Problema Solution
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 ?
Answer provided by our tutors
We have three fractions with different denominators: sixths, thirds, and halves.
The first step is to make all the denominators equal for 1/6, 1/3, 1/2.
In this case we want sixths since LCM(2, 3, 6) = 6
click here to see the LCM calculation:
since 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 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.
2/6 = 1/3
3/6 = 1/2
4/6 = 2/3
6/6 = 1
The answers are: 1/3, 1/2, 2/3, and 1.