MathsICSE Class 9 · Compound Interest

Compound Interest Formula with Yearly Compounding: Amount, Principal, Rate and Time (ICSE Class 9)

Updated 17 October 2026 · Fact-checked

With interest compounded yearly, the amount after n years is A = P(1 + r/100)^n and the compound interest is CI = A − P. If the rate changes each year, multiply the factors (1 + r₁/100)(1 + r₂/100)… instead of using a power. To find P, r or n, set up the formula and solve, often by recognising a perfect square or cube.

Each year the interest is added to the principal, so the interest earned grows (1,000, 1,100, 1,210) and the amount follows A = P(1 + r/100)ⁿ.

Illustration drawn for QuestionBuddy.

Understand Compound Interest Formula: Yearly Compounding, Finding P, Rate and Time

Idea. In compound interest the interest earned each period is added to the principal, so the next period earns interest on the larger sum. Simple interest, by contrast, is always on the original principal.

Year-by-year working. Principal at the start of year 1 is $P$ . After one year the amount is $P (1 + \frac{r}{100})$ . This is the principal for year 2, so after two years $P (1 + \frac{r}{100})^{2}$ , and so on.

Formula.
$ $A = P (1 + \frac{r}{100})^{n}, C I = A - P,$ $
where $r$ is the annual rate per cent and $n$ the number of years.

Different rates in different years. If the rates are $r_{1}, r_{2}, r_{3}$ in successive years,
$ $A = P (1 + \frac{r _{1}}{100}) (1 + \frac{r _{2}}{100}) (1 + \frac{r _{3}}{100}) .$ $

Finding the unknown.

Principal: $P = \frac{A}{( 1 + r /100 ) ^{n}}$ . If only $C I$ is given, $P = \frac{C I}{( 1 + r /100 ) ^{n} - 1}$ .
Rate: compute $A / P = (1 + r /100)^{n}$ and recognise a perfect square or cube (for example $1.21 = 1. 1^{2}$ , $1.331 = 1. 1^{3}$ , $1.1025 = 1.0 5^{2}$ ).
Time: write $A / P$ as a power of $(1 + r /100)$ and read $n$ .

Compare with simple interest. For the first year, $C I = S I$ . From the second year, compound interest is larger. For comparing plans, compute each plan's interest in rupees and then decide; do not assume compounding is always better, because a higher simple-interest rate can still win over a short period.

Exam care. Do not round intermediate factors; keep the decimal exactly ( $1.0 5^{3} = 1.157625$ ). Give amounts and interest in rupees, and state which one you found.

Level. Class IX is a school-exam class. The unequal-rates, reverse-finding and plan-comparison items here are at the hardest school-exam level and a few touch foundation-olympiad style.

Compound Interest Formula: Yearly Compounding, Finding P, Rate and Time: formulas to remember

Amount: $A = P (1 + \frac{r}{100})^{n}$; Annual compounding.
Compound interest: $C I = A - P$; Interest earned over n years.
Varying rates: $A = P (1 + \frac{r _{1}}{100}) (1 + \frac{r _{2}}{100}) \dots$; One factor per year.
Principal: $P = \frac{A}{( 1 + r /100 ) ^{n}}$; Or P = CI divided by (factor − 1).
Common factors: $1. 1^{2} = 1.21, 1. 1^{3} = 1.331, 1.0 5^{2} = 1.1025$; Recognise these when finding r or n.

How to solve Compound Interest Formula: Yearly Compounding, Finding P, Rate and Time questions

Write the formula with the numbers in place, then evaluate the factor carefully.

1Note P, r, n (or A) and the compounding period (here yearly).
2Write $A = P (1 + r /100)^{n}$ with the values substituted.
3Evaluate the power exactly; for unequal rates multiply the factors year by year.
4Subtract P to get $C I$ if interest is asked.
5To find an unknown, rearrange and match the ratio $A / P$ with a known power.
6Write the answer with the rupee sign and state whether it is amount or interest.

Compound Interest Formula: Yearly Compounding, Finding P, Rate and Time 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.

A savings plan that grows on itself

Suppose a family puts ₹5,000 in a deposit that pays 10% a year, compounded yearly, and leaves it for 2 years.

Find the amount and the compound interest.

Try it first, then show the solutionSolution

$A = 5000 \times 1. 1^{2} = 5000 \times 1.21 = 6050$ .
$C I = 6050 - 5000 = 1050$ .

Answer: Amount ₹6,050; compound interest ₹1,050 (simple interest would be ₹1,000).

Memory hook: Year 2 earns interest on year 1 interest too.

Choosing between two plans

Imagine a shop owner choosing between compound interest at 8% a year and simple interest at 9% a year on ₹15,000 for 2 years.

Which plan gives more interest and by how much?

Try it first, then show the solutionSolution

CI plan: $15000 \times (1.0 8^{2} - 1) = 15000 \times 0.1664 = 2496$ .
SI plan: $15000 \times 0.09 \times 2 = 2700$ .
Difference $= 2700 - 2496 = 204$ .

Answer: The 9% simple-interest plan gives ₹204 more over 2 years.

Memory hook: Compare in rupees, not in labels.

Tricks to remember Compound Interest Formula: Yearly Compounding, Finding P, Rate and Time

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

Power, not product
n years means a power of the factor, not n times r.
Use it when: Stops the simple-interest habit.
Perfect powers
Learn 1.1, 1.21, 1.331; 1.05, 1.1025, 1.157625; 1.08, 1.1664.
Use it when: Quick rate and time questions.
A minus P
CI is the amount minus the principal; do not forget the subtraction.
Use it when: Final line of every CI question.

Compound Interest Formula: Yearly Compounding, Finding P, Rate and Time: solved examples

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

Example 1: Compound interest in two years · 1 mark · MCQBoard level

The compound interest on ₹5,000 for 2 years at 10% per annum, compounded yearly, is:
(a) ₹1,000
(b) ₹1,050
(c) ₹1,100
(d) ₹6,050

Try it first, then show the solutionSolution

$A = 5000 \times 1. 1^{2} = 6050$ .
$C I = 6050 - 5000 = 1050$ .

Answer: (b) ₹1,050

Example 2: Three years at one rate · 3 marks · Short answerBoard level

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

Try it first, then show the solutionSolution

$A = 8000 \times (1.05)^{3} = 8000 \times 1.157625 = 9261$ .
$C I = 9261 - 8000 = 1261$ .

Answer: Amount ₹9,261; CI ₹1,261.

Example 3: Different rates in different years · 3 marks · Short answerHarder board level

A sum of ₹10,000 is invested at 10% for the first year and 12% for the second year, compounded yearly. Find the amount and the interest.

Try it first, then show the solutionSolution

$A = 10000 \times 1.10 \times 1.12 = 10000 \times 1.232 = 12320$ .
$C I = 12320 - 10000 = 2320$ .

Answer: Amount ₹12,320; interest ₹2,320.

Example 4: Principal from the interest · 4 marks · Short answerHarder board level

The compound interest on a sum for 2 years at 10% per annum is ₹2,100. Find the sum, and the amount after 2 years.

Try it first, then show the solutionSolution

$C I = P (1. 1^{2} - 1) = P \times 0.21 = 2100$ .
$P = \frac{2100}{0.21} = 10000$ .
Amount $= 10000 + 2100 = 12100$ (check: $10000 \times 1.21 = 12100$ ).

Answer: Sum ₹10,000; amount ₹12,100.

Example 5: Rate and time from the amount · 5 marks · Long answerHOTS / competency

₹4,000 amounts to ₹4,840 in 2 years at compound interest compounded yearly. (a) Find the rate per cent. (b) Using the same rate, find in how many years ₹4,000 will amount to ₹5,324.

Try it first, then show the solutionSolution

(a) $\frac{4840}{4000} = 1.21 = (1 + \frac{r}{100})^{2}$ , so $1 + \frac{r}{100} = 1.1$ and $r = 10%$ .
(b) $\frac{5324}{4000} = 1.331 = 1. 1^{3}$ , so $n = 3$ years.
Check: $4000 \times 1. 1^{3} = 5324$ .

Answer: (a) 10% per annum (b) 3 years.

Where students lose marks in Compound Interest Formula: Yearly Compounding, Finding P, Rate and Time

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

Mistake: Using simple interest by habit
Compound interest is a power, not n times the rate.
Wrong working
$A = 5000 (1 + 2 \times 0.1) = 6000$ .
Right working
$A = 5000 \times 1. 1^{2} = 6050$ .
Do this: Write the power in the formula first.
Usually costs: Whole question.
Mistake: Giving the amount as the interest
CI is A minus P.
Wrong working
$C I = ₹6, 050$ .
Right working
$C I = 6050 - 5000 = ₹1, 050$ .
Do this: Read what is asked: amount or interest.
Usually costs: Final mark.
Mistake: Rounding the factor too early
Rounding 1.157625 to 1.16 changes the answer.
Wrong working
$8000 \times 1.16 = 9280$ .
Right working
$8000 \times 1.157625 = 9261$ .
Do this: Keep the exact decimal.
Usually costs: Accuracy mark.

Exam tips

Class IX is examined in school papers; plan-comparison and reverse-finding items here are at the hardest school-exam level.
Keep the exact powers; they are short (1.1² = 1.21, 1.1³ = 1.331, 1.05³ = 1.157625), so do not round them.
In reverse problems, divide A by P first; a recognisable number (1.21, 1.331) gives the rate immediately.
Always finish with the unit: rupees, per cent or years.

Practice questions on Compound Interest Formula: Yearly Compounding, Finding P, Rate and Time

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 of ₹2,000 for 2 years at 5% per annum compounded yearly is:
(a) ₹2,200
(b) ₹2,205
(c) ₹2,210
(d) ₹2,250
Hint
$2000 \times 1.0 5^{2}$ .
Answer
(b). $2000 \times 1.1025 = 2205$ .
Q2 · 3 marks · Short answerHarder board level
A sum amounts to ₹5,832 in 2 years at 8% per annum compounded yearly. Find the sum.
Hint
$P = 5832/1.0 8^{2}$ .
Answer
$1.0 8^{2} = 1.1664$ , so $P = 5832/1.1664 = ₹5, 000$ .
Q3 · 4 marks · Short answerHarder board level
Find the compound interest earned on ₹6,000 in each of the three years at 10% per annum, and the total. Verify using the formula.
Hint
Year-by-year principal: 6000, 6600, 7260.
Answer
Year 1: ₹600; year 2: ₹660; year 3: ₹726; total ₹1,986. Formula: $6000 \times (1.331 - 1) = 1986$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): For a deposit held for one year, compound interest and simple interest at the same rate are equal.
Reason (R): In the first year, interest is calculated on the original principal.
(a) Both (A) and (R) are true and (R) explains (A).
(b) Both are true but (R) does not explain (A).
(c) (A) is true, (R) is false.
(d) (A) is false, (R) is true.
Hint
Compounding adds interest to the principal only after a year.
Answer
(a)
Q5 · 4 marks · Case-basedHOTS / competency
Case study: Meera deposits ₹15,000 in a bank at 8% per annum compounded yearly for 2 years. Her cousin invests the same sum for 2 years at 9% per annum simple interest.
(a) Find Meera's amount.
(b) Find the simple interest earned by her cousin.
(c) Whose interest is greater, and by how much?
Hint
Compound: $15000 \times 1.1664$ .
Answer
(a) ₹17,496 (b) ₹2,700 (c) The cousin earns ₹2,700 against Meera's ₹2,496, which is ₹204 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.A principal P earns r% per annum, compounded yearly, for n years. What is the amount?Show answer
A = P(1 + r/100)^n.
2.How do you get compound interest from the amount?Show answer
CI = A − P.
3.What is 1.1 cubed?Show answer
1.331.
4.When are CI and SI equal?Show answer
For one year, at the same rate.

Compound Interest Formula: Yearly Compounding, Finding P, Rate and Time: frequently asked questions

Is compound interest always more than simple interest?

At the same rate and for more than one year, yes; but a plan with a higher simple rate can still earn more over a short time.

What if the rate is different each year?

Multiply the yearly factors one by one.

Can I use the formula for a fraction of a year?

For ICSE Class IX, time is taken in whole years (and half-years, covered in the next topic).