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algebra: work problem.. so hard.?
a tank is filled with 2 pipes. the first pipe can fill the tank in 10 hours. But after it has been opened for 3 and 1/3 hours, the second pipe is opened and the tank is filled up in 4 hours more. how long would it take the second pipe alone to fill the tank? the two pipes have different diameters.
2 Answers
- Ed ILv 79 years agoFavorite Answer
The first pipe fills the tank in 1/10 hr.
Let h = time require for second pipe to fill it.
(1/10)(10/3 + 4) + (1/h)(4) = 1
1/3 + 2/5 + 4/h = 1
4/h = 1 - 1/3 - 2/5 = 4/15
h = 15
It would take the second pipe 15 hr to fill it.
- 9 years ago
so if we call a full tank "10"
then 10 times pipe a would be 10 or 10a=10 .
in your example the tank is, in this case, filled up to a third (three hours and 20 minutes is a third of 10 hours)
The two thirds that are left are done in 4 hours using the second pipe, that's a third per two hours that's a whole tank in 6 hours
