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Chapter003

Chapter003 - The Theory of Interest Solutions Manual...

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The Theory of Interest - Solutions Manual 23 Chapter 3 1. The equation of value using a comparison date at time 20 t = is 20 10 50,000 1000 at 7%. s Xs = + Thus, 20 10 50,000 1000 50,000 40,995.49 \$651.72. 13.81645 s X s = = = 2. The down payment ( D ) plus the amount of the loan ( L ) must equal the total price paid for the automobile. The monthly rate of interest is .18/12 .015 j = = and the amount of the loan ( L ) is the present value of the payments, i.e. ( ) 48 .015 250 250 34.04255 8510.64. L a = = = Thus, the down payment needed will be 10,000 8510.64 \$1489.36. D = = 3. The monthly interest rate on the first loan ( L 1 ) is 1 .06/12 .005 j = = and ( )( ) 1 48 .005 500 500 42.58032 21,290.16. L a = = = The monthly interest rate on the second loan ( L 2 ) is 2 .075/12 .00625 j = = and 2 1 25,000 25,000 21,290.16 3709.84. L L = = = The payment on the second loan ( R ) can be determined from 12 .00625 3709.84 Ra = giving 3709.84 \$321.86. 11.52639 R = = 4. A’s loan: 8 .085 20,000 Ra = 20,000 3546.61 5.639183 R = = so that the total interest would be ( )( ) 8 3546.61 20,000 8372.88. = B’s loan: The annual interest is ( ) ( ) .085 20,000 1700.00 = so that the total interest would be ( )( ) 8 1700.00 13,600.00. = Thus, the difference is 13,600.00 8372.88 \$5227.12. =

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