2440.51 – Race to the Bus

Dino, Kyle, and Nina race the 100 meters to the school bus each day. It develops that Dino and Kyle will reach the bus together if Dino is given a head start of 20 meters. Kyle and Nina will arrive together if Kyle is given a head start of 25 meters. If Dino and Nancy want to arrive at the bus at the same time, who should start where?

Solution

Let $D$ , $K$ , and $N$ be Dino's, Kyle's, and Nancy's rates of speed, respectively.

Dino runs 80m when Kyle runs 100m, so $D = \left(\dfrac{4}{5}\right)K$ . Nina runs 100m when Kyle runs 75m, so $K = \left(\dfrac{3}{4}\right)N$ .

So $D = \left(\dfrac{4}{5}\right)\left(\dfrac{3}{4}\right)N = \left(\dfrac{3}{5}\right)N$ . This means that when Nancy runs 100m, Dino runs 60m. So give Dino a 40m head start.