A loan of Rs 400000 at the interest rate of 6.75% p.a. compounded monthly is to be amortized by equal payments at the end of each month for 10 years. Find the total interest paid.

(Given $(1.005625)^{12}1201.9603, (1.005625)^{60}=1.4001)$

₹151160

The correct answer is option (1) : ₹151160

Given $P= ₹400000$

$n=120$

$i=\frac{6.75}{1200}=0.005625$

$∴EMI=\frac{400000×0.005625×(1.005625)^{100}}{(1.005625)^{120}-1}$

$=\frac{400000×0.005625×1.9603}{0.9603}$

$=₹45493$

Principal outstanding at the beginning of 61 months

$=\frac{EMI[(1+i)^{n-K+1}-1]}{i(1+i)^{n-K+1}}$

$=\frac{4593[(1.005625)^{120-61+1}-1}{0.005625(1.005625)^{120-61+1}}$

$=\frac{4593(1.4001-1)}{0.005625×1.4001}$

$= ₹233336.89$

Interest paid in 61st payment $=\frac{EMI[(1+i)^{n-K+1}-1]}{(1+i)^{n-K+1}}$

$=\frac{4593×0.4001}{1.4001}$

$=₹1312.52$

Principal paid in 61st payment $=EMI-$Interest paid in 61st period

$=₹4593-₹1312.52$

$=₹3280.48$

Total interest paid $=n×EMI-P$

$=120×4593-400000$

$=₹151160$