Miguel is making an obstacle course for field day. At the end of every sixth of the course, 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?

Question

Mathematics

Nicholas Lawson

3 November, 06:23

Miguel is making an obstacle course for field day. At the end of every sixth of the course, 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?

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Answers (1)

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Jamie Barber · Answer 1

So,

We have three fractions with different denominators: sixths, thirds, and halves. The first step is to make all the denominators equal. In this case we want sixths.

1/3 = 2/6, and 1/2 = 3/6. Now we can start solving.

1. There are six tires at the following: 1/6, 2/6, 3/6, 4/6, 5/6, and 6/6.

2. There are three cones at the following (G. C. F.) : 2/6 (or 1/3), 4/6 (or 2/3), and 6/6 (or 3/3).

3. There are two hurdles at the following (G. C. F.) : 3/6 (or 1/2) and 6/6 (or 2/2).

We simply look for common numbers.

1. At 2/6, there are two obstacles: a tire and a cone.

2. At 3/6, there are two obstacles: a tire and a hurdle.

3. At 4/6, there are two obstacles: a tire and a cone.

4. At 6/6, there are three obstacles: a tire, cone, and a hurdle. (hard!)

The answers are: 2/6, 3/6, 4/6, and 6/6.

S = {2/6, 3/6, 4/6, 6/6}