Question 193776: A business invests $8,000 in a savings account for two years. At the beginning of the second year, an additional $2,500 is invested. At the end of the second year, the account balance is $11,445. What was the annual interest rate?
Found 2 solutions by Mathtut, RAY100:
Answer by Mathtut(3670)   (Show Source):
You can put this solution on YOUR website! the annual interest rate = r
$8,000 in a savings account for two years.--> 8000*(1+r)^2
beginning of the second year, an additional $2,500 is invested-->2500*(100+r)
8000*(1+r)^2 + 2500*(1+r) = 11,445
let 100+r = A
:
8000*A^2 + 2500*A = 11,445:
8000*A^2 + 2500*A - 11,445 = 0:
A = 1.05
:
r = 0.05 or 5 % Ans.
Answer by RAY100(1637) (Show Source):
You can put this solution on YOUR website! assume simple interest ???????,,,I=PRT
,
yr,,,, invested,,,,,,interest,,,,,,invested,,,,,interest
0,,,,,,8000
1,,,,,,,,,,,,,,,,,,,,,,, 8000R(1),,,,2500
2,,,,,,,,,,,,,,,,,,,,,,, 8000R(1),,,,,,,,,,,,,,,,,,, 2500R(1)
total 8000 +2500+2*8000R +2500R =11,445
,
10,500 + 16000R +2500R=11,445
18500R=945
R= 945/18500 = .05108 = 5.108 %
,
check , 8000 + 2500 + 2*8000*.05108 + 2500* .0518 =11,445
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