Question

Punam opened a recurring deposit amount with Bank of Baroda for \[1\dfrac{1}{2}years\]. If the rate of interest is \[6 \% \] per annum and the bank pays \[Rs.11313\] on maturity, find how much Punam is deposited each month?

Answer 1

Hint: Use the maturity value formula, where maturity value is the amount to be received on the due date.

Maturity value is the sum of the amount loaned (taken from the bank) and the interest given. Mathematically, \[M = P + I\]

Where simple interest is the accumulated interest on the principal amount while the total amount accumulated in a defined time period is the sum of the principal amount and the interest accumulated on the principal amount, the formula used for the calculation of the simple interest is

\[Interest = {\text{Principle}} \times \dfrac{{n\left( {n + 1} \right)}}{2} \times \dfrac{{r \times t}}{{100}}\].

Here, in this question we need to calculate the amount to be deposited in the bank by Punam so that at the end of 1.5 years, she will get Rs. 11313 as the maturity amount from the bank. For this we need to first evaluate the value of the interest in terms of the principal amount by using the formula

\[Interest = {\text{Principle}} \times \dfrac{{n\left( {n + 1} \right)}}{2} \times \dfrac{{r \times

t}}{{100}}\] and use the formula $M = P + I$ to calculate the depositing amount.

Complete step by step answer:

Given, Time period is \[1\dfrac{1}{2}years\] and rate of interest =\[6\% \]

Maturity amount=\[Rs.11313\]

Now find the amount deposited by Punam deposited each month

Where the number of months \[n = 1\dfrac{1}{2} \times 12 = 18 months\]

Since Interest Amount is given as \[Interest = {\text{Principle}} \times \dfrac{{n\left( {n + 1} \right)}}{2} \times \dfrac{{r \times t}}{{100}}\]

Substituting n as 18, r as 6 and t as $\dfrac{1}{{12}}$ in the equation \[Interest = {\text{Principle}} \times \dfrac{{n\left( {n + 1} \right)}}{2} \times \dfrac{{r \times t}}{{100}}\] to determine the relation between Interest and the principal amount.

\[Interest = {\text{Principle}} \times \dfrac{{n\left( {n + 1} \right)}}{2} \times \dfrac{{r \times t}}{{100}}\]

\[= P \times \dfrac{{18\left( {18 + 1} \right)}}{2} \times \dfrac{6}{{100}} \times \dfrac{1}{{12}}\]

\[= P \times \dfrac{{18 \times 19}}{2} \times \dfrac{1}{{2 \times 100}}\]

\[= P \times \dfrac{{342}}{{400}}\]

\[I = 0.86P - - - - (i)\]

Hence the simple interest \[I = 0.86P\]

As maturity value is the sum of the amount loaned and the interest, given by the formula

\[M = P + I\]

where P is the amount deposited per month.

For 18 months,

\[Maturity = P \times 18 + Interest\]

Now substitute the value of the interest in the maturity value formula from equation (i) as:

\[M = P + I\]

\[= 18P + 0.86P - - - - (ii)\]

According to the question, the maturity value received by Punam is \[Rs.11313\], so substituting $M = 11313$ in the equation (ii), we get:

\[M = 18P + 0.86P\]

\[11313 = 18.86P\]

\[P = \dfrac{{11313}}{{18.86}}\]

\[= 599.98\]

\[= Rs.600\]

Hence, the amount deposited by Punam each month is \[Rs.600\].

Note: A very important point to be noted down here that, we cannot use the formula for the interest as $I = \dfrac{{prt}}{{100}}$ here in this case as the maturity rate is fixed here and if Punam misses any of the depositing amount any month then, the maturity amount would be varied which is not the case here.