Simple and compound interest

Leave ₹15,000 in a bank at 8% per annum and after three years you will have earned either ₹3,600 or ₹3,895.68 — and which one depends entirely on whether the bank pays simple interest or compound interest. Same principal, same rate, same time, different answer, because the two methods disagree about one thing only: whether interest earned in one period is itself allowed to earn interest in the next. Simple interest treats the principal as fixed forever — every year's interest is calculated on the same original sum. Compound interest keeps adding each period's interest back into the principal, so the base on which interest is charged grows a little every time. That single difference is why compound interest overtakes simple interest as soon as more than one period has passed, and why banks, loans, and growth problems in the real world almost always use compounding. This article builds both formulas from first principles, works through yearly and half-yearly compounding side by side with simple interest, and flags the traps — like forgetting to halve the rate for half-yearly compounding — that cost easy marks in this ICSE Class 8 chapter.

The core insight

Simple interest is charged only on the original principal each period, while compound interest is charged on principal plus all previously earned interest, so $A = P(1+\frac{R}{100})^n$ for yearly compounding always gives a result at least as large as $SI = \frac{PRT}{100}$ for the same $P$ , $R$ , $T$ .

The core idea

Simple interest (SI) is the plainest case: interest is calculated only on the original principal $P$ , at rate $R\%$ per annum, for time $T$ years. It never changes from period to period, so the total interest simply scales with time.

SI = \frac{P \times R \times T}{100}

where: $P$ is the principal (the sum originally lent or invested), $R$ is the rate of interest per annum as a percentage, $T$ is the time in years, and $SI$ is the simple interest earned. The amount owed or received after time $T$ is just principal plus interest:

\text{Amount} = P + SI

Compound interest (CI) works differently: at the end of each period, the interest earned is added to the principal, and the next period's interest is calculated on this new, larger sum. For yearly compounding over $n$ years at rate $R\%$ per annum:

A = P\left(1 + \frac{R}{100}\right)^{n}

where: $A$ is the final amount after $n$ years, and the compound interest itself is what has been added on top of the principal, $CI = A - P$ . For half-yearly compounding, each compounding period is six months, so the rate charged per period is half the annual rate, and the number of periods in $n$ years is double the number of years:

A = P\left(1 + \frac{R}{200}\right)^{2n}

where: $\frac{R}{200}$ is the half-yearly rate (half of $\frac{R}{100}$ ) and $2n$ is the number of half-year periods. At this level, yearly compounding is used for up to 3 years and half-yearly compounding for up to 3 half-year steps only — quarterly compounding and fractional time beyond that are not part of this chapter. Because compound interest keeps earning interest on interest already earned, while simple interest never does, $CI \geq SI$ for the same $P$ , $R$ , $T$ whenever more than one compounding period has passed, with equality only in the first period.

Figure

Two growth curves for the same principal showing simple interest rising as a straight line and compound interest rising as a curve that overtakes it after the first year

Derivation ABuilding the compound interest formula from year-by-year growth

Starting from a principal $P$ growing at rate $R\%$ per annum, show step by step why adding each year's interest back into the principal before computing the next year's interest leads directly to the closed-form formula $A = P(1+\frac{R}{100})^n$ , rather than requiring a separate year-by-year calculation every time.

Step 1. After Year 1, the interest earned is $P \times \frac{R}{100}$ , so the new principal becomes $P + P\cdot\frac{R}{100} = P\left(1+\frac{R}{100}\right)$ . Call this new principal $P_1$ .

Step 2. In Year 2, interest is charged on $P_1$ , not on the original $P$ . So the amount after Year 2 is $P_1\left(1+\frac{R}{100}\right) = P\left(1+\frac{R}{100}\right)\left(1+\frac{R}{100}\right) = P\left(1+\frac{R}{100}\right)^2$ .

Step 3. The same multiplication by $\left(1+\frac{R}{100}\right)$ happens again in Year 3, giving $P\left(1+\frac{R}{100}\right)^3$ , and in general, after $n$ years of repeating this same multiplication, the amount is $A = P\left(1+\frac{R}{100}\right)^n$ .

Step 4. For half-yearly compounding, each "period" is six months long, so the interest rate applied per period is half the annual rate, $\frac{R}{2}\%$ , and the same reasoning from Steps 1 to 3 applies with $\frac{R}{200}$ in place of $\frac{R}{100}$ , and with the number of periods doubled to $2n$ , giving $A = P\left(1+\frac{R}{200}\right)^{2n}$ .

Worked example

Find the simple interest and amount on a principal of ₹15,000 at 8% per annum for 3 years, then find the compound interest on the same principal, rate and time with yearly compounding, and compare the two.

Step 1. Calculate simple interest using $SI = \frac{P \times R \times T}{100}$ .

SI = \frac{15000 \times 8 \times 3}{100} = \frac{360000}{100} = ₹3600

\text{Amount (SI)} = 15000 + 3600 = ₹18{,}600

Step 2. Calculate compound interest year by year. In Year 1, interest is $15000 \times \frac{8}{100} = ₹1200$ , giving a new principal of $15000 + 1200 = ₹16{,}200$ . In Year 2, interest is $16200 \times \frac{8}{100} = ₹1296$ , giving a new principal of $16200 + 1296 = ₹17{,}496$ . In Year 3, interest is $17496 \times \frac{8}{100} = ₹1399.68$ , giving a final amount of $17496 + 1399.68 = ₹18{,}895.68$ .

Step 3. Find the compound interest by subtracting the principal from the final amount.

CI = 18895.68 - 15000 = ₹3895.68

Step 4. Compare the two results. $CI - SI = 3895.68 - 3600 = ₹295.68$ . The compound interest is larger because from Year 2 onward, interest is earned on the previous year's interest as well as on the original principal — the amount by which CI exceeds SI is exactly the "interest on interest" accumulated after Year 1.

Competitive stretch

A sum of ₹8,000 is invested at 10% per annum, compounded half-yearly, for $1\tfrac12$ years. Since the rate per half-year is $\frac{10}{2} = 5\%$ and the number of half-year steps is $2 \times 1.5 = 3$ , the amount is $A = 8000\left(1+\frac{10}{200}\right)^3 = 8000(1.05)^3 = 8000 \times 1.157625 = ₹9261$ , so $CI = 9261 - 8000 =$ ₹1,261.

Common traps

Using the annual rate directly in half-yearly compounding. The rate must be halved to $\frac{R}{2}$ and the number of periods doubled to $2n$ — using the full annual rate badly overstates the interest.
Reporting the amount $A$ as if it were the compound interest. Compound interest is always found by subtracting the principal, $CI = A - P$ , never read off directly from the formula.
Assuming CI grows by the same amount every year, like SI does. In compound interest, the rupee amount of interest increases each period because it is calculated on a growing principal, not a fixed one.
Compounding beyond the prescribed limit. This chapter covers yearly compounding up to 3 years and half-yearly compounding up to 3 half-year steps only — quarterly compounding and fractional time periods beyond that are out of scope.

Ask the tutor: if the jump from year-by-year compounding to the direct formula, or the half-yearly rate-and-period adjustment, still feels shaky, ask the AI tutor to walk through one more worked example with you.

Why is compound interest always the same as simple interest in the very first year but different afterward?

In the first period, there has been no earlier interest yet to add to the principal, so both methods are charging interest on the same untouched sum $P$ — they must give identical results. The difference only appears from the second period onward, once compound interest has folded Year 1's interest into a larger principal while simple interest keeps using the original, smaller $P$ .

Do I have to compute compound interest year by year, or can I just use the formula every time?

Either method gives the same answer, and it is worth being comfortable with both. The year-by-year method is safer when a question asks you to show intermediate amounts or when the compounding is irregular, while the formula $A = P(1+\frac{R}{100})^n$ is faster once you trust it, especially for half-yearly problems where tracking each half-year by hand is more error-prone.

If half-yearly compounding uses a smaller rate per period, why does it give more interest than yearly compounding at the same annual rate?

It is true the per-period rate is smaller, $\frac{R}{2}$ instead of $R$ , but there are also more compounding periods, $2n$ instead of $n$ , and interest gets added to the principal twice as often. That extra frequency of "interest earning interest" more than makes up for the smaller per-period rate, so half-yearly compounding always yields a slightly higher amount than yearly compounding for the same nominal annual rate.

Simple and compound interest

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads