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If it takes Mike and Joshua 2 hours to do a certain job, and it takes Mike and Kelvin 3 hours to do the same job, how long would it take for Mike, Joshua and Kelvin to do the job if all 3 of them worked together?

Jeffreymars16  May 31, 2017
edited by Jeffreymars16  May 31, 2017
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3+0 Answers

 #1
avatar+90645 
+2

If it takes Mike and Joshua 2 hours to do a certain job, and it takes Mike and Kelvin 3 hours to do the same job, how long would it take for Mike, Joshua and Kelvin to do the job if all 3 of them worked together?

 

I got in a mess trying to solve this alsgebraically so I decided to takle it with an example.

Let mike and Josh both do  1/4 of the job in an hours so together they do 1/2 the job in an hour which is 1 job in 2hours.

Now Mike and Kevin together do 1/3 of a job in an hour.   so 1/4 + K = 1/3       K=1/12 of a job in an hour

so together 

M+J+K = 1/4 +1/4 + 1/12 = 7/12 jobs in an hour   = 12/7 hours for a job = aprrox  1 hour and 43 minutes

Melody  May 31, 2017
 #3
avatar+90645 
+1

I am going to try a different example.

 

Say Mike does 1/6 job in an hours, then then Josh does 2/6 of a job in an hour which adds to  1/2 a job in an hour.

Since mike does 1/6 of a job in an hour, Kelvin must do 1/6 of a job in an hour too totalling 1/3 of a job in an hour.

So together they do   1/6 + 2/6 +  1/6 = 4/6 = 2/3 job in an hour  = 3/2 hours for a job

 

So this time they take  1.5 hours to do the job which is different from my first answer.

So either my logic is up the creek or there is not just one definite answer.

 

I see CPhill is answering, he is better at these then me so we will see what he says :)

Melody  May 31, 2017
 #4
avatar+77243 
+2

 

I believe that we need to know something about at least one of the individual's rate of work per minute to have a definite solution for this.

 

Note that  Mike and Joshua   must do   1/120 of the job in one minute [since it takes them 120 minutes for the whole job]

And Mike and Kelvin must do 1/180 of the job in one minute [ since it takes them 180 minutes for the whole job ]

 

Let M be the part of the job that Mike can do in one minute , J the part of the job that Joshua can do in one minute and K the part of the job that Kelvin can do in one minute

 

Then we have the following system

 

M +  J  =  1/120

M + K  =  1/180

 

Subtracting the second equation from the first, we have that J - K = 1/360....i.e., Joshua can do 1/360  more of the job per minute than Kelvin can

 

Suppose that Mike can do   1/200 of the job each minute

Then Joshua can do  1/300 of the job each minute and Kelvin can  do 1/1800 of the job each minute......and  1/300 - 1/1800  =  1/360

 

Or......suppose that Mike can do  1/300 of the job each minute

Then Joshua can do 1/200 of the job each minute and Kelvin can do 1/450 of the job each minute

And  1/200 - 1.450  =  1/360

 

Thus....there are infinite solutions to this......all we definitely know is that each person must do less than 1/180  of the job every minute

 

 

cool cool cool

CPhill  May 31, 2017
edited by CPhill  May 31, 2017

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