The Theory of Interest - Solutions Manual 23 Chapter 31. The equation of value using a comparison date at time 20t=is 201050,0001000at 7%.sXs=+Thus, 201050,000100050,00040,995.49$622.214.171.124645sXs−−===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 .18/12.015j==and the amount of the loan (L) is the present value of the payments, i.e. ()48 .015250250 34.042558510.64.La===Thus, the down payment needed will be 10,0008510.64$1489.36.D=−=3. The monthly interest rate on the first loan (L1) is 1.06/12.005j==and ()()148 .00550050042.5803221,290.16.La===The monthly interest rate on the second loan (L2) is 2.075/12.00625j==and 2125,00025,00021,290.163709.84.LL=−=−=The payment on the second loan (R) can be determined from 12 .006253709.84Ra=giving 3709.84$3126.96.36.199639R==4. A’s loan: 8 .08520,000Ra=20,0003546.615.639183R==so that the total interest would be ( )()83546.6120,0008372.88.−=B’s loan: The annual interest is ()().08520,0001700.00=so that the total interest would be ( )()81700.0013,600.00.=Thus, the difference is 13,600.008372.88$5227.12.−=
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