Ch.8_Solution_Manual_Ed.1_v6_

# Ch.8_Solution_Manual_Ed.1_v6_ - Exercises 8.2 Exercise 8.2...

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Exercises 8.2 Exercise 8.2, Solution 1: a. January 01, 2011: 1 st day of the year (using the table) February 19, 2011: 50 th day of the year (using the table) Difference = 50 1 = 49 days Therefore, the number of days in the given time period = 49 days. In terms of years, t = years b. February 26, 2010: 57 th day of the year (using the table) December 02, 2010: 336 th day of the year (using the table) Difference = 336 57 = 279 days Therefore, the number of days in the given time period = 279 days. In terms of years, t = years c. November 23, 2010: 327 th day of the year (using the table) December 31, 2010: 365 th day of the year (end of the year) Therefore, in 2010 we have 365 327 = 38 days April 04, 2011: 94 th day of the year (using the table) Therefore, in 2011 we have 94 days Therefore, the number of days in the given time period = 38 + 94 = 132 days. In terms of years, t = years d. August 25, 2011: 237 th day of the year (using the table) December 31, 2011: 365 th day of the year (end of the year) Therefore, in 2011 we have 365 237 = 128 days September 06, 2012: 249 th day of the year (using the table) But 2012 is a leap year, therefore September 06, 2012 will be 249 + 1 = 250 th day of the year since February 29 will be day 60. Therefore, in 2012 we have 250 days. Therefore, the number of days in the given time period = 128 + 250 = 378 days. In terms of years, t = years Exercise 8.2, Solution 3 : a. I = Prt = (200) (0.10) () = \$10.958904. .. = \$10.96 Therefore, amount of interest earned on the given investment = \$10.96. b. From the days table, April 15: 105 th day November 1: 305 th day Difference = 305 105 = 200 days Therefore t = 200 days = years I = Prt = (2200) (0.04) () = \$48.219178. .. = \$48.22 Therefore, amount of interest earned on the given investment = \$48.22 c. From the days table, January 24: 24 th day of the year

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December 31: 365 th day of the year Difference = 365 24 = 341 days Therefore, in 2010 we have 341 days February 15: 46 th day of the year Therefore, in 2011 we have 46 days Total days in the given time period = 341 + 46 = 387 days Therefore, t = × 12 months I = Prt = (5605) (0.012) ( × 12) = \$855.768328. .. = \$855.77 Therefore, amount of interest earned in the given period = \$855.77. d. t = 1 year and 9 months = × 12 months = 21 months I = Prt = (150) (0.008) (21) = \$25.20 Therefore, the amount of interest earned in the given period = \$25.20 Exercise 8.2, Solution 5: t = 1 year and 3 months = 1 + = 1 + = years I = Prt = (5000)(0.05)() = \$312.50 Therefore, he will receive an interest of \$312.50 at the end of the time period. Exercise 8.2, Solution 7: t = 11 months = years, r = 5.5% = 0.055 I = Prt = (750) (0.055) () = \$37.8125 = \$37.81 Therefore, Nathan would have to pay an interest of \$37.81. Exercise 8.2, Solution 9:
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Ch.8_Solution_Manual_Ed.1_v6_ - Exercises 8.2 Exercise 8.2...

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