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 Post subject: Solving for unknown VariablesPosted: Tue, 6 Sep 2011 23:41:58 UTC
 S.O.S. Oldtimer

Joined: Tue, 30 Dec 2008 20:37:56 UTC
Posts: 205
Location: Accra,Ghana
How best can I approach this question.

Martin and Henry, working together, can finish a job in 120 minutes. Martin starts to a job, 160 minutes into the task, Henry joins him and they finish the job spending an additional 50 minutes. How long would it have taken each man, separately, to do this job?

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 Post subject: Re: Solving for unknown VariablesPosted: Wed, 7 Sep 2011 00:00:59 UTC
 Member of the 'S.O.S. Math' Hall of Fame

Joined: Mon, 23 Feb 2009 23:20:33 UTC
Posts: 1049
Suppose the job is handing out flyers. If there are 100 flyers, and Martin hands out 2 a minute, it will take 100/2 minutes to hand out all of them. If Henry hands out 5 a minute, it will take 100/5 minutes to hand out all of them. And working together, it will take 100/(2+5) minutes to hand out all of them.

Martin starts 1 task, and part of the way through (the fraction p), he is joined by Henry, who helps him for the fraction 1-p of the task.

You also have

and you now have enough information to find 1/m and 1/h and solve the problem.

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 Post subject: Re: Solving for unknown VariablesPosted: Wed, 7 Sep 2011 05:01:31 UTC

Joined: Sat, 26 Apr 2003 22:14:40 UTC
Posts: 2063
Location: El Paso TX (USA)
Very nice explanation, aswoods!

_________________
The greater danger for most of us lies not in setting our aim too high and falling short; but in setting our aim too low, and achieving our mark. - Michelangelo Buonarroti

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 Post subject: Re: Solving for unknown VariablesPosted: Wed, 7 Sep 2011 12:20:25 UTC

Joined: Wed, 3 Dec 2008 11:17:17 UTC
Posts: 6
120/M + 120/H = 1 ……….(1

210/M + 50/H = 1 …………(2

120/M + 120/H = 210/M + 50/H

70/H = 90/M

M = (9/7)H

Substitute in (1

120/(9/7)H + 120/H = 1

120/(9/7) + 120 = H

H = 213 1/3 minutes

M = 9/7 x 213 1/3 = 274 2/7 minutes

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