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29 September, 13:51

Eric deposits X into a savings account at time 0, which pays interest at a nominal rate of i, compounded semiannually. Mike deposits 2X into a different savings account at time 0, which pays simple interest at an annual rate of i. Eric and Mike earn the same amount of interest during the last 6 months of the 8th year. Calculate i.

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  1. 29 September, 16:19
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    i = 9.46%

    Step-by-step explanation:

    Money deposited by Eric is X.

    Now formula for compound interest is;

    A = P (1 + i) ⁿ

    Where n = kt and k is the number of compounding per annum.

    Since Eric compounded interest semi annually, we have;

    A = X (1 + i/2) ^ (2t)

    After 7.5 years, Eric will earn;

    A (7.5) = X (1 + i/2) ^ (2 * 7.5)

    A_7.5 = X (1 + i/2) ^ (15)

    For the last half year, where n = 1, Eric earns; (A_7.5) (1 + i/2) ¹

    Thus, the interest eric earns = (A_7.5) (1 + i/2) - A_7.5

    Interest = A_7.5 + (A_7.5) * (i/2) - A_7.5

    Interest = (A_7.5) * (i/2)

    We initially got A_7.5 = X (1 + i/2) ^ (15)

    Thus, interest Eric earns is now;

    X (1 + i/2) ^ (15) * (i/2)

    Which gives;

    0.5Xi (1 + i/2) ^ (15)

    Now to mike. He deposited 2X

    Simple interest earned by mike will be;

    A = 2X (1 + it)

    During the last half year of any yeat, Mike earns; 2X * i/2 = Xi

    Equating last half year interests of both Eric and Mike gives us;

    0.5Xi (1 + i/2) ^ (15) = Xi

    Divide both sides by 0.5Xi to give;

    (1 + i/2) ^ (15) = 2

    So,

    1 + i/2 = 2^ (1/15)

    i/2 = 1.04729 - 1

    i/2 = 0.04729

    Multiply both sides by 2 to give;

    i = 0.09458

    Thus, interest is approximately 0.0946 or 9.46%
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