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Worked Examples · Example 14

Q.Find the least number of years for which an ordinary annuity of ₹1,500 per annum must run in order that its amount just exceeds ₹30,000 at 9% compound annually.

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✓ Free question

Set the annuity future-value formula greater than ₹30,000, solve for nn using logarithms, then verify with the nearest integers.

FV=R⋅(1+i)n−1iFV=R\cdot\dfrac{(1+i)^{n}-1}{i}, where RR = annual deposit, ii = interest rate per period, nn = number of years (periods).

Given: R=₹1,500R=₹1{,}500, i=9%=0.09i=9\%=0.09; require FV>₹30,000FV>₹30{,}000.

  1. Set up the inequality:

1500⋅(1.09)n−10.09>300001500\cdot\dfrac{(1.09)^n-1}{0.09}>30000

  1. Divide both sides by 1500:

(1.09)n−10.09>20 ⇒ (1.09)n−1>1.8 ⇒ (1.09)n>2.8\dfrac{(1.09)^n-1}{0.09}>20\ \Rightarrow\ (1.09)^n-1>1.8\ \Rightarrow\ (1.09)^n>2.8

  1. Take logarithms:

n>log⁡2.8log⁡1.09=1.0296190.086178≈11.95n>\dfrac{\log 2.8}{\log 1.09}=\dfrac{1.029619}{0.086178}\approx11.95

  1. Since nn must be a whole number of years and 11.9511.95 is not an integer, the least integer satisfying it is n=12n=12. Verify both sides:
  • At n=11n=11: (1.09)11=2.580426(1.09)^{11}=2.580426, FV=1500×1.5804260.09=1500×17.5603=₹26,340.44FV=1500\times\dfrac{1.580426}{0.09}=1500\times17.5603=₹26{,}340.44 (< ₹30,000 — not enough).
  • At n=12n=12: (1.09)12=2.812665(1.09)^{12}=2.812665, FV=1500×1.8126650.09=1500×20.1407=₹30,211.08FV=1500\times\dfrac{1.812665}{0.09}=1500\times20.1407=₹30{,}211.08 (> ₹30,000 ✓).
  1. Self-check: n=11n=11 falls short, n=12n=12 clears ₹30,000 — confirms 12 is the least sufficient number of years.
✓Final answer

The annuity must run for at least 1212 years (amount then ≈₹30,211.08\approx₹30{,}211.08, which just exceeds ₹30,000).

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