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### Combining Rates of Work

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Date: 10/27/98 at 21:09:06
From: Sen
Subject: Sergeant Marco's Maintenance Check

I don't get this question. Can you help me?

Corporal Jones takes 12 hours to complete a planned maintenance check
on an anti-submarine helicopter. Jones had been working on a helicopter
for 4 hours when Sergeant Marco arrived to help him finish the job.
They complete the check in 2 hours. How long would it have taken
Sergeant Marco to complete the job alone?
```

```
Date: 10/28/98 at 09:04:21
From: Doctor Rick
Subject: Re: Sergeant Marco's Maintenance Check

Hi, Sen. This kind of problem is always tricky to think about. The
trick is to make it into a rate problem, like "distance = rate *
time". But the Sergeant doesn't run a certain distance in an hour. He
does a certain amount of work in an hour.

If Corporal Jones takes 12 hours to check a helicopter, then he does
1/12 of the job in one hour. That's his rate: 1/12 helicopter/hour.

In 4 hours, how much of the job had he finished? "Distance" = rate *
time:

1/12 helicopter/hour * 4 hours = 1/3 helicopter

So he has 2/3 of the job left to do.

When Sergeant Marco gets to work, you have to assume that they work
separately - they don't get in each other's way, and they don't help
each other. (Sometimes just a little help can make a job go a lot
faster - I hope I'm doing that for you!)

If you make this assumption, then Jones and Marco together work at a
rate equal to Jones' rate plus Marco's rate. So if you can find the
rate at which they worked together, you can just subtract Jones' rate
to find Marco's rate.

See if you can solve the problem now.

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