Simple Interest — notes, question types & practice · Interest (SI & CI)

01

The four words

Principal (P): the money lent, borrowed or invested.
Rate (R): the yearly charge, written as a per cent per annum.
Time (T): the period, always counted in years.
Interest (SI): the charge for the whole period.

Borrow Rs 100 for a year at 8% and you pay Rs 8 extra. Keep it three years and you pay Rs 24 — the same Rs 8 three times over.

Rule: At simple interest, every year costs the same fixed slice of the original principal. Interest already paid never earns interest.

02

The formula and its four faces

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

Cover the one you want and the formula solves itself: $P = \dfrac{100 \times SI}{RT}$ , $R = \dfrac{100 \times SI}{PT}$ , $T = \dfrac{100 \times SI}{PR}$ .

SI on Rs 6,500 at 8% for 3 years: $\dfrac{6500 \times 8 \times 3}{100} = 1560$ .

Tip: Keep T in years. Nine months is $\dfrac{9}{12} = \dfrac{3}{4}$ of a year, never 9.

03

Amount questions

Amount (A) is the total handback: $A = P + SI$ . When a question says "amounts to", strip the principal first: $SI = A - P$ .

A sum amounts to Rs 1,540 in 2 years at 5%. Then $SI = 1540 - P$ and also $SI = \dfrac{P \times 5 \times 2}{100} = \dfrac{P}{10}$ . So $1.1P = 1540$ and $P = 1400$ .

04

Growth is a straight line

The amount climbs by the same rupees every year, so the yearly amounts sit on a straight line. That single fact gives the n-times rule:

R = \frac{100(n-1)}{T}

"Becomes 3 times in 16 years" means interest $= 2P$ over 16 years: $R = \dfrac{200}{16} = 12.5\%$ .

Check the same sum doubles in 8 years: $\dfrac{100}{8} = 12.5\%$ . Same line, same slope. And "5 times at 12%" needs $T = \dfrac{400}{12} = 33\dfrac{1}{3}$ years.

Watch: "Amounts to n times" includes the principal. Only $(n-1)P$ is interest. Using n instead of n−1 picks the trap option.

05

Interest as a fraction of the sum

"The SI is $\dfrac{1}{5}$ of the sum at 4%." Write $\dfrac{P \times 4 \times T}{100} = \dfrac{P}{5}$ and cancel P: $T = \dfrac{100}{5 \times 4} = 5$ years.

The principal never enters the answer. Same shape: SI $= \dfrac{2}{9}$ of P at 8% gives $T = \dfrac{200}{9 \times 8} = 2\dfrac{7}{9}$ years.

06

Comparing two situations

SI is proportional to each of P, R and T. Double the sum and the interest doubles; double the time and it doubles too. "Double P, half R" leaves SI unchanged.

For rupee answers, compute both and subtract: $\dfrac{12000 \times 8 \times 3}{100} - \dfrac{10000 \times 7.5 \times 3}{100} = 2880 - 2250 = 630$ .

07

One loan, two rates

A sum split into two parts at different rates is one linear equation. Split Rs 13,000 so both parts earn equal yearly interest, at 8% and 12%: $8a = 12(13000 - a)$ , so $a = 7800$ . Check: $8\%$ of 7,800 $= 624 = 12\%$ of 5,200.

Note: Equal yearly interest means the smaller rate must hold the larger part.

08

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

Direct simple interest

How to spot it:

P, R and T are given and the SI, or the amount, is asked straight out.

SI = \frac{PRT}{100}, \qquad A = P + SI

Method

Convert T to years (months divided by 12).
Multiply P, R, T and divide by 100.
For the amount, add P back.

Why it works:

Each year charges the same fixed slice of the principal.

Try this

Find the simple interest on Rs 6,500 at 8% per annum for 3 years.

Show solution

$SI = \dfrac{6500 \times 8 \times 3}{100} = 1560$ .
Amount $= 6500 + 1560 = 8060$ .

Answer

Rs 1,560

Type 2very common2 practice Q

Find P, R or T from the interest or amount

How to spot it:

Three of P, R, T, SI (or the amount) are given; the fourth is asked.

P = \frac{100 \times SI}{RT}, \quad R = \frac{100 \times SI}{PT}, \quad T = \frac{100 \times SI}{PR}

Method

If an amount is given, first take SI = A − P.
Rearrange the formula for the missing letter.
Sanity check the size: doubling money in 12 years is under 10%.

Why it works:

One formula links all four quantities, so any three fix the fourth.

Try this

At what rate per annum will Rs 1,250 amount to Rs 2,000 in 12 years at simple interest?

Show solution

$SI = 2000 - 1250 = 750$ .
$R = \dfrac{750 \times 100}{1250 \times 12} = 5$ .

Answer

5% per annum

Type 3very common2 practice Q

n times in T years

How to spot it:

'A sum doubles, triples or becomes n times itself in T years — find the rate or the time.'

R = \frac{100(n-1)}{T}, \qquad T = \frac{100(n-1)}{R}

Method

Interest earned is (n − 1)P — subtract the principal once.
Rate: 100(n−1) over T. Time: 100(n−1) over R.
Chains stay linear: doubling in T years means tripling in 2T.

Why it works:

'Becomes n times' fixes the interest as a multiple of P, and P cancels out.

Try this

A sum of money triples itself in 16 years at simple interest. The rate per annum is:

Show solution

Interest $= (3-1)P = 2P$ in 16 years.
$R = \dfrac{100 \times 2}{16} = 12.5$ .

Answer

12.5%

Type 4common2 practice Q

Interest as a fraction of the principal

How to spot it:

'The SI is 1/5 (or 2/9 ...) of the sum after T years at R% — find T or R.'

\frac{PRT}{100} = \frac{P}{k} \ \Rightarrow\ T = \frac{100}{kR}

Method

Set the fraction equal to PRT over 100.
Cancel P from both sides.
Solve for the missing letter.

Why it works:

The principal appears on both sides, so it never enters the answer.

Try this

The simple interest on a sum at 4% per annum is one-fifth of the sum. The number of years is:

Show solution

$\dfrac{4T}{100} = \dfrac{1}{5}$ .
$T = \dfrac{100}{5 \times 4} = 5$ years.

Answer

5 years

Type 5common2 practice Q

Comparing two SI situations

How to spot it:

Two principals, rates or times are compared — 'how much more interest', or interest on a scaled sum.

\frac{SI_1}{SI_2} = \frac{P_1 R_1 T_1}{P_2 R_2 T_2}

Method

SI is proportional to each of P, R and T — build the ratio.
For a rupee difference, compute both SIs and subtract.
Scaling checks: double P and half R leaves SI unchanged.

Why it works:

The formula is a product, so scaling factors multiply.

Try this

The SI on Rs 12,000 at 8% for 3 years exceeds the SI on Rs 10,000 at 7.5% for 3 years by:

Show solution

$\dfrac{12000 \times 8 \times 3}{100} = 2880$ .
$\dfrac{10000 \times 7.5 \times 3}{100} = 2250$ .
$2880 - 2250 = 630$ .

Answer

Rs 630

Type 6common

One sum split at two rates

How to spot it:

A sum is lent in two parts at different rates; the total or the equal yearly interest is given.

r_1 a = r_2 (P - a) \ \text{(equal interest)}, \quad \frac{r_1 a + r_2 (P-a)}{100} = \text{yearly interest}

Method

Name the first part a and the second P − a.
Write each part's yearly interest: part × rate ÷ 100.
Equate to the given total, or to each other for equal interest.
Solve and check the two interests.

Why it works:

Yearly interest of each part is linear in its size, so the split solves one linear equation.

Try this

Rs 13,000 is lent in two parts, one at 8% and the other at 12% simple interest. The yearly interest from the two parts is equal. Find the part lent at 8%.

Show solution

$8a = 12(13000 - a)$ .
$20a = 156000$ , so $a = 7800$ .
Check: $8\%$ of $7800 = 624 = 12\%$ of $5200$ .

Answer

Rs 7,800

09

Formula sheet

Simple interest

SI = \frac{PRT}{100}

Any one of the four recovers from the other three.

Amount

A = P + SI = P\left(1 + \frac{RT}{100}\right)

'Amounts to' includes the principal.

Recovering inputs

P = \frac{100\,SI}{RT},\quad R = \frac{100\,SI}{PT},\quad T = \frac{100\,SI}{PR}

n-times in T years

R = \frac{100(n-1)}{T}

Interest is only (n−1)P.

Equal yearly interest

SI_{\text{per year}} = \frac{SI_{\text{total}}}{T}

Same rupees every year.

10

Shortcuts that save time

⚡ Cover the unknown

Write $SI = \dfrac{PRT}{100}$ and cover the letter you want. That covered formula is the whole equation.

Example

Find the simple interest on Rs 5,000 at 8% per annum for 3 years.

Show solution

$SI = \dfrac{5000 \times 8 \times 3}{100}$ .
$= 1200$ .

Answer

Rs 1,200

⚡ n-times to rate

'Becomes n times in T years' means the interest earned is $(n-1)P$ . The P cancels.

Example

At what rate per annum will a sum double itself in 10 years at simple interest?

Show solution

Interest $= (2-1)P = P$ in 10 years.
$R = \dfrac{100 \times 1}{10} = 10$ .

Answer

10% p.a.

⚡ One year at a time

Divide the total interest by the years; every year carries exactly that much.

Example

The simple interest on a sum at 4% per annum is two-fifths of the sum. Find the time.

Show solution

$\dfrac{P \times 4 \times T}{100} = \dfrac{2P}{5}$ .
$T = \dfrac{100 \times 2}{5 \times 4} = 10$ years.

Answer

10 years

11

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Plugging the amount A into the SI formula as the principal.

SI is always on the original P; from an amount take SI = A − P first.

Mistake 02

Using n instead of n − 1 in 'becomes n times'.

The interest part is only (n − 1)P, since n times includes the principal.

Mistake 03

Leaving months as months in T.

R is per annum, so T must be in years: 9 months = 3/4 year.

Mistake 04

Letting old interest earn interest.

At simple interest each year costs the same slice of the original principal only.

Mistake 05

Reporting the amount when the question asks for the interest.

A = P + SI; read which of the two the question wants.

12

Quick revision

Read this the night before the exam.

SI $=$ PRT/100; A $=$ P $+$ SI.
Any one of P, R, T, SI from the other three.
n times in T years: R $= \dfrac{100(n-1)}{T}$ .
Growth is linear: equal rupees every year.
SI as a fraction of P: T $=$ 100 $\times$ fraction $\div$ R.
Months to years: divide by 12.
Split loan, equal interest: rate $\times$ part equal on both sides.

13

Practice: 15 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 10 questions

Suggested time 4 min · wrong answers go to your mistake notebook automatically.