Skip to content
Question

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 life time when its scrap realizes 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. [Use (1.05)12=1.7958(1.05)^{12} = 1.7958]

CBSECBSE Class XII Board 2025Subjective· 5mImportance★★★★★
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Need ₹1,00,000−₹5,000=₹95,000₹1{,}00{,}000 - ₹5{,}000 = ₹95{,}000 in 12 years at 5%; from A=R⋅(1.05)12−10.05A = R\cdot\frac{(1.05)^{12}-1}{0.05}, R=95000×0.050.7958≈₹5,968.84R = \frac{95000\times0.05}{0.7958} \approx ₹5{,}968.84.

Future value of an ordinary annuity (sinking fund): A=R[(1+i)n−1i]A = R\left[\dfrac{(1+i)^{n} - 1}{i}\right], where AA = amount to accumulate, RR = yearly deposit, ii = annual rate, nn = number of years.

  1. Amount needed for replacement =cost−scrap=1,00,000−5,000=₹95,000= \text{cost} - \text{scrap} = 1{,}00{,}000 - 5{,}000 = ₹95{,}000.
  2. Here i=0.05i = 0.05, n=12n = 12, and (1.05)12=1.7958(1.05)^{12} = 1.7958 (given). …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.