Simple interest

Every time you borrow money, you pay a little extra for the privilege of using it before you have earned it yourself; every time you deposit money in a bank, the bank pays you the same courtesy. That extra amount is interest, and the simplest way to calculate it is called simple interest (SI) — a fixed slice of the original sum, charged fresh every year, for as long as the money stays borrowed. This chapter builds directly on percentages: the "rate" in every interest problem is nothing but a percentage applied year after year to a fixed Principal (P). A Class 8 student is expected to compute the SI and the Amount (A) to be repaid for any Principal, Rate and Time — including when the time is given in months rather than years — and, just as importantly, to work backwards: given any three of P, R, T and SI, find the missing fourth quantity. Simple interest is also the launchpad for the next big idea in the chapter, compound interest, so getting comfortable with it now pays off later.

The core insight

Simple interest is calculated only on the original principal every year, so $SI = \dfrac{P \times R \times T}{100}$ grows in a straight line with time, and the Amount owed is always $A = P + SI$ .

The core idea

1. Principal, Rate, Time, Amount. Four quantities run through every simple interest problem: the Principal (P), the original sum of money; the Rate (R), the percent charged per year (per annum); the Time (T), the duration in years; and the Amount (A), the total sum finally repaid — principal plus interest.

2. The interest is fixed each year. Under simple interest, interest is charged only on the original principal, never on interest already earned. That is exactly what makes it "simple": the interest for year 1, year 2, year 3, ... is identical every single time.

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

where: $P$ is the principal (in rupees), $R$ is the rate of interest per annum, and $T$ is the time in years.

3. Amount is principal plus interest.

A = P + SI

Since $SI = A - P$ as well, knowing any two of $A$ , $P$ , $SI$ instantly gives the third by simple addition or subtraction — no formula needed.

4. Months must become years. Because $R$ is a rate per annum, any time stated in months has to be converted to years by dividing by 12 before it goes into the formula — for instance, 8 months becomes $\frac{8}{12} = \frac{2}{3}$ year. Skipping this step is the single most common slip in the whole chapter.

5. Reading the straight line. Because the yearly interest amount, $\frac{PR}{100}$ , never changes, plotting Amount owed against Time gives a straight line — it rises by the same step every year. This "always the same step" behaviour is the visual signature of simple interest, and it is exactly what distinguishes it from compound interest, where each year's step gets a little bigger.

Figure

Amount owed rising in equal straight-line steps year after year under simple interest, contrasted with a curve that bends upward for compound interest

Derivation AWhy simple interest is PRT over 100

Show that the simple interest earned over $T$ years is always $T$ times the interest earned in a single year, which is why the amount owed rises by an identical step every year and why $SI = \dfrac{PRT}{100}$ correctly captures that behaviour for any principal, rate or time — including fractional years.

Step 1. Start with one year. If a Rate of $R$ percent is charged on a Principal $P$ for one year, the interest for that single year is $R$ percent of $P$ , which is $\dfrac{P \times R}{100}$ .

Step 2. Recognise that every later year repeats the same charge. Under simple interest, the principal used for the calculation never changes — it stays $P$ , not $P$ plus whatever interest has already accumulated. So the interest for year 2 is again $\dfrac{PR}{100}$ , and likewise for year 3, year 4, and so on.

Step 3. Add up $T$ identical yearly interests. For $T$ complete years, the total interest is simply the one-year interest added to itself $T$ times: $SI = T \times \dfrac{PR}{100} = \dfrac{P \times R \times T}{100}$ .

Step 4. Check that this also works for fractional time. If $T$ is not a whole number of years — say $2.5$ years — the same formula still applies, because interest accrues continuously at the steady rate $\dfrac{PR}{100}$ per year; half a year simply contributes half of one year's interest. This is why time in months must first be converted to a fraction of a year before substitution, and why the formula $SI = \dfrac{PRT}{100}$ is written once and used for every case.

Worked example

Find the simple interest and the Amount on a Principal of ₹12,000 lent at 9% per annum for 3 years, then use the same rate to find how much a friend borrowed if she owed ₹6,900 after 2 years 6 months.

Step 1. Identify the given values for the first part. $P = 12{,}000$ , $R = 9$ , $T = 3$ .

Step 2. Apply the SI formula.

SI = \dfrac{P \times R \times T}{100} = \dfrac{12{,}000 \times 9 \times 3}{100} = \dfrac{324{,}000}{100} = 3{,}240

Step 3. Find the Amount.

A = P + SI = 12{,}000 + 3{,}240 = 15{,}240

So the simple interest is ₹3,240, and ₹15,240 must be repaid in total after 3 years.

Step 4. Now reverse the problem. Rohan borrowed a sum at 8% per annum simple interest. After 2 years 6 months he repaid ₹6,900. Convert the time first: $2$ years $6$ months $= 2\tfrac{6}{12} = 2.5$ years.

Step 5. Express SI in terms of the unknown principal. Let the principal be $P$ .

SI = \dfrac{P \times 8 \times 2.5}{100} = \dfrac{20P}{100} = \dfrac{P}{5}

Step 6. Use the Amount relation to solve.

6{,}900 = P + \dfrac{P}{5} = \dfrac{6P}{5} \implies 6P = 34{,}500 \implies P = 5{,}750

So Rohan borrowed ₹5,750. Check: $SI = \dfrac{5{,}750 \times 8 \times 2.5}{100} = 1{,}150$ , and $5{,}750 + 1{,}150 = 6{,}900$ — matches exactly.

Competitive stretch

Suppose the rate itself changes mid-way: ₹12,000 earns 9% for the first 2 years and then 6% for the final 1 year. Because each year's interest is still computed on the same original ₹12,000 (never on an updated principal), the correct method is to work out each period separately and add: $\frac{12{,}000 \times 9 \times 2}{100} = 2{,}160$ for the first two years, plus $\frac{12{,}000 \times 6 \times 1}{100} = 720$ for the third year, giving a total SI of ₹2,880 — not the single-rate formula applied blindly across all 3 years.

Common traps

Forgetting to convert months to years. Plugging in "9" for a time of "9 months" instead of $\frac{9}{12}$ gives an answer 12 times too large.
Confusing Amount with Interest. A question may ask for the total Amount owed or just the extra Interest charged; always check which one is being asked before writing the final answer.
Compounding by accident. A student used to compound interest might add last year's interest to the principal before computing the next year's interest — under simple interest, every year's interest is calculated on the same original principal, never on a running total.
Applying one rate across changing-rate periods. When different rates apply to different stretches of time, each stretch must be calculated separately on the original principal and then added, as in the competitive stretch above.

Ask the tutor: if rearranging the SI formula to solve for the rate or the time feels shaky, or you want extra practice mixing months-and-years word problems, ask the AI tutor to walk through a few with you step by step.

Why is it called simple interest if the formula has three letters multiplied together

The name refers to how the interest behaves over time, not to how many symbols are in the formula. Because the same original principal is used every single year, the yearly interest never changes — it is a constant, simple, repeating amount. Compound interest, by contrast, recalculates a new principal every year, so its formula and its behaviour are genuinely more complex.

If a problem gives days instead of months, how should I convert the time

Convert days to years the same way as months, but using 365 (or sometimes 360, if the problem specifies a banking convention) as the number of days in a year, so $T = \frac{\text{number of days}}{365}$ . ICSE Class 8 problems almost always use months rather than days, but the underlying idea is identical: express whatever time unit you are given as a fraction of one year before substituting into the SI formula.

Can the Rate or the Time ever come out negative or be found using the same formula rearranged

Yes — since $SI = \frac{PRT}{100}$ connects four quantities, knowing any three lets you solve for the fourth by rearranging: $R = \frac{100 \times SI}{P \times T}$ or $T = \frac{100 \times SI}{P \times R}$ . A negative or nonsensical result usually signals an arithmetic slip or a wrongly converted time unit, since in real interest problems P, R, T and SI are always positive.

Simple interest

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads