Mathematics for FinanceMTP June 24 Series IIQuestion 1413 of 512
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A sum of money, lent out at simple interest, doubles itself in 8 years. Find in how many years will the sum become 3 times itself.

Options

A16 years
B15 years
C20 years
DNone of these
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Correct Answer

✅ Option a — 16 years

All Options:

  • A16 years
  • B15 years
  • C20 years
  • DNone of these

Detailed Solution & Explanation

Let the principal be P\displaystyle P and the rate of simple interest be r%\displaystyle r\% p.a. The formula for simple interest is: SI=P×r×t100SI = \frac{P \times r \times t}{100} Given that the sum doubles itself in 8\displaystyle 8 years, the simple interest earned is equal to the principal: SI=P for t=8SI = P \text{ for } t = 8 P=P×r×8100P = \frac{P \times r \times 8}{100} 1=8r100  ⟹  r=1008=12.5% p.a.1 = \frac{8r}{100} \implies r = \frac{100}{8} = 12.5\% \text{ p.a.} We want to find the number of years t\displaystyle t for the sum to become 3\displaystyle 3 times itself, which means the interest earned will be 2P\displaystyle 2P: 2P=P×12.5×t1002P = \frac{P \times 12.5 \times t}{100} 2=0.125×t2 = 0.125 \times t Solving for t\displaystyle t: t=20.125=16 yearst = \frac{2}{0.125} = 16 \text{ years} Thus, the sum will become 3\displaystyle 3 times itself in 16\displaystyle 16 years. Hence, **Option A** is the correct answer.

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