Mathematics for FinanceMTP Dec 23 Series IQuestion 3945 of 507
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The sum required to earn a monthly interest of Rs.1200\displaystyle 1200 at 18%\displaystyle 18\% P.a Simple Interest is -

Options

ARs.80,000\displaystyle 80,000
BRs.90,000\displaystyle 90,000
CRs.80,000\displaystyle 80,000
DNone of these
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Correct Answer

✅ Option c — Rs.80,000\displaystyle 80,000

All Options:

  • ARs.80,000\displaystyle 80,000
  • BRs.90,000\displaystyle 90,000
  • CRs.80,000\displaystyle 80,000
  • DNone of these

Detailed Solution & Explanation

Let the required sum of money (principal) be P\displaystyle P. Given parameters: * Monthly Interest (Imonthly\displaystyle I_{\text{monthly}}) = Rs. 1,200\displaystyle \text{Rs. }1,200 * Annual Interest (SI\displaystyle SI) = 1,200×12=Rs. 14,400\displaystyle 1,200 \times 12 = \text{Rs. }14,400 * Rate of Simple Interest (r\displaystyle r) = 18%\displaystyle 18\% p.a. * Time (t\displaystyle t) = 1\displaystyle 1 year The formula for Simple Interest is: SI=P×r×t100SI = \frac{P \times r \times t}{100} Substituting the values: 14,400=P×18×110014,400 = \frac{P \times 18 \times 1}{100} 14,400=0.18P14,400 = 0.18P Solving for P\displaystyle P: P=14,4000.18=80,000P = \frac{14,400}{0.18} = 80,000 Thus, the required sum of money is Rs. 80,000\displaystyle \text{Rs. }80,000. Hence, **Option C** is the correct answer.

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