Three runners

Anik, Banik and Zanik agree to a simple 100 yard race.

Here’s what happened. Anik finished the race 10 yards ahead of Banik. Subsequently Banik finished the race 10 yards ahead of Zanik.

The question is, how far ahead was Anik over Zanik when he finished the race?

Assume all three ran at constant speeds throughout the race.





Three-runnersThe tempting answer is 20 yards, but that is not correct. The reason is that Zanik runs at a slower pace than Banik. In the time it takes Banik to run the final 10 yards, Zanik would end up running less than 10 yards. So we know that Zanik is definitely closer than 20 yards when Anik finishes the race.

How close is he?

It’s best to solve this problem in terms of relative speeds. Banik runs 90 yards in the time it takes Anik to run 100 yards. Therefore, Banik runs at a speed 90 percent as fast as Anik (B = 0.9 A).

Similarly, in the time it takes Zanik to run 90 yards, Banik can run 100 yards. Hence Zanik runs at 90 percent the speed of Banik (Z = 0.9 B).

We can combine these equations to find that:
Z = 0.9 B = 0.9(0.9 A) = 0.81 A

That is, Zanik runs at 81 percent the speed of Anik. When Anik runs 100 yards, Zanik would run 81 yards. This means Zanik ends up 19 yards behind Anik when he finishes.

Share this post
  • Share to Facebook
  • Share to Twitter
  • Share to Google+
  • Share to Pinterest
  • Share to Stumble Upon
  • Share to Evernote
  • Share to Blogger
  • Share to Email
  • Share to Yahoo Messenger
  • More...


Related Posts Plugin for WordPress, Blogger...

Mental Math for Faster Calculation

Mental Math for Faster Calculation
© 2013-16 Burning Math
Posts RSSComments RSS