. It is a percentage of the insured value. A vehicle's value falls over time — this is depreciation.
\(\text{Premium} = \dfrac{R}{100} \times \text{Insured value}\)
Example: Sameena's car (Rs 500 000) is insured at \(4\%\). In year 4 it is worth Rs 475 000.
Years 1–3: \(\dfrac{4}{100}\times 500000 = \) Rs 20 000 each
Year 4: \(\dfrac{4}{100}\times 475000 = \) Rs 19 000
Total \(= 20000\times 3 + 19000 = \) Rs 79 000
Years 1–3: \(\dfrac{4}{100}\times 500000 = \) Rs 20 000 each
Year 4: \(\dfrac{4}{100}\times 475000 = \) Rs 19 000
Total \(= 20000\times 3 + 19000 = \) Rs 79 000
Inheritance (Islamic law)
How is an inheritance distributed?
The widow's share is \(\tfrac{1}{8}\) of the estate; the rest is divided among the children so that a son inherits twice a daughter's share. If the widow is already deceased, no widow's share is taken out.
Example: A man leaves Rs 50 000 to his widow, 2 sons and a daughter.
Widow \(= \dfrac{1}{8}\times 50000 = \) Rs 6250; left \(= 43750\)
Ratio sons : daughter \(= (2\times 2):1 = 4:1\), sum \(=5\)
2 sons \(= \dfrac{4}{5}\times 43750 = 35000\) (Rs 17 500 each)
Daughter \(= \dfrac{1}{5}\times 43750 = \) Rs 8750
Widow \(= \dfrac{1}{8}\times 50000 = \) Rs 6250; left \(= 43750\)
Ratio sons : daughter \(= (2\times 2):1 = 4:1\), sum \(=5\)
2 sons \(= \dfrac{4}{5}\times 43750 = 35000\) (Rs 17 500 each)
Daughter \(= \dfrac{1}{5}\times 43750 = \) Rs 8750
Partnership
How is partnership profit shared?
If partners invest equal amounts for the same time, profit is divided equally. If they invest different amounts, profit is divided in the ratio of their investments.
Example: A, B, C invest Rs 120 000, 135 000, 150 000; annual profit Rs 56 700.
Ratio \(= 8:9:10\), sum \(=27\)
A \(= \dfrac{8}{27}\times 56700 = \) Rs 16 800
B \(= \dfrac{9}{27}\times 56700 = \) Rs 18 900
C \(= \dfrac{10}{27}\times 56700 = \) Rs 21 000
Ratio \(= 8:9:10\), sum \(=27\)
A \(= \dfrac{8}{27}\times 56700 = \) Rs 16 800
B \(= \dfrac{9}{27}\times 56700 = \) Rs 18 900
C \(= \dfrac{10}{27}\times 56700 = \) Rs 21 000
Vocabulary
Key Terms
Quick reference for the important words in this chapter.
| Term | Meaning |
|---|---|
| Principal \((P)\) | The original money borrowed or lent. |
| Rate \((R)\) | The percentage of interest charged per year (per annum). |
| Amount \((A)\) | The total to be repaid \(= P + I\) (simple) or \(P\left(1+\frac{R}{100}\right)^n\) (compound). |
| Hire purchase | Buying by a deposit plus instalments; costs more than the cash price. |
| Exchange rate | How much of one currency you get for another. |
| Premium | The amount paid for an insurance policy. |
| Depreciation | The fall in the market value of a vehicle over time. |
| Partnership | A business run by two or more people who share the profit. |
Chapter Summary
Formulas at a Glance
Every formula you need for this chapter in one place.
| Topic | Formula |
|---|---|
| Simple interest | \(I = \dfrac{P \times R \times T}{100}\), \(A = P + I\) |
| Compound interest | \(A = P\left(1+\dfrac{R}{100}\right)^{n}\), \(\text{C.I.} = A - P\) |
| Hire purchase | \(\text{HP price} = \text{Downpayment} + (\text{Instalment} \times \text{No.})\) |
| Insurance | \(\text{Premium} = \dfrac{R}{100} \times \text{Insured value}\) |
| Inheritance | Widow \(= \tfrac{1}{8}\); a son inherits twice a daughter's share |
| Partnership | Equal invest → equal profit; else profit in the ratio of investments |
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