| 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: | ||
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| 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! | ||
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So the "set up" on this type of problem is: |
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| 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. | ||
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| 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? | ||
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| So working together, it takes Fred and Fred Jr | ||
| 2 hours to mow the lawn. | ||
| Example: | ||
| 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? | ||
| OK FREEZE, STOP, HALT, TIME OUT ... | ||
| 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: | ||
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| 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 ... | ||
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3 × 2 × 2 × 2 = 24 |
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| 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) | ||
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| 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 ... | ||
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| Multiply each side by 24 and simplify ... | ||
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| Now divide by 13 to get X by itself ... | ||
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| Now use a calculator to find | ||
| a decimal equivalent to 24/13 ... | ||
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X = 1.846154 hours |
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| Convert fractional hours to minutes ... | ||
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| 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: | ||
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| 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 ... | ||
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| 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. | ||
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| Multiply both sides by 4X, and simplify ... | ||
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| Multiply both sides by 2.4 and simplify. | ||
| A calculator really helps here ... | ||
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| Subtract 2.4X from each side ... | ||
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| Divide both sides by 1.6 and simplify ... | ||
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| So it would take Steve 6 hours | ||
| to paint the barn by himself. | ||
|
copyright 2005 Bruce Kirkpatrick |
