A machine costs a factory Rs. 6,00,000 and its useful life is 6 years, after which its scrap will sell for Rs. 40,000. To replace it by a machine of the same cost, the factory sets aside an equal amount at the end of every year in a sinking fund earning 6% per annum compounded annually. The yearly amount is (Given \((1.06)^6 = 1.4185\))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Rs. 80,286.74
The scrap money will be available at that time, so the fund needs only
\(A = 600000 - 40000 = 560000\)
The yearly deposit R in a sinking fund that must grow to A in n years is
\( \displaystyle R = \frac{A\,i}{(1+i)^n - 1} \)
With i = 0.06 and n = 6:
\( \displaystyle R = \frac{560000 \times 0.06}{1.4185 - 1} \)
\( \displaystyle = \frac{33600}{0.4185} \)
\(\approx 80286.74\)
Why the other options are incorrect
Option (2) Rs. 86,021.51 — Incorrect. This is 36000 ÷ 0.4185, which builds up the full Rs. 6,00,000 and ignores the scrap value.
Option (3) Rs. 93,333.33 — Incorrect. This is 560000 ÷ 6, which ignores the interest earned by the fund.
Option (4) Rs. 1,13,886.74 — Incorrect. This is \(\frac{33600 \times 1.4185}{0.4185}\), the loan instalment formula, which treats Rs. 5,60,000 as a loan taken today.