Question

A mutual fund manager wishes to purchase a property that's been valued at $1.5 m. She has $200,000 in cash to use as a deposit, and she will require a mortgage for the rest, which is to be paid monthly. The annual interest rate on the loan is 2.45% and the loan is for 25 years. Calculate the total interest payable, giving your answer in dollars to the nearest hundred dollars. Do not include commas or the dollar sign in your answer.

Answer 1

To calculate the total interest payable on a mortgage for a property valued at $1.5 million, with a $200,000 deposit and an annual interest rate of 2.45% for 25 years, we apply a formula to determine the monthly payment, multiply by the total number of payments to get the total amount paid, then subtract the principal to find the total interest.

Step-by-step explanation:

The question asks us to calculate the total interest payable on a mortgage for a property valued at $1.5 million with a deposit made of $200,000. The loan amount hence is $1.3 million ($1,500,000 - $200,000), with an annual interest rate of 2.45% over a period of 25 years. The mortgage will be paid in monthly installments.

To find the total interest, we can use the formula for calculating the total payment for an annuity (which in this case, is the monthly mortgage payment) and then subtracting the principal from the total amount paid over the life of the loan. The formula for the monthly payment, M, for a loan amount, P, with an annual interest rate, r, converted to a monthly interest rate (r/12), over n months, is:

M = P [r(1+r)^n] / [(1+r)^n - 1]

Where:

• P = Principal loan amount ($1.3 million)
• r = Monthly interest rate (2.45% per year or 0.02045/12 per month)
• n = Total number of payments (25 years x 12 months/year)

After calculating M, multiply M by n to get the total amount paid. Subtract the principal (P) from this to get the total interest paid.

Answer 2

Answer: 439, 800 is the correct answer

Step-by-step explanation:

$439800$

We note that the manager requires a mortgage of $1,500,000−$200,000=$1,300,000. We must apply the formula for P0 and solve for d, that is,

P0=d(1−(1+rk)−Nk)(rk).

We have P0=$1,300,000,r=0.0245,k=12,N=25, so substituting in the numbers into the formula gives

$1,300,000=d(1−(1+0.024512)−25⋅12)(0.024512),

that is,

$1,300,000=224.163565d⟹d=$5,799.34.

So the total interest payable is

I=$5,799.34×25×12−$1,300,000=$439,802

which is $439,800 to the nearest $100.