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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(1.4775)^{1/8} = 1.05]

Solution

✅ Correct Option: 4

The company needs to buy a new machine for ₹4,00,000 but will receive ₹50,000 by selling the old machine as scrap.

Amount needed = ₹4,00,000 - ₹50,000

Amount needed = ₹3,50,000


The future value of a sinking fund with annual deposits is given by:

Future Value = Annual Deposit × (1+r)n−1r\dfrac{(1 + r)^n - 1}{r}

where Future Value = ₹3,50,000, Annual Deposit = ₹36,650, and r=0.05r = 0.05


Substituting the values:

3,50,000=36,650×(1.05)n−10.053,50,000 = 36,650 \times \dfrac{(1.05)^n - 1}{0.05}

3,50,00036,650=(1.05)n−10.05\dfrac{3,50,000}{36,650} = \dfrac{(1.05)^n - 1}{0.05}

9.55=(1.05)n−10.059.55 = \dfrac{(1.05)^n - 1}{0.05}


9.55×0.05=(1.05)n−19.55 \times 0.05 = (1.05)^n - 1

0.4775=(1.05)n−10.4775 = (1.05)^n - 1

0.4775+1=(1.05)n0.4775 + 1 = (1.05)^n

1.4775=(1.05)n1.4775 = (1.05)^n


The given information states that (1.4775)1/8=1.05(1.4775)^{1/8} = 1.05

Raising both sides to the power of 8:

[(1.4775)1/8]8=(1.05)8[(1.4775)^{1/8}]^8 = (1.05)^8

1.4775=(1.05)81.4775 = (1.05)^8


Comparing 1.4775=(1.05)n1.4775 = (1.05)^n with 1.4775=(1.05)81.4775 = (1.05)^8:

n=8n = 8

Therefore, the useful life of the machine is approximately 8 years.

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