There are three pipes in the pool. With the first pipe, the pool can be filled in 10 hours, with the second

There are three pipes in the pool. With the first pipe, the pool can be filled in 10 hours, with the second pipe in 8 hours, and with the third pipe, all the water from the pool can be poured out in 5 hours. What part of the pool will be filled in 1 hour if all three pipes are in operation?

Let us introduce the notation:

Vob is the volume of the pool;

v∑ is the total rate of filling (pouring out) of the pool when all 3 pipes are in operation;

v1 is the speed with which the pipe fills the pool I;

v2 is the velocity for pipe II;

v3 is the speed at which water flows out of the III pipe.

Let’s write the formula for the volume of the pool through the filling rate:

Vob = v∑ ​​* t, where t is time.

From the condition of the problem by the formula v = Vob / t we express:

v1 = Vob / 10;

v2 = Vob / 8;

v3 = Vob / 5.

Let us express v∑:

v∑ = v1 + v2 – v3.

Let’s designate the part of the pool, which will have time to fill in 1 hour, – “X”. Substitute all expressions into the formula Vob:

(Vob / 10 + Vob / 8 – Vob / 5) * 1 = Vob * X;

(4 Vob + 5 Vob – 8 Vob) / 40 = Vob * X;

Vob / 40 = Vob * X;

X = 1/40.

Answer: 1/40 of the pool will be filled in 1 hour.