MathsICSE Class 9 · Chapter 2

Compound Interest ICSE Class 9: Formula, Half-Yearly Interest, Growth and Depreciation

Updated 15 October 2026 · Fact-checked

Compound interest is interest on the principal plus interest already earned, so the principal grows each period. ICSE Class 9 asks you to work it year by year for 2 or 3 years, use A = P(1 + r/100)ⁿ with CI = A − P, handle half-yearly compounding, compare CI with SI, and solve growth and depreciation.

What Compound Interest is about

With simple interest (SI) the interest each year is worked out on the same original sum. With compound interest (CI) the interest of each period is added to the principal, and the next period's interest is worked out on this bigger amount. So CI is simple interest computed again and again on a growing principal.

The CISCE 2028 Class IX syllabus covers two ways of working: (a) year by year for 2 or 3 years, as repeated simple interest, and (b) the formula $A = P (1 + \frac{r}{100})^{n}$ , with $C I = A - P$ . You must also handle interest compounded half-yearly, find any one of $A$ , $P$ , $r$ , $n$ or CI from the others, solve difference between CI and SI questions, and use the same idea for rate of growth (population, bacteria) and depreciation (value of a machine or car).

Paying back a loan in equal instalments is not in the syllabus.

Class IX is examined by your school (80 marks, 3 hours). Expect 3-mark year-by-year sums, 4-mark 'find the rate and the sum' questions and 4 to 5-mark growth or depreciation problems.

Weightage: CISCE publishes no chapter weightage and no question-paper layout for ICSE Class 9 Mathematics. Class IX is examined by your school: one 80-mark paper of 3 hours plus 20 marks of internal assessment. Compound Interest is in the 2. Commercial Mathematics part of the 2028 syllabus.Source: CISCE ICSE 2028 Mathematics syllabus, Class IX

Key concepts

Compound interest as repeated simple interest

Work one year at a time:
1. Interest for the year $= \frac{P \times r \times 1}{100}$ .
2. Amount at the end of the year $=$ principal $+$ interest.
3. This amount is the principal for the next year.

After the last year, $C I =$ final amount $-$ original principal. This method is the one to use when the question says 'without using the formula', or when the rate changes from year to year and you want to show each year clearly.

Worked example

Find the compound interest on ₹8,000 for 3 years at 10% per annum, working year by year.

Try it first, then show the solutionSolution

Year 1: interest $= 10%$ of 8000 $= 800$ ; amount $= 8800$ .
Year 2: interest $= 10%$ of 8800 $= 880$ ; amount $= 9680$ .
Year 3: interest $= 10%$ of 9680 $= 968$ ; amount $= 10648$ .
$C I = 10648 - 8000 = ₹2, 648$ .

Answer: ₹2,648

The compound interest formula

Each year the money is multiplied by $(1 + \frac{r}{100})$ . After $n$ years:

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

Here $P$ is the principal, $r$ the rate per cent per annum and $n$ the number of years. If the rates for successive years are $r_{1}$ , $r_{2}$ , $r_{3}$ , then $A = P (1 + \frac{r _{1}}{100}) (1 + \frac{r _{2}}{100}) (1 + \frac{r _{3}}{100})$ .

A useful habit: write the multiplier as a fraction, for example $1 + \frac{10}{100} = \frac{11}{10}$ , and cancel before multiplying.

Worked example

Find the amount and CI on ₹15,000 for 2 years if the rate is 8% for the first year and 10% for the second year.

Try it first, then show the solutionSolution

$A = 15000 \times \frac{108}{100} \times \frac{110}{100}$ .
$= 15000 \times 1.08 \times 1.1 = 17820$ .
$C I = 17820 - 15000 = 2820$ .

Answer: A = ₹17,820; CI = ₹2,820

Interest compounded half-yearly

When interest is added every six months, there are two conversion periods in a year. So:

rate per period $= \frac{r}{2} %$ ;
number of periods $= 2 n$ .

$A = P (1 + \frac{r}{200})^{2 n}$ .

For the same yearly rate, half-yearly compounding gives slightly more interest, because interest starts earning interest sooner.

Worked example

Find the CI on ₹12,000 for $1 \frac{1}{2}$ years at 10% per annum compounded half-yearly.

Try it first, then show the solutionSolution

Rate per half-year $= 5%$ ; periods $= 3$ .
$A = 12000 \times (1.05)^{3} = 12000 \times 1.157625 = 13891.50$ .
$C I = 13891.50 - 12000 = ₹1, 891.50$ .

Answer: ₹1,891.50

Finding P, r or n

Put the known values into $A = P (1 + \frac{r}{100})^{n}$ and solve.

Principal: $P = \frac{A}{( 1 + r /100 ) ^{n}}$ . If only CI is given, use $C I = P [(1 + \frac{r}{100})^{n} - 1]$ .
Rate: find $\frac{A}{P}$ , recognise it as a square or cube (for example $\frac{7290}{6250} = \frac{729}{625} = (\frac{27}{25})^{2}$ ), then $1 + \frac{r}{100} = \frac{27}{25}$ , so $r = 8$ .
Time: find $\frac{A}{P}$ and see which power of the multiplier it is.

When the amounts after two consecutive years are known, their ratio is $1 + \frac{r}{100}$ . When the interests for two consecutive years are known, the difference is one year's interest on the earlier interest.

Worked example

In how many years will ₹5,000 amount to ₹6,655 at 10% per annum compound interest?

Try it first, then show the solutionSolution

$\frac{6655}{5000} = \frac{1331}{1000} = (\frac{11}{10})^{3}$ .
Multiplier per year $= \frac{11}{10}$ , so $n = 3$ .

Answer: 3 years

Difference between CI and SI

For the first year, CI and SI are equal. The difference appears from the second year, because CI also earns interest on the first year's interest.

For the same $P$ and $r$ :

2 years: $C I - S I = P (\frac{r}{100})^{2}$ .
3 years: $C I - S I = P (\frac{r}{100})^{2} (3 + \frac{r}{100})$ .

You can always find the difference by working CI and SI separately; the short formulas are a check.

Worked example

The difference between CI and SI on a sum for 2 years at 8% per annum is ₹48. Find the sum.

Try it first, then show the solutionSolution

$P \times (\frac{8}{100})^{2} = 48$ .
$P \times 0.0064 = 48$ , so $P = 7500$ .

Answer: ₹7,500

Rate of growth and depreciation

The compound interest idea works for anything that grows or shrinks by a fixed percentage each year.

Growth (population, bacteria, prices): $V_{n} = V_{0} (1 + \frac{r}{100})^{n}$ .
Depreciation (value of a machine, car or phone): $V_{n} = V_{0} (1 - \frac{r}{100})^{n}$ .

If the rate changes, multiply the year factors: a 10% rise followed by a 10% fall gives $\frac{11}{10} \times \frac{9}{10} = 0.99$ , a net fall of 1%, not zero.

Worked example

A machine bought for ₹50,000 depreciates at 10% per annum. Find its value after 2 years.

Try it first, then show the solutionSolution

$V = 50000 \times (1 - \frac{10}{100})^{2} = 50000 \times 0.81$ .
$V = ₹40, 500$ .

Answer: ₹40,500

Maths in real life: 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.

Fixed deposit with yearly compounding

Suppose you put your ₹20,000 savings in a deposit that pays 8% per annum compounded annually, and you leave it for 2 years.

How much will you get back, and how much more is it than simple interest would give?

Try it first, then show the solutionSolution

$A = 20000 \times (1.08)^{2} = 20000 \times 1.1664 = 23328$ .
SI for 2 years $= \frac{20000 \times 8 \times 2}{100} = 3200$ ; CI $= 3328$ .
Extra $= 3328 - 3200 = 128$ , which equals $20000 \times (0.08)^{2}$ .

Answer: ₹23,328; ₹128 more than with simple interest.

Memory hook: The extra is interest on the first year's interest.

Depreciation of a scooter

Imagine a family buys a scooter for ₹80,000 and its value falls by 15% every year.

What would it sell for after 2 years?

Try it first, then show the solutionSolution

$V = 80000 \times (0.85)^{2}$ .
$= 80000 \times 0.7225 = 57800$ .

Answer: ₹57,800

Memory hook: Each year you keep 85%, not lose 15% of the original.

Growth then fall

Suppose a village has 20,000 people. Its population rises by 10% in one year and then, after families move to the city, falls by 10% the next year.

Is the population back to 20,000?

Try it first, then show the solutionSolution

After year 1: $20000 \times 1.1 = 22000$ .
After year 2: $22000 \times 0.9 = 19800$ .

Answer: No; it is 19,800, a fall of 1%.

Memory hook: +10% then −10% is never a return to the start.

Topics in this chapter

Compound Interest formula sheet

Amount (annual compounding): $A = P (1 + \frac{r}{100})^{n}$; r % per annum, n years.
Compound interest: $C I = A - P$; Always subtract the original principal.
Half-yearly compounding: $A = P (1 + \frac{r}{200})^{2 n}$; Half the rate, twice the periods.
Different rates each year: $A = P (1 + \frac{r _{1}}{100}) (1 + \frac{r _{2}}{100})$; Extend for a third year.
CI − SI for 2 years: $C I - S I = P (\frac{r}{100})^{2}$; Same P and r.
CI − SI for 3 years: $C I - S I = P (\frac{r}{100})^{2} (3 + \frac{r}{100})$; Check by working both separately.
Depreciation: $V_{n} = V_{0} (1 - \frac{r}{100})^{n}$; Growth uses 1 + r/100.

How the school exam asks Compound Interest

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

Question type	How it is asked	Method
Year-by-year CI (2 or 3 years)	Find the CI without using the formula, often with a table of principal, interest and amount for each year.	Interest on the current principal; add; carry forward. Show each year's line.
Direct formula	Find the amount and CI for given P, r, n, sometimes with different rates for each year.	Write the multipliers as fractions, cancel, then multiply.
Half-yearly	Interest compounded every six months for 1, 1½ or 2 years.	Halve the rate, double the number of periods.
Find P, r or n	Given the amount or CI, or amounts in two consecutive years, find the sum, the rate or the time.	Ratio of amounts gives the multiplier; recognise squares and cubes such as 1.21, 1.331, 1.1664.
CI − SI	Given the difference for 2 or 3 years, find the sum or the rate.	Use P(r/100)² for 2 years; for 3 years work both or use the 3-year formula.
Growth and depreciation	Population of a town, value of a car, number of bacteria, with one rate or with a rise and a fall.	Multiply by (1 + r/100) for each year of growth and (1 − r/100) for each year of fall.

Tricks to remember Compound Interest

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

Interest earns interest
SI: same base every year. CI: the base snowballs.
Use it when: Explaining why CI > SI after year 1.
Half the rate, double the time
Half-yearly? r becomes r/2 and n becomes 2n.
Use it when: Half-yearly compounding.
Multiplier cubes
1.1² = 1.21, 1.1³ = 1.331; 1.05² = 1.1025; 1.08² = 1.1664. Spot them in A/P.
Use it when: Finding r or n quickly.
Up is plus, down is minus
Growth: 1 + r/100. Depreciation: 1 − r/100.
Use it when: Population and value problems.

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: CI equals SI when · 1 mark · MCQExam level

On the same sum at the same rate, compound interest compounded annually equals simple interest for:
(a) 1 year
(b) 2 years
(c) 3 years
(d) every period

Try it first, then show the solutionSolution

In the first year both are calculated on the same principal.
From the second year, CI also earns interest on interest.

Answer: (a)

Example 2: Year by year with changing rates · 3 marks · Short answerExam level

Without using the formula, find the compound interest on ₹16,000 for 3 years if the rates are 5%, 10% and 15% for the first, second and third years respectively.

Try it first, then show the solutionSolution

Year 1: interest $= 5%$ of 16000 $= 800$ ; amount $= 16800$ .
Year 2: interest $= 10%$ of 16800 $= 1680$ ; amount $= 18480$ .
Year 3: interest $= 15%$ of 18480 $= 2772$ ; amount $= 21252$ .
$C I = 21252 - 16000 = ₹5, 252$ .

Answer: ₹5,252

Example 3: Rate from principal and amount · 3 marks · Short answerExam level

At what rate per cent per annum compound interest will ₹6,250 amount to ₹7,290 in 2 years?

Try it first, then show the solutionSolution

$(1 + \frac{r}{100})^{2} = \frac{7290}{6250} = \frac{729}{625}$ .
$1 + \frac{r}{100} = \frac{27}{25}$ , so $\frac{r}{100} = \frac{2}{25}$ .
$r = 8$ .

Answer: 8% per annum

Example 4: Half-yearly versus yearly · 3 marks · Short answerHarder exam level

Find the difference between the CI on ₹20,000 for 1 year at 8% per annum when compounded (i) yearly (ii) half-yearly.

Try it first, then show the solutionSolution

(i) Yearly: $C I = 8%$ of 20000 $= 1600$ .
(ii) Half-yearly: $A = 20000 \times (1.04)^{2} = 21632$ , so $C I = 1632$ .
Difference $= 1632 - 1600 = ₹32$ .

Answer: ₹32

Example 5: Two consecutive amounts · 4 marks · Short answerHarder exam level

A sum of money amounts to ₹13,230 in 2 years and to ₹13,891.50 in 3 years at compound interest. Find the rate and the sum.

Try it first, then show the solutionSolution

One year's interest on 13230 $= 13891.50 - 13230 = 661.50$ .
$r = \frac{661.50}{13230} \times 100 = 5%$ .
$P \times (1.05)^{2} = 13230$ , so $P = \frac{13230}{1.1025} = 12000$ .

Answer: Rate 5%; sum ₹12,000

Example 6: Interest of the second and third years · 4 marks · Long answerHarder exam level

The compound interest on a sum for the second year is ₹1,320 and for the third year is ₹1,452. Find the rate of interest and the sum.

Try it first, then show the solutionSolution

Third-year interest exceeds second-year interest by $1452 - 1320 = 132$ ; this is interest for one year on ₹1,320.
$r = \frac{132}{1320} \times 100 = 10%$ .
Second-year interest $= 10%$ of the amount after 1 year, so that amount $= 13200$ .
$P \times 1.1 = 13200$ , so $P = 12000$ .

Answer: 10% per annum; ₹12,000

Example 7: CI − SI for 3 years, find the rate · 4 marks · Long answerHOTS / competency

On ₹10,000, the difference between compound interest and simple interest for 3 years is ₹310. Find the rate, given that it is a whole number.

Try it first, then show the solutionSolution

$C I - S I = P (\frac{r}{100})^{2} (3 + \frac{r}{100})$ .
$10000 \times \frac{r ^{2}}{10000} (3 + \frac{r}{100}) = 310$ , so $r^{2} (3 + \frac{r}{100}) = 310$ .
Try $r = 10$ : $100 \times 3.1 = 310$ . ✓
Check: CI $= 10000 (1.331 - 1) = 3310$ , SI $= 3000$ , difference 310.

Answer: 10% per annum

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.

A school library fund

A school sets aside ₹50,000 for its library. The bank offers two plans for 2 years. Plan A: 10% per annum compounded annually. Plan B: 9% per annum compounded half-yearly. The library's existing books, worth ₹40,000, lose 20% of their value every year.

(i) · 3 marks
Find the amount under Plan A after 2 years.
Answer
Answer: ₹60,500
$50000 \times (1.1)^{2} = 50000 \times 1.21 = 60500$ .
(ii) · 3 marks
Find the amount under Plan B after 2 years, to the nearest rupee, and state which plan is better.
Answer
Answer: ₹59,626; Plan A
Rate per half-year 4.5%, 4 periods: $50000 \times (1.045)^{4} = 50000 \times 1.19251860 \dots = 59625.93$ , about ₹59,626. Plan A gives more.
(iii) · 4 marks
Find the value of the existing books after 2 years and the total value of books plus fund under the better plan.
Answer
Answer: ₹25,600; ₹86,100
Books: $40000 \times (0.8)^{2} = 25600$ . Total $= 60500 + 25600 = ₹86, 100$ .

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.

Find the amount and compound interest on ₹12,500 for 2 years at 12% per annum compounded annually. (3 marks)

Step	Marks
$A = 12500 \times (1.12)^{2}$	1
$A = 12500 \times 1.2544 = ₹15, 680$	1
$C I = 15680 - 12500 = ₹3, 180$	1

A sum amounts to ₹9,680 in 2 years and to ₹10,648 in 3 years at compound interest. Find the rate and the sum. (4 marks)

Step	Marks
Interest for the third year on ₹9,680 $= 10648 - 9680 = 968$	1
$r = \frac{968}{9680} \times 100 = 10%$	1
$P (1.1)^{2} = 9680$	1
$P = \frac{9680}{1.21} = ₹8, 000$	1

The population of a town is 64,000. It increases by 5% in the first year, by 10% in the second year and decreases by 5% in the third year. Find the population after 3 years. (5 marks)

Step	Marks
Multipliers: $1.05$ , $1.10$ and $0.95$	1
After year 1: $64000 \times 1.05 = 67200$	1
After year 2: $67200 \times 1.1 = 73920$	1
After year 3: $73920 \times 0.95 = 70224$	1
Population after 3 years $= 70, 224$	1

Where students lose marks in Compound Interest

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

Mistake: Giving the amount when CI is asked
The formula gives $A$ ; the interest is $A - P$ . Students stop one line early.
Wrong working
CI $= 12500 \times (1.12)^{2} = 15680$
Right working
CI $= 15680 - 12500 = 3180$
Do this: Underline 'interest' or 'amount' in the question before starting.
Usually costs: 1 mark.
Mistake: Not halving the rate (or not doubling the time) for half-yearly
Students change one but not the other, for example $(1.10)^{3}$ for 1½ years.
Do this: Write 'rate per half-year = r/2 %, number of half-years = 2n' as a separate line.
Usually costs: Most of the question.
Mistake: Using the original principal every year in year-by-year working
That is simple interest. CI needs the new amount as the next principal.
Do this: Draw a three-column table: principal, interest, amount; carry the amount down.
Usually costs: Every year after the first.
Mistake: Subtracting percentages for successive changes
A 10% rise and 10% fall is treated as 0% change.
Wrong working
$20000 + 10% - 10% = 20000$
Right working
$20000 \times 1.1 \times 0.9 = 19800$
Do this: Multiply year factors; never add or subtract percentages across years.
Usually costs: The final answer.
Mistake: Rounding in the middle of the working
Rounding 1.157625 to 1.16 changes the answer by several rupees.
Do this: Keep exact multipliers (fractions or full decimals) until the last line; round money to 2 decimal places only at the end.
Usually costs: 1 mark for accuracy.

Practice questions on 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 compound interest on ₹10,000 for 2 years at 10% per annum is:
(a) ₹2,000
(b) ₹2,100
(c) ₹2,200
(d) ₹12,100
Hint
1.1² = 1.21.
Answer
(b): $A = 12100$ , $C I = 2100$ .
Q2 · 2 marks · Short answerExam level
Find the amount on ₹4,000 for 1 year at 10% per annum compounded half-yearly.
Hint
5% for 2 periods.
Answer
$4000 \times 1.1025 = ₹4, 410$ .
Q3 · 3 marks · Short answerExam level
Find the sum of money which amounts to ₹9,261 in 3 years at 5% per annum compound interest.
Hint
1.05³ = 1.157625.
Answer
$P = \frac{9261}{1.157625} = ₹8, 000$ .
Q4 · 1 mark · Assertion-reasonHarder exam level
Assertion (A): The CI on a sum for 2 years at 10% exceeds the SI by 1% of the sum.
Reason (R): For 2 years, $C I - S I = P (\frac{r}{100})^{2}$ .
(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
Compute (0.1)².
Answer
(c): $P (0.1)^{2} = 0.01 P$ , which is 1% of the sum.
Q5 · 3 marks · Short answerHarder exam level
The CI on a certain sum for 2 years at 5% per annum is ₹1,025. Find the sum.
Hint
CI = P[(1.05)² − 1].
Answer
$P \times 0.1025 = 1025$ , so $P = ₹10, 000$ .
Q6 · 3 marks · Short answerHarder exam level
Without using the formula, find the CI on ₹25,000 for 2 years at 12% per annum, compounded annually, and the interest for the second year.
Hint
Table of principal, interest, amount.
Answer
Year 1: 3000, amount 28000; year 2: 3360, amount 31360. CI $= ₹6, 360$ ; second-year interest $= ₹3, 360$ .
Q7 · 4 marks · Short answerHarder exam level
A sum is invested for $1 \frac{1}{2}$ years at 8% per annum compounded half-yearly. If the interest for the first half-year is ₹1,000, find the sum and the total interest.
Hint
4% per half-year.
Answer
$P \times 4% = 1000$ , so $P = 25000$ . $A = 25000 \times (1.04)^{3} = 28121.60$ ; CI $= ₹3, 121.60$ .
Q8 · 4 marks · Long answerHarder exam level
The value of a car depreciates by 12% in the first year and 10% in each later year. It was bought for ₹6,00,000. Find its value after 3 years.
Hint
Multiply 0.88 × 0.9 × 0.9.
Answer
$600000 \times 0.88 \times 0.81 = ₹4, 27, 680$ .
Q9 · 4 marks · Long answerHarder exam level
The difference between the CI and SI on a sum for 2 years at 6% per annum is ₹54. Find the sum, and the CI for 2 years.
Hint
P × 0.0036 = 54.
Answer
$P = 15000$ ; $A = 15000 \times 1.1236 = 16854$ , CI $= ₹1, 854$ .
Q10 · 4 marks · Long answerHarder exam level
A bacteria culture grows by 20% every hour. If there are 4,32,000 bacteria now, how many were there 2 hours ago, and how many will there be after 2 hours?
Hint
Divide by 1.2² for the past.
Answer
Two hours ago: $\frac{432000}{1.44} = 3, 00, 000$ . After 2 hours: $432000 \times 1.44 = 6, 22, 080$ .
Q11 · 4 marks · Long answerHOTS / competency
Ravi lends ₹20,000 at 10% per annum compound interest and Meena lends the same sum at $10 \frac{1}{2} %$ per annum simple interest, both for 2 years. Who earns more, and by how much?
Hint
Work both interests fully.
Answer
Ravi: $20000 \times 0.21 = 4200$ . Meena: $\frac{20000 \times 10.5 \times 2}{100} = 4200$ . They earn the same: ₹4,200 each.
Q12 · 4 marks · Long answerHOTS / competency
At what rate per cent per annum compounded annually will a sum become 1.44 times itself in 2 years? At this rate, after how many years will it first exceed double itself?
Hint
1.2² = 1.44; try powers of 1.2.
Answer
$1 + \frac{r}{100} = 1.2$ , so $r = 20%$ . $1. 2^{3} = 1.728$ , $1. 2^{4} = 2.0736 > 2$ : after 4 years.

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 the amount at compound interest.Show answer
A = P(1 + r/100)ⁿ.
2.How do you change r and n for half-yearly compounding?Show answer
Use r/2 % per period and 2n periods.
3.For which year are CI and SI equal?Show answer
The first year.
4.Difference between CI and SI for 2 years?Show answer
P(r/100)².
5.Formula for depreciation?Show answer
V = V₀(1 − r/100)ⁿ.
6.What is (1.1)³?Show answer
1.331.
7.If amounts after 2 and 3 years are A₂ and A₃, what is 1 + r/100?Show answer
A₃/A₂.
8.Is paying back in equal instalments in the Class IX syllabus?Show answer
No, it is excluded.

Best order to study it

Revise simple interest: SI = PRT/100.
Practise year-by-year CI for 2 and 3 years, including changing rates.
Learn the formula and practise half-yearly compounding.
Practise finding P, r and n, and consecutive-year questions.
Finish with CI − SI, growth and depreciation problems.

Quick revision checklist

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

Can you build a year-by-year table for 3 years?
Do you subtract P to get CI?
Do you halve the rate and double the periods for half-yearly?
Can you recognise 1.21, 1.331, 1.1025 and 1.1664 as powers?
Can you find the rate from two consecutive amounts or interests?
Do you use P(r/100)² for the 2-year CI − SI difference?
Do you use (1 − r/100) for depreciation?

Compound Interest: frequently asked questions

When should I use the year-by-year method instead of the formula?

When the question says 'without using the formula', or asks for interest for a particular year (such as the second year). Otherwise the formula is quicker.

Are quarterly or monthly compounding in the Class IX syllabus?

The syllabus names annual and half-yearly compounding. Practise those two.

Do I need logarithms to find n?

No. Class IX questions are set so that A/P is a recognisable power such as 1.331 = 1.1³.

Is depreciation the same as negative interest?

It uses the same formula with a minus sign: the value is multiplied by (1 − r/100) every year.

Are instalment problems in the syllabus?

No. CISCE explicitly leaves out paying back in equal instalments.

Based on: CISCE ICSE 2028 Mathematics (51) syllabus (Classes IX and X); CISCE Regulations for ICSE Examination Year 2028