- 29 Jul '06 08:59

First hose has a rate of 5/12 litres per second and the other hose has a rate of 5/25 litres per second.*Originally posted by Knight Square***Two hoses have a rate of 12s per 5L and 25s per 5L how long will it take to fill up a 150m^3 pool?**

So the combined rate is 5/12+5/25 litres per second.

There 150*1000 litres to fill, so the total time needed is

150*1000/(5/12+5/25) = 9000000/37 seconds. - 29 Jul '06 15:04

You have to keep them still so they keep Peeing in the pool. Also you have to keep them from pooping in the pool which would throw off the liquid calculation*Originally posted by deriver69***If like me you read horses instead of hoses it makes the idea of them filling a pool far more entertaining.** - 29 Jul '06 15:35 / 1 edit

If you divide out each contibutor, you get 5/12 of a liter per second or 0.41666 L/sec. The other one comes out at 5/25 or 0.2 L/sec. That makes 0.61666 L/sec filling the pool. Inverting that gives 1.6216 seconds per liter, times 1000 gives 1621.6 seconds to fill one cubic meter. Times 150 gives 243,245 seconds to fill the 150 M^3 pool.*Originally posted by SPMars***First hose has a rate of 5/12 litres per second and the other hose has a rate of 5/25 litres per second.**

So the combined rate is 5/12+5/25 litres per second.

There 150*1000 litres to fill, so the total time needed is

150*1000/(5/12+5/25) = 9000000/37 seconds.

The same as SP, but made the final division.

That problem can be restated as the current flowing in parallel in two resistors where they are charging a battery to X amount of joules of energy. It could be stated as 5/12 amp in one and 5/25 amp in the other. Same kind of problem. - 30 Jul '06 12:47

in abc minutes tap A fills bc baths, B ac baths and C ab baths.*Originally posted by ThudanBlunder***If**

tap A fills a pool in a minutes

tap B fills a pool in b minutes

tap C fills a pool in c minutes

how long will they need to fill the pool when used together?

so a total of ab + bc + ac baths are filled in abc minutes.

so one bath is filled in abc/(ab + bc + ac) - 30 Jul '06 13:17

Why would you do that? The words surrounding the numbers in these types of problems don't make a lick of difference.*Originally posted by sonhouse*

That problem can be restated as the current flowing in parallel in two resistors where they are charging a battery to X amount of joules of energy. It could be stated as 5/12 amp in one and 5/25 amp in the other. Same kind of problem.