Three Sprinters

Algebra Level 4

Allan, Bill, and Carl had to sprint from points $P$ to $Q$ , which are $55$ meters apart, and back again (starting in that order). The time interval between their starting times was $5$ seconds each. Carl started $10$ seconds after Allan, while Bill started $5$ seconds after Allan. They passed a certain point $R$ , which is somewhere between $P$ and $Q$ , simultaneously (none of them having reached point $Q$ yet). Having reached $Q$ and reversed the direction, the third sprinter met the second one $9$ m short of $Q$ and met the first sprinter $15$ m short of $Q$ .

If the sum of their speeds, in meters per second, can be expressed as $\frac{m}{n}$ , where $m$ and $n$ are coprime positive integers, find $m + n$ .

This is not my original problem.

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The answer is 19.

1 solution

John Angelo Buscado
Jun 21, 2019

Let $A$ , $B$ , and $C$ be the speed of the Allan, Bill and Carl respectively.

i. $C$ would overtake $B$ and at the same time $B$ would overtake $A$ as they passed a certain point $R$ .

$A$ started $10$ seconds ahead of $C$ while $B$ started $5$ seconds ahead of $C$ . $PR = \frac{10AC}{C-A} = \frac{5BC}{C-B}$ $2AC = BC+AB (1)$

ii. $C$ met $B$ $9$ m short of $Q$ .

Time of $C$ $=$ Time of $B$ $-$ $5$ seconds $\frac{55 + 9}{C} = \frac{55 - 9}{B} - 5$ $B = \frac{46C}{64 + 5C} (2)$

iii. $C$ met $A$ $15$ m short of $Q$ .

Time of $C$ $=$ Time of $A$ $-$ $10$ seconds $\frac{55 + 15}{C} = \frac{55 - 15}{A} - 10$ $A = \frac{4C}{7 + C} (3)$

Substitute $(2)$ and $(3)$ to $(1)$ and you'll get $C = 1$ $m/s$ . It follows that $B = \dfrac{2}{3}$ $m/s$ and $A = \dfrac{1}{2}$ $m/s$ . $A + B + C = \frac{13}{6}$

Hence, $m + n = \boxed{19}$ .

Three Sprinters

The answer is 19.

1 solution

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