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 principal contained in 61st payment.
(Given $(1.005625)^{12}1201.9603, (1.005625)^{60}=1.4001)$
Answer & explanation
Correct answer: option 4
The correct answer is option (4) : ₹3280.48
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
$=₹4593-₹1312.52$
$=₹3280.48$