Question

A page from a savings bank account passbook is given.

Date	Particulars	Amount Withdrawal (Rs.)	Amount Deposited (Rs.)	Balance (Rs.)
Jan 7, 2016	B/F			3000.00
Jan 10, 2016	By Cheque		2600.00	5600.00
Feb 8, 2016	To self	1500.00		4100.00
April 6, 2016	By cheque	2100.00		2000.00
May 4, 2016	By Cash		6500.00	8500.00
May 27, 2016	By Cheque		1500.00	10000.00

(a) Calculate the interest for the 6 months from January to June 2016, at 6% per annum.
(b) If the account is closed on $ {{1}^{st}} $ July 2016, find the amount received by the account holder.

Answer 1

Hint: In order to solve this problem, we need to assume that the interest is calculated by a simple interest formula. The formula for simple interest is $ \text{Simple interest}=\dfrac{P\times N\times R}{100} $ where, P = principal amount, N = number of years, R = rate of interest. Also, we need to assume that the bank calculates the interest according to the minimum balance in the account

Complete step-by-step answer:
We need to find the interest after 6 months but as the principal amount is changing every month we need to calculate the simple interest each month and then add them up for 6 months.
We are assuming that the interest is calculated based on the minimum balance each month.
Therefore, we can make a table of minimum balance each month
The table is as follows,

Month	Minimum balance
January	3000.00
February	4100.00
March	4000.00
April	2000.00
May	8500.00
June	10000.00

Now, the formula for simple interest is $ \text{Simple interest}=\dfrac{P\times N\times R}{100} $
Where, P = principal amount.
N = number of years.
R = rate of interest.
The rate of interest in every month is 6%.
The time period for every month is $ \dfrac{1}{12} $ .
Let’s consider the month of January.
P = Rs. 3000
Substituting in the formula we get,
 $ S{{I}_{1}}=\dfrac{3000\times 1\times 6}{100\times 12} $
Solving this we get,
 $ S{{I}_{1}}=\dfrac{18000}{1200}=15 $ .
Let’s consider the month of February.
P = Rs. 4100.
Substituting in the formula we get,
 $ S{{I}_{2}}=\dfrac{4100\times 1\times 6}{100\times 12} $
Solving this we get,
 $ S{{I}_{2}}=\dfrac{24600}{1200}=20.5 $ .
As the principal amount is the same for March, simple interest will also be the same.
 $ S{{I}_{3}}=20.5 $
Let’s consider the month of April.
P = Rs. 2000
Substituting in the formula we get,
 $ S{{I}_{4}}=\dfrac{2000\times 1\times 6}{100\times 12} $
Solving this we get,
 $ S{{I}_{4}}=\dfrac{12000}{1200}=10 $ .
Let’s consider the month of May.
P = Rs. 8500
Substituting in the formula we get,
 $ S{{I}_{5}}=\dfrac{8500\times 1\times 6}{100\times 12} $
Solving this we get,
 $ S{{I}_{5}}=\dfrac{51000}{1200}=42.5 $ .
Let’s consider the month of June.
P = Rs. 10000
Substituting in the formula we get,
 $ S{{I}_{6}}=\dfrac{10000\times 1\times 6}{100\times 12} $
Solving this we get,
 $ S{{I}_{6}}=\dfrac{60000}{1200}=50 $ .
Adding all the simple interest we get,
 $ SI=15+20.5+20.5+10+42.5+50=158.50 $
Therefore, the total interest at the end of 6 months is Rs. 158.50.
Now, let’s move to the second part.
We are asked to find the amount received by the account holder before closing the account.
The total account will be the balance from the last month + simple interest.
The balance at the end of June is Rs. 10000.00.
Therefore, the total amount received by the account holder = 10000 + 158.50 = Rs. 10158.50.

Note: Here, we have assumed that the bank gives interest on the minimum balance in that particular month. Also, we are calculating the interest every month so we need to modify the number of years. The total number of years will be $ \dfrac{1}{12} $ .