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Year 10 Maths - 10 Indices and Surds

Comparing Simple and Compound Interest

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Theory

Simple interest is worked out on the original amount only, using \(I = Prn\). Compound interest is simple interest worked out again each period on the growing balance. This Year 10 maths page shows how to find simple interest and the total amount, solve for the principal, rate or time, read simple interest graphs, and compare simple interest with compound interest, built up as repeated simple interest.

The principal (\(P\)) is the amount invested or borrowed. Interest (\(I\)) is the extra money earned on an investment or paid on a loan.

With simple interest, the interest is worked out on the principal only, so the same amount is added every period. The rate is usually per annum (p.a.), which means per year.

The amount or future value at the end is the principal plus the interest: \(A = P + I\). A flat-rate loan charges simple interest on the amount borrowed.

The simple interest formulaFour boxes: I, the interest, equals P, the principal, times r, the rate per period, times n, the number of periods. The rate and the time must use the same period, for example 6 percent per annum for 3 months gives r equals 0.06 and n equals 0.25. I interest P principal r rate per period n number of periods = Γ— Γ— rate and time must use the same period 6% p.a. for 3 months: r = 0.06, n = 0.25
Simple interest is the principal times the rate times the time. Change months to years (or the rate to the same period) first.
Simple and compound interest comparedGraph of the value of 1000 dollars at 12 percent per annum over 6 years. Simple interest grows in a straight line to 1720 dollars. Compound interest curves upward and reaches about 1974 dollars, pulling further ahead each year. 0 1 2 3 4 5 6 1000 1500 2000 compound simple years
Compound interest earns interest on interest, so its graph curves upward and pulls away from simple interest.
\[I = Prn \qquad A = P + I\]
I=Prn,A=P+I

\(r\) is the rate per period as a decimal (\(4.5\% = 0.045\)) and \(n\) is the number of periods.

\[P = \dfrac{I}{rn} \qquad r = \dfrac{I}{Pn} \qquad n = \dfrac{I}{Pr}\]
P=Irn,r=IPn,n=IPr
Match the periods. A yearly rate needs time in years: \(3\) months \(= 0.25\) years. A rate per quarter needs time in quarters: \(2\) years \(= 8\) quarters.
Compound interest is repeated simple interest. Each year, work out one year of simple interest on the balance at the start of that year and add it on. Adding \(r\) of the balance is multiplying by \(1 + r\), so after \(n\) years the value is \(P(1 + r)^n\).
Comparing the two. Simple: \(A = P(1 + rn)\), a straight-line graph. Compound: \(A = P(1 + r)^n\), a curve that bends upward. At the same rate they agree after one year and compound is ahead after that: \(\$2000\) at \(5\%\) p.a. for \(3\) years gives \(\$2300\) simple but \(\$2315.25\) compound.

How to solve a simple interest problem

  1. List \(P\), \(r\) and \(n\), with the rate as a decimal and the time in the same period as the rate.
  2. Substitute into \(I = Prn\). If \(I\) is known, solve the equation for the unknown.
  3. Add the interest to the principal if the total amount, value or amount owed is asked for.

Other questions you will meet

  1. Value known, principal unknown: find the total percentage added, rate \(\times\) time, then divide the value by \(1 +\) that decimal (Example 3).
  2. Value known, time unknown: interest \(=\) value \(-\) principal, then divide by one year's interest.
  3. Days: write the time as a fraction of a year, \(n = \dfrac{\text{days}}{365}\).
  4. Simple against compound: work out each final value (or each total interest) and subtract. If the rates differ, tabulate both values year by year to see when compound first gets ahead.
  5. Comparing offers: change rates to the same period. \(1.2\%\) per quarter is \(4 \times 1.2\% = 4.8\%\) p.a.
  6. Short loans: to give a rate for a few months as a yearly rate, multiply by the number of those periods in a year.
  7. Doubling: the money has doubled when the interest equals the principal, so \(n = 1 \div r\).
  8. Money split between two accounts: let \(x\) be the amount in one account and the rest be (total \(- x\)), then write total interest as an equation.
  9. Graphs: the interest or value grows by the same amount each year. Read values from the graph; the rate is one year's interest \(\div\) principal. Where two value lines cross, the investments are worth the same.
Example 1 β€” Interest and amount over months
Mia invests \(\$8400\) at \(4.5\%\) p.a. simple interest for \(30\) months. Find
(i) the interest earned
(ii) the value of the investment at the end.
Solution

The rate is per year, so write the time in years: \(30\) months \(= \dfrac{30}{12} = 2.5\) years.

\(I\)\(=\)\(Prn\)
\(\)\(=\)\(8400 \times 0.045 \times 2.5\)
\(\)\(=\)\(\mathbf{\$945.00}\)
I=8400×0.045×2.5=945

(ii) The value at the end is the principal plus the interest.

\(A\)\(=\)\(P + I\)
\(\)\(=\)\(8400 + 945\)
\(\)\(=\)\(\mathbf{\$9345.00}\)
Example 2 β€” Finding the rate or the time
(i) \(\$2500\) earns \(\$337.50\) simple interest in \(3\) years. Find the interest rate per annum.
(ii) How long will it take \(\$6200\) at \(3.5\%\) p.a. to earn \(\$542.50\) simple interest? Answer in months.
Solution

Substitute what you know into \(I = Prn\) and solve for the unknown.

(i)

\(337.50\)\(=\)\(2500 \times r \times 3\)
\(337.50\)\(=\)\(7500r\)
\(r\)\(=\)\(337.50 \div 7500\)
\(\)\(=\)\(0.045\)
\(\)\(=\)\(\mathbf{4.5\%} \text{ p.a.}\)

(ii) One year's interest is \(6200 \times 0.035 = \$217.00\).

\(542.50\)\(=\)\(217 \times n\)
\(n\)\(=\)\(542.50 \div 217\)
\(\)\(=\)\(2.5 \text{ years}\)
\(\)\(=\)\(2.5 \times 12\)
\(\)\(=\)\(\mathbf{30} \text{ months}\)
n=IPr=542.50217=2.5
Example 3 β€” Working back to the principal
Sam borrowed some money at \(8.4\%\) p.a. simple interest. After \(15\) months he owed \(\$7735\) in total. How much did he borrow?
Solution

Find the total percentage added over the whole loan, then undo it.

\(n\)\(=\)\(\dfrac{15}{12}\)
\(\)\(=\)\(1.25 \text{ years}\)
\(\text{total rate}\)\(=\)\(8.4\% \times 1.25\)
\(\)\(=\)\(10.5\%\)

So \(\$7735\) is \(110.5\%\) of the amount borrowed.

\(1.105 \times P\)\(=\)\(7735\)
\(P\)\(=\)\(7735 \div 1.105\)
\(\)\(=\)\(\mathbf{\$7000.00}\)
P=77351.105=7000

Check: \(7000 \times 0.084 \times 1.25 = 735\), and \(7000 + 735 = 7735\).

Example 4 β€” Compound as repeated simple interest
\(\$3500\) is invested at \(4\%\) p.a. for \(3\) years.
(i) Find the value each year if the interest is compounded annually, by working out one year of simple interest at a time. Check with \(P(1 + r)^n\).
(ii) How much more interest is earned than with simple interest?
Solution

(i) Each year's interest is \(4\%\) of the balance at the start of that year.

YearStartInterestEnd
\(1\)\(\$3500.00\)\(\$140.00\)\(\$3640.00\)
\(2\)\(\$3640.00\)\(\$145.60\)\(\$3785.60\)
\(3\)\(\$3785.60\)\(\$151.42\)\(\$3937.02\)
\(3500 \times 1.04^3\)\(=\)\(3937.024\)
\(\)\(\approx\)\(\mathbf{\$3937.02}\)

(ii) Compare the total interest:

\(\text{compound } I\)\(=\)\(3937.02 - 3500\)
\(\)\(=\)\(\$437.02\)
\(\text{simple } I\)\(=\)\(3500 \times 0.04 \times 3\)
\(\)\(=\)\(\$420.00\)
\(\text{difference}\)\(=\)\(\mathbf{\$17.02}\)

Simple interest adds \(\$140\) every year; compound interest adds a little more each year because it earns interest on interest.

Common pitfalls

Months are not years. For \(8\) months at a yearly rate, \(n = \dfrac{8}{12}\), not \(8\). Change the time to match the rate.
Percent to decimal. \(3.25\%\) is \(0.0325\), not \(0.325\) or \(3.25\).
Interest or amount? \(I = Prn\) gives the interest only. If the question asks how much is owed or what the investment is worth, add the principal.
Undo a percentage by dividing. If a value already includes the interest, divide by \(1 +\) the total rate. Taking the percentage off the value gives the wrong principal.
Same rate, different totals. At the same rate, compound interest earns more than simple interest after the first year. Do not use \(I = Prn\) for a compound account.

Frequently asked questions

What is the formula for simple interest?

The formula is \(I = Prn\), where \(I\) is the interest, \(P\) is the principal, \(r\) is the rate per period as a decimal and \(n\) is the number of periods.

How do you calculate simple interest for months?

Write the months as a fraction of a year by dividing by \(12\), then use that value for \(n\) with the yearly rate. For example, \(21\) months is \(1.75\) years.

How do you find the interest rate in simple interest?

Rearrange \(I = Prn\) to get \(r = \dfrac{I}{Pn}\). Multiply the decimal by \(100\) to write it as a percentage per annum.

What does p.a. mean?

p.a. stands for per annum, which means per year. A rate of \(5\%\) p.a. adds \(5\%\) of the principal each year with simple interest.

Why is compound interest more than simple interest?

Simple interest is always worked out on the original principal. Compound interest is worked out on the balance, which includes earlier interest, so from the second year on it earns interest on interest.