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Simple and Compound Interest

15 minintermediateModule 2: Quantitative ReasoningContinues from hat-profit-loss-discount
15 min · Module 2: Quantitative Reasoning · Applied arithmetic

Why This Matters

Interest is the percentage change that repeats. Simple interest applies the rate to the original sum every year, so it grows in a straight line. Compound interest applies the rate to the running balance, so each year earns interest on the previous interest, and the growth curves upward.

The test cares about one thing above all: the difference between the two over two years. It has a short formula, it appears constantly, and it separates candidates who understand the compounding from those who only memorised a rate.

A Simple Model

Three quantities and one multiplier drive everything.

simple grows by addition  ·  compound grows by repeated multiplication

The model assumes the rate is fixed and, for compound interest, that the interest is left in the account. If the question says the interest is withdrawn each year, the account behaves like simple interest even when the wording says compound.

The Actual Detail

Simple interest. The rate is charged once per year on the original principal only.

SI = P × R × T ÷ 100    amount = P + SI

Compound interest. The amount is the principal multiplied by the growth factor once per year.

A = P × (1 + r ÷ 100)n    CI = A − P

The two-year difference. Over exactly two years the gap between compound and simple interest has a tidy closed form worth memorising.

CI − SI over two years = P × (r ÷ 100)2

Half-yearly compounding. When interest is compounded every six months, halve the annual rate and double the number of periods. A rate of 10 percent over one year becomes 5 percent over two periods.

TermP = 10000, r = 10%, n = 2 years
Simple interest10000 × 10 × 2 ÷ 100 = 2000
Compound interest10000 × 1.102 − 10000 = 2100
Difference2100 − 2000 = 100, and P × (r ÷ 100)2 = 100

Growth and decay. The same multiplier handles a rising population and a falling price. A quantity growing at r percent per year for n years is multiplied by (1 + r ÷ 100) n times; a quantity shrinking uses (1 − r ÷ 100).

The fastest check on any interest question: for a whole number of years, compound interest is always greater than simple interest over the same principal, rate and time. If your answer has them reversed, the arithmetic is wrong somewhere.
Two common slips. Applying the rate to P for every year of a compound question, which quietly turns it into simple interest; and using the two-year difference formula for a term that is not two years.

A Worked Example

A sum of 10000 is invested at 10 percent per annum for 2 years. Find the simple interest, the compound interest and the difference between them.

Simple interest applies the rate to the original principal each year.

SI = 10000 × 10 × 2 ÷ 100 = 2000

Compound interest multiplies by the growth factor 1.10 twice.

A = 10000 × 1.10 × 1.10 = 12100   so   CI = 12100 − 10000 = 2100

The difference is 2100 − 2000 = 100. The two-year shortcut confirms it: P × (r ÷ 100)2 = 10000 × 0.01 = 100. The extra 100 is the interest earned on the first year interest during the second year.

Try It

What is the simple interest on 5000 at 8 percent per annum for 3 years?

Find the compound interest on 1000 at 10 percent per annum for 2 years.

For a principal of 20000 at 5 percent per annum, what is the difference between compound and simple interest over 2 years?

Find the amount when 10000 is invested at 10 percent per annum compounded half-yearly for 1 year.

Check Your Understanding

Without looking back: write the simple interest formula, and state what happens to the rate and the number of periods under half-yearly compounding.

The answers: SI equals P times R times T divided by 100; and for half-yearly compounding the rate is halved while the number of periods is doubled.

Simple interest divides P × R × T by . For two years, CI − SI equals P × (r ÷ 100) squared, which for P = 10000 and r = 10 gives . Under half-yearly compounding the annual rate is halved and the number of periods is . Compound interest exceeds simple interest over the same term because it earns interest on the itself.

Apply It

A sum of 8000 is invested at 10 percent per annum compound interest for 2 years. Find the amount and the compound interest, then confirm the two-year difference formula.

Show the reasoning

The growth factor is 1.10, applied twice.

A = 8000 × 1.10 × 1.10 = 8000 × 1.21 = 9680

Compound interest is the amount minus the principal: CI = 9680 − 8000 = 1680.

Simple interest would be 8000 × 10 × 2 ÷ 100 = 1600, so the difference is 1680 − 1600 = 80. The shortcut agrees: P × (r ÷ 100)2 = 8000 × 0.01 = 80.

The 80 is the interest on the first year interest of 800, which is 10 percent of 800. This connects the two-year difference formula to the intuition behind compounding, instead of leaving it as a memorised fact.

Connect

That completes the finance-arithmetic family. The next lesson stops teaching and starts training: a timed mixed set of twenty-five quantitative questions, which forces you to choose the right technique under a clock — the only skill that matters on the day.