At 3 p.m , four runners all leave the starting line, running laps around the indoor track.If the runners maintain their pace, at what time will Sue,Drew, and Ste finish a lap together? At what time will all 4 runners finish a lap together? Explain your reasoning
Sue lap 1 time- 1.30
Drew lap 1 time- 2.00
Stu lap 1 time- 1.12
Marylou lap 1 time- 1.20
Note that
Sue lap 1 time- 1.30 → runs 2/3 of a lap in one minute
Drew lap 1 time- 2.00 → runs 1/2 of a lap in one minute
Stu lap 1 time- 1.12 → runs 60/72 = 5/6 of a lap in one minute
Marylou lap 1 time- 1.20 → runs 3/4 of a lap in one minute
So ...getting a common denominator between their rates, we have that
Sue = 4/6 laps in one minute = 8/ 12 of a lap in one minute
Drew = 3/6 laps in one minute = 6/12 of a lap in one minute
Stu = 5/6 laps in one minute = 10/12 of a lap in one minute
Marylou = 3/4 lap in one minute = 9/12 of a lap in one minute
So...in 6 minutes [at 3:06 PM ]
Sue will have run 4 laps, Drew will have run 3 laps and Stu will have run 5 laps...and they are all back at the starting line
And in 12 minutes [ at 3:12 PM]
Sue will have run 8 laps, drew will have run 6 laps, Stu will have run 10 laps and Marylou will have run 9 laps...and they are all back at the starting line
Note that
Sue lap 1 time- 1.30 → runs 2/3 of a lap in one minute
Drew lap 1 time- 2.00 → runs 1/2 of a lap in one minute
Stu lap 1 time- 1.12 → runs 60/72 = 5/6 of a lap in one minute
Marylou lap 1 time- 1.20 → runs 3/4 of a lap in one minute
So ...getting a common denominator between their rates, we have that
Sue = 4/6 laps in one minute = 8/ 12 of a lap in one minute
Drew = 3/6 laps in one minute = 6/12 of a lap in one minute
Stu = 5/6 laps in one minute = 10/12 of a lap in one minute
Marylou = 3/4 lap in one minute = 9/12 of a lap in one minute
So...in 6 minutes [at 3:06 PM ]
Sue will have run 4 laps, Drew will have run 3 laps and Stu will have run 5 laps...and they are all back at the starting line
And in 12 minutes [ at 3:12 PM]
Sue will have run 8 laps, drew will have run 6 laps, Stu will have run 10 laps and Marylou will have run 9 laps...and they are all back at the starting line
Sue lap 1 time- 1.30
Drew lap 1 time- 2.00
Stu lap 1 time- 1.12
Marylou lap 1 time- 1.20
I assume the times of their laps are in minutes and seconds. If that is so, then we have:
LCM{90, 120, 72, 80} =720 seconds, or:
720 / 60 =12 minutes - or 3:12 pm when they will finish together.