MathsICSE Class 9 · Compound Interest

Half-Yearly Compounding, Growth and Depreciation, and CI − SI for Two Years (ICSE Class 9)

Updated 17 October 2026 · Fact-checked

For half-yearly compounding use A = P(1 + r/200)^(2n): half the rate, twice the periods. Growth and depreciation use the same power with a plus or minus sign: A = P(1 ± r/100)^n. For 2 years, the difference between compound and simple interest is CI − SI = P(r/100)².

Half-yearly compounding uses ×1.05 twice, not ×1.10 once; the extra ₹50 is interest earned on the first half-year's interest.

Illustration drawn for QuestionBuddy.

Understand Half-Yearly Compounding, Growth and Depreciation, and the Difference Between CI and SI

Half-yearly compounding. When interest is added every six months, the rate per period is $\frac{r}{2} %$ and the number of periods is $2 n$ :
$ $A = P (1 + \frac{r}{200})^{2 n} .$ $
For 18 months, $2 n = 3$ periods. The annual rate $r$ is always quoted per year; never apply the full $r$ to a half-year.

Why more frequent compounding gives more. Compare ₹20,000 for 1 year at 10% per annum: yearly gives $20000 \times 1.1 = 22000$ ; half-yearly gives $20000 \times 1.0 5^{2} = 22050$ . The extra ₹50 is the interest earned on the first half-year's interest.

Growth and depreciation. A quantity that grows by $r %$ each year follows $A = P (1 + r /100)^{n}$ (population, bacteria in a school lab model, a rising price). A quantity that loses $r %$ of its current value each year follows $A = P (1 - r /100)^{n}$ (a machine, a vehicle). If the rate differs year by year, multiply the factors, using $(1 - r /100)$ for a fall. Note that a 10% rise followed by a 10% fall does not return to the start: $1.1 \times 0.9 = 0.99$ .

Difference between CI and SI (2 years).
$ $C I - S I = P (\frac{r}{100})^{2} .$ $
Derivation: $C I = P [(1 + i)^{2} - 1] = P (2 i + i^{2})$ with $i = \frac{r}{100}$ , while $S I = 2 P i$ . The difference is $P i^{2}$ , which is the interest on the first year's interest. Given the difference, $P = \frac{difference}{( r /100 ) ^{2}}$ .

Working tips.

Decide the factor first: half-yearly rate, number of periods, then power.
For percentage change over several years, compare final with initial value: percentage change $= \frac{A - P}{P} \times 100$ .
Keep exact decimals as long as possible; round money to paise only at the end.

Level. In school papers you meet half-yearly and CI − SI in 3 to 4 mark questions. These are typical ICSE Class 9 questions of 1 to 5 marks; the comparison and mixed-rate items are the more demanding ones.

Half-Yearly Compounding, Growth and Depreciation, and the Difference Between CI and SI: formulas to remember

Half-yearly amount: $A = P (1 + \frac{r}{200})^{2 n}$; Rate halved, periods doubled.
Growth: $A = P (1 + \frac{r}{100})^{n}$; Population, price rise.
Depreciation: $A = P (1 - \frac{r}{100})^{n}$; Machine or vehicle value.
CI − SI (2 years): $C I - S I = P (\frac{r}{100})^{2}$; Interest on the first year interest.
Percentage change: $\frac{A - P}{P} \times 100$; Positive growth, negative fall.

How to solve Half-Yearly Compounding, Growth and Depreciation, and the Difference Between CI and SI questions

Fix the rate per period and the number of periods before touching the formula.

1For half-yearly compounding write rate per half-year $= r /2$ and periods $= 2 n$ .
2For growth use plus, for depreciation use minus in the factor.
3Evaluate the power exactly, then multiply by P.
4For CI − SI over 2 years use $P (r /100)^{2}$ directly, or compute both interests and subtract to check.
5Subtract P to get interest, or divide by P for percentage change.
6Round money only at the end and state the unit.

Half-Yearly Compounding, Growth and Depreciation, and the Difference Between CI and SI 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.

Machine losing value

Suppose a workshop buys a machine for ₹80,000 and its value falls by 10% of the previous year's value each year.

What is its value after 2 years and how much has been lost?

Try it first, then show the solutionSolution

$A = 80000 \times 0. 9^{2} = 80000 \times 0.81 = 64800$ .
Loss $= 80000 - 64800 = 15200$ .

Answer: Value ₹64,800; loss ₹15,200 (not ₹16,000).

Memory hook: Each year falls on the new, smaller value.

Town growth

Imagine a town with 50,000 people whose population is modelled to grow 4% each year.

Find the population after 2 years.

Try it first, then show the solutionSolution

$A = 50000 \times 1.0 4^{2} = 50000 \times 1.0816 = 54080$ .

Answer: 54,080 people.

Memory hook: Growth multiplies; it does not add the same amount each year.

Tricks to remember Half-Yearly Compounding, Growth and Depreciation, and the Difference Between CI and SI

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

Halve and double
Half-yearly: halve the rate, double the years.
Use it when: Setting up every half-yearly problem.
Plus up, minus down
Growth factor 1 + r/100; fall factor 1 − r/100.
Use it when: Depreciation questions.
Interest on interest
CI − SI over 2 years is just the interest on one year's interest.
Use it when: Remembering P(r/100)².

Half-Yearly Compounding, Growth and Depreciation, and the Difference Between CI and SI: solved examples

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

Example 1: Half-yearly amount · 1 mark · MCQBoard level

₹4,000 is invested for 1 year at 10% per annum compounded half-yearly. The amount is:
(a) ₹4,400
(b) ₹4,410
(c) ₹4,420
(d) ₹4,440

Try it first, then show the solutionSolution

Rate per half-year $= 5%$ , periods $= 2$ .
$A = 4000 \times 1.0 5^{2} = 4000 \times 1.1025 = 4410$ .

Answer: (b) ₹4,410

Example 2: Eighteen months · 3 marks · Short answerBoard level

A sum of ₹25,000 is lent for $1 \frac{1}{2}$ years at 12% per annum compounded half-yearly. Find the compound interest.

Try it first, then show the solutionSolution

Rate per half-year $= 6%$ ; periods $= 3$ .
$A = 25000 \times 1.0 6^{3} = 25000 \times 1.191016 = 29775.40$ .
$C I = 29775.40 - 25000 = 4775.40$ .

Answer: ₹4,775.40

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

The difference between the compound interest and the simple interest on a sum for 2 years at 8% per annum is ₹32. Find the sum, and verify by finding CI and SI separately.

Try it first, then show the solutionSolution

$P (0.08)^{2} = 32$ , so $P = \frac{32}{0.0064} = 5000$ .
$C I = 5000 \times (1.1664 - 1) = 832$ .
$S I = 5000 \times 0.08 \times 2 = 800$ .
$C I - S I = 832 - 800 = 32$ . ✓

Answer: Sum ₹5,000 (CI ₹832, SI ₹800)

Example 4: Depreciation and growth together · 4 marks · Short answerHarder board level

A machine bought for ₹80,000 depreciates 10% per annum. The town where it is used has a population of 50,000 growing at 4% per annum. Find the value of the machine and the population after 2 years, and the loss in the machine's value.

Try it first, then show the solutionSolution

Value $= 80000 \times 0. 9^{2} = 64800$ .
Population $= 50000 \times 1.0 4^{2} = 54080$ .
Loss $= 80000 - 64800 = 15200$ .

Answer: ₹64,800; 54,080; loss ₹15,200.

Example 5: Yearly against half-yearly · 5 marks · Long answerHOTS / competency

₹20,000 is invested for 2 years at 10% per annum. Find how much more it earns when compounded half-yearly than when compounded yearly.

Try it first, then show the solutionSolution

Yearly: $20000 \times 1. 1^{2} = 20000 \times 1.21 = 24200$ .
Half-yearly: rate $5%$ , 4 periods: $20000 \times 1.0 5^{4} = 20000 \times 1.21550625 = 24310.125$ .
Difference $= 24310.125 - 24200 = 110.125$ , that is ₹110.13 (to the nearest paisa).

Answer: About ₹110.13 more.

Where students lose marks in Half-Yearly Compounding, Growth and Depreciation, and the Difference Between CI and SI

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

Mistake: Using the annual rate for half a year
The rate must be halved when interest is added every six months.
Wrong working
$A = 4000 \times 1. 1^{2}$ for half-yearly.
Right working
$A = 4000 \times 1.0 5^{2} = 4410$ .
Do this: Write "rate per half-year =" first.
Usually costs: Whole answer.
Mistake: Adding percentages in depreciation
Successive falls multiply.
Wrong working
Two falls of 10% leave 80%.
Right working
$0.9 \times 0.9 = 0.81$ , so 81%.
Do this: Multiply the factors.
Usually costs: Final answer.
Mistake: Using the wrong power for CI − SI
The shortcut works for exactly 2 years.
Wrong working
$C I - S I = P (r /100)$ for 2 years.
Right working
$C I - S I = P (r /100)^{2}$ for 2 years.
Do this: Square the rate fraction.
Usually costs: Final answer.

Exam tips

These are typical ICSE Class 9 questions of 1 to 5 marks; the comparison and mixed-rate items are the more demanding ones.
Write "r per period" and "number of periods" on separate lines before substituting.
Percentage change compares with the original value, not with the final value.
When a question gives a difference, use P(r/100)² directly and then verify with separate CI and SI.

Practice questions on Half-Yearly Compounding, Growth and Depreciation, and the Difference Between CI and SI

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
At 12% per annum compounded half-yearly, the rate per period and the number of periods for 2 years are:
(a) 12%, 2
(b) 6%, 4
(c) 12%, 4
(d) 6%, 2
Hint
Halve the rate, double the years.
Answer
(b)
Q2 · 3 marks · Short answerHarder board level
Find the amount and the compound interest on ₹16,000 for $1 \frac{1}{2}$ years at 10% per annum compounded half-yearly.
Hint
$16000 \times 1.0 5^{3}$ .
Answer
$A = 16000 \times 1.157625 = ₹18, 522$ ; $C I = ₹2, 522$ .
Q3 · 4 marks · Short answerHarder board level
For 2 years at 5% per annum, the compound interest on a sum exceeds its simple interest by ₹60. Find the sum, the CI and the SI.
Hint
$P \times 0.0025 = 60$ .
Answer
$P = 24000$ ; $C I = 24000 \times 0.1025 = ₹2, 460$ and $S I = ₹2, 400$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): A machine whose value falls by 10% of the previous year's value each year is worth 81% of its original value after 2 years.
Reason (R): The value is multiplied by 0.9 in each year, and $0.9 \times 0.9 = 0.81$ .
(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
$0. 9^{2} = 0.81$ .
Answer
(a)
Q5 · 4 marks · Case-basedHOTS / competency
Case study: The population of a town is 40,000. It grows by 5% in the first year and falls by 10% in the second year because of migration.
(a) Find the population after one year.
(b) Find the population after two years.
(c) Find the overall percentage change over the two years.
Hint
Multiply 1.05 and 0.9.
Answer
(a) 42,000 (b) 37,800 (c) $\frac{37800 - 40000}{40000} \times 100 = - 5.5%$ , a fall of 5.5%.

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.What changes in the formula for half-yearly compounding?Show answer
The rate is halved and the number of periods is doubled.
2.What is the depreciation factor for r% per year?Show answer
1 − r/100.
3.What is CI − SI for 2 years?Show answer
P(r/100)².
4.How do you find the percentage change?Show answer
(final − initial) divided by initial, times 100.

Half-Yearly Compounding, Growth and Depreciation, and the Difference Between CI and SI: frequently asked questions

Does half-yearly compounding always give more than yearly?

Yes, for the same annual rate and time, because interest is earned on interest sooner.

Is the growth formula part of ICSE Class IX?

The chapter covers growth and depreciation as applications of the compound interest formula.

What if the time is not a whole number of half-years?

Class IX questions use whole numbers of periods, such as 1, 1.5 or 2 years.