A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to pay the balance in annual....

Question

A farmer buys a used tractor for Rs $12000$. He pays Rs $6000$ cash and agrees to pay the balance in annual installments of Rs $500$ plus $12 \%$ interest on the unpaid amount. How much will be the tractor cost him?

Answer

It is given that the farmer pays Rs $6000$ in cash. Therefore, unpaid amount = Rs $12000$ – Rs $6000$ = Rs $6000$

According to the given condition, the interest paid annually is

$12 \%$ of $6000$, $12 \%$ of $5500$, $12 \%$ of $5000$, ..., $12 \%$ of $500$ ( I am unable to understand this step.. please suggest thanks.... )

Answer 1

The tractor is worth Rs 12000 but the farmer pays an upfront amount of Rs 6000 and the rest as per the arrangement. In the end, total cost will be Rs 12000 + interest payments.

Answer 2

If I'm understanding correctly, the question states that the interest paid annually is 12% of 6000, 12% of 5500, 12% of 5000, ..., 12% of 500. From this we infer that either the interest payable each year is calculated before the Rs $500$ for that year is subtracted from the amount left to pay, or that annual payments of Rs $500$ don't start until the second year. Either way, the answer is the same.

The farmer pays Rs 6000 to begin with, leaving Rs 6000 to pay, so the interest payable in the first year is 12% of 6000.

In year 2, he pays Rs 500, leaving Rs $6000-500=$ Rs $5500$ to pay, so the interest for year 2 is 12% of 5500.

In year 3, he pays Rs 500, leaving Rs $6000-500(2)=$ Rs $5000$ to pay, so the interest for year 3 is 12% of 5000.

,...,

In year 12, he pays Rs 500, leaving $6000-500(11)=$ Rs $500$ to pay, so the interest for year 12 is 12% of 500.

The interest payments form an arithmetic progression with $a_1 = 6000\frac{12}{100} = 720$ and $d = -500\frac{12}{100} = -60$, so the total interest payments are $s_{12} = \frac{12}{2}(720+60) =$ Rs $4680$.

As Ankush pointed out, the total paid is $12000 +$ total interest $= $Rs $16680.