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 (3)
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.
At 2/6, there are two obstacles: a tire and a cone.
At 3/6, there are two obstacles: a tire and a hurdle.
At 4/6, there are two obstacles: a tire and a cone.
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}
People will need to go through more than one obstacle at the following locations of the course: every 6 1 of the course (i.e., 6 1 , 6 2 = 3 1 , 6 3 = 2 1 , 6 4 = 3 2 , 6 5 , 6 6 = 1 ).[/tex]
To determine the locations of the course where participants will encounter more than one obstacle, we need to find the common multiples of the fractions representing where the obstacles are placed.
The obstacles are placed as follows:
Tire: every \frac{1}{ of the course
Cone: every [tex] 3 1 of the course
Hurdle: every 2 1 of the course
We need to find the least common multiple (LCM) of the denominators of these fractions: 6, 3, and 2.
First, we find the LCM of 6, 3, and 2.
The prime factorizations are:
6 = \( 2 \times 3 \)
3 = \( 3 \)
2 = \( 2 \)
The LCM is the highest power of each prime factor appearing in the factorizations:
The highest power of 2 is 2 1
The highest power of 3 is 3 1
So, the LCM is: 2 1 × 3 1 = 6
Therefore, every 6 1 of the course, all three obstacles (tire, cone, and hurdle) will coincide.
To find other locations where two or more obstacles overlap, let's identify the least common multiple of the pairs:
Tires (\( \frac{1}{6} \)) and Cones (\( \frac{1}{3} \)): LCM of 6 and 3 is 6.
Tires (\( \frac{1}{6} \)) and Hurdles (\( \frac{1}{2} \)): LCM of 6 and 2 is 6.
Cones (\( \frac{1}{3} \)) and Hurdles (\( \frac{1}{2} \)) : LCM of 3 and 2 is 6.
Thus, the only common location for multiple obstacles (more than one) is at the end of every 6 1 of the course, where there will be a tire, cone, and hurdle.
Participants will encounter more than one obstacle at these points in the course: 2/6, 3/6, 4/6, and 6/6. At these locations, obstacles like tires, cones, and hurdles overlap. This setup creates a challenging experience for the participants.
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