A person invests 9500 dollars in a bank. The bank pays 5.75% interest compounded annually. To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 22600 dollars?

Solution

Solution

Time : 15.5

Solution steps

Translate the problem into an equation:

22600 = 9500 (1 + \frac{0.0575}{1})^{1 \cdot t}

Solve for

t, 22600 = 9500 (1 + \frac{0.0575}{1})^{1 \cdot t} : t = 15.5

Time : 15.5

Popular Examples

Prompted when I was 6 years old my sister was half of my age, and my mother was 10:times older than my sister. Now I'm 60. How old is my mother now?Maggie has 30 in an account. The interest rate is 10% compounded annually.To the nearest cent, how much will she have in 1 year?Ian invested 90,000 in an account paying an interest rate of 2.6% compounded quarterly. Assuming no deposits or withdrawals are made, how much money, to the nearest dollar, would be in the account after 7 years?The sum of two consecutive odd numbers equal 96. Find the numbers The sum of two numbers is 31. Half the larger number increased by four is same as the smaller number. Find the numbers.