Mr. Jain wishes to purchase a shop for Rs. 25,00,000 with down payment of Rs. 2,00,000. If he can amortize the balance at 9% per annum compounded monthly for 25 years, what will be his monthly installment if it is given that \(\left(a_{\overline{300}|0.0075}=119.1616\right)\)?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Rs. 19301.52
The loan to be repaid is the price minus the down payment:
\(P = 25{,}00{,}000-2{,}00{,}000 = 23{,}00{,}000\)
The rate per month is \(i = \frac{9\%}{12} = 0.0075\) and the number of payments is \(n = 25 \times 12 = 300\). By the EMI (amortization) formula, P is the present value of the 300 instalments:
\(P = \text{EMI} \times a_{\overline{300}|0.0075}\)
\(\displaystyle \text{EMI} = \frac{23{,}00{,}000}{119.1616}\)
\(\approx 19301.52\)
Why the other options are incorrect
Option (1) Rs. 12140.52 — Incorrect. \(12140.52 \times 119.1616 \approx\) Rs. 14.47 lakh, not the loan of Rs. 23 lakh.
Option (2) Rs. 12607.57 — Incorrect. \(12607.57 \times 119.1616 \approx\) Rs. 15.02 lakh, not Rs. 23 lakh.
Option (3) Rs. 18302.52 — Incorrect. \(18302.52 \times 119.1616 \approx\) Rs. 21.81 lakh, not Rs. 23 lakh.
CBSE Applied Mathematics (Class XII):
Methods of Calculation of EMI or Instalment
