SOLUTION: Question 3 (15 marks) John has just purchased an apartment at a price of $5,000,000. He made a down-payment of $2,000,000 and financed the remaining with a 30-year mortgage at APR

Question 1201043: Question 3 (15 marks) John has just purchased an apartment at a price of $5,000,000. He made a down-payment of $2,000,000 and financed the remaining with a 30-year mortgage at APR 12%, compounded monthly. (a) Determine the size of the fixed month-end payments. (5 marks)
(b) Calculate the amount John still owes the bank right after the 120th payment was made. (5 marks)
(c) Calculate the interest payment and the amount of principal paid in the 121st loan repayment. (5 marks)
Found 3 solutions by mananth, ikleyn, math_tutor2020:
Answer by mananth(16946)   (Show Source): You can put this solution on YOUR website!
Thanks to to Tutor Ikleyn.
I deleted the solution

Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.

Calculations in the post by @mananth all are INCORRECT.

Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

Answers
(a) $30,858.38
(b) $2,802,539.87
(c) interest payment = $28,025.40; principal payment = $2,832.98

====================================================================================================

Work Shown for part (a)

Price = $5 million
Down-payment = $2 million
Loan amount = 5-2 = $3 million

We will use this monthly payment formula
P = (L*i)/( 1-(1+i)^(-n) )
where,
P = monthly payment
L = loan amount
i = monthly interest rate in decimal form
n = number of months

In this case
L = 3,000,000
i = 0.12/12 = 0.01
n = 30*12 = 360 months

P = (L*i)/( 1-(1+i)^(-n) )
P = (3,000,000*0.01)/( 1-(1+0.01)^(-360) )
P = 30,858.3779077651
P = 30,858.38

The answer can be confirmed through the use of a loan calculator such as this one
https://www.calculator.net/loan-calculator.html

===============================================

Work Shown for part (b)

Refer to the 2nd formula mentioned on this page
https://www.mtgprofessor.com/formulas.htm
It calculates the remaining balance at any given month.

That formula looks rather messy.
The numerator is L*( (1+c)^n - (1+c)^p )
The denominator is (1+c)^n - 1

L = loan amount
c = monthly interest rate in decimal form
n = number of months of the entire loan
p = current month number

In this case we have
L = 3,000,000
c = 0.01 calculated earlier
n = 360 months total
p = 120

numerator = L*( (1+c)^n - (1+c)^p )
numerator = 3,000,000*( (1+0.01)^360 - (1+0.01)^120 )
numerator = 97,947,763.299334

denominator = (1+c)^n - 1
denominator = (1+0.01)^360 - 1
denominator = 34.949641327685

Divide the results:
97,947,763.299334/34.949641327685 = 2,802,539.87103856

John still owes $2,802,539.87 after the 120th payment was made. This is approximately 2.8 million dollars.

The first link I mentioned earlier (the loan calculator) provides a handy amortization table to show the balance for any given month.
Scroll down to month 120 to confirm that $2,802,539.87 is the ending balance.
120 months = 120/12 = 10 years

At first a big chunk of the monthly payment is composed of interest.
As time goes on, more of the monthly payment is devoted to the principal.

===============================================

Work Shown for part (c)

The balance at the end of month 120 is $2,802,539.87 as mentioned in part (b)

Multiply this with c = 0.01 mentioned in the previous part.

0.01*(2,802,539.87) = 28,025.3987 = $28,025.40 is the interest part of the payment

Subtract the interest from the monthly payment to determine the principal.
30,858.38 - 28,025.40 = $2,832.98

For month 121 we have
interest = $28,025.40
principal = $2,832.98

The amortization table can be used to confirm these results.

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