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

#1**+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**+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**+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

CPhill
May 31, 2017