Question

Dave currently makes monthly payments towards a 25-year mortgage of $290000 with an interest rate of 13.2% compounded monthly.

After making the 65th payment, Dave refinanced his mortgage at an interest rate of 8.4% compounded monthly and renegotiated his term. This resulted in Dave's monthly payments being reduced by $400. How large will Dave's final drop payment be?

Answer 1

years	25.00
Month (n)	300.00
Principal (pv)	2,90,000.00
Monthly Interest rate - i	13.2%/12=1.1%
Monthly Interest rate - i	0.01
(1+i)	1.01
(1+i)^n	26.63
Month left after 65th payment (n1)	300-65=235
(1+i)^n1	13.08

Formula to calculate equal Monthly Payment ( EMP) is :

PV = EMP*( ((1+i)^n-1)/ (i*(1+i)^n))

290000 = EMP*((26.63-1)/0.01*26.63

290000=EMP*25.63/0.29

EMP= 290000*0.29/25.63

EMP= 3314.47

After the 65th Payment , he will owe money to the bank , the amount is

PV = EMP*( ((1+i)^n1-1)/ (i*(1+i)^n1))

PV = 3314.47*(13.08-1)/0.01*13.08

PV = 3314.47*12.08/0.143851

PV = 278274.53

		After refinance
	years	n
	Month (n2)	n
	Principal (pv)	2,78,274.53
	Monthly Interest rate - i	8.4%/12=0.7%
	Monthly Interest rate - i	0.0070
	(1+i)	1.0070
	EMP	3,314.47
	EMP reduced by 400	3314.47-400 = 2914.47

PV = EMP*( ((1+i)^n2-1)/ (i*(1+i)^n2))

278274.53 = 2914.47*( ((1+i)^n2-1)/ (i*(1+i)^n2))

n2 =158.22 months

now till 158 month EMI PAID PV will be as under:

PV = EMP*( ((1+i)^n2-1)/ (i*(1+i)^n2))

2914.47*((1.007)^158)-1))/ (0.007*(1.007)^158))

PV = 2,78,058.26

Therefore Dave final drop payment will be 278274.53- 2,78,058.26 = 216.27/- ANS