1. ## Running laps

Dionne is running clockwise. She can run around a circular track in 120 seconds. Basha, running counter-clockwise, meets Dionne every 48 seconds. Sandra runs clockwise and passes Basha every 240 seconds. How often does Sandra meet Dionne in 10 minutes?

I feel like I'm going in circles myself trying to solve this and not getting anywhere. Any advice?

2. Originally Posted by amolabola
Dionne is running clockwise. She can run around a circular track in 120 seconds. Basha, running counter-clockwise, meets Dionne every 48 seconds. Sandra runs clockwise and passes Basha every 240 seconds. How often does Sandra meet Dionne in 10 minutes?
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Hello,

I hope this post doesn't come too late.

You already know that a constant speed is defined by: $speed = \frac{distance}{time}$

Let $v_D, \ v_B \ and \ v_S$ be the speeds with which Dionne, Basha and Sandra are running. The running distance is measured in laps (L).

You know that $v_D = \frac{L}{120}$

The relative speed between two runners is calculated by the difference of their speeds and because $distance=speed \cdot time$ you can extract from the text 2 linear equations:

$\left \{ \begin{array}{l}(v_D-(-v_B)) \cdot 48 = 1L \\(v_S - v_B) \cdot 240 = 1L\end{array}\right.$ . Multiply the first equation by 5 and add columnwise:

$\underbrace{240 v_D}_{\text{equals 2 L}} + 240 v_S = 6L~\implies~v_S=\frac{L}{60}$

That means Sandra is running twice as fast as Dionne.
Sandra and Dionne are running in the same direction. The difference of their speeds is the relative speed between them. When Sandra overtakes Dionne she has covered 1 lap more than Dionne:

$(v_S - v_D) \cdot x = 1L~\implies~\left(\frac{L}{60} - \frac{L}{120}\right) \cdot x = 1L$ Divide both sides by L and simplify the value in the bracket:

$\frac1{120} \cdot x = 1~\iff~x = 120$ . That means: Every 120 s = 2 minutes Sandra overtakes Dionne. If her speed is constant she'll meet Dionne 5 times in 10 minutes.

3. ## WHat if

Sandra and Basha are running in the opposite direction of Dionne?