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A man deposited a certain amount in a bank that would repay double the amount after a year. At the beginning of the second year, the man took out Rs 8000 and deposited the rest in the same bank. Again, at the beginning of the third year, he took out Rs 8000 and deposited the rest in the same bank. At the beginning of the fourth year, he took out Rs 8000 as before but was not left with any balance in the bank. What was his initial deposit?A. 6000B. 9000C. 8000D. 7000

Last updated date: 23rd Jul 2024
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Hint: We first assume the initial deposit. We follow the continuous process of doubling the money in the bank at the end of year and withdrawing 8000. This happens thrice in total. We find the algebraic form and find the solution.

Complete step by step solution:
Let us assume the man has an initial deposit of $x$. We know that the bank would repay double the amount after a year.
At the beginning of the second year, he would get $2x$, the man took out Rs 8000 and deposited the rest in the same bank which is $2x-8000$.
At the beginning of the third year, he would again get a double which is $2\left( 2x-8000 \right)=4x-16000$. The man again took out Rs 8000 and deposited the rest in the same bank which is equal to the amount $4x-16000-8000=4x-24000$.
At the beginning of the fourth year, he took out Rs 8000 as before but was not left with any balance.
The money would get doubled and become 8000 in all to withdraw.
So, $2\left( 4x-24000 \right)-8000=0$. Simplifying we get
\begin{align} & 8x-48000-8000=0 \\ & \Rightarrow 8x=56000 \\ & \Rightarrow x=\dfrac{56000}{8}=7000 \\ \end{align}
The initial deposit of the man was Rs. 7000.

Note:
We need to double the amount that is remaining after the withdrawal. The amount takes 12 year to double. The equation will be formed with the last remaining amount being 8000. We can also perform the back calculation to find the solution.