Simple interest

Borrow money and you pay for the privilege; lend or invest it and you get paid for the wait. That price of waiting is interest, and the simplest way to charge it is to fix a yearly rate and apply it to the same original sum every single year, never letting the interest itself earn more interest. This is simple interest, and it is the natural next step after profit, loss and discount in the chapter Fractions in Disguise — the same percentage skill, now applied across time instead of a single transaction. A bank loan, a fixed deposit, or a friend lending you ₹500 "at 10% a year" all use this idea. Master the one formula here, the amount it produces, and how to run the formula backwards, and every simple-interest question in the chapter falls into place.

The core insight

Simple interest is a fixed percentage of the original principal, charged the same way every year, so it grows in a straight, unchanging line rather than compounding.

The core idea

Three quantities decide any simple-interest problem: the Principal ( $P$ ), the sum originally borrowed or invested; the annual Rate ( $R$ ), a percentage understood as "per year" unless told otherwise; and the Time ( $T$ ), always measured in years for the formula below to work.

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

where: $SI$ is the simple interest earned or owed (in rupees), $P$ is the principal, $R$ is the rate per cent per annum, and $T$ is the time in years. The Amount ( $A$ ) is simply what must finally be repaid or received — principal plus interest:

A = P + SI

Because $SI$ is calculated on the original $P$ every time, not on a growing balance, the same rupee amount of interest is added each year. Three years at 8% on ₹15,000 add exactly ₹1,200 in year one, ₹1,200 in year two, and ₹1,200 in year three — a straight line, unlike compound interest, where each year's interest is computed on the previous year's larger amount. Units matter as much as the arithmetic. $R$ is always a yearly rate, and $T$ must be in years — a period given in months is converted by dividing by 12 (8 months becomes $T = \tfrac{8}{12}$ years), and a period in days is generally converted using 365 days to a year. Skipping this conversion is the single most common source of wrong answers in this topic.

Figure

A bar chart showing yearly interest of 1200 rupees added in year one year two and year three with each bar the same height and a growing total line above marking 1200 2400 and 3600 rupees

Derivation ABuilding the simple interest formula from a rate per year

The goal is to see why $SI = \dfrac{P \times R \times T}{100}$ is the correct formula, by building it up from what "a rate of $R\%$ per annum" actually means, rather than memorising the expression — this makes it easy to rearrange the formula later for $P$ , $R$ , or $T$ when any one of them is the unknown.

Step 1. Read what $R\%$ per annum means. A rate of $R\%$ per annum means that for every ₹100 of principal, ₹ $R$ is charged as interest for one full year. So the interest for one year on a principal $P$ is $R$ rupees for every 100 rupees of $P$ , which is the fraction $\dfrac{R}{100}$ of $P$ : one year's interest is $P \times \dfrac{R}{100}$ .

Step 2. Extend from one year to $T$ years. Simple interest never lets the interest itself earn more interest — every year adds exactly the same amount, the one-year interest found in Step 1. Over $T$ years, that one-year interest is simply repeated $T$ times, so the total interest is the one-year interest multiplied by $T$ .

Step 3. Combine the two steps into a single formula. Multiplying the one-year interest by $T$ gives $SI = P \times \dfrac{R}{100} \times T$ , which is the same expression as $SI = \dfrac{P \times R \times T}{100}$ once the multiplication is rearranged — division by 100 can be done at any point without changing the answer.

Step 4. Confirm the formula behaves as expected. If $T = 1$ , the formula reduces to $SI = \dfrac{PR}{100}$ , matching Step 1 exactly. If $R = 0$ , no interest is charged at all, as expected. Doubling $T$ exactly doubles $SI$ , consistent with the same fixed amount being added every year — this straight-line growth is the defining feature of simple interest.

Worked example

Find the simple interest on ₹15,000 borrowed for 3 years at an annual rate of 8%, and find the total amount to be repaid.

Step 1. Identify the known values: $P = 15000$ , $R = 8$ , $T = 3$ .

Step 2. Apply the formula: $SI = \dfrac{P \times R \times T}{100} = \dfrac{15000 \times 8 \times 3}{100}$ .

Step 3. Compute in stages: $15000 \times 8 = 120000$ , then $120000 \times 3 = 360000$ , then $SI = \dfrac{360000}{100} = 3600$ . The simple interest is ₹3,600.

Step 4. Find the amount: $A = P + SI = 15000 + 3600 = 18600$ . The borrower must repay a total of ₹18,600.

Step 5. Check using the one-year method from the derivation: one year's interest is $\dfrac{15000 \times 8 \times 1}{100} = 1200$ , and $1200 \times 3 = 3600$ , matching Step 3.

Competitive stretch

A sum of money amounts to ₹13,000 in 3 years at 10% per annum simple interest. Find the original principal.

Step 1. Write the amount in terms of $P$ alone: $A = P + SI = P\left(1 + \dfrac{R \times T}{100}\right)$ .

Step 2. Substitute the known values: $13000 = P\left(1 + \dfrac{10 \times 3}{100}\right) = P(1.3)$ .

Step 3. Solve: $P = \dfrac{13000}{1.3} = 10000$ . The original principal was ₹10,000 — this "express $A$ in terms of $P$ first" trick handles any reverse problem cleanly.

Common traps

Forgetting to divide by 100 — the rate is a percentage, so $P \times R \times T$ alone is not the interest.
Mixing up Amount and Simple Interest — the amount includes the principal; SI is only the extra money.
Using the time in months or days without converting to years first (6 months means $T = 0.5$ , not $T = 6$ ).
Assuming the interest changes each year — under simple interest it stays exactly the same every year, since it is always computed on the original, unchanging principal.

Ask the tutor: stuck on rearranging the simple interest formula to solve for rate or time instead of interest — ask your AI tutor to walk through isolating the unknown step by step.

Why do we divide by 100 in the simple interest formula?

Because the rate $R$ is given as a percentage, meaning " $R$ out of every 100." To turn a percentage into the actual fraction of the principal it represents, you divide by 100 — the same move used everywhere else with percentages, like profit or discount. Skipping it gives an answer 100 times too large, a very common slip under exam pressure.

Why is simple interest called simple if compound interest also uses a similar formula?

It is "simple" because the interest is always calculated on the same fixed principal, so it forms a straight, unchanging pattern year after year — easy to predict and easy to check by hand, as shown in the worked example. Compound interest instead adds each year's interest back into the principal, so the next year's interest is computed on a larger base and the growth speeds up over time. The two ideas share a family resemblance but grow in very different ways.

How do I find the rate or the time if the interest and principal are already known?

Rearrange the same formula. Since $SI = \dfrac{P \times R \times T}{100}$ , multiply both sides by 100 to get $P \times R \times T = 100 \times SI$ , then divide by whichever two quantities you already know to isolate the third — for example, $R = \dfrac{100 \times SI}{P \times T}$ . The derivation above shows exactly why this formula holds, which makes it much easier to rearrange correctly than trying to memorise several separate formulas.

Simple interest

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads