MathsICSE Class 8 · Chapter 8

Simple and Compound Interest ICSE Class 8: Formulae and Solved Problems

Updated 15 October 2026 · Fact-checked

Simple interest is charged only on the original sum: SI = P × R × T / 100. Compound interest is charged on the principal plus the interest already added, so the amount after n years is A = P(1 + r/100)ⁿ. ICSE Class 8 compounds yearly for up to 3 years or half-yearly for up to 3 steps.

What Simple and Compound Interest is about

When you borrow or deposit money, the extra paid for using it is the interest. The money borrowed or deposited is the principal ( $P$ ), the yearly percentage is the rate ( $R$ or $r$ ), and the amount ( $A$ ) is principal plus interest.

The CISCE Class VIII curriculum asks you to:

use simple interest (SI), where interest is always on the original principal,
find compound interest (CI) step by step, where each year's interest is added to the principal before the next year's interest is worked out,
see the difference between SI and CI,
arrive at the formula $A = P (1 + \frac{r}{100})^{n}$ from the pattern, and use it for simple problems,
compound yearly for up to 3 years or half-yearly for up to 3 steps (1½ years).

In a school exam you will compute SI and CI, find a missing principal, rate or time, compare SI with CI, and handle half-yearly compounding.

Weightage: CISCE publishes no question paper, marking scheme or chapter weightage for Class VIII Mathematics: your school sets and marks the exam. Simple and Compound Interest sits in the Ratio and Proportion theme of the CISCE Curriculum for Upper Primary Classes (VI-VIII).Source: CISCE Curriculum for Upper Primary Classes (VI-VIII), RDCD 2016

Key concepts

Simple interest

For simple interest the principal never changes, so the interest is the same every year:
$SI = \frac{P \times R \times T}{100}$ , and $A = P + SI$ .

Here $T$ is in years: 6 months is $T = \frac{1}{2}$ .

Rearranging gives the missing quantity:

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

Worked example

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

Try it first, then show the solutionSolution

SI $= \frac{5000 \times 8 \times 3}{100} = ₹1200$ .
Amount $= 5000 + 1200 = ₹6200$ .

Answer: SI ₹1,200; amount ₹6,200

Compound interest step by step

With compound interest, interest is added to the principal at the end of each period. The next period's interest is on this larger sum, so the interest grows each year.

Method (yearly):
1. Interest for year 1 on $P$ ; add it to get the principal for year 2.
2. Interest for year 2 on the new principal; add it.
3. Repeat; CI $=$ final amount $-$ original principal.

Set the working out as a small table: principal, interest, amount for each year.

Worked example

Find the compound interest on ₹10,000 at 10% per annum for 2 years, compounded yearly.

Try it first, then show the solutionSolution

Year 1: interest $= 10%$ of 10000 $= ₹1000$ ; amount $= ₹11000$ .
Year 2: interest $= 10%$ of 11000 $= ₹1100$ ; amount $= ₹12100$ .
CI $= 12100 - 10000 = ₹2100$ .

Answer: ₹2,100

How simple and compound interest differ

In the first year (yearly compounding), SI and CI are equal, because both are on $P$ .
From the second year, CI is larger: it also earns "interest on interest".

For 2 years, the difference is exactly the interest on the first year's interest:
$CI - SI = P (\frac{r}{100})^{2}$ .

On ₹10,000 at 10% for 2 years: SI $= ₹2000$ , CI $= ₹2100$ , difference $= ₹100$ (10% of the first year's ₹1000 interest).

Worked example

Find the difference between CI and SI on ₹10,000 at 10% per annum for 2 years.

Try it first, then show the solutionSolution

SI $= \frac{10000 \times 10 \times 2}{100} = ₹2000$ .
CI $= ₹2100$ (from the step-by-step working).
Difference $= ₹100$ ; check: $10000 \times (\frac{10}{100})^{2} = 100$ ✓.

Answer: ₹100

The compound interest formula

Each year the amount is multiplied by the same factor $(1 + \frac{r}{100})$ :

after 1 year: $P (1 + \frac{r}{100})$
after 2 years: $P (1 + \frac{r}{100})^{2}$
after 3 years: $P (1 + \frac{r}{100})^{3}$

This pattern gives $A = P (1 + \frac{r}{100})^{n}$ and CI $= A - P$ .

If the rate changes from year to year, multiply by each year's factor: $A = P (1 + \frac{r _{1}}{100}) (1 + \frac{r _{2}}{100})$ .

Worked example

Use the formula to find the amount and CI on ₹8,000 for 3 years at 5% per annum, compounded yearly.

Try it first, then show the solutionSolution

$A = 8000 \times (\frac{21}{20})^{3} = 8000 \times \frac{9261}{8000}$ .
$A = ₹9261$ .
CI $= 9261 - 8000 = ₹1261$ .

Answer: Amount ₹9,261; CI ₹1,261

Half-yearly compounding

When interest is compounded half-yearly, it is added every 6 months. So:

the rate per step is half the yearly rate: $\frac{r}{2} %$ ,
the number of steps is twice the number of years: 1 year = 2 steps, $1 \frac{1}{2}$ years = 3 steps.

$A = P (1 + \frac{r}{200})^{2 n}$ . In Class VIII you need at most 3 half-yearly steps.

Half-yearly compounding gives slightly more interest than yearly compounding at the same yearly rate, because interest starts earning interest sooner.

Worked example

Find the CI on ₹4,000 for 1 year at 10% per annum, compounded half-yearly.

Try it first, then show the solutionSolution

Rate per half-year $= 5%$ ; number of steps $= 2$ .
After 6 months: $4000 + 200 = ₹4200$ . After 1 year: $4200 + 210 = ₹4410$ .
CI $= 4410 - 4000 = ₹410$ (yearly compounding would give ₹400).

Answer: ₹410

Rates that change from year to year

Sometimes the rate is different each year, for example 8% in the first year and 10% in the second. Work year by year, or multiply by each factor in turn:
$A = P \times \frac{100 + r _{1}}{100} \times \frac{100 + r _{2}}{100}$ .

The order of the rates does not change the final amount, because multiplication is commutative.

Worked example

Find the amount on ₹5,000 for 2 years if the rate is 8% in the first year and 10% in the second, compounded yearly.

Try it first, then show the solutionSolution

After year 1: $5000 \times \frac{108}{100} = ₹5400$ .
After year 2: $5400 \times \frac{110}{100} = ₹5940$ .

Answer: ₹5,940

Maths in real life: Simple and Compound Interest

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Simple interest

Suppose you keep ₹2,000 in a savings scheme that pays 6% simple interest per year for 3 years.

How much interest do you earn?

Try it first, then show the solutionSolution

SI $= \frac{2000 \times 6 \times 3}{100}$ .
SI $= ₹360$ , so you get ₹2,360 back.

Answer: ₹360

Memory hook: Simple interest: same interest every year.

Compound growth

Imagine a town of 20,000 people whose population grows by 5% every year.

What is the population after 2 years?

Try it first, then show the solutionSolution

Growth works like compound interest: $20000 \times (\frac{105}{100})^{2}$ .
$= 20000 \times 1.1025 = 22050$ .

Answer: 22,050 people

Memory hook: Each year's growth is on the new, larger number.

Compound decrease

Suppose a scooter bought for ₹50,000 loses 10% of its value every year.

What is it worth after 2 years?

Try it first, then show the solutionSolution

Each year multiply by $\frac{90}{100}$ .
$50000 \times 0.9 \times 0.9 = ₹40500$ .

Answer: ₹40,500

Memory hook: A yearly fall uses the factor (1 − r/100).

Topics in this chapter

Simple and Compound Interest formula sheet

Simple interest: $SI = \frac{P \times R \times T}{100}$; T in years.
Amount: $A = P + I$; For SI or CI.
Compound amount (yearly): $A = P (1 + \frac{r}{100})^{n}$; n = 1, 2 or 3 in Class VIII.
Compound interest: $CI = A - P$; Subtract the original principal.
Half-yearly: $A = P (1 + \frac{r}{200})^{2 n}$; Half the rate, twice the steps (up to 3).
CI − SI for 2 years: $CI - SI = P (\frac{r}{100})^{2}$; Interest on the first year's interest.

How the school exam asks Simple and Compound Interest

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Simple interest	Find SI and amount, or find P, R or T from the others.	Use SI = PRT/100 and rearrange.
CI by steps	Find the CI on ₹2,500 at 12% for 2 years, compounded annually (without the formula).	Year-by-year table: principal, interest, amount.
CI by formula	Find the amount and CI on ₹12,500 at 4% for 3 years.	A = P(1 + r/100)ⁿ, then CI = A − P.
Half-yearly	Find the CI on ₹16,000 at 10% p.a. for 1½ years, compounded half-yearly.	Halve the rate, double the time: 3 steps at 5%.
Compare or reverse	Difference between CI and SI; find the principal or rate from two amounts.	Use P(r/100)² for 2-year differences; divide consecutive amounts to find the growth factor.

Tricks to remember Simple and Compound Interest

Short memory aids. Say each one aloud once, then use it on a question the same day.

Simple stays, compound grows
Simple interest is the same every year. Compound interest gets bigger every year.
Use it when: Choosing the method.
Half the rate, double the time
For half-yearly compounding, halve the rate and double the number of periods.
Use it when: Half-yearly questions.
Year 1 is a tie
In the first year, yearly CI and SI are the same.
Use it when: Quick checks and assertion questions.
Subtract P at the end
The formula gives the amount. CI is what is left after taking away P.
Use it when: Not giving A as the CI.

Simple and Compound Interest: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Simple interest · 1 mark · MCQExam level

The simple interest on ₹2,000 at 6% per annum for 2 years is:
(a) ₹120
(b) ₹240
(c) ₹247.20
(d) ₹2,240

Try it first, then show the solutionSolution

SI $= \frac{2000 \times 6 \times 2}{100} = ₹240$ .
(c) is the compound interest and (d) is the amount.

Answer: (b)

Example 2: CI year by year · 3 marks · Short answerExam level

Find the compound interest on ₹5,000 for 2 years at 8% per annum, compounded yearly, without using the formula.

Try it first, then show the solutionSolution

Year 1: interest $= ₹400$ ; amount $= ₹5400$ .
Year 2: interest $= 8%$ of 5400 $= ₹432$ ; amount $= ₹5832$ .
CI $= 5832 - 5000 = ₹832$ .

Answer: ₹832

Example 3: Finding the time · 3 marks · Short answerExam level

In how many years will ₹4,000 earn ₹960 as simple interest at 6% per annum?

Try it first, then show the solutionSolution

$T = \frac{100 \times SI}{P \times R}$ .
$T = \frac{100 \times 960}{4000 \times 6} = \frac{96000}{24000}$ .
$T = 4$ years.

Answer: 4 years

Example 4: CI minus SI · 3 marks · Short answerHarder exam level

Find the difference between the compound interest and the simple interest on ₹15,000 for 2 years at 6% per annum.

Try it first, then show the solutionSolution

SI $= \frac{15000 \times 6 \times 2}{100} = ₹1800$ .
CI: $15000 \to 15900 \to 16854$ , so CI $= ₹1854$ .
Difference $= ₹54$ ; check: $15000 \times (0.06)^{2} = 54$ ✓.

Answer: ₹54

Example 5: Half-yearly, 3 steps · 4 marks · Long answerHarder exam level

Find the compound interest on ₹16,000 for $1 \frac{1}{2}$ years at 10% per annum, compounded half-yearly.

Try it first, then show the solutionSolution

Rate per half-year $= 5%$ ; steps $= 3$ .
After 6 months: $₹16800$ ; after 1 year: $₹17640$ .
After $1 \frac{1}{2}$ years: $17640 + 882 = ₹18522$ .
CI $= 18522 - 16000 = ₹2522$ .

Answer: ₹2,522

Example 6: Using the formula · 4 marks · Long answerHarder exam level

Using the formula, find the amount and the compound interest on ₹12,500 for 3 years at 4% per annum, compounded yearly.

Try it first, then show the solutionSolution

$A = 12500 \times (\frac{104}{100})^{3} = 12500 \times (\frac{26}{25})^{3}$ .
$= 12500 \times \frac{17576}{15625} = \frac{4}{5} \times 17576$ .
$A = ₹14060.80$ .
CI $= 14060.80 - 12500 = ₹1560.80$ .

Answer: Amount ₹14,060.80; CI ₹1,560.80

Example 7: From two amounts to the rate · 5 marks · Long answerHOTS / competency

A sum of money at compound interest (compounded yearly) amounts to ₹6,050 in 2 years and to ₹6,655 in 3 years. Find the rate of interest and the sum.

Try it first, then show the solutionSolution

From year 2 to year 3 the amount is multiplied by $1 + \frac{r}{100}$ .
$\frac{6655}{6050} = 1.1$ , so $\frac{r}{100} = 0.1$ and $r = 10%$ .
$P \times (1.1)^{2} = 6050$ , so $P = \frac{6050}{1.21}$ .
$P = ₹5000$ .
Check: $5000 \times 1.331 = 6655$ ✓.

Answer: Rate 10%; sum ₹5,000

Case-study practice (Section B style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 10-mark question split into 3 + 3 + 4 marks.

Saving for college

Suppose Riya's parents deposit ₹6,000 in a bank that pays 10% per annum compound interest, compounded yearly, for 3 years.

(i) · 3 marks
Find the amount at the end of the first and the second year.
Answer
Answer: ₹6,600 and ₹7,260
Year 1: $6000 + 600 = ₹6600$ . Year 2: $6600 + 660 = ₹7260$ .
(ii) · 3 marks
Find the amount after 3 years and the compound interest earned.
Answer
Answer: ₹7,986; CI ₹1,986
Year 3: $7260 + 726 = ₹7986$ (also $6000 \times 1. 1^{3}$ ). CI $= 7986 - 6000 = ₹1986$ .
(iii) · 4 marks
How much more is this than the simple interest for the same 3 years? Explain where the extra comes from.
Answer
Answer: ₹186 more
SI $= \frac{6000 \times 10 \times 3}{100} = ₹1800$ . Extra $= 1986 - 1800 = ₹186$ . It is interest earned on the interest added in earlier years (₹60 in year 2, and ₹66 + ₹60 = ₹126 in year 3).

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

The simple interest on a sum for 3 years at 9% per annum is ₹1,350. Find the sum. (3 marks)

Step	Marks
$P = \frac{100 \times SI}{R \times T}$	1
$P = \frac{100 \times 1350}{9 \times 3}$	1
$P = ₹5000$	1

Find the compound interest on ₹10,000 for 3 years at 10% per annum, compounded yearly, working year by year. (4 marks)

Step	Marks
Year 1: interest ₹1000; amount ₹11000	1
Year 2: interest ₹1100; amount ₹12100	1
Year 3: interest ₹1210; amount ₹13310	1
CI $= 13310 - 10000 = ₹3310$	1

Find the compound interest on ₹8,000 for $1 \frac{1}{2}$ years at 10% per annum, compounded half-yearly. How much more is it than the simple interest? (5 marks)

Step	Marks
Rate per half-year 5%, 3 steps	1
Amounts: ₹8400, ₹8820, ₹9261	1
CI $= 9261 - 8000 = ₹1261$	1
SI $= \frac{8000 \times 10 \times 1.5}{100} = ₹1200$	1
Difference $= ₹61$	1

Where students lose marks in Simple and Compound Interest

The slips examiners see most often, and what to write instead.

Mistake: Giving the amount as the compound interest
The formula gives A, not CI.
Wrong working
CI $= 8000 \times 1.0 5^{3} = ₹9261$
Right working
CI $= 9261 - 8000 = ₹1261$
Do this: Always finish with CI = A − P.
Usually costs: 1 mark.
Mistake: Using the full yearly rate for half-yearly steps
Each half-year earns only half the yearly rate.
Wrong working
Half-yearly at 10%: $4000 \times 1. 1^{2}$
Right working
$4000 \times 1.0 5^{2} = ₹4410$
Do this: Halve the rate and double the number of periods.
Usually costs: Most of the question.
Mistake: Working out every year's interest on P
That is simple interest, not compound interest.
Do this: Each year, find the interest on the previous year's amount.
Usually costs: Most of the question.
Mistake: Taking months as years in SI
T must be in years.
Wrong working
6 months: $T = 6$
Right working
6 months: $T = \frac{1}{2}$
Do this: Convert months to years by dividing by 12.
Usually costs: The answer.
Mistake: Rounding in the middle of a calculation
Rounding each year's amount can make the final CI wrong by a few paise or rupees.
Do this: Keep exact values (paise) until the last line.
Usually costs: 1 mark for accuracy.

Practice questions on Simple and Compound Interest

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
The amount on ₹1,000 at 10% per annum compound interest for 2 years, compounded yearly, is:
(a) ₹1,200
(b) ₹1,210
(c) ₹1,100
(d) ₹1,331
Hint
Multiply by 1.1 twice.
Answer
(b)
Q2 · 2 marks · Short answerExam level
Find the simple interest and the amount on ₹7,500 at 9% per annum for 3 years.
Hint
SI = PRT/100.
Answer
SI $= ₹2025$ ; amount $= ₹9525$ .
Q3 · 3 marks · Short answerExam level
Find the compound interest on ₹2,500 for 2 years at 12% per annum, compounded yearly.
Hint
Year-by-year table.
Answer
Amounts ₹2800 and ₹3136; CI $= ₹636$ .
Q4 · 2 marks · Short answerHarder exam level
At what rate of simple interest will ₹2,400 amount to ₹3,000 in 5 years?
Hint
SI = 600.
Answer
$R = \frac{100 \times 600}{2400 \times 5} = 5%$ .
Q5 · 1 mark · Assertion-reasonHarder exam level
Assertion (A): For the first year, compound interest (compounded yearly) equals simple interest on the same sum at the same rate.
Reason (R): In the first year, interest is calculated on the principal only.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
What is the principal in year 1 in each case?
Answer
(c)
Q6 · 3 marks · Short answerHarder exam level
Find the compound interest on ₹25,000 for 2 years if the rate is 10% for the first year and 12% for the second year.
Hint
Multiply by 1.10, then by 1.12.
Answer
$25000 \times 1.1 \times 1.12 = ₹30800$ ; CI $= ₹5800$ .
Q7 · 3 marks · Short answerHarder exam level
Find the compound interest on ₹6,250 for 1 year at 8% per annum, compounded half-yearly.
Hint
Two steps of 4%.
Answer
$6250 \to 6500 \to 6760$ ; CI $= ₹510$ .
Q8 · 2 marks · Short answerHarder exam level
Find the difference between the compound interest and simple interest on ₹8,000 for 2 years at 5% per annum.
Hint
$P (\frac{r}{100})^{2}$ .
Answer
$8000 \times (0.05)^{2} = ₹20$ .
Q9 · 3 marks · Short answerHarder exam level
What sum of money will amount to ₹9,261 in 3 years at 5% per annum, compounded yearly?
Hint
$P = A \div (1.05)^{3}$ .
Answer
$P = 9261 \times (\frac{20}{21})^{3} = ₹8000$ .
Q10 · 3 marks · Short answerHarder exam level
Using the formula, find the compound interest on ₹20,000 for 3 years at 10% per annum, compounded yearly.
Hint
$1. 1^{3} = 1.331$ .
Answer
$A = 20000 \times 1.331 = ₹26620$ ; CI $= ₹6620$ .
Q11 · 4 marks · Long answerHOTS / competency
The simple interest on a sum for 2 years at 10% per annum is ₹1,200. Find the compound interest on the same sum for the same time and rate, compounded yearly.
Hint
Find P first.
Answer
$P = \frac{100 \times 1200}{10 \times 2} = ₹6000$ . CI $= 6000 \times (1.21 - 1) = ₹1260$ .
Q12 · 3 marks · Short answerHOTS / competency
Which earns more on ₹10,000 for 1 year: (A) 10% per annum compounded half-yearly, or (B) 10.2% per annum simple interest? By how much?
Hint
Work out both interests.
Answer
(A): $10000 \times 1.0 5^{2} - 10000 = ₹1025$ . (B): $₹1020$ . Option (A) earns ₹5 more.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Write the formula for simple interest.Show answer
SI = P × R × T / 100.
2.Write the formula for the amount under yearly compounding.Show answer
A = P(1 + r/100)ⁿ.
3.For half-yearly compounding, what happens to the rate and the time?Show answer
The rate is halved and the number of periods is doubled.
4.Are SI and CI equal for the first year (yearly compounding)?Show answer
Yes.
5.₹1,000 at 10% CI for 2 years: what is the amount?Show answer
₹1,210.
6.What is the difference between CI and SI on ₹1,000 at 10% for 2 years?Show answer
₹10.
7.How many half-years make 1½ years?Show answer
3.
8.What is P in SI = PRT/100 called?Show answer
The principal.

Best order to study it

Revise simple interest and practise finding P, R and T.
Learn compound interest year by year with a table.
Compare SI and CI and spot the 'interest on interest'.
Arrive at A = P(1 + r/100)ⁿ from the pattern and use it for up to 3 years.
Finish with half-yearly compounding (up to 3 steps) and changing rates.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Did you convert the time into years for SI?
Did you work each year's CI on the new amount?
Did you subtract P to get CI?
Did you halve the rate and double the periods for half-yearly?
Can you show why CI − SI = P(r/100)² for 2 years?
Did you keep exact values until the final answer?

Simple and Compound Interest: frequently asked questions

Why is compound interest more than simple interest?

After the first year, compound interest is also charged on the interest already added, so each year's interest is bigger. Simple interest is always on the original principal.

Must I use the formula for compound interest?

No. The year-by-year method always works and is required when the question says 'without using the formula'. The formula is quicker for 3 years.

What does 'compounded half-yearly' mean?

Interest is added to the principal every 6 months, at half the yearly rate each time.

Does compound interest apply to anything other than money?

The same calculation describes anything that grows (or shrinks) by a fixed percentage each period, such as a population or the value of a machine.

How long can compounding run in Class VIII questions?

Up to 3 years when compounded yearly, or up to 3 half-yearly steps (1½ years) when compounded half-yearly.

Based on: CISCE Curriculum for Upper Primary Classes (VI-VIII), Mathematics (Class VIII); CISCE RDCD: Curriculum Pre-School to Class VIII