Skilled and Semi-Skilled Workers

Date: 09/04/2002 at 04:42:54
From: Riya Bajaj
Subject: Work and time problem

Hi Dr. Math,

This is regarding a problem on "work and time" in the Dr. Math FAQ:

   Working Together
   http://mathforum.org/dr.math/faq/faq.working.together.html 

Jack can paint a house in 5 days, and Richard can paint the same house 
in 7 days. Working together, how long will it take them to finish the 
job? 

A very easy method was used:

Jack and Richard =  1 house  -  5*7 / 5+7
                             =  35/12 days

I tried to apply the same formula to the following question but was 
unable to get the answer.

Four skilled workers do a job in 5 days, and five semi-skilled workers 
do the same job in 6 days. How many days will it take for two skilled 
and one semi-skilled worker to do that job?

Here is what I did:

4 skilled workers   5 days
1 skilled worker    5/4 days
2 skilled workers   5*2/4 = 5/2 days.

Then,

5 semi-skilled      6 days
1 semi-skilled      6/5 days

Now applying the same method used in the first question, here we say:

1 skilled and 1 semi-skilled   (5/2 * 6/5)
                              -------------
                                5/2 + 6/5

                                = 30/37

But the right answer is 60/7.

Why can't we use the same method for harder word problems? Could you 
please suggest a common method or way to solve such problems?

Thank you,
Riya Bajaj

Date: 09/04/2002 at 10:29:01
From: Doctor Ian
Subject: Re: Work and time problem

You can't just divide the time by the number of workers - which is 
exactly what makes these problems so hard.

Think of it this way:

  four skilled workers can do 1 job in 5 days. 

We could just have them repeat what they did, so 

  four skilled workers can do 4 jobs in 20 days. 

Now, if four workers are doing 4 jobs, each of them is essentially 
doing a job by himself. So we can just eliminate any number of jobs by 
eliminating the same number of workers:

  1 skilled worker can do 1 job in 20 days. 

> 2 skilled workers   5*2/4 = 5/2 days.
> 
> Then,
> 5 semi-skilled      6 days
> 1 semi-skilled      6/5 days

Doing the same thing as before:

  5 semi-skilled workers can do 1 job  in  6 days.
  5 semi-skilled workers can do 5 jobs in 30 days. 
  1 semi-skilled worker  can do 1 job  in 30 days. 

Now, what can two skilled workers and one semi-skilled worker do 
together? Suppose we give them 60 days. Each of the skilled workers 
can do 3 jobs, for a total of 6 jobs. The semi-skilled worker can do 2 
jobs. So in 60 days, the three of them could do 8 jobs, which means 
that they can do one job in 60/8 days.  

>The right answer is 60/7.

I think it's 60/8, or 15/2.  

Can we check that another way?  Let's start from the beginning:

   4 skilled      workers can do 1 job in 5 days
   5 semi-skilled workers can do 1 job in 6 days

We'd like to get a ratio of 2 skilled workers to 1 semi-skilled 
worker:
    
    4 skilled workers can do 1 job in 5 days
   20 skilled workers can do 5 jobs in 5 days
     
    5 semi-skilled workers can do 1 job in 6 days 
   10 semi-skilled workers can do 2 jobs in 6 days

And we'd like to consider the same number of days:

    4 skilled workers can do 1 job in 5 days
   20 skilled workers can do 5 jobs in 5 days
   20 skilled workers can do 30 jobs in 30 days
     
    5 semi-skilled workers can do 1 job in 6 days
   10 semi-skilled workers can do 2 jobs in 6 days
   10 semi-skilled workers can do 10 jobs in 30 days 

Now, working together, 20 skilled workers and 10 semi-skilled workers 
can do 40 jobs in 30 days. We can divide them into 10 work crews, each 
with 2 skilled workers and 1 semi-skilled worker. Each of those crews 
can do 4 of the jobs in those 30 days.  

So two skilled and one semi-skilled worker can do 4 jobs in 30 days,
or one job in 30/4 days, or 15/2 days, which is what we got before.  

> Why can't we use the same method for harder word problems? Could 
> you please suggest a common method or way to solve such problems?

I think I've shown that we _can_ use a single method for all the 
various kinds of problems. The key is to avoid the temptation to use 
formulas without understanding why they work, and instead use methods 
where each step makes sense.  

I hope this helps.  Write back if you'd like to talk more about this, 
or anything else.

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 09/04/2002 at 15:22:40
From: Riya Bajaj
Subject: Thank you (Work and time problem)

Hi Dr. Math,

Thank you so much for your response. Things seem to be clearer than 
before!

Regards,
Riya