Solving for unknown Variables

Kodwo · **Posted:** Tue, 6 Sep 2011 23:41:58 UTC

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?

aswoods · **Posted:** Wed, 7 Sep 2011 00:00:59 UTC

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.

Now instead of 100 flyers, call it 1 task. You have
$$\frac{1}{m+h}=120$

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.
$$\frac{p}{m}+\frac{1-p}{m+h}=210$

You also have
$$\frac{p}{m}=160,\quad\frac{1-p}{m+h}=50$
and you now have enough information to find 1/m and 1/h and solve the problem.

helmut · **Posted:** Wed, 7 Sep 2011 05:01:31 UTC

Very nice explanation, aswoods!

hoblygobly · **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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Solving for unknown Variables

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