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# algebra 2

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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

Guest Jun 6, 2017

#1
+77165
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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

CPhill  Jun 6, 2017
Sort:

#1
+77165
+1

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

CPhill  Jun 6, 2017
#2
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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.

Guest Jun 6, 2017

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