Explanation Given information: Present value = $ 50 , 000 i = 8 % n = 15 Calculation: At the end of first month, interest on $ 50 , 000 is $ 50 , 000 × 0.08 12 = $ 333.33 The remainder of first month payment is P M T − P V = $ 50 , 000 × 0.08 12 1 − ( 1 + 0.08 12 ) − 180 − $ 50 , 000 = $ 144.50 It will go on reducing the principal. Thus, in second month the outstanding principal will be, $ 50 , 000 − $ 144.50 ≃ $ 49856 Part of second monthly payment will be for the interest on this amount, which is $ 49856 × 0.08 12 ≃ $ 145.46 After continuing this calculation for the 180 months for fifteen years, it can achieve the mortgage’s amortization table. Add the payments on principal and interest payments for each year, Year Interest Payment on Principal 1 $ 3 , 934.99 $ 1 , 798.97 2 $ 3 , 785.67 $ 1 , 948.29 3 $ 3 , 623.97 $ 2 , 109

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ISBN: 9781337291262

Author: Stefan Waner, Steven Costenoble

Publisher: Brooks Cole

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Chapter 2.3, Problem 81E

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To calculate: An amortization table showing interest each year, which is to be paid on a 15 year mortgage for $50,000 at 8%, to be paid off monthly.

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Chapter 2.3, Problem 80E

Chapter 2.3, Problem 82E

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An amortization table showing interest each year, which is to be paid on a 15 year mortgage for $ 50 , 000 at 8 % , to be paid off monthly.

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To calculate: An amortization table showing interest each year, which is to be paid on a 15 year mortgage for $50,000 at 8%, to be paid off monthly.

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