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Exercise 7.1 · Q7

Q.To save for child's education, a sinking fund is created to have ₹1,00,000 at the end of 25 years. How much money should be retained out of the profit each year for the sinking fund, if the investment can earn interest at the rate 4% per annum.

Arunachal CbseNCERTSubjective· 3mImportance★★★★★
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The problem asks for the annual payment into a sinking fund that accumulates to ₹1,00,000 in 25 years at 4% p.a. interest. This is a future value of an annuity problem — the annual payment (sinking fund installment) is found using the formula for the future value of an ordinary annuity. The required annual payment is ₹2,401.20.

The core idea here is that a sinking fund is simply an annuity — a series of equal payments made at regular intervals, earning compound interest, with the goal of reaching a specific future sum. Instead of thinking about "how much to set aside," think: "If I deposit the same amount every year into an account that grows at 4%, what must that amount be so that after 25 years I have exactly ₹1,00,000?"

This is the future value of an ordinary annuity (payments at the end of each year). The formula is:

FV=P×(1+r)n−1rFV = P \times \frac{(1 + r)^n - 1}{r}

where FVFV = future value, PP = annual payment, rr = annual interest rate (as decimal), nn = number of years.

We know FV=1,00,000FV = 1,00,000, r=0.04r = 0.04, n=25n = 25. We need to solve for PP.


  1. Write the formula with the known values.

1,00,000=P×(1+0.04)25−10.041,00,000 = P \times \frac{(1 + 0.04)^{25} - 1}{0.04}

  1. Compute the compound factor (1.04)25(1.04)^{25}.

    This is the only heavy calculation. You can do it stepwise or use a calculator:

    (1.04)5≈1.21665(1.04)^5 \approx 1.21665, then square twice:

    (1.04)10≈(1.21665)2=1.48024(1.04)^{10} \approx (1.21665)^2 = 1.48024

    (1.04)20≈(1.48024)2=2.19112(1.04)^{20} \approx (1.48024)^2 = 2.19112

    Then multiply by (1.04)5(1.04)^5: 2.19112×1.21665≈2.665842.19112 \times 1.21665 \approx 2.66584

    So (1.04)25≈2.66584(1.04)^{25} \approx 2.66584.

    Tip

    For exam speed, remember that (1.04)25(1.04)^{25} is very close to e0.04×25=e1≈2.71828e^{0.04 \times 25} = e^1 \approx 2.71828, but the exact value is slightly less. Using 2.66584 is accurate enough.

  2. Subtract 1 and divide by the rate.

    2.66584−10.04=1.665840.04=41.646\frac{2.66584 - 1}{0.04} = \frac{1.66584}{0.04} = 41.646 …

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