Question

Jon and Steve ride their bicycles on a path that parallels two side-by-side train tracks running in the east/west direction. Jon rides east at $20$ miles per hour, and Steve rides west at $20$ miles per hour. Two trains of equal length, traveling in opposite directions at constant but different speeds, each pass the two riders. Each train takes exactly $1$ minute to go past Jon. The westbound train takes $10$ times as long as the eastbound train to go past Steve. The length of each train is $\frac{m}{n}$ miles, where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Answer

Solution:

Let the eastbound and westbound trains have speeds $v_1$ and $v_2$ miles per hour and common length $L$ miles. A train passes a rider in time $L$ divided by their relative speed. Passing Jon (riding east at $20$ ) in $\frac{1}{60}$ hour gives $\frac{L}{v_1 - 20} = \frac{L}{v_2 + 20} = \frac{1}{60},$ so $v_1 = 60L + 20$ and $v_2 = 60L - 20.$

Relative to Steve (riding west at $20$ ), the speeds are $v_1 + 20$ and $v_2 - 20,$ and the westbound train takes $10$ times as long: $\frac{L}{v_2 - 20} = \frac{10L}{v_1 + 20},$ so $v_1 + 20 = 10(v_2 - 20).$ Substituting, $60L + 40 = 600L - 400,$ so $540L = 440$ and $L = \frac{22}{27}.$

Since $\gcd(22, 27) = 1,$ the answer is $22 + 27 = 49.$

2014 AIME I Problem 4

Solution:

Problem 4 in Other Years