Skip to main contentSkip to solution

A company purchased a machine for ₹ 15,00,000 and its effective life is estimated to be 10 years. A sinking fund is created for replacing the machine at the end of its effective life when its scrap value is ₹ 2,42,000. What amount company should provide, at the end of every year out of profits for the sinking fund if it accumulates an interest of 5% per annum? [Given(1.05)¹⁰=1.629]

Solution

✅ Correct Option: 2

The company wants to save money every year so that after 10 years, they have enough to buy a new machine.

Given:

  • New machine cost: ₹15,00,000
  • Scrap value: ₹2,42,000
  • Years to save: 10 years
  • Interest rate: 5% per annum
  • (1.05)10=1.629(1.05)^{10} = 1.629

The company will get some money by selling the old machine (scrap value), so they don't need to save the full ₹15,00,000.

Amount needed from savings == New machine cost −- Scrap value

=₹15,00,000−₹2,42,000= ₹15,00,000 - ₹2,42,000

=₹12,58,000= ₹12,58,000


When depositing the same amount every year with interest, the future value formula is:

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

Where:

  • Future Value =₹12,58,000= ₹12,58,000
  • Annual Payment =P= P (to be found)
  • r=0.05r = 0.05
  • n=10n = 10 years

₹12,58,000=P×(1.05)10−10.05₹12,58,000 = P \times \dfrac{(1.05)^{10} - 1}{0.05}

₹12,58,000=P×1.629−10.05₹12,58,000 = P \times \dfrac{1.629 - 1}{0.05}

₹12,58,000=P×0.6290.05₹12,58,000 = P \times \dfrac{0.629}{0.05}

₹12,58,000=P×12.58₹12,58,000 = P \times 12.58

P=₹12,58,00012.58P = \dfrac{₹12,58,000}{12.58}

P=₹1,00,000P = ₹1,00,000


Therefore, the company should provide ₹1,00,000 at the end of every year.

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question