How many years does it take for a deposit of $1000 to reach $2700 with an annual effective interest rate of 11.1%? |
Amount investment = $ 1000 | ||||
Interest rate per year = 11.1% per annum | ||||
Period = 15 Years | ||||
Years | Investment | Interest @ 11.1% Per Year | Running Bal | |
1 | $1,000.00 | $111.00 | $1,111.00 | |
2 | $123.32 | $1,234.32 | ||
3 | $137.01 | $1,371.33 | ||
4 | $152.22 | $1,523.55 | ||
5 | $169.11 | $1,692.66 | ||
6 | $187.89 | $1,880.55 | ||
7 | $208.74 | $2,089.29 | ||
8 | $231.91 | $2,321.20 | ||
9 | $257.65 | $2,578.85 | ||
10 | $286.25 | $2,865.11 | ||
In the 10 th year we reach the $ 2700 but we not required whole 10th year | ||||
So exact period calculaiton is as below, | ||||
Amount required in the 10th year = $ 2865.11 - $ 2,700 | ||||
Amount required in the 10th year = $ 165.11 | ||||
So period required = $ 165.11 / $ 286.25 | ||||
So period required = 0.58 Years | ||||
So , period required for reach $ 2,700 = 9.58 Years | ||||
Answer = 9.58 Years | ||||
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