From Parade (24 Feb 2008):

In today's column, Marilyn adds an additional hypothesis needed to find a mathematical solution. Instead, she should have educated her readers on the need to properly formulate a problem before it can be solved. She missed an opportunity to be helpful to her readers. To wit ...

Say a plane departs from City A at 10 a.m., flies due east and arrives at City B at 7 p.m. Another plane takes off from City B at noon and lands at City A at 3 p.m. What is the difference in their time zones? (All times are local. Also, exclude the effect of the jet stream on flying time.)

Marilyn's Wrong Answer: 3 hours. (Let x = hours of time difference, and y = hours of flight time. Then:y + x = 9, and y – x = 3. Solve for x.)

Bernard's Correct Answer: The problem, as stated, does not provide sufficient information - we need to know the speed of each plane. Marilyn improperly assumes that both planes require the same amount of time, y. More properly, let D = distance between A and B. If plane 1 flies with speed V1 and plane 2 with speed V2, the amount of time each plane requires is T1 = D/V1 and T2 = D/V2. Therefore, T1 + x = 9 and T2 - x = 3. Since there are 3 variables and 2 equations, there is no unique solution. QED.

Page last modified by Bernard Roth Mon Feb 25/2008 06:47