Daily Math Puzzle — Two Trains

Sunday, July 13, 2014

Taken from Project Gutenberg’s Amusements in Mathematics, by Henry Ernest Dudeney, originally published ca. 1917 … #68 — “The Two Trains” …

Two trains start at the same time, one from Baltimore to Philadelphia, the other from Philadelphia to Baltimore. If each train is traveling at a constant speed and the trains arrive at their destinations one hour and four hours respectively after passing one another, how much faster is one train running than the other?

Solution coming soon … Or post your solution in a comment. Be sure to explain how you found your answer …

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1 COMMENT

Paul Katula Wednesday, July 16, 2014 At 2:11 am

Let’s assume the train leaving Baltimore is the faster train (at velocity v_b).

The key to understanding the problem is knowing what it means for the two trains to “meet.” If one is traveling from Baltimore to Philly and the other from Philly to Baltimore, they meet at some point in the middle. That meeting point happens at the same time for both trains, so we can set the times equal to each other in an algebraic equation ( $t_{b}=t_{p}$ ).

Now, to figure out an expression that represents $t_{b}$ and $t_{p}$ so we can actually get an expression for one velocity ( $v_{b}$ ) in terms of the other ( $v_{p}$ ).

We know that velocity is distance divided by time, which means

$t = \frac{d}{v}$

Then, since distance = velocity × time ( $v_{p}(4)$ for the train leaving Baltimore since the train from Philly met it this far from Philly or $v_{b}(1)$ for the train leaving Philly since the train from Baltimore met it this far from Baltimore), we have

$\frac{(1) v_{b}}{v_{p}}=\frac{(4) v_{p}}{v_{b}}$

$v_{b} = \frac{(4) v_{p}^{2}}{v_{b}}$

$v_{b}^{2} = (4) v_{p}^{2}$

$v_{b} = \sqrt{(4) v_{p}^{2}}$

$v_{b} = 2 v_{p}$

Therefore, the train from Baltimore is traveling exactly twice as fast as the train from Philadelphia.