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 outstanding at the beginning of 40th month.

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

₹1055326.2

The correct answer is option (2) : ₹1055326.2

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

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

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

$⇒₹1055326.2$