Homework Help: Math Help Please!!!

Posted by Nieda on Thursday, September 15, 2011 at 9:42am.

A couple purchased a house and signed a mortgage contract for $110,000 to be paid in “every second week “ installments over 25 years, at 3.0 %. The contract stipulates that after 3 years the mortgage will be renegotiated at the new prevailing rate of interest. Calculate:
a) The every second week payment for the initial 3-year period
b) The outstanding principal after 3 years.
c) The new payment ( now, once a month )after the 3 years, at 4.5 %

NOTE: mortgages rates in Canada are always compounded twice a year.

Canadian Math help needed - bobpursley, Thursday, September 15, 2011 at 10:08am
Help!
Math Help Please!!! - Reiny, Thursday, September 15, 2011 at 11:14am
first we must find the rate for the "every second week" period.
let that rate be i
(1+i)^26 = 1.015 , (semi-annual rate is 1.5%)
1+i = 1.015^(1/26) = 1.000572803
i = .000572803 ( I stored the full decimal in memory of calculator)

then if P is the payment
110000 = P[ 1 - 1.000572803^-650 ]/.000572803
P = 202.73

b) This continues for 3 years, or 78 payments
Balance after 3 years
= 110000(1.000572803)^78 - 202.73[ 1.000572803^78 - 1]/.000572803
= 115024.61 - 16167.01
= 98857.60

c) Now we have to find the equivalent monthly rate for 4.5%
let it be j
(1+j)^6 = 1.0225
j = .00371532
new montly payment M , 22 years left or 264 payments
98857.60 = M [1 - 1.00371532^-264]/.00371532
M = 588.30 per month

I suggest you check my "arithmetic" on this one.
LET'S TRY THIS AGAIN - Reiny, Thursday, September 15, 2011 at 12:14pm
Totally missed the part that our rates are assumed to be compounded semi-annually !!

So we first have to find the equilavalent semi-annual rate equal to 3% compounded annually.
let that rate be j
(1+j)^2 = 1.03
1+j = √1.03
j = .014889157 (I stored that in calculator memory)

Now we must find the rate for the "every second week" period.
let that rate be i
(1+i)^26 = 1.014889157 , (semi-annual rate is 1.014889157 %)
same steps as above ...
i = .0005686

then if P is the payment
110000 = P[ 1 - 1.0005686^-650 ]/.0005686
P = 202.47

b) This continues for 3 years, or 78 payments
Balance after 3 years
= 110000(1.0005686)^78 - 202.73[ 1.0005686^78 - 1]/.0005686
= 114986.94 - 16143.68
= 98843.26 ---> balance after 3 years (78 payments)

c) Equivalent rate compounded semi-annual ...
let that rate be k
(1+k)^2 = 1.045
k = .0022252415
Now we have to find the equivalent monthly rate for the above rate
let it be j
(1+j)^6 = 1.0022252415
j = .003674809
new montly payment M , 22 years left or 264 payments
98857.60 = M [1 - 1.003674809^-264]/.003674809
M = 585.57 per month
Final correction - Math Help Please!!! - Reiny, Thursday, September 15, 2011 at 12:17pm
This is the problem when "cutting and pasting"
It is so easy to miss a change that should be made

2nd last line should say

98843.26 = M [1 - 1.003674809^-264]/.003674809

Answer this Question

Related Questions

Accounting - I need to enter the following into Journal Enteries a) Borrowed $12...
accounting - You have just purchased a house and have obtained a 30-year, $200,...
accounting - House mortgage You have just purchased a house and have obtained a...
accounting - The following transactions occurred during March 2009 for the ...
infinite math - Hey guys! really need help with this one!!! On December 31, 1995...
math - Ever wonder how much a house “actually” costs? Consider Alex ...
Financial Accounting - Prepare the general journal entries for the following ...
accounting - how would the table of analysis be set up with these transactions: ...
math - Lauren and mark obtained a 20 year 120,000 conventional mortgage at 10.5...
Finance - Say that you purchase a house for $270,000 by getting a mortgage for $...

For Further Reading

First Name:
School Subject:
Answer: