Math 618 Answer to Selected Problems in Section 5.2 Autumn 2005 Problem 5.2.1 100,000 ?8,000 X 103,992 2005a27 a45 2006a27 a45 In the picture above, X is the account value at the end of 2005. Dollar-weighted method for 2005 gives 100000(1 + x)?8000 parenleftbigg 1 + 34x parenrightbigg = X. Time-weighted method for 2006 gives X(1 + x) = 103992, or X = 1039921 + x . So, (the two red X?s are equal) 100000(1 + x)?8000 parenleftbigg 1 + 34x parenrightbigg = 1039921 + x . Multiplying both sides of the last equation by 1 + x and simplifying, we get 94000x2 + 186000x?11992 = 0. Solving this quadratic equation (by quadratic formula for example), we get x1 = 0.06249905183, x2 = ?2.041222456. Since x > 0 (the money has actually grown), x = 0.06249905183. Answer: x = 6.25% Note that the value 130000 on April 1, 2005 is not useful here.