Respuesta :
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 1) the value of the fund has increased to 62,000 and 10,000 is withdrawn.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,
[tex]$\frac{7000}{\frac{50000+16000}{3-10000(1-t)}}= \frac{52}{50} \frac{62}{60} \frac{55}{52} - 1$[/tex]
= 0.13667
7000 = 0.13667(55,333.33 - 10000 + 10000t)
[tex]$t=\frac{7000-0.13667(45333.33)}{1366.7}$[/tex]
= 0.5885
Thus, time - weighted rate of the return = 0.5885
