Rates of Work Combined, and Staggered

Date: 08/24/2010 at 19:03:38
From: David
Subject: rate of work

Corporal Jones takes 12 hours to complete a planned maintenance check on
an anti-submarine helicopter. Jones had been working on the helicopter for
4 hours when Sergeant Marco arrived to help him finish the job. Together,
they completed the check in 2 hours.

How long would it have taken Sergeant Marco to do the entire maintenance
check alone?

The formula I have is

   1/a + 1/b = 1/c,

where a and b are each person's time, and c is their time together.

I don't know how to use these numbers in my formula for rate of work
because of the second guy arriving later. Together they finished in 2
hours, but do I use the 4 hours somehow with that 2?

I really don't know what goes where!

Date: 08/25/2010 at 10:20:43
From: Doctor Ian
Subject: Re: rate of work

Hi David,

The most fundamental principle here is one they don't always emphasize
enough in math classes, which is this:

   When you have a problem that you don't know how to solve,
   try to turn it into a simpler problem that you do know how to solve!

Here, you have Jones needing 12 hours to do a job, and he's 4 hours into
it.

... or one-third done. What if you thought of the maintenance check as
divisible into 3 parts, each of which takes 4 hours? Then Marko shows up
at the time when Jones has completed one of those three jobs.

By re-scaling the problem, it now looks a little different:

  Jones needs 8 hours to complete 2 jobs. 
  Marko and Jones working together complete those jobs in 2 hours. 
  How long would Marko need to do the jobs by himself?

And in fact, we can make this simpler again, by cutting everything in
half:

  Jones needs 4 hours to complete 1 job. 
  Marko and Jones working together complete the job in 1 hour.
  How long would Marko need to do the job by himself?

Now you have the standard "working together problem," which you can solve
in the usual way:

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

The tricky bit is remembering that we broke up this unit of work (one
"maintenance check") into three pieces. That means that solving this
simpler version of the problem reveals how long it takes Marko to do
one-third of the original job. So you need to remember to triple the
result.

Does that make sense?  

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