Question

On January 1, an investment account is worth 50,000. On May 1, the value has increased to 52,000 and 8,000 of new principal is deposited. At time t, in years, (4/12

Answer 1

The question is incomplete. The complete question is :

On January 1, an investment account is worth 50,000. On May 1, the value has increased to 52,000 and 8,000 of new principal is deposited. At time t, in years, (4/12 < t < 1) the value of the fund has increased to 62,000 and 10,000 is withdrawn. On January 1 of the next year, the investment account is worth 55,000. The approximate dollar-weighted rate of return (using the simple interest approximation) is equal to the time-weighted rate of return for the year. Calculate t.

Solution :

It is given that :

Worth of investment account on 1st Jan = 50,000

Worth of investment account on 1st Jan next year = 55,000

New principal deposited = 8000

Therefore the interest earned = 55,000 - 50,000 - 8,000 + 10,000

                                                  = 7,000

Therefore,

$\frac{7000}{\frac{50000+16000}{3-10000(1-t)}}= \frac{52}{50} \frac{62}{60} \frac{55}{52} - 1$

                   = 0.13667

7000 = 0.13667(55,333.33 - 10000 + 10000t)

$t=\frac{7000-0.13667(45333.33)}{1366.7}$

    = 0.5885

Thus, time - weighted rate of the return = 0.5885