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29 May, 10:59

Hector invests $800 in an account that earns 6.98% annual interest compounded semiannually. Rebecca invests $1000 in an account that earns 5.43% annual interest compounded monthly. Find when the value of Hector's investment equals the value of Rebecca's investment and find the common value of the investments at that time.

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  1. 29 May, 11:37
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    A=P (1+r/n) ^nt

    A = Total amount invested, P=principal amount, r=Interest rate, n=number of time in a year when the interest is earned (for annual, n=1; for semi-annual, n=2, ...), t = time in years

    In the current scenario, case 1, n=2; case 2, n=1 and A1=A2, t1=t2

    Therefore,

    800 (1+0.0698/2) ^2t = 1000 (1+0.0543/1) t

    Dividing by 800 on both sides;

    (1+0.0349) ^2t = 1.25 (1+0.02715) ^t

    (1.0349) ^2t = 1.25 (1.02715) ^t

    Taking ln on both sides of the above equation;

    2t*ln (1.0349) = ln 1.25 + t*ln (1.02715)

    2t*0.0343 = 0.2231 + t*0.0268

    0.0686 t = 0.2231+0.0268 t

    (0.0686-0.0268) t = 0.2231

    0.0418t=0.2231

    t=5.337 years

    Therefore, after 5.337 years or 5 years and approximately 4 months, their value of investments will be equal.

    This value will be,

    A=800 (1+0.0698/2) ^2*5.337 = $1,153.76
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