SOLUTION: Susan places $6300 in three investments at rates of 9%, 10.8% and 16.2% per annum, respectively. The total income after ine year is $784.80. If the amount placed in the third inves
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-> SOLUTION: Susan places $6300 in three investments at rates of 9%, 10.8% and 16.2% per annum, respectively. The total income after ine year is $784.80. If the amount placed in the third inves
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Question 481532: Susan places $6300 in three investments at rates of 9%, 10.8% and 16.2% per annum, respectively. The total income after ine year is $784.80. If the amount placed in the third investment id $900 more than the amount placed in the second, find the amount of each investment. Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Write an equation for each statement:
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Susan places $6300 in three investments at rates of 9%, 10.8% and 16.2% per annum, respectively.
a + b + c = 6300
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The total income after one year is $784.80.
.09a + .108b + .162c = 784.80
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If the amount placed in the third investment is $900 more than the amount placed in the second,
c = (b+900)
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In the 1st equation, replace c with (b+900)
a + b + (b+900) = 6300
a + 2b = 6300 - 900
a + 2b = 5400
a = (5400-2b)
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Do the same in the 2nd equation
.09a + .108b + .162(b+900) = 784.80
.09a + .108b + .162b + 145.8 = 784.80
.09a + .27b = 784.80 - 145.8
.09a + .27b = 639
replace a with (5400-2b)
.09(5400-2b) + .27b = 639
486 - .18b + .27b = 639
.09b = 639 - 486
.09b = 153
b =
b = $1700 invested at 10.8%
then
a = 5400 - 2(1700))
a = 5400 - 3400
a = $2000 invested at 9%
and
c = 1700 + 900
c = $2600 invested at 16.2%
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You can check these amts in the original equations
