Question

Ajit opens a bank account by depositing Rs $20,000$. At the end of each year, he withdraws a sum of money that makes the balance exactly half of what it was in the beginning of the year. If the bank pays $10\% $interest, what is the total amount withdrawn by Ajit at the end of three years?
(A) $Rs{\text{ 17,856}}$
(B) $Rs{\text{ 20,377}}$
(C) $Rs{\text{ 18,330}}$
(D) $Rs{\text{ 19,400}}$

Answer 1

Hint: If $P$, $R$and $T$ be the principal amount, rate of interest, and time in years respectively then the total amount at the end of $T$ years will be the sum of simple interest and principal amount. As he is withdrawing half of the total amount at the end of each year, so the total amount withdrawn at the end of $3$ years will be the sum of individual amounts withdrawn at the end of each year. So, first we need to calculate the total amount at the end of each year using the formula, total amount at the end of first year = $PI + SI$, Where $PI$ is the principal amount and $SI$ is the simple interest.
$SI = \dfrac{{P \times R \times T}}{{100}}$
And then we need to add all the amounts withdrawn at the end of each year to get the total amount withdrawn.

Complete Step by Step Solution:
Principal amount = $P$= $Rs$$20,000$
Rate of interest = $R$= $10\% $
Simple interest at the end of first year = $SI = \dfrac{{P \times R \times T}}{{100}}$= $\dfrac{{20,000 \times 10 \times 1}}{{100}}$= $Rs$$2000$
$\therefore $total amount at the end of first year = $PI + SI$= $Rs{\text{ 20,000 + }}Rs{\text{ 2000}}$ = $Rs{\text{ }}22,000$
Amount withdrawn at the end of first year = amount left = ${\text{22,000}} \times {\text{0}}{\text{.5}}$ = $Rs{\text{ 11,000}}$
Now, the amount left at the end of first year will be the principal amount for second year.
Principal amount = $P$= $Rs$$11,000$
Rate of interest = $R$= $10\% $
Simple interest at the end of second year = $SI = \dfrac{{P \times R \times T}}{{100}}$= $\dfrac{{11,000 \times 10 \times 1}}{{100}}$= $Rs{\text{ 1100}}$
$\therefore $total amount at the end of second year = $PI + SI$= $Rs{\text{ 11,000 + }}Rs{\text{ 1100}}$ = $Rs{\text{ 1}}2100$
Amount withdrawn at the end of second year = amount left = ${\text{12100}} \times {\text{0}}{\text{.5}}$ = $Rs{\text{ 6050}}$
Now, the amount left at the end of second year will be the principal amount for third year.
Principal amount = $P$= $Rs$$6050$
Rate of interest = $R$= $10\% $
Simple interest at the end of third year = $SI = \dfrac{{P \times R \times T}}{{100}}$= $\dfrac{{6050 \times 10 \times 1}}{{100}}$= $Rs{\text{ 605}}$
$\therefore $total amount at the end of third year = $PI + SI$= $Rs{\text{ 6050 + }}Rs{\text{ 605}}$ = $Rs{\text{ 6655}}$
Amount withdrawn at the end of third year = amount left = $6655 \times {\text{0}}{\text{.5}}$ = $Rs{\text{ 3327}}{\text{.5}}$
$\therefore $total amount withdrawn = Amount withdrawn at the end of first year $ + $total amount at the end of second year $ + $ Amount withdrawn at the end of third year
=$Rs{\text{ 11000 + }}Rs{\text{ 6050 + }}Rs{\text{ }}3327.5$
= $Rs{\text{ 20377}}{\text{.5}}$
Hence, total amount withdrawn = $Rs{\text{ 20377}}{\text{.5}}$

So, the correct option is (B) which is nearest to the obtained result.

Note:
Here, we are calculating the total amount withdrawn at the end of each year. Always be careful regarding the time in years. Suppose we are calculating the total amount at the end of second year where principal amount will be the amount left at the end of first year, then don’t put the value of time as $2$years because we are not calculating for two year, we are calculating for each year.