A person invests 1000 dollars in a bank. The bank pays 6% interest compounded annually. To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 2100 dollars?

Solution

Solution

Time : 12.73

Solution steps

Translate the problem into an equation:

2100 = 1000 (1 + \frac{0.06}{1})^{1 \cdot t}

Solve for

t, 2100 = 1000 (1 + \frac{0.06}{1})^{1 \cdot t} : t = 12.73

Time : 12.73

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