Banking (Recurring Deposit)

ICSE Class 10 Mathematics • Chapter 02

1. Introduction

Recurring Deposit (RD): A savings scheme where a fixed amount is deposited every month for a predetermined period. At maturity, the depositor receives the principal plus interest.

Interest Type: Simple Interest (SI) is used for RD calculations in ICSE.

Key Terminology

Symbol	Meaning	Unit
P	Monthly installment (Principal per month)	₹
n	Number of months (time period)	months
r	Rate of interest per annum	% p.a.
I	Total interest earned	₹
MV	Maturity Value (Total amount received)	₹

2. Understanding How Interest Works

How Interest is Calculated:

Each monthly installment earns interest for a different duration:

1st installment earns interest for n months
2nd installment earns interest for (n-1) months
3rd installment earns interest for (n-2) months
... and so on ...
Last (nth) installment earns interest for 1 month

Total months of interest = n + (n-1) + (n-2) + ... + 1

This is an AP sum = $\frac{n(n+1)}{2}$

3. Master Formulas

Interest on Recurring Deposit:

$I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}$

or simplified as:

$I = P \times \frac{n(n+1)}{24} \times \frac{r}{100}$

Total Principal Deposited:

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

Maturity Value (MV):

$MV = P \times n + I$

or in combined form:

$MV = P \times n + P \times \frac{n(n+1)}{24} \times \frac{r}{100}$

Remember: The fraction $\frac{n(n+1)}{24}$ comes from $\frac{n(n+1)}{2 \times 12}$ where:

$\frac{n(n+1)}{2}$ = Sum of months for interest calculation
Division by 12 = Conversion from months to years (for annual rate)

4. Types of Problems

Type 1: Finding Interest and Maturity Value

Given: P, n, r → Find: I, MV

Calculate Interest: $I = P \times \frac{n(n+1)}{24} \times \frac{r}{100}$
Calculate Principal: $P \times n$
Calculate MV: $MV = P \times n + I$

Example 1: Anita deposits ₹500 per month in an RD account for 2 years at 10% p.a. Find the interest and maturity value.

Solution:

Given	Value
P (monthly deposit)	₹500
n (months)	2 years = 24 months
r (rate)	10% p.a.

Step 1: Calculate Interest

$I = 500 \times \frac{24 \times 25}{24} \times \frac{10}{100}$

$I = 500 \times 25 \times 0.1$

$I = 500 \times 2.5 = ₹1,250$

Step 2: Calculate Principal

Principal = $500 \times 24 = ₹12,000$

Step 3: Calculate MV

MV = 12,000 + 1,250 = ₹13,250

Type 2: Finding Monthly Installment (P)

Given: MV, n, r → Find: P

Rearrange: $MV = P \times n + P \times \frac{n(n+1)}{24} \times \frac{r}{100}$

$MV = P \left[ n + \frac{n(n+1) \times r}{2400} \right]$

$P = \frac{MV}{n + \frac{n(n+1) \times r}{2400}}$

Example 2: What monthly deposit is required to get ₹51,000 after 2 years at 10% p.a.?

Solution:

MV = ₹51,000, n = 24, r = 10%

Let monthly installment = P

$51000 = P \times 24 + P \times \frac{24 \times 25}{24} \times \frac{10}{100}$

$51000 = 24P + 25P \times 0.1$

$51000 = 24P + 2.5P$

$51000 = 26.5P$

$P = \frac{51000}{26.5} = ₹1,924.53 \approx$ ₹1,925

Type 3: Finding Rate of Interest (r)

Given: P, n, MV (or I) → Find: r

Find Interest: $I = MV - P \times n$
Substitute in formula: $I = P \times \frac{n(n+1)}{24} \times \frac{r}{100}$
Solve for r

Example 3: Ramesh deposits ₹600 per month for 18 months and receives ₹11,232 on maturity. Find the rate of interest.

Solution:

P = ₹600, n = 18, MV = ₹11,232

Step 1: Principal = 600 × 18 = ₹10,800

Step 2: Interest = 11,232 − 10,800 = ₹432

Step 3: Using formula:

$432 = 600 \times \frac{18 \times 19}{24} \times \frac{r}{100}$

$432 = 600 \times \frac{342}{24} \times \frac{r}{100}$

$432 = 600 \times 14.25 \times \frac{r}{100}$

$432 = 85.5r$

$r = \frac{432}{85.5} = $ 5.05% ≈ 5% p.a.

Type 4: Finding Time Period (n)

Given: P, r, I (or MV) → Find: n

This often leads to a quadratic equation in n.

Example 4: Sheela deposits ₹2,000 per month at 8% p.a. and receives ₹1,020 as interest. Find the time period.

Solution:

P = ₹2,000, r = 8%, I = ₹1,020

$1020 = 2000 \times \frac{n(n+1)}{24} \times \frac{8}{100}$

$1020 = 2000 \times \frac{n(n+1)}{24} \times 0.08$

$1020 = \frac{160 \times n(n+1)}{24}$

$1020 \times 24 = 160 \times n(n+1)$

$24480 = 160n(n+1)$

$n(n+1) = 153$

$n^2 + n - 153 = 0$

$(n - 12)(n + 13) = 0$

n = 12 (rejecting negative)

Time = 12 months = 1 year

5. Quick Reference Formula Table

To Find	Formula
Interest (I)	$P \times \frac{n(n+1)}{24} \times \frac{r}{100}$
Principal	$P \times n$
Maturity Value	$P \times n + I$
Monthly Installment	$\frac{MV}{n + \frac{n(n+1)r}{2400}}$
Rate of Interest	$\frac{I \times 2400}{P \times n(n+1)}$
Equivalent Principal	$P \times \frac{n(n+1)}{2}$ (in months)

6. Important Points to Remember

Always convert years to months: 1 year = 12 months, 2 years = 24 months
The rate is always given per annum but interest is calculated monthly
Interest is calculated on each installment from the date of deposit to maturity
For finding n, you may get a quadratic equation - always reject negative value
Quick check: Interest should always be less than Principal for reasonable rates and periods

Exam Practice Questions (PYQ Trends)

PYQ: 2023

BOARD A man deposited ₹1,000 per month in a recurring deposit account for 2 years at 8% simple interest per annum. Find: (i) The total interest earned (ii) The maturity value

PYQ: 2022

BOARD 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 the rate of interest per annum.

Additional Practice

HOTS The maturity value of a recurring deposit is ₹16,176. If monthly installment is ₹400 and the rate of interest is 8% p.a., find the time.