Mr. Jain wishes to purchase a shop for ₹25,00,000 with down payment of ₹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 $\left(a_{\overline{300} / 0.0075}=119.1616\right)$?

₹19301.52

The correct answer is Option (4) → ₹19301.52

$\text{Cost of shop} = 25,00,000,\quad \text{Down payment} = 2,00,000$

$\text{Loan amount } P = 23,00,000$

$\text{Rate per month } r = \frac{9}{12 \times 100} = 0.0075$

$n = 25 \times 12 = 300$

$\text{EMI} = P \cdot \frac{r(1+r)^n}{(1+r)^n - 1}$

$\text{EMI} = 2300000 \cdot \frac{0.0075(1.0075)^{300}}{(1.0075)^{300} - 1}$

$(1.0075)^{300} \approx 9.40$

$\text{EMI} = 2300000 \cdot \frac{0.0075 \times 9.40}{9.40 - 1}$

$= 2300000 \cdot \frac{0.0705}{8.40}$

$= 2300000 \cdot 0.008392$

$\approx 19302$

$\text{Monthly installment} \approx ₹19,300$