Work Problem 2 Solution

This is the solution to the Algebra work problem # 2.

The key to solving any work problem is to figure out how much work can be done in 1 hour, minute, day- whatever time classification your using. In this problem it's talking hours. Since pump A takes 10 hours to finish then in 1 hours times 1/10th of the work would be complete and with the 8 hour pump B, 1/8th of the work would be done.

Now the way to understand this is to look at just 1 of the pumps. Take pump A for example. It takes 10 hours to fill the pool and 1/10 of the pool is finished in 1 hour. So ask yourself how many hours would it take a pump that fills 1/10th of a pool an hour to fill the entire pool? Or X/10 = 1 --- Without Algebra you know its 10 but using algebra you can see it works out the same way.

So all we do when were dealing with more than just 1 pump is to add their respective ratios together and set them equal to 1. We add because it's the added effect of both pumps that fills the pool.

So we have algebraically:

X/10 + X/8 = 1

multiply by 40 ----- the common denominator....

40X/10 + 40X/8 = 40

4X + 5X = 40

9X = 40

X = 40/9

X = 40/9 or 4 and 4/9ths

so almost 4 and a half hours. To get the exact minutes take 60 * 4/9 = 240/9 = 26 and 2/3rds

so we got 4 hours 26 minutes and if u want to be real anal 2/3's of a minute is 40 seconds (60 * 2/3 = 120/3 = 40)

So, it takes 4 hours 26 minutes and 40 seconds for pumps A and B to fill up the pool if they work together.

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