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Mr. R.K. Nair gets ₹ 6455 at the end of one year at the rate of 14% per annum recurring deposit account. Find the monthly instalment.

Mr. R.K. Nair gets ₹ 64556455 at the end of one year at the rate of 14% 14 \% per annum recurring deposit account. Find the monthly instalment.

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Q. Mr. R.K. Nair gets ₹ 64556455 at the end of one year at the rate of 14% 14 \% per annum recurring deposit account. Find the monthly instalment.
  1. Identify Given Values: Step 11: Identify the given values and the formula to use.\newlineMr. R.K. Nair receives ₹ 64556455 after one year with an interest rate of 1414% per annum on a recurring deposit. We need to find the monthly installment.\newlineUsing the formula for the future value of a recurring deposit: A=P×(1+r)n−1r A = P \times \frac{(1 + r)^n - 1}{r} \newlineWhere:\newline- A A is the future value (₹ 64556455)\newline- P P is the monthly installment (unknown)\newline- r r is the monthly interest rate (1414% per annum, so 1412% \frac{14}{12} \% per month)\newline- n n is the total number of deposits (1212 deposits in one year)
  2. Convert Annual Interest Rate: Step 22: Convert the annual interest rate to a monthly interest rate and substitute the values into the formula.\newlineThe monthly interest rate r r is 14100×12=0.01167 \frac{14}{100 \times 12} = 0.01167 \newlineSubstitute into the formula:\newline6455=P×(1+0.01167)12−10.01167 6455 = P \times \frac{(1 + 0.01167)^{12} - 1}{0.01167}
  3. Calculate Denominator: Step 33: Calculate the denominator (1+0.01167)12−1 (1 + 0.01167)^{12} - 1 \newlineUsing a calculator,\newline(1+0.01167)12≈1.1508 (1 + 0.01167)^{12} \approx 1.1508 \newlineSo, 1.1508−1=0.1508 1.1508 - 1 = 0.1508 \newlineNow, substitute back:\newline6455=P×0.15080.01167 6455 = P \times \frac{0.1508}{0.01167}
  4. Solve for P: Step 44: Solve for P P (monthly installment).\newline6455=P×1292.5 6455 = P \times 1292.5 \newlineP=64551292.5≈4.995 P = \frac{6455}{1292.5} \approx 4.995

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