Assuming that the value of a property in a Toronto suburb would double over 25 years, Morgan would purchase a house worth $600,000 by making a down-payment of $30,000 and obtaining a mortgage for the balance amount from a local bank at an interest rate of 4% compounded semi-annually for 25 years. If the interest rate is constant over the 25-year period, calculate the month-end payments for the mortgage. What would be his total investment in the house over the term?

Step-by-step explanation:

When the interest is compounded a different number of times per year than payments are made, it is necessary to find an equivalent interest rate to apply to each payment.

For this purpose, we can use the formula ...

  r = (1 +R/n)^(n/p) -1

where R is the nominal annual rate, n is the number of times interest is compounded per year, p is the number of payments made per year, and r is the effective monthly interest rate.

Putting the given numbers into this formula, we find ...

  r = (1 +.04/2)^(2/12) -1 ≈ 0.00330589032

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We can use this value of r in the amortization formula to find the monthly payment for a loan of $600,000 -30,000 = $570,000. That formula is ...

  A = Pr/(1 -(1+r)^-n)

where P is the principal amount of the loan, r is the monthly interest rate, and n is the number of months of the loan.

Using the values for this loan, we find the payment to be ...

  A = $570,000·0.00330589032/(1 -1.00330589033^-300) ≈ $2998.32

The monthly payment on the loan is $2998.32.

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The total amount of the 300 monthly payments is ...

  loan repayment amount = 300·$2998.32 = 899,496

This amount is added to the down payment to find the total investment:

  total investment = $30,000 +899,496 = $929,496