Question

In: Accounting

Mortgages are annuities in that a fixed monthly payment is made to the lender (assume end...

Mortgages are annuities in that a fixed monthly payment is made to the lender (assume end of month payments and an interest rate that compounds semi-annually). Sara is planning to take on a mortgage of $100 000 and believes she can afford monthly payments up to $700. How much interest would she save if she decided to pay off her mortgage over 20 years, rather than over 25 years? Her mortgage is at five percent interest calculated semi-annually.

Solutions

Expert Solution

Her savings in Interest in case she decides to pay off her mortgage over 20 years rather than 25 years would be $74,210.

Calculation:

Particulars 20 years   25 years Savings in Interest if loan repaid in 20 years rather than 25 years
Formula Amount Formula Amount
Loan Amount 100000 100000 = Interest in 25 years - Interest in 20 years
= 242710 - 168500 = $74,210
Interest Rate (r) compounded semi-annually 5% 5%
Therefore, Semi-annual rate of Interest (5%/2) 2.50% (5%/2) 2.50%
No. of terms (No. of years X 2)
since semi-annual
40 (No. of years X 2)
since semi-annual
50
Compound Interest [P (1 + r)^n] - P = [100000 (1 + 2.5%)^40] - 100000 [P (1 + r)^n] - P = [100000 (1 + 2.5%)^50] - 100000
= [100000 (1.025)^ 40] - 100000 = [100000 (1.025)^ 50] - 100000
= [100000 X 2.6850] -100000 = [100000 X 3.4271] -100000
= 268500 - 100000 = 342710 - 100000
= 168500 = 242710

Further, the calculation of monthly installment in each case i.e. 20 years and 25 years would be as under:

Particulars 20 years   25 years
Formula Amount Formula Amount
Monthly Interest Rate 5%/12 0.42% 5%/12 0.42%
No. of Monthly Installments No. of Years X 12                                                                                                       240 No. of Years X 12                                                                                                       300
EMI P (r(1+r)^n / ((1+r)^n-1) = 100000 (0.0042(1+0.0042)^240 / ((1+0.0042)^240-1)) P (r(1+r)^n / ((1+r)^n-1) = 100000 (0.0042(1+0.0042)^300 / ((1+0.0042)^300-1))
= 100000 (0.0042 X 2.7343) / (2.7343 - 1) = 100000 (0.0042 X 3.5161) / (3.5161 - 1)
= 100000 (0.01148) / 1.7343 = 100000 (0.01476) / 2.5161
= 1148 / 1.7343 = 1476 / 2.5161
= $661.93 = $586.62

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