Algebra 1 Word Problems - Work
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Get to Work
Word Problems - Work

 Here's a real cute type of problem that you might see:
 Fred can mow the yard in 3 hours.
 Fred Jr can mow the yard in 6 hours.
 How long will it take if they both work on it?
 Well in real life there are all kinds of questions like
 "Will they get in each other's way?" and
 "Do they have two mowers?" and stuff like that.
 For these problems, we just figure that they
 have all of the equipment that they need,
 and that they don't get in each other's way.
 Here's how we solve it.
 It takes Fred 3 hours to mow the entire yard
 so in 1 hour, he will be able to mow 1/3 of the yard.
 It takes Fred Jr. 6 hours to mow the entire yard
 so in 1 hour, he will be able to mow 1/6 of the yard.
 So in 1 hour together, they mow:


 So here comes the big trick of these problems ...
 If they do 1/X of the job in 1 hour,
 then the whole job takes X hours!

 So the "set up" on this type of problem is:


 X is how long it takes to do the entire job
 with both of them working on it.
 The first thing we need to do to solve this,
 is to get a common denominator on the left side.

 We want to wind up with "X = STUFF" 
 so the first thing to do is get X out of the denominator.
 If we multiply both sides by 2X (the product of the 2 denominators), 
 we get rid of both denominators at the same time.
 Tricky, eh?

 So working together, it takes Fred and Fred Jr
 2 hours to mow the lawn.
 Pump 1 can fill the pool in 6 hours by itself.
 Pump 2 can fill the pool in 4 hours by itself.
 Pump 3 can fill the pool in 8 hours by itself.
 How long does it take if all 3 pumps are used?
 Let's think about this for a second.
 How long would you guess it will take?
 1 hour? 10 hours? what?
 Well, if pump 2 did all the work 
 and the other two pumps just stood around,
 it would take 4 hours.
 So as long as pump 2 is on the job,
 it can't take any longer than that..
 If there was  one other pump and it was as fast as pump 2
 they would both do half the job and be done in half the time.
 That would be 2 hours.
 Here we have 2 other pumps besides pump 2.
 They are both slower than pump 2,
 but maybe between the two of them they are as good
 as another pump 2 would be.
 That means that a guess of 2 hours for the 3 pumps
 might be close to right..
 Lets see ...
 Set up the problem:

 Find a common denominator.
 6 can be broken into 3 x 2
 4 can be broken into 2 x 2
 8 can be broken into 2 x 2 x 2
 The least common denominator is made up 
 of the most times any factor appears in any denominator
 so we have one 3 in the first denominator
 and three 2's in the third denominator.
 That is ...

 3 2 2 2 = 24

 So we multiply each fraction by another name for 1
 that contains the factors it's denominator is missing.
 For example, 6 contains a 2 and a 3, 
 so it's missing two of the 2' (2 x 2 = 4)

 Now get this to look like X = stuff
 We need to multiply by X and by 24
 to get rid of the denominators.
 We can do it all at once like we did last time.
 Or take them one at a time.
 Let's do that this time.
 Multiply each side by X and simplify ...

 Multiply each side by 24 and simplify ...

 Now divide by 13 to get X by itself ...

 Now use a calculator to find
 a decimal equivalent to 24/13 ...

X = 1.846154 hours

 Convert fractional hours to minutes ...

 Last Example:
 Joe can paint the barn in 4 hours by himself.
 Steve has never timed himself painting a barn before 
 so we don't know how fast he is.
 Working together, the two finish the job in 2.4 hours.
 How long would it take Steve to paint the barn by himself?
 The formula for doing this type of problem is:

 In the other problems we've done
 we did not know the total time.
 Here we do know the total time for the job,
 but we don't know the "time for the second person alone."
 We have ...

 This will take a bit longer to solve,
 but it's really no big deal.
 First get a common denominator on the left.
 The factors are 2 x 2 x X = 4X.

 Multiply both sides by 4X, and simplify ...

 Multiply both sides by 2.4 and simplify.
 A calculator really helps here ...

 Subtract 2.4X from each side ...

 Divide both sides by 1.6 and simplify ...

 So it would take Steve 6 hours 
 to paint the barn by himself.

   copyright 2005 Bruce Kirkpatrick

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