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How do I solve this

+2 votes
Miguel is making an obstacle course for field day at the end of every sixth of the corse there is a tire. At the end of every third of the course there is a cone. At the end of every half of the course there is a hurdle. At which locations of the course will people need to go through more than one obstacle?
asked Jan 15, 2014 in Fractions by anonymous
   

2 Answers

+3 votes
You solve this by determining which fractions can have common denominators.

1/3 = 2/6 and 2/3 = 4/6, so at 1/3 and 2/3 there will be tire and a cone.

1/2 = 3/6 so there will be a tire and a hurdle.

6/6 = 3/3 = 2/2 so at the end of the course there will be all three: a tire, a cone and a hurdle.
answered Jan 15, 2014 by stg1.stewart Inquisitive Expert (21,460 points)
+3 votes
2/6 3/6 4/6 6/6
answered Feb 6, 2017 by anonymous

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