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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View Full DocumentDecember 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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 Winter '12
 Kit
 1981, 1973, 1985, 1955, 1988

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