A person amortizes a loan of Rs 1500000 for renovation of his house by 8 years mortgage at the rate of 12% p.a. compounded monthly. Find the principal contained in 40th payment.

(Given $(1.01)^{96}=2.5993, (1.01)^{57}=1.7633)$

₹12825.84

The correct answer is option (4) : ₹13825.84

Given $P=₹1500000, n=12×8=96\, months $

i.e $i=\frac{12}{1200}=0.01$

$EMI=\frac{1500000×0.01(1.01)^{96}}{(1.01)^{96}-1}$

$=₹24379.10$

Principal outstanding at beginning of 40th month

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

$⇒\frac{24379.10[(1.01)^{57}-1]}{0.01(1.01)^{57}}$

$⇒₹1055326.2$

Interest paid in 40th payment

$=EMI[(1+i)^{90-40+1}-1]$

$=(1+i)^{96-40+1}$

$=\frac{24379.10×0.7633}{1.7633}$

$=₹10553.26$

Principal contained in 40th payment $= EMI-$ Interest paid in 40th payment

$= ₹24379.10-₹10553.26$

$=₹13825.84$