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Exercise 7.1 · Q8

Q.A machine costs ₹1,00,000 and its effective life is estimated to be 12 years. A sinking fund is created for replacing the machine by a new model at the end of its lifetime when its scrap realises a sum of ₹5,000 only. Find what amount should be set aside at the end of each year, out of the profits, for the sinking fund if it accumulates at 5% effective.

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The key idea is to treat the replacement cost (₹95,000) as the future value of an annuity of equal annual deposits into a sinking fund earning 5% per year. The required annual deposit is approximately ₹5,968.40.

  1. Understand the problem. A machine costs ₹1,00,000 today. After 12 years, it will be sold as scrap for ₹5,000. To replace it with a new machine (assuming the same cost), we need ₹1,00,000 at that time. But we already have ₹5,000 from scrap, so the net additional amount needed at the end of 12 years is:

Required future sum=1,00,000−5,000=₹95,000\text{Required future sum} = 1,00,000 - 5,000 = ₹95,000

This ₹95,000 must be accumulated by making equal annual deposits into a sinking fund that earns 5% per annum, compounded annually. The deposits are made at the end of each year.

  1. Recall the sinking fund formula. When equal payments AA are made at the end of each year for nn years into an account earning interest rate ii per year (compounded annually), the accumulated amount SS (future value of the annuity) is:

S=A⋅(1+i)n−1iS = A \cdot \frac{(1+i)^n - 1}{i}

Here, S=95,000S = 95,000, i=0.05i = 0.05, n=12n = 12. We need to solve for AA.

  1. Plug in the numbers.

95,000=A⋅(1.05)12−10.0595,000 = A \cdot \frac{(1.05)^{12} - 1}{0.05}

First compute (1.05)12(1.05)^{12}. You can do this step by step:

  • (1.05)2=1.1025(1.05)^2 = 1.1025
  • (1.05)4=(1.1025)2=1.21550625(1.05)^4 = (1.1025)^2 = 1.21550625
  • (1.05)8=(1.21550625)2≈1.477455(1.05)^8 = (1.21550625)^2 \approx 1.477455
  • Then (1.05)12=(1.05)8×(1.05)4≈1.477455×1.215506≈1.795856(1.05)^{12} = (1.05)^8 \times (1.05)^4 \approx 1.477455 \times 1.215506 \approx 1.795856

More precisely, using a calculator: (1.05)12≈1.795856326(1.05)^{12} \approx 1.795856326. So:

(1.05)12−10.05=1.795856326−10.05=0.7958563260.05=15.91712652\frac{(1.05)^{12} - 1}{0.05} = \frac{1.795856326 - 1}{0.05} = \frac{0.795856326}{0.05} = 15.91712652

  1. Solve for AA.

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