ICSE Class 10 Maths Chapter 02 Banking

Banking

Recurring Deposit Account

2.1 Introduction

The business of receiving, safeguarding and lending of money is called banking.

In general, people who have some spare money do not keep it with them to avoid the risk of losing it by theft, etc. They deposit this spare money in a bank. In the bank, the money is safe as well as it fetches interest on it. On the other hand, some people need money to start a business or to expand their business. So, they borrow money from the bank at a nominal interest on it.

Thus, a bank is an institution which carries on the business of taking deposits and lending money. The rate of interest charged by the bank from its borrowers is usually higher than what it pays to depositors.

In addition to money-taking and money-lending, the banks also perform various other functions and almost every individual and every section of society deals with banks in one way or the other. Some of the main functions of a bank are:

1. Receiving money from depositors

Different banks have different types of schemes to attract people who keep their money in the bank of their choice or the bank which gives them maximum return on their deposits.

2. Lending money on demand

Of course, banks take deposits (money) at a lower rate of interest and lend it at a higher rate; but they also perform certain other functions for the benefit of the needy. For example, they give loans at concessional rates to small farmers, petty shopkeepers, educated unemployed persons to start business, handicapped persons, widows to help them earn their livelihood, etc.

3. Providing other useful services to society

The general public, the Government, etc.

Now-a-days, a maximum number of salaried persons get their salaries through banks. Banks help in transferring money from one place to the other. In big cities, the school fees, different types of bills, Government loan instalments, income tax, etc. are paid through banks. Banks provide lockers for the safe custody of valuables. They provide traveller's cheques, foreign currency, etc. to benefit tourists and travellers. ATM cards, debit cards, credit cards, etc. are some other facilities provided by banks which are highly beneficial to the public.

Teacher's Note

Banks are essential in modern life - from depositing your allowance safely to making purchases with a debit card, banking services simplify how we manage money daily.

2.2 Types of Accounts

Out of the various types of accounts (deposit schemes) provided by the banks, we shall confine our discussion to the popular and most commonly used bank account, namely:

Recurring Deposit Accounts.

2.3 Recurring Deposit Account (R.D. Account)

Under this scheme, a depositor chooses a specified amount and deposits that amount every month for a fixed period, chosen by him or her at the time of opening this account. This period may vary from 3 months to 10 years. At the expiry of this period (called the maturity period) the depositor is paid a lumpsum amount (called the maturity value). The maturity value of a R.D. Account includes the amount deposited by the account holder together with interest compounded quarterly at a fixed rate. This rate is fixed by the Reserve Bank of India and is revised from time to time.

2.4 Computing maturity value of a Recurring Deposit Account

Let a sum of \(\text{\u20b9}\ P\) be deposited every month in a bank for \(n\) months. If the rate of interest be \(r\%\) per year, the interest on the whole deposit is calculated by using the formula:

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

Since, the total sum deposited in \(n\) months

\(= \text{Sum deposited every month} \times \text{Number of months}\)

\(= P \times n\)

Maturity value (M.V.) of the recurring deposit

\(= \text{Total sum deposited} + \text{Interest on it}\)

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

M.V. = \(P \times n + I\)

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

Problem 1

Kiran deposited \(\text{\u20b9}\ 200\) 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. [2012]

Solution

Given: \(P = \text{\u20b9}\ 200\), \(n = 36\) months and \(r = 11\%\)

\(\therefore\ I = \text{\u20b9}\ 200 \times \frac{36(36 + 1)}{2 \times 12} \times \frac{11}{100} = \text{\u20b9}\ 1,221\)

\[\left[\therefore\ I = P \times \frac{n(n + 1)}{2 \times 12} \times \frac{r}{100}\right]\]

Since, the sum deposited \(= P \times n = \text{\u20b9}\ 200 \times 36 = \text{\u20b9}\ 7,200\)

The amount that Kiran will get at the time of maturity

\(= \text{\u20b9}\ 7,200 + \text{\u20b9}\ 1,221 = \text{\u20b9}\ 8,421\)

Ans.

Teacher's Note

Regular deposits with compound interest teach the power of consistent saving - even small monthly amounts grow significantly over time, like how pocket money saved monthly becomes substantial.

Problem 2

Mohan deposited \(\text{\u20b9}\ 80\) per month in a cumulative (recurring) deposit account for six years. Find the amount payable to him on maturity, if the rate of interest is 6% per annum. [2006]

Solution

Since, money deposited \(= \text{\u20b9}\ 80\) per month i.e. \(P = \text{\u20b9}\ 80\)

and, number of months \(= 6 \times 12 = 72\) i.e. \(n = 72\)

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

\(= \text{\u20b9}\ 80 \times \frac{72 \times 73}{2 \times 12} \times \frac{6}{100} = \text{\u20b9}\ 1,051.20\)

Amount payable to him on maturity

\(= \text{Sum deposited} + \text{Interest}\)

\(= \text{\u20b9}\ 80 \times 72 + \text{\u20b9}\ 1,051.20\)

\(= \text{\u20b9}\ 6,811.20\)

Ans.

Problem 3

Mr. R.K. Nair gets \(\text{\u20b9}\ 6,455\) at the end of one year at the rate of 14% per annum in a Recurring Deposit Account. Find the monthly instalment. [2005]

Solution

Suppose Mr. Nair deposited \(\text{\u20b9}\ 100\) per month i.e. \(P = \text{\u20b9}\ 100\)

Since, number of months \((n) = 12\) and rate of interest \((r) = 14\%\)

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

\(= \text{\u20b9}\ 100 \times \frac{12 \times 13}{2 \times 12} \times \frac{14}{100} = \text{\u20b9}\ 91\)

As the money deposited in 12 months \(= 12 \times \text{\u20b9}\ 100 = \text{\u20b9}\ 1,200\)

\(\therefore\) Maturity value (M.V.) \(= \text{\u20b9}\ 1,200 + \text{\u20b9}\ 91\) \(= \text{\u20b9}\ 1,291\)

Now applying Unitary Method, we get:

When M.V. is \(\text{\u20b9}\ 1,291\); the monthly instalment \(= \text{\u20b9}\ 100\)

\(\Rightarrow\) When M.V. is \(\text{\u20b9}\ 6,455\); the monthly instalment \(= \text{\u20b9}\ \frac{100}{1,291} \times 6,455\)

\(= \text{\u20b9}\ 500\)

Ans.

Alternative method:

Let Mr. Nair deposits \(\text{\u20b9}\ x\) per month i.e. \(P = \text{\u20b9}\ x\)

Since, \(n = 12\) and \(r = 14\%\)

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

\(= \text{\u20b9}\ x \times \frac{12 \times 13}{2 \times 12} \times \frac{14}{100} = \text{\u20b9}\ 0.91\ x\)

As the money deposited in 12 months \(= \text{\u20b9}\ 12x\)

\(\therefore\) Maturity value \(= \text{\u20b9}\ 12x + \text{\u20b9}\ 0.91\ x = \text{\u20b9}\ 12.91\ x\)

Given, M.V. is \(\text{\u20b9}\ 6,455\) \(\Rightarrow\) \(\text{\u20b9}\ 12.91\ x = \text{\u20b9}\ 6,455\)

\(\Rightarrow\) \(x = \text{\u20b9}\ \frac{6,455}{12.91} = \text{\u20b9}\ 500\)

The monthly instalment = \(\text{\u20b9}\ 500\)

Ans.

Teacher's Note

Understanding how to reverse-engineer a savings goal teaches financial planning - knowing your target maturity value helps you determine how much to deposit monthly.

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