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13 January, 21:31

A person invests 10000 dollars in a bank. The bank pays 6.75% interest compounded monthly. To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 37900 dollars?

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  1. 13 January, 21:43
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    Answer: it will take 20.6 years

    Step-by-step explanation:

    We would apply the formula for determining compound interest which is expressed as

    A = P (1 + r/n) ^nt

    Where

    A = total amount in the account at the end of t years

    r represents the interest rate.

    n represents the periodic interval at which it was compounded.

    P represents the principal or initial amount deposited

    From the information given,

    A = $37900

    P = $10000

    r = 6.75% = 6.75/100 = 0.0675

    n = 12 because it was compounded 12 times in a year.

    Therefore,

    37900 = 10000 (1 + 0.065/12) ^12 * t

    37900/10000 = (1 + 0.00542) ^12t

    3.79 = (1.00542) ^12t

    Taking log of both sides, it becomes

    Log 3.79 = 12t * log 1.00542

    0.579 = 12t * 0.00234752004

    0.579 = 0.02817t

    t = 0.579/0.02817

    t = 20.6 years
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