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Kumar buys a Maruti car for $Rs.2,40,000$ at a down payment of $Rs.1,20,000$ and the rest annual installments of $Rs.10,000$ plus $12\%$ interest on the unpaid amount. Eventually what will the car cost him?

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Last updated date: 13th Jun 2024
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Answer
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Hint: For answering this question we will use the given information that is Kumar buys a Maruti car for $Rs.2,40,000$ at a down payment of $Rs.1,20,000$ and the rest annual installments of $Rs.10,000$ plus $12\%$ interest on the unpaid amount and calculate the remaining amount that Kumar needs to pay and then apply interest on it and get the sum of them.

Complete step-by-step solution
Now let us consider from the question the given information is Kumar buys a Maruti car for $Rs.2,40,000$ at a down payment of $Rs.1,20,000$ and the rest annual installments of $Rs.10,000$ plus $12\%$ interest on the unpaid amount.
So we have the remaining amount that Kumar needs to pay. It is $Rs.2,40,000-Rs.1,20,000=Rs.1,20,000$ .
Now as we know that the rest amount $Rs.1,20,000$ Kumar pays in annual installments of $Rs.10,000$ plus $12\%$ interest on it.
According to the given condition, the interest paid annually is $12\%$ of $Rs.1,20,000$ , $12\%$ of $Rs.1,10,000$ , $12\%$ of $Rs.1,00,000$ ,………………., $12\%$ of $Rs.10,000$ .
The total interest to be paid is $\begin{align}
  & 12\%\text{ of Rs}\text{.1,20,000 + 12 }\!\!\%\!\!\text{ of Rs}\text{.1,10,000 + 12 }\!\!\%\!\!\text{ of Rs}\text{.1,00,000 + }................\text{ + 12 }\!\!\%\!\!\text{ of Rs}\text{.10,000} \\
 & \Rightarrow \text{12 }\!\!\%\!\!\text{ of }\left( \text{Rs}\text{.1,20,000 + Rs}\text{.1,10,000 + Rs}\text{.1,00,000 + }................\text{ + Rs}\text{.10,000} \right) \\
 & \Rightarrow 12\%\text{ of }\left( Rs.10,000+Rs.20,000+................+Rs.1,20,000 \right) \\
\end{align}$. Now, we see an arithmetic series $10000,20000,.........,120000$ having first term and a common difference of $10,000$ .
Let the number of terms be $n$.
Here we have the last term $1,20,000$ which implies
$\begin{align}
  & 1,20,000=10,000+\left( n-1 \right)10,000 \\
 & \Rightarrow n=12 \\
\end{align}$ .
We know from the basic concept that the sum of $n$ terms in an Arithmetic progression is ${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$ where $a$ is the first term and $d$ is the common difference.
Sum of this series will be
 $\begin{align}
  & \dfrac{12}{2}\left( 2\left( 10,000 \right)+\left( 12-1 \right)\left( 10,000 \right) \right) \\
 & \Rightarrow 6\left( 1,30,000 \right) \\
 & \Rightarrow 7,80,000 \\
\end{align}$.
So, the total interest to be paid will be $12\%\text{ of }\left( Rs.7,80,000 \right)=93,600$.
Thus the total amount spent by Kumar to buy the car is $1,20,000+1,20,000+93,600=3,33,600$ .

Note: While answering questions of this type we need to care of the calculations that we perform and be sure with them. Here while applying the interest on the remaining amount that we need to pay in annual installments we should be clear and damn sure. Here we can use another way to find the sum of the arithmetic progression of the annual installments. By applying the basic concept, that the sum of arithmetic series is given by ${{S}_{n}}=\dfrac{n}{2}\left( a+l \right)$ where $l$ is the last term. We will have the total interest to be paid will be $\begin{align}
  & 12\%\text{ of }\left( \dfrac{12}{2}\left( 10,000+1,20,000 \right) \right) \\
 & \Rightarrow 12\%\text{ of }\left( 7,80,000 \right)=93,600 \\
\end{align}$ .
So the answer will be the same in both cases.