Question

14) To purchase studio space, on April 3, 2025 a pianist takes out a 30-year loan of $120,000 at an annual interest rate of 11%. The first monthly payment is to be made on May 3, 2025; the final payment is scheduled to be made April 3, 2055. The monthly payment is $1,142.79 . [Note: If you use the table you’ll get that the monthly payment is $1,142.40, but the table produces an approximation. $1,142.79 is the precise payment.]

a) Calculate the total interest charged over the course of the loan, assuming that nothing but the standard payment is made each month.

b) How much will she owe immediately after making the payment on June 3, 2025? (= the second payment) Round appropriately to the nearest penny.

c) Suppose that as part of that June 3, 2025 payment she pays an extra $300 as an overpayment. Calculate the dollar amount of interest which this saves in the future. Round your answer down to the nearest $10.

Answer 1

Since the first monthly payment is to be made on May 3, 2025 and since the final payment is scheduled to be made April 3, 2055, the period of the loan is 130*12-1 = 359 months.

a). The formula for calculating the fixed monthly payment (P) required to fully amortize a loan of L dollars over a term of n months at a monthly interest rate of r is

P = L[r(1 + r)ⁿ]/[(1 + r)ⁿ - 1].

Here, P = $ 120000, n = 359 months and r = 11/1200. Therefore, P = 120000* (11/1200)[ (1+11/1200)³⁵⁹]/ [ (1+11/1200)³⁵⁹ -1] = 1100 (26.46549722)/( 25.46549722) = $ 1143. 195701 = $1143.20 ( on rounding off to the nearest cent). The total repayment made by the pianist is 359*$1143.20 = $ 410408.80. Therefore, the interest paid by the pianist is $ 410408.80- $ 120000 = $ 290408.80.

b). The formula used to calculate the remaining loan balance (B), of a fixed payment loan after p months is

B = L[(1 + r)ⁿ - (1 + r)^p]/[(1 + r)ⁿ - 1] , where L, r and n are as above.

Here, p = 2 so that B = 120000[(1+11/1200)³⁵⁹ –(1+11/1200)²]/ [ (1+11/1200)³⁵⁹ -1] = 120000(26.46549722-1.018417361)/( 25.46549722)= 120000*25.44707987/ ( 25.46549722) = $ 119913.21( on rounding off to the nearest cent).

c). If the pianist makes an extra payment of $ 300 on June,3, 2025, the loan amount as on June,3, 2025 gets reduced to $ 119913.21 - $ 300 = $ 119613.21. This is to be repaid in 359-2=357 monthly installments. Further, the monthly installment now =119613.21(11/1200)[(1+11/1200)³⁵⁷]/[ (1+11/1200)³⁵⁷ -1]= (1096.454425)(25.98688734) /(24.98688734)= $ 1140.335618 = $ 1140.34 ( on rounding off to the nearest cent). Now, the total repayment made by the pianist is 2*$1143.20 +$ 300 + 357*$1140.34 = $ 409687.78. Therefore, the interest paid by the pianist is $409687.78 - $ 120000 = $ 289687.78. The saving in interest is $ 290408.80-$289687.78 = $ 721.02 = $ 720 ( on rounding off to the nearest $ 10).