Financial Maths · Ordinary Level
Compound Interest
Ordinary Level · Part 4 · Tap NEXT to begin
Section 1 of 3
Compound Interest — Future Value
With compound interest, each year’s interest is added on, and the next year earns interest on the new total. One formula does it all.
The formula
$F = P(1 + i)^t$. $P$ = present value (start with), $F$ = future value (end with), $t$ = time in years, $i$ = interest rate as a decimal. Round money to 2 d.p.
Worked example — find the final value
Find the value of €600 invested for $3$ years at $2\%$ per annum.
$P = 600,\ t = 3,\ i = 0.02$
$F = 600(1.02)^3$
€636.72
Worked example — value and interest
€5200 is invested for $2$ years at $3.1\%$. Find the total value and the interest gained.
$F = 5200(1.031)^2 = 5527.397 = $ €$5527.40$
Interest $= 5527.40 - 5200 = $ €$327.40$
€5527.40; interest €327.40
You try
€5000 is borrowed for $3$ years at $8.2\%$ APR. How much is owed at the end of year 3?
$F = P(1+i)^t$ with $P=5000,\ i=0.082,\ t=3$.
$F = 5000(1.082)^3 = 6333.616$
€6333.62
Section 2 of 3
Finding the Present Value
Work backwards
Given the final amount, find what was invested/borrowed with $P = \dfrac{F}{(1+i)^t}$.
Worked example — how much was borrowed
A sum is borrowed for $3$ years at $8.6\%$ APR and €5243.51 is paid back. How much was borrowed?
$F = 5243.51,\ t = 3,\ i = 0.086$
$P = \dfrac{5243.51}{(1.086)^3}$
€4093.86
You try
A sum is invested for $3$ years at $2\%$ AER and is worth €4321.12 at the end. How much was invested?
$P = \dfrac{F}{(1+i)^t}$ with $F=4321.12,\ i=0.02,\ t=3$.
$P = \dfrac{4321.12}{(1.02)^3} = 4071.887$
€4071.89
Section 3 of 3
APR, AER & Changing Rates
Two rates & year-by-year
APR = annual percentage rate = the rate you borrow at. AER = annual equivalent rate = the rate you invest at. If the rate changes each year, do one year at a time — the end of one year is the start of the next.
Worked example — a different rate each year
€2000 is invested for $2$ years: $2\%$ in year 1, $2.5\%$ in year 2. Find the value at the end of each year.
Year 1: $F = 2000(1.02) = $ €$2040$
Year 2: $F = 2040(1.025) = $ €$2091$
€2040 then €2091
You try
€2600 is borrowed for $2$ years at $45.2\%$. Find the total repayment and the interest paid.
$F = 2600(1.452)^2$; interest = $F -$ the amount borrowed.
$F = 2600(1.452)^2 = $ €$5481.59$
Interest $= 5481.59 - 2600 = $ €$2881.59$
€5481.59; interest €2881.59
That’s Part 4.
Compound interest both ways — future value $F = P(1+i)^t$ and present value $P = F/(1+i)^t$ — plus APR/AER and a changing rate year by year. Next: income tax, then foreign exchange.