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numerical - The Theory of Interest Solutions Manual Chapter...

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The Theory of Interest - Solutions Manual 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. ( 29 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 ( 29 ( 29 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 ( 29 ( 29 8 3546.61 20,000 8372.88. - = B’s loan: The annual interest is ( 29 ( 29 .085 20,000 1700.00 = so that the total interest would be ( 29 ( 29 8 1700.00 13,600.00. = Thus, the difference is 13,600.00 8372.88 \$5227.12. - = 23

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The Theory of Interest - Solutions Manual Chapter 3 5. Using formula (3.2), the present value is ( 29 1 1 1 where . n n n i na i i n - - + = = This expression then becomes 2 1 1 1 . 1 1 n n n n n n n n n - + - ÷ = - ÷ + 6. We are given 1 , n n v a x i - = = so that 1 . n v ix = - Also, we are given 2 2 1 , n n v a y i - = = so that 2 1 . n v iy = - But ( 29 2 2 n n v v = so that ( 29 2 1 1 . iy ix - = - This equation is the quadratic ( 29 2 2 2 0 x i x y i - - = so that 2 2 . x y i x - = Then applying formula (1.15a), we have 2 2 . 1 2 i x y d i x x y - = = + + - 7. We know that 1 , d v = - and directly applying formula (3.8), we have ( 29 ( 29 8 8 8 8 1 1 1 1 .9 5.695. .1 v d a d d - - - - = = = = 8. The semiannual interest rate is .06/ 2 .03. j = = The present value of the payments is ( 29 ( 29 21 9 100 100 15.87747 8.01969 \$2389.72. a a + = + = 9. We will use a comparison date at the point where the interest rate changes. The equation of value at age 65 is 25 .08 15 .07 3000 s Ra = so that 25 .08 15 .07 3000 236,863.25 \$24,305 9.74547 s R a = = = to the nearest dollar. 10. ( a ) Using formulas (3.1) and (3.7) ( 29 ( 29 2 1 2 1 1 1 . n n n n n n n n a v v v v v v v v v a v - = + + + + + - = + + + + - = + - L L 24
The Theory of Interest - Solutions Manual Chapter 3 ( b ) Using formulas (3.3) and (3.9) ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 1 1 1 1 1 1 1 1 1 1 1 1 1 1 . n

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numerical - The Theory of Interest Solutions Manual Chapter...

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