Chapter 2: Banking (Recurring Deposit Account)

The business of receiving, safeguarding, and lending money is called banking. In modern commercial banking, a Recurring Deposit (R.D.) Account allows a depositor to save a fixed monthly sum for a specified tenure (maturity period). Interest is paid on a cumulative basis calculated per month. This master note covers all definitions, formula derivations, equivalent principal calculations, solved textbook examples, and past ICSE board examination problems covering all 4 question types.

Official ICSE Class 10 Question Types Master Checklist

✔ Banking Fundamentals & Accounts: Functions of banks, comparison of Savings, Fixed Deposit, and Cumulative/Recurring Deposit Accounts.
✔ Equivalent Principal Derivation: Deriving 1-month equivalent principal $P_{\text{eq}} = P \times \frac{n(n+1)}{2}$.
✔ Interest Formula: Calculating interest $I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}$.
✔ Maturity Value Formula: Calculating Maturity Value $\text{M.V.} = (P \times n) + I$.
✔ Type 1: Finding Maturity Value ($\text{M.V.}$) — Given monthly deposit $P$, time $n$ (months/years), and rate $r\%$.
✔ Type 2: Finding Monthly Instalment ($P$) — Given Maturity Value $\text{M.V.}$ (or Interest $I$), time $n$, and rate $r\%$.
✔ Type 3: Finding Rate of Interest ($r\%$) — Given monthly deposit $P$, time $n$, and Maturity Value $\text{M.V.}$ (or Interest $I$).
✔ Type 4: Finding Time Period ($n$ in months/years) — Leads to Quadratic Equation $a n^2 + b n - c = 0$. Solving for $n$ and discarding negative roots.

1. Fundamental Banking Concepts & Account Types

Functions of Commercial Banks

Receiving Money from Depositors: Banks accept spare funds from individuals and institutions, offering security and interest returns.
Lending Money on Demand: Banks grant loans to businesses, farmers, traders, and individuals at interest rates higher than deposit interest rates.
Public Utility Services: Transferring money (drafts/online transfers), safe custody of valuables (lockers), bill payments, traveller's cheques, and ATM/Debit/Credit card facilities.

Comparison of Popular Bank Deposit Accounts

Account Type	Deposit Pattern	Withdrawal Rules	Interest & Purpose
Savings Bank Account	Flexible amounts deposited at any time	Withdrawal permitted by cheque/ATM anytime	Low interest rate; promotes saving habits among public.
Fixed Deposit (F.D.) Account	One-time lump sum deposited for a fixed period	Withdrawn only after completion of maturity period	Highest interest rate; suitable for long-term investments.
Recurring Deposit (R.D.) Account	Fixed monthly instalment ($P$) deposited every month	Lump sum (Maturity Value) paid at the end of tenure	Compounded monthly interest; tailored for salaried & regular earners.

2. Recurring Deposit (R.D.) Account & Interest Derivation

Why Simple Interest Formula Requires Equivalent Principal

In a Recurring Deposit Account, a fixed sum ₹$P$ is deposited every month for $n$ months:

The 1st instalment remains in the bank for $n$ months.
The 2nd instalment remains in the bank for $(n-1)$ months.
The 3rd instalment remains in the bank for $(n-2)$ months.
...
The $n$-th (last) instalment remains in the bank for $1$ month.

Total equivalent principal for 1 month is the sum of the first $n$ natural numbers multiplied by $P$:

$$\text{Equivalent Principal for 1 month} = P \times [n + (n-1) + (n-2) + \dots + 1] = P \times \frac{n(n+1)}{2}$$

MASTER FORMULAS FOR RECURRING DEPOSIT ACCOUNT

$$\text{Interest (I)} = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} = \frac{P \cdot n(n+1) \cdot r}{2400}$$ $$\text{Total Sum Deposited} = P \times n$$ $$\text{Maturity Value (M.V.)} = (P \times n) + I = P \times n + \left[\frac{P \cdot n(n+1) \cdot r}{2400}\right]$$

where $P = \text{Monthly Instalment}$, $n = \text{Number of Months } (\text{Years} \times 12)$, $r = \text{Rate of Interest per annum}$.

3. Formula Quick Reference Grid

Monthly Deposit

$$P = \text{Instalment Amount}$$

Time in Months

$$n = \text{Years} \times 12$$

Total Deposited

$$\text{Total } P = P \times n$$

Interest Amount

$$I = \frac{P \cdot n(n+1) \cdot r}{2400}$$

Maturity Value

$$\text{M.V.} = (P \times n) + I$$

Time Quadratic

$$a n^2 + b n - c = 0$$

4. Solved Master Examples Covering All 4 ICSE Question Types

Type 1 — Finding Maturity Value (Given P, n, r) [ICSE 2012]

Problem: Kiran deposited ₹200 per month for 36 months in a bank's recurring deposit account. If the bank pays interest at the rate of 11% per annum, find the amount she gets on maturity.

Solution:

Monthly deposit ($P$) = ₹200
Number of months ($n$) = 36
Rate of interest ($r$) = 11% p.a.
Interest ($I$) = $P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} = 200 \times \frac{36 \times 37}{24} \times \frac{11}{100} = 200 \times 55.5 \times 0.11 =$ ₹1,221.
Total sum deposited = $P \times n = 200 \times 36 =$ ₹7,200.
Maturity Value (M.V.) = $\text{Total Deposited} + I = 7200 + 1221 =$ ₹8,421.

Type 1 — Finding Maturity Value (Time Given in Years) [ICSE 2006]

Problem: Mohan deposited ₹80 per month in a cumulative (recurring) deposit account for 6 years. Find the amount payable to him on maturity, if the rate of interest is 6% per annum.

Solution:

Monthly deposit ($P$) = ₹80
Number of months ($n$) = $6 \times 12 = 72$ months
Rate of interest ($r$) = 6% p.a.
Interest ($I$) = $80 \times \frac{72 \times 73}{24} \times \frac{6}{100} = 80 \times 219 \times 0.06 =$ ₹1,051.20.
Total sum deposited = $80 \times 72 =$ ₹5,760.
Maturity Value (M.V.) = $5760 + 1051.20 =$ ₹6,811.20.

Type 2 — Finding Monthly Instalment (P) [ICSE 2005]

Problem: Mr. R.K. Nair gets ₹6,455 at the end of 1 year at the rate of 14% per annum in a Recurring Deposit Account. Find the monthly instalment.

Solution:

Let monthly instalment = ₹$x$.
Number of months ($n$) = $1 \times 12 = 12$ months, Rate ($r$) = 14%.
Interest ($I$) = $x \times \frac{12 \times 13}{24} \times \frac{14}{100} = \frac{13 \times 14}{200} x = 0.91 x$.
Total sum deposited = $x \times 12 = 12x$.
Maturity Value ($\text{M.V.}$) = $12x + 0.91x = 12.91x$.
Given $\text{M.V.} = 6455 \implies 12.91x = 6455 \implies x = \frac{6455}{12.91} = 500$.
Monthly Instalment (P) = ₹500.

Type 3 — Finding Rate of Interest (r%) [ICSE 2011]

Problem: Ahmed has a recurring deposit account in a bank. He deposits ₹2,500 per month for 2 years. If he gets ₹66,250 at the time of maturity, find: (i) the interest paid by the bank, (ii) the rate of interest.

Solution:

$P = \text{₹}2500$, $n = 2 \times 12 = 24$ months.
Total money deposited = $24 \times 2500 =$ ₹60,000.
(i) Interest paid by bank = $\text{M.V.} - \text{Total Deposited} = 66250 - 60000 =$ ₹6,250.
(ii) Using Interest formula: $$6250 = 2500 \times \frac{24 \times 25}{24} \times \frac{r}{100}$$ $$6250 = 2500 \times 25 \times \frac{r}{100} = 625 r$$ $$r = \frac{6250}{625} = 10\%$$
Rate of Interest (r) = 10% per annum.

Type 4 — Finding Time Period (n) via Quadratic Equation [Textbook Example 5]

Problem: Monica had a R.D. Account in the Union Bank of India and deposited ₹600 per month. If the maturity value of this account was ₹24,930 and the rate of interest was 10% per annum, find the time (in years) for which the account was held.

Solution:

Let the account be held for $n$ months. $P = \text{₹}600$, $r = 10\%$.
Interest ($I$) = $600 \times \frac{n(n+1)}{24} \times \frac{10}{100} = \frac{600 \times 10}{2400} \cdot n(n+1) = \frac{5 n(n+1)}{2}$.
Total deposited = $600 \times n = 600n$.
$\text{M.V.} = 600n + \frac{5n(n+1)}{2} = 24930$.
Multiply by 2: $1200n + 5n^2 + 5n = 49860 \implies 5n^2 + 1205n - 49860 = 0$.
Divide by 5: $n^2 + 241n - 9972 = 0$.
Factorizing: $(n + 277)(n - 36) = 0 \implies n = 36 \text{ or } n = -277$.
Since time cannot be negative, $n = 36$ months.
Time in years = $\frac{36}{12} =$ 3 years.

Type 4 — Finding Time Period (Quadratic Case 2) [Textbook Example 6]

Problem: A recurring deposit account of ₹1,200 per month has a maturity value of ₹12,440. If the rate of interest is 8% per annum, find the time (in months) of this Recurring Deposit Account.

Solution:

$P = \text{₹}1200$, $r = 8\%$. Let time = $n$ months.
Interest ($I$) = $1200 \times \frac{n(n+1)}{24} \times \frac{8}{100} = 4n(n+1) = 4n^2 + 4n$.
$\text{M.V.} = 1200n + 4n^2 + 4n = 12440 \implies 4n^2 + 1204n - 12440 = 0$.
Divide by 4: $n^2 + 301n - 3110 = 0$.
Factorizing: $(n + 311)(n - 10) = 0 \implies n = 10 \text{ or } n = -311$.
Time period = 10 months.

5. Comprehensive ICSE Board Practice Problems

Textbook Exercises & Past Board Questions

Q1 (ICSE 2001, 2007). Amit deposited ₹150 per month in a bank for 8 months under the Recurring Deposit Scheme. What will be the maturity value of his deposits, if the rate of interest is 8% per annum?

Solution:
$P = 150$, $n = 8$, $r = 8\%$.
Interest = $150 \times \frac{8 \times 9}{24} \times \frac{8}{100} = 150 \times 3 \times 0.08 =$ ₹36.
Total sum deposited = $150 \times 8 =$ ₹1,200.
Maturity Value = $1200 + 36 =$ ₹1,236.

Q2 (ICSE 2008). David opened a Recurring Deposit Account in a bank and deposited ₹300 per month for two years. If he received ₹7,725 at the time of maturity, find the rate of interest per annum.

Solution:
$P = 300$, $n = 24$, $\text{M.V.} = 7725$.
Total deposited = $300 \times 24 =$ ₹7,200 → Interest = $7725 - 7200 =$ ₹525.
$525 = 300 \times \frac{24 \times 25}{24} \times \frac{r}{100} = 75r \implies r = \frac{525}{75} =$ 7% p.a.

Q3 (ICSE 2010). Mr. Gupta opened a recurring deposit account in a bank. He deposited ₹2,500 per month for two years. At the time of maturity he got ₹67,500. Find: (i) the total interest earned, (ii) the rate of interest per annum.

Solution:
$P = 2500$, $n = 24$, $\text{M.V.} = 67500$.
Total deposited = $2500 \times 24 =$ ₹60,000.
(i) Interest earned = $67500 - 60000 =$ ₹7,500.
(ii) $7500 = 2500 \times \frac{24 \times 25}{24} \times \frac{r}{100} = 625r \implies r = \frac{7500}{625} =$ 12% p.a.

Q4 (ICSE 2013). Mr. Britto deposits a certain sum of money each month in a Recurring Deposit Account of a bank. If the rate of interest is 8% per annum and Mr. Britto gets ₹8,088 from the bank after 3 years, find the value of his monthly instalment.

Solution:
Let monthly instalment = ₹$P$. $n = 3 \times 12 = 36$, $r = 8\%$.
Interest = $P \times \frac{36 \times 37}{24} \times \frac{8}{100} = \frac{111 \times 8}{200} P = 4.44 P$.
Total deposited = $36P$.
$\text{M.V.} = 36P + 4.44P = 40.44P = 8088 \implies P = \frac{8088}{40.44} =$ ₹200.

Q5 (ICSE 2014). Shahrukh opened a Recurring Deposit Account in a bank and deposited ₹800 per month for $1\frac{1}{2}$ years. If he received ₹15,084 at the time of maturity, find the rate of interest per annum.

Solution:
$P = 800$, $n = 1.5 \times 12 = 18$ months.
Total deposited = $800 \times 18 =$ ₹14,400.
Interest = $15084 - 14400 =$ ₹684.
$684 = 800 \times \frac{18 \times 19}{24} \times \frac{r}{100} = 8 \times 14.25 \times r = 114r \implies r = \frac{684}{114} =$ 6% p.a.

Q6 (ICSE 2015). Katrina opened a recurring deposit account of ₹1,000 per month with a Nationalised Bank for 2 years. If the bank pays interest at the rate of 6% per annum, find: (i) interest earned in 2 years, (ii) maturity value.

Solution:
$P = 1000$, $n = 24$, $r = 6\%$.
(i) Interest = $1000 \times \frac{24 \times 25}{24} \times \frac{6}{100} = 1000 \times 25 \times 0.06 =$ ₹1,500.
Total deposited = $1000 \times 24 =$ ₹24,000.
(ii) Maturity Value = $24000 + 1500 =$ ₹25,500.

Q7 (ICSE 2018). Priyanka has a Recurring Deposit Account in a bank. She deposits ₹1,000 per month for 2 years. If she gets ₹26,000 on maturity, find the rate of interest per annum.

Solution:
$P = 1000$, $n = 24$, $\text{M.V.} = 26000$.
Total deposited = $1000 \times 24 =$ ₹24,000 → Interest = $26000 - 24000 =$ ₹2,000.
$2000 = 1000 \times \frac{24 \times 25}{24} \times \frac{r}{100} = 250r \implies r = \frac{2000}{250} =$ 8% p.a.

Q8 (ICSE 2020). Mrs. Geeta deposited ₹350 per month in a bank for 1 year and 3 months under the Recurring Deposit Scheme. If the maturity value of her deposits is ₹5,565, find the rate of interest per annum.

Solution:
$P = 350$, $n = 15$ months ($1 \text{ year } 3 \text{ months}$).
Total deposited = $350 \times 15 =$ ₹5,250 → Interest = $5565 - 5250 =$ ₹315.
$315 = 350 \times \frac{15 \times 16}{24} \times \frac{r}{100} = 350 \times 10 \times \frac{r}{100} = 35r \implies r = \frac{315}{35} =$ 9% p.a.