A machine costing ₹3,00,000 will have its scrap value of ₹50,000. The company at present plans to put ₹36,650 per annum at the end of each year in a sinking fund at the rate 5% per annum for the replacement of the machine after its useful life. Suppose the new machine will cost ₹4,00,000 at that time, then the useful life (approx.) of the machine is:

[Given: $(1.4775)^{1/8}≈ 1.05$]

8 yrs

The correct answer is Option (4) → 8 yrs **

Required accumulation = ₹300000 − ₹50000 = ₹250000.

Annual deposit = ₹36650, rate = 5% = 0.05.

Sinking-fund formula:

$36650 \cdot \frac{(1.05)^{n}-1}{0.05} = 250000$

So

$(1.05)^{n}-1 = \frac{250000 \cdot 0.05}{36650}$

$(1.05)^{n} = 1 + \frac{12500}{36650}$

$(1.05)^{n} = 1.34125$

Given that $(1.4775)^{\frac{1}{8}} \approx 1.05$, so $1.05^{8} \approx 1.4775$.

Since $1.34125$ is close to $1.4775$, the nearest integer solution is:

$n = 8$