To the pool three pipes are brought. The first pipe fills the pool with water for 6 h, and the second - for 8 h. On the third pipe water from the filled pool flows for 4 h completely. What part of volume of the pool is filled with water for an hour if water flows at the same time on three pipes?

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We THINK Tasks about pipes it is possible to compare to tasks about a way. Formulas are approximately identical. And = p*t is work. Formulas r = A: t - pipe performance - t speed = A: p - time the Pool at us ONE - And = 1 - should be filled ONE pool. t1 = 6 t2 = 8 t3 = 4 Pc = is GIVEN to p1+p2 - p3 - the joint performance of pipes - two pour, the third - merges. To FIND t4(1) = - a part of the pool for 1 hour of work of three pipes. SOLUTION 1) of p1 = A/t1 = 1/6 (1/h) - the performance of the first pipe. Similarly 2) r2 = 1/8 and r3 = 1/4. We write the main equation of a task. at t = 1 hour 3) And * (1/6 + 1/8 - 1/4) = = A*(4+3-6)/24 = A*1/24 the ANSWER In 1 hour will be filled (HOK=24) 1/24 part of the pool. IN ADDITION Time of filling of the pool completely. T = A: 1/24 = 24 h - will be filled the pool.
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