Compound Interest Class 9 ICSE — Bansal Chapter 3
This chapter explains compound interest class 9 ICSE using one formula instead of repeated simple interest. Chapter 3 of the ICSE Mathematics textbook covers the amount formula, inverse problems, half-yearly compounding, fractional time, and applications like growth and depreciation. Each section below follows the book’s order, with formulas and a worked pattern you can try yourself. For the official syllabus, visit the CISCE board website.
ICSE Class 9 Mathematics Chapter 3 Compound Interest Using Formula PDF
What is inside Chapter 3
15 pages · 7 sections · 15 worked examples · 40 exercise questions
| Section | Title |
|---|---|
| 3.1 | INTRODUCTION |
| 3.2 | USING FORMULA |
| 3.3 | INVERSE PROBLEMS |
| 3.4 | MISCELLANEOUS PROBLEMS |
| 3.5 | WHEN THE INTEREST IS COMPOUNDED HALF-YEARLY (TWO TIMES IN A YEAR) |
| 3.6 | WHEN THE TIME IS NOT AN EXACT NUMBER OF YEARS AND THE INTEREST IS COMPOUNDED YEARLY |
| 3.7 | OTHER APPLICATIONS OF THE FORMULA |
- EXERCISE 3 (B): 10 questions
- EXERCISE 3 (C): 10 questions
- EXERCISE 3 (D): 10 questions
- EXERCISE 3 (E): 10 questions
3.1 INTRODUCTION
The previous chapter computed compound interest as repeated simple interest on a growing principal. Each year, the new principal is the previous amount, and interest is calculated on it again. This works, but becomes tedious as conversion periods increase.
Conversion periods mean the number of times interest is added in a year. Years, half-years, and quarters are all conversion periods. More periods mean more steps, and each step adds room for error. The full set of Class 9 ICSE books covers this topic across subjects.
This chapter replaces the step-by-step process with a single formula. You substitute the principal, rate, and time into one expression and evaluate it once. The formula does the repeated work in one calculation, which is faster and easier to check.
This section exists to explain why the formula approach is needed. It sets up the formula that section 3.2 introduces. Students who understand this motivation find the rest of the chapter easier to follow.
3.2 USING FORMULA
This section gives the core amount formula. When interest is compounded yearly, the amount after n years is:
A = P \left(1 + \frac{r}{100}\right)^n
Here, A is the amount, P is the principal, r is the yearly rate, and n is the number of years. The bracket \left(1 + \frac{r}{100}\right) is the growth factor for one year. Raising it to the power n applies that growth n times, which is why the formula replaces repeated calculation.
Compound interest is the amount minus the principal. The direct form computes interest without finding the amount first:
\text{C.I.} = P \left[katex display="true"]\left(1 + \frac{r}{100}\right)^n - 1 \right[/katex]
When rates differ across successive years, the formula becomes:
A = P \left(1 + \frac{r_1}{100}\right) \left(1 + \frac{r_2}{100}\right) \left(1 + \frac{r_3}{100}\right)
Here r_1\%, r_2\%, r_3\% are the rates for successive years. Each bracket handles one year’s growth, so the product applies all three rates in sequence.
Worked pattern: Find the amount and compound interest on ₹10,000 at 8% for 2 years.
Step 1: Write P = 10{,}000, r = 8, n = 2.
A = 10{,}000 \left(1 + \frac{8}{100}\right)^2 = 10{,}000 \times (1.08)^2
Step 2: Evaluate (1.08)^2 = 1.1664 and multiply.
A = 10{,}000 \times 1.1664 = ₹11{,}664
Step 3: Subtract the principal to get the interest.
\text{C.I.} = 11{,}664 - 10{,}000 = ₹1{,}664
Final answer: Amount = ₹11,664 and C.I. = ₹1,664.
3.3 INVERSE PROBLEMS
Inverse problems give you the amount or interest and ask you to find the principal, rate, or time. You solve them by rearranging the amount formula.
Finding the principal: Divide both sides of A = P \left(1 + \frac{r}{100}\right)^n by the growth factor. If the question gives compound interest instead of amount, add the principal back first and use the C.I. formula.
Finding the rate: Divide the amount by the principal, then match the result to a known power. If \left(1 + \frac{r}{100}\right)^n equals a fraction like \frac{21}{20}, the rate is 5%.
Finding the time: Express the ratio of amount to principal as a power of the growth factor. When both sides have the same base, the exponents are equal.
Worked pattern: What sum amounts to ₹5,445 at 10% for 2 years?
Step 1: Substitute into the amount formula.
5{,}445 = P \left(1 + \frac{10}{100}\right)^2 = P \times (1.1)^2
Step 2: Divide both sides by (1.1)^2 = 1.21.
P = \frac{5{,}445}{1.21} = ₹4{,}500
Final answer: The principal is ₹4,500.
The section also covers a two-person share problem. A sum is split so both receive equal amounts at a future age. Write each person’s amount with their own principal and time, then equate the two expressions and solve.
3.4 MISCELLANEOUS PROBLEMS
This section combines ideas from earlier parts of the chapter. The two main problem types are the difference between compound interest and simple interest, and finding both principal and rate from two known amounts.
C.I. minus S.I. for two years: Write compound interest as P \left[katex display="true"]\left(1 + \frac{r}{100}\right)^2 - 1 \right[/katex] and simple interest as \frac{P \times r \times 2}{100}. Subtract the second from the first. The terms simplify, leaving an equation you can solve for the principal.
Two known amounts at different times: Write the amount at year 1 and the amount at year 3 using the same principal and rate. Divide one equation by the other. The principal cancels, leaving an equation for the rate. Then substitute back to find the principal.
Worked pattern: The difference between C.I. and S.I. on a sum at 5% for 2 years is ₹10. Find the sum.
Step 1: Write C.I. and S.I. in terms of P.
\text{C.I.} = P \left(1.05\right)^2 - P = 0.1025P
\text{S.I.} = \frac{P \times 5 \times 2}{100} = 0.1P
Step 2: Subtract and solve for P.
0.1025P - 0.1P = 10 \Rightarrow 0.0025P = 10 \Rightarrow P = ₹4{,}000
Final answer: The sum is ₹4,000.
3.5 WHEN THE INTEREST IS COMPOUNDED HALF-YEARLY (TWO TIMES IN A YEAR)
When interest is compounded half-yearly, the rate is divided by 2 and the number of years is multiplied by 2. The formula becomes:
A = P \left(1 + \frac{r}{2 \times 100}\right)^{n \times 2}
Why halve the rate? Each half-year earns half the yearly rate. Why double the years? There are two conversion periods per year. So n years become 2n half-years.
Common mistake: Students double the rate instead of halving it. That inflates the growth factor and gives a wrong amount. Always divide the rate by 2 and multiply the years by 2.
The quarterly version A = P \left(1 + \frac{r}{4 \times 100}\right)^{n \times 4} is mentioned in the book but is outside the ICSE syllabus.
Worked pattern: Find the amount on ₹5,000 at 10% for 1 year, compounded half-yearly.
Step 1: Halve the rate to 5% and double the years to 2.
A = 5{,}000 \left(1 + \frac{10}{200}\right)^2 = 5{,}000 \times (1.05)^2
Step 2: Evaluate and subtract the principal.
A = 5{,}000 \times 1.1025 = ₹5{,}512.50
\text{C.I.} = 5{,}512.50 - 5{,}000 = ₹512.50
Final answer: Amount = ₹5,512.50 and C.I. = ₹512.50.
3.6 WHEN THE TIME IS NOT AN EXACT NUMBER OF YEARS AND THE INTEREST IS COMPOUNDED YEARLY
When the time is fractional, like 2\frac{1}{2} years, use a two-step method. First, find the amount for the whole number of years using the yearly formula. Then treat that amount as the new principal for the remaining fraction of a year.
Worked pattern: Find the amount on ₹8,000 at 10% for 2\frac{1}{2} years, compounded yearly.
Step 1: Find the amount for 2 full years.
A_2 = 8{,}000 \left(1.1\right)^2 = 8{,}000 \times 1.21 = ₹9{,}680
Step 2: Use ₹9,680 as the principal for the remaining half-year.
A = 9{,}680 \left(1 + \frac{10}{200}\right)^1 = 9{,}680 \times 1.05 = ₹10{,}164
Final answer: The amount after 2\frac{1}{2} years is ₹10,164.
The section includes a comparison table for n = 1, 1\frac{1}{2}, 2, and 2\frac{1}{2} years under yearly and half-yearly compounding:
| Time (n) | Compounded yearly | Compounded half-yearly |
|---|---|---|
| 1 year | P \left(1 + \frac{r}{100}\right)^1 | P \left(1 + \frac{r}{200}\right)^2 |
| 1\frac{1}{2} years | P \left(1 + \frac{r}{100}\right)^1 \left(1 + \frac{r}{200}\right) | P \left(1 + \frac{r}{200}\right)^3 |
| 2 years | P \left(1 + \frac{r}{100}\right)^2 | P \left(1 + \frac{r}{200}\right)^4 |
| 2\frac{1}{2} years | P \left(1 + \frac{r}{100}\right)^2 \left(1 + \frac{r}{200}\right) | P \left(1 + \frac{r}{200}\right)^5 |
For the same rate, half-yearly compounding always gives a larger amount because interest is added more often.
3.7 OTHER APPLICATIONS OF THE FORMULA
The amount formula works beyond money. It applies to growth, depreciation, population change, doubling time, and loan repayment. The sign in the bracket decides whether the value grows or shrinks.
| Application | Formula | Rate sign |
|---|---|---|
| Growth (production, population increase) | \text{Future value} = \text{Initial} \times \left(1 + \frac{r}{100}\right)^n | Add r |
| Depreciation (machine value) | \text{Future value} = \text{Present value} \times \left(1 - \frac{r}{100}\right)^n | Subtract r |
Growth: Production or population after n years equals the initial value times \left(1 + \frac{r}{100}\right)^n. Add the rate because the value increases. For population, the earlier population takes the place of P and the later population takes the place of A.
Depreciation: Value after n years equals present value times \left(1 - \frac{r}{100}\right)^n. Subtract the rate because the value decreases. This minus sign is the key difference from growth.
Doubling time: If a sum doubles in 5 years, then 2 = \left(1 + \frac{r}{100}\right)^5. To find when it becomes 8 times, notice 8 = 2^3, so n = 3 \times 5 = 15 years.
Loan repayment: A borrowed sum repaid in two installments is the sum of the present values of each installment. Find the principal for each payment at its own time, then add.
Worked pattern: A machine bought for ₹3,20,000 depreciates at 20% yearly. Find its value after 3 years.
Step 1: Use the depreciation formula with a minus sign.
\text{Value} = 3{,}20{,}000 \left(1 - \frac{20}{100}\right)^3 = 3{,}20{,}000 \times (0.8)^3
Step 2: Evaluate the power and multiply.
= 3{,}20{,}000 \times 0.512 = ₹1{,}63{,}840
Final answer: The machine is worth ₹1,63,840 after 3 years. For more practice across the subject, browse the other ICSE Class 9 Maths chapters.
What to practise from this chapter
- EXERCISE 3 (B) — 10 questions. Practises miscellaneous problems from section 3.4: the C.I. versus S.I. difference and finding principal plus rate from two amounts.
- EXERCISE 3 (C) — 10 questions. Practises half-yearly compounding from section 3.5, with rate halved and years doubled.
- EXERCISE 3 (D) — 10 questions. Practises applications from section 3.7: growth, depreciation, and population change.
- EXERCISE 3 (E) — 10 questions. Mixed review covering all methods in the chapter.
Frequently asked questions
What is the formula to find the amount when interest is compounded annually?
Use A = P \left(1 + \frac{r}{100}\right)^n. Here P is the principal, r is the yearly rate, and n is the number of years. Compound interest is A - P, or directly P \left[katex display="true"]\left(1 + \frac{r}{100}\right)^n - 1 \right[/katex].
How do I find the principal when the amount and rate are given?
Rearrange the amount formula. Divide the amount by \left(1 + \frac{r}{100}\right)^n. For example, if a sum amounts to ₹5,445 at 10% for 2 years, the principal is \frac{5{,}445}{1.21} = ₹4{,}500.
What changes when interest is compounded half yearly?
Halve the rate and double the years. Use A = P \left(1 + \frac{r}{200}\right)^{2n}. The common mistake is doubling the rate instead of halving it.
How do I find the amount when the time is two and a half years?
Use two steps. First find the amount for 2 years with the yearly formula. Then treat that amount as the principal for the remaining half-year at half the yearly rate.
What is the difference between compound interest and simple interest for two years?
The difference is P \left( \frac{r}{100} \right)^2. Compound interest is higher because each year’s interest earns interest in the following year.
Reference: ICSE Class 9 Mathematics textbook material, chapter 3: Compound Interest Using Formula — CISCE-aligned.