MathsICSE Class 8 · Simple and Compound Interest

Compound Interest with Yearly Compounding up to 3 Years (ICSE Class 8)

Updated 17 October 2026 · Fact-checked

When interest is compounded yearly, each year's interest is added to the principal for the next year. The amount after $n$ years is $A = P\left(1 + \frac{R}{100}\right)^n$ and the compound interest is $CI = A - P$. The curriculum limit is yearly compounding up to 3 years.

Each year the orange interest joins the teal principal, so the next year's interest is larger and the amount is multiplied by (1 + R/100) every year.

Illustration drawn for QuestionBuddy.

Understand Compound Interest Formula: Yearly Compounding

Simple interest is $S I = \frac{P R T}{100}$ and always works on the original principal. Compound interest works on a principal that grows: after each year the interest earned is added to the principal.

Year-by-year method. Year 1: interest $= \frac{P R}{100}$ ; new principal $= P + \frac{P R}{100} = P (1 + \frac{R}{100})$ . Year 2 starts from this new principal, so after 2 years the amount is $P (1 + \frac{R}{100})^{2}$ , and so on. The CISCE curriculum asks you to arrive at the formula through this pattern and use it for simple problems.

$ $A = P (1 + \frac{R}{100})^{n}, C I = A - P .$ $

Rates that change. If the rates are $R_{1} %, R_{2} %, R_{3} %$ in successive years, then $A = P (1 + \frac{R _{1}}{100}) (1 + \frac{R _{2}}{100}) (1 + \frac{R _{3}}{100})$ .

CI versus SI. For the first year they are equal. For 2 years at the same rate, the difference is $C I - S I = P (\frac{R}{100})^{2}$ , the interest earned on the first year's interest.

Finding the rate or principal. If $A$ , $P$ and $n$ are known, solve $(1 + \frac{R}{100})^{n} = \frac{A}{P}$ by recognising a perfect power, such as $1.1025 = 1.0 5^{2}$ . Successive amounts give the rate directly: $\frac{A _{n + 1}}{A _{n}} = 1 + \frac{R}{100}$ .

Keep the compounding unit consistent: here $R$ is per annum and $n$ in years.

Compound Interest Formula: Yearly Compounding: formulas to remember

Amount (yearly compounding): $A = P (1 + \frac{R}{100})^{n}$; $n$ in years, up to 3.
Compound interest: $C I = A - P$; Interest only, not the total.
Rates changing each year: $A = P (1 + \frac{R _{1}}{100}) (1 + \frac{R _{2}}{100}) (1 + \frac{R _{3}}{100})$; One factor per year.
CI minus SI for 2 years: $C I - S I = P (\frac{R}{100})^{2}$; Same rate both years.

How to solve Compound Interest Formula: Yearly Compounding questions

Find the yearly growth factor, raise it to the number of years, then multiply by P.

1Write $P$ , $R$ and $n$ with units (per annum, years).
2Find the growth factor $1 + \frac{R}{100}$ as a decimal.
3Raise it to the power $n$ (or multiply the separate year factors).
4Multiply by $P$ to get $A$ , and subtract $P$ for $C I$ .
5Check against the year-by-year method for small cases.

Compound Interest Formula: Yearly Compounding in real life

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.

Deposit that earns on its interest

Suppose Rs 20,000 is deposited at 10% per annum compounded yearly and left untouched for 3 years.

Find the amount and the compound interest.

Try it first, then show the solutionSolution

$1. 1^{3} = 1.331$ .
$A = 20000 \times 1.331 = 26620$ ; $C I = 6620$ .

Answer: Amount Rs 26,620; CI Rs 6,620.

Memory hook: Interest earns interest: multiply the same factor each year.

Tricks to remember Compound Interest Formula: Yearly Compounding

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

Factor first
Turn R% into 1 + R/100 and just multiply that factor n times.
Use it when: Any compounding sum.
Subtract at the end
The formula gives the amount; subtract P last to get the CI.
Use it when: Avoid answering with A.

Compound Interest Formula: Yearly Compounding: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Two-year compound interest · 1 mark · MCQBoard level

The compound interest on Rs 10,000 at 10% per annum for 2 years is:
(a) Rs 2,000 (b) Rs 2,010 (c) Rs 2,100 (d) Rs 2,200

Try it first, then show the solutionSolution

$A = 10000 \times 1. 1^{2} = 12100$ ; $C I = 2100$ .

Answer: (c)

Example 2: Three years · 3 marks · Short answerBoard level

Find the amount and compound interest on Rs 8,000 at 5% per annum compounded yearly for 3 years.

Try it first, then show the solutionSolution

$1.0 5^{3} = 1.157625$ .
$A = 8000 \times 1.157625 = 9261$ .
$C I = 9261 - 8000 = 1261$ .

Answer: Amount Rs 9,261; CI Rs 1,261

Example 3: Rates changing · 3 marks · Short answerHarder board level

A sum of Rs 15,000 is invested for 2 years, at 10% per annum for the first year and 20% for the second, compounded yearly. Find the amount and the compound interest.

Try it first, then show the solutionSolution

$A = 15000 \times 1.10 \times 1.20$ .
$1.10 \times 1.20 = 1.32$ ; $A = 19800$ .
$C I = 19800 - 15000 = 4800$ .

Answer: Amount Rs 19,800; CI Rs 4,800

Example 4: Difference between CI and SI · 4 marks · Short answerHarder board level

Find the difference between the compound interest and the simple interest on Rs 12,500 at 8% per annum for 2 years. Verify by calculating both.

Try it first, then show the solutionSolution

$S I = \frac{12500 \times 8 \times 2}{100} = 2000$ .
$A = 12500 \times 1.0 8^{2} = 12500 \times 1.1664 = 14580$ ; $C I = 2080$ .
Difference $= 2080 - 2000 = 80$ .
Check: $P (\frac{R}{100})^{2} = 12500 \times 0.0064 = 80$ .

Answer: Rs 80

Example 5: Finding the rate · 4 marks · Long answerHOTS / competency

Rs 3,200 amounts to Rs 3,528 in 2 years at compound interest compounded yearly. Find the rate of interest per annum, and the extra interest compared with simple interest at the same rate.

Try it first, then show the solutionSolution

$(1 + \frac{R}{100})^{2} = \frac{3528}{3200} = 1.1025 = 1.0 5^{2}$ .
So $\frac{R}{100} = 0.05$ and $R = 5%$ .
$C I = 328$ ; $S I = \frac{3200 \times 5 \times 2}{100} = 320$ .
Extra $= 8$ (check: $3200 \times 0.0 5^{2} = 8$ ).

Answer: $R = 5%$ ; CI exceeds SI by Rs 8.

Where students lose marks in Compound Interest Formula: Yearly Compounding

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

Mistake: Giving the amount as the compound interest
$A$ includes the principal.
Wrong working
$C I = 12100$ for Rs 10,000 at 10% for 2 years.
Right working
$C I = 12100 - 10000 = 2100$ .
Do this: Subtract P as the last line.
Usually costs: Answer mark.
Mistake: Using n x R in the exponent or multiplying the rate by years
The factor is raised to the power n; the rate is not multiplied by n.
Wrong working
$A = 8000 \times (1 + 0.05 \times 3)$ .
Right working
$A = 8000 \times 1.0 5^{3} = 9261$ .
Do this: Write the formula with the exponent before substituting.
Usually costs: Method and answer marks.

Exam tips

Write the formula first, then substitute.
Round only at the end; keep exact decimals such as 1.157625.
For "difference between CI and SI", show both calculations or the formula.

Practice questions on Compound Interest Formula: Yearly Compounding

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 · MCQBoard level
The amount on Rs 2,000 at 10% per annum compounded yearly for 1 year is:
(a) Rs 2,100 (b) Rs 2,200 (c) Rs 2,020 (d) Rs 2,400
Hint
For one year, compound and simple interest are equal.
Answer
(b)
Q2 · 3 marks · Short answerBoard level
Find the compound interest on Rs 6,000 at 5% per annum for 2 years, compounded yearly.
Hint
$1.0 5^{2} = 1.1025$ .
Answer
$A = 6000 \times 1.1025 = 6615$ ; $C I = 615$ .
Q3 · 3 marks · Short answerHarder board level
Find the amount and compound interest on Rs 20,000 at 10% per annum for 3 years, compounded yearly.
Hint
$1. 1^{3} = 1.331$ .
Answer
$A = 20000 \times 1.331 = 26620$ ; $C I = 6620$ .
Q4 · 3 marks · Short answerHarder board level
A sum of Rs 10,000 earns interest compounded yearly at 4%, 5% and 6% in the first, second and third year respectively. Find the amount after 3 years.
Hint
Multiply the three factors.
Answer
$1.04 \times 1.05 = 1.092$ ; $1.092 \times 1.06 = 1.15752$ . $A = 10000 \times 1.15752 = 11575.20$ .
Q5 · 4 marks · Long answerHOTS / competency
A sum amounts to Rs 4,410 in 2 years and Rs 4,630.50 in 3 years at compound interest. Find the rate and the sum.
Hint
Divide successive amounts.
Answer
$\frac{4630.50}{4410} = 1.05$ , so $R = 5%$ . $P = \frac{4410}{1.0 5 ^{2}} = \frac{4410}{1.1025} = 4000$ . The sum is Rs 4,000.

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.Formula for amount with yearly compounding?Show answer
A = P(1 + R/100)^n.
2.CI minus SI for 2 years?Show answer
P(R/100)^2.
3.When are SI and CI equal?Show answer
For the first year.
4.How do you find CI from A?Show answer
CI = A - P.

Compound Interest Formula: Yearly Compounding: frequently asked questions

Why is CI more than SI after one year?

From the second year, interest is earned on the earlier interest as well.

Can the rate differ each year?

Yes; multiply the separate yearly factors.