What amount would you have if you deposited 1800 a year for 30 years at 8% compounded annually
1 answer:
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Answer:
<em>It will take me </em>= <em>30.99 years</em> to pay-off the borrowed amount paying $60.00 every month
Explanation:
SOLUTION:
Using the Formula: A = P(1+r/n)
Where:
A= $60.00;
P = $3,200.00;
r = 12.9%;
n = 12
t =?
Substituting the values into the Formula =
$60 = $3,200(1 + 12.9%/12)
= $60 = $3,200(1 + 12.9%/12)
= $60.00 = $3,200.00 (1 + 0.129/12)
= $60.00 = $3,200.00 (1 + 0.01075 )
= $60.00 = $3,200.00 (1.01075 )
= $60.00/$3,200.00 = $3,200.00 (1.01075 )/$3200.00
= 0.01875 = (1.01075)
= Using the law of logarithm
= log = Nlog A
=12tlog(1.01075) = log0.01875
= 12log(1.01075)/(1.01075) = log0.01875/log(1.01075)
= 12t = log(0.01875)/log1.01075
= 12t = 1.7270/0.04644
Divide 12 by both sides
=12t/12 = 371.88774/12
t= 30.9898
<em>∴ 30.99 years</em>
<em>It will take me </em>= 30.99 years to pay-off the borrowed amount paying $60.00 every month