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ch 4 sols

# ch 4 sols - Solutions to Chapter 4 The Time Value of Money...

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Solutions to Chapter 4 The Time Value of Money 1. a. \$100/(1.08) 10 = \$46.32 b.\$100/(1.08) 20 = \$21.45 c. \$100/(1.04) 10 = \$67.56 d.\$100/(1.04) 20 = \$45.64 2. a. \$100 × (1.08) 10 = \$215.89 b.\$100 × (1.08) 20 = \$466.10 c. \$100 × (1.04) 10 = \$148.02 d.\$100 × (1.04) 20 = \$219.11 3. \$100 × (1.04) 113 = \$8,409.45 \$100 × (1.08) 113 = \$598,252.29 4. With simple interest, you earn 4% of \$1,000 or \$40 each year. There is no interest on interest. After 10 years, you earn total interest of \$400, and your account accumulates to \$1,400. With compound interest, your account grows to: \$1,000 × (1.04) 10 = \$1480.24 Therefore \$80.24 is interest on interest. 5. PV = \$700/(1.05) 5 = \$548.47 6. Present Value Years Future Value Interest Rate a. \$400 11 \$684 % 00 . 5 1 400 684 ) 11 / 1 ( = - b. \$183 4 \$249 % 00 . 8 1 183 249 ) 4 / 1 ( = - c. \$300 7 \$300 % 0 1 300 300 ) 7 / 1 ( = - To find the interest rate, we rearrange the basic future value equation as follows: FV = PV × (1 + r) t r = 1 PV FV ) t / 1 ( - 7. You should compare the present values of the two annuities. 4-1

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a. 73 . 721 , 7 \$ (1.05) 0.05 1 0.05 1 \$1,000 PV 10 = × - × = 73 . 303 , 8 \$ (1.05) 0.05 1 0.05 1 \$800 PV 15 = × - × = b. 47 . 192 , 4 \$ (1.20) 0.20 1 0.20 1 \$1,000 PV 10 = × - × = 38 . 740 , 3 \$ (1.20) 0.20 1 0.20 1 \$800 PV 15 = × - × = c. When the interest rate is low, as in part (a), the longer (i.e., 15-year) but smaller annuity is more valuable because the impact of discounting on the present value of future payments is less significant. 8. \$100 × (1 + r) 3 = \$115.76 r = 5.00% \$200 × (1 + r) 4 = \$262.16 r = 7.00% \$100 × (1 + r) 5 = \$110.41 r = 2.00% 9. PV = (\$200/1.06) + (\$400/1.06 2 ) + (\$300/1.06 3 ) = \$188.68 + \$356.00 + \$251.89 = \$796.57 10. In these problems, you can either solve the equation provided directly, or you can use your financial calculator, setting: PV = ( - )400, FV = 1000, PMT = 0, i as specified by the problem. Then compute n on the calculator. a. \$400 × (1.04) t = \$1,000 t = 23.36 periods b.\$400 × (1.08) t = \$1,000 t = 11.91 periods c. \$400 × (1.16) t = \$1,000 t = 6.17 periods 11. APR Compounding period Effective annual rate a. 12% 1 month (m = 12/yr) 1.01 12 - 1 = 0.1268 = 12.68% b. 8% 3 months (m = 4/yr) 1.02 4 - 1 = 0.0824 = 8.24% c. 10% 6 months (m = 2/yr) 1.05 2 - 1 = 0.1025 = 10.25% 12. Effective Ra te Compounding period Per period rate APR a. 10.00% 1 month 1.10 (1/ 12) - 1 = 0.0080 0.096 = 9.6% 4-2
(m = 12/yr) b. 6.09% 6 months (m = 2/yr) 1.0609 (1/ 2) - 1 = 0.0300 0.060 = 6.0% c. 8.24% 3 months (m = 4/yr) 1.0824 (1/ 4) - 1 = 0.0200 0.080 = 8.0% 13. Solve the following for t: 1.08 t = 2 t = 11.9 years On a financial calculator, enter: PV = ( - )1, FV = 2, PMT = 0, i = 6 and then compute n. 14.

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ch 4 sols - Solutions to Chapter 4 The Time Value of Money...

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