Slow and Steady?

Algebra Level 4

Alex and Drew are running a long race of unknown distance.
Alex runs at a steady rate of 2 meters per second for the entire race.
Drew sprints at a steady rate of 3 meters per second for 300 meters but then has to rest for 50 seconds before sprinting again at the same rate.

Assuming that they do not tie, who finishes the race first?

Alex Drew Depends on the exact distance

View wiki
Chung Kevin by Chung Kevin , 46, Australia

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Nihar Mahajan
Nov 28, 2016

Suppose they start at t=0. Note that Alex and Drew come together every 150 seconds. So if the race time is an integral multiple of 150 it ends in a tie, else Drew is always ahead of Alex as his speed is more. Since, the question eliminates the possibility of a tie, Drew should win the race.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

great observation!

Agnishom Chattopadhyay - 4 years, 6 months ago

Say the length of the track is x x long.

How much time does Alex take? x 2 \frac{x}{2}

How much time does Drew take? x 3 + 50 ⌊ x 300 ⌋ \frac{x}{3} + 50 \lfloor \frac{x}{300} \rfloor

So, let's see ... can Alex do better than Drew? That is, can the following hold?

x 2 < x 3 + 50 ⌊ x 300 ⌋ ⟹ x 2 − x 3 = x 6 < 50 ⌊ x 300 ⌋ ⟹ x 300 < ⌊ x 300 ⌋ \frac{x}{2} < \frac{x}{3} + 50 \lfloor \frac{x}{300} \rfloor \\ \implies \frac{x}{2} - \frac{x}{3} = \frac{x}{6} < 50 \lfloor \frac{x}{300} \rfloor \\ \implies \frac{x}{300} < \lfloor \frac{x}{300} \rfloor

We have ended up with an equation of the form n < ⌊ n ⌋ n < \lfloor n \rfloor But this cannot happen, unless n n is negative. So, Alex can never do better.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...