Jan, Febby, and March are participating a 1kilometer dash-run. Jan completes a lap in 60 minutes, Febby completes a lap in 120 minutes and March completes a lap in 80 minutes. If they all started at the same time, then how long will they meet again?
A. 2 hours
B. 3 hours
C. 3.5 hours
D. 4 hours
My answer is D please check this
Answer D is correct.
You can check that.
60 min = 1 h
120 min = 2 h
80 min = 80 min / 60 = 20 ∙ 4 / 20 ∙ 3 = 4 / 3 h
1 km = 1 lap
Jan´s speed = 1 km / h = 1 lap / h
Febby´s speed = 1 km / 2 h = 0.5 lap / h
March´s speed = 1 km / ( 4 / 3 ) h = 3 / 4 km / h = 0.75 lap / h
After 1 h
Jan complete 1 lap , Febby complete 0.5 lap , March complete 0.75 lap
After 2 h
Jan complete 2 laps , Febby complete 2 ∙ 0.5 = 1 lap , March complete 2 ∙ 0.75 = 1.5 laps
After 3 h
Jan complete 3 laps , Febby complete 3 ∙ 0.5 = 1.5 laps , March complete 3 ∙ 0.75 = 2.25 laps
After 3.5 h
Jan complete 3.5 laps , Febby complete 3.5 ∙ 0.5 = 1.75 laps , March complete 3.5 ∙ 0.75 = 2.625 laps
After 4 h
Jan complete 4 laps , Febby complete 4 ∙ 0.5 = 2 laps , March complete 4 ∙ 0.75 = 3 laps
They will meet again when each of them runs the whole number of laps, in this case after 4 hours.
To figure out when Jan, Febby, and March will meet again, we need to find the least common multiple (LCM) of their lap times.
First, let's find the factors of each lap time:
- Jan: 60 minutes = 2 × 2 × 3 × 5
- Febby: 120 minutes = 2 × 2 × 2 × 3 × 5
- March: 80 minutes = 2 × 2 × 2 × 2 × 5
Next, we identify the highest power of each prime factor that appears in any of the lap times:
- 2^3
- 3^1
- 5^1
The LCM is then calculated by multiplying these highest powers:
LCM = 2^3 × 3^1 × 5^1 = 8 × 3 × 5 = 120 minutes
Therefore, Jan, Febby, and March will meet again after 120 minutes, which is equal to 2 hours.
So, the correct answer is A. 2 hours.