The pool is filled with water by two pipes operating simultaneously in 6 hours.

The pool is filled with water by two pipes operating simultaneously in 6 hours. One first pipe fills it 5 hours faster than one second. How many hours does it take to fill a pool through one second pipe?

Let us denote the number of hours for which the pool is filled with the first pipe after x, the second pipe after y.

The first pipe fills the pool 5 hours faster than the second, therefore:

y – x = 5.

In 1 hour, the first pipe fills the pool by 1 / x volume, the second pipe by 1 / y volume.

Since if the pool is filled through both pipes at the same time in 6 hours, then in 1 hour it is filled by 1/6 of the volume, that is:

1 / x + 1 / y = 1/6.

Let’s solve the resulting system of equations:

y = x + 5;

1 / x + 1 / (x + 5) = 1/6;

x + 5 + x = 1/6 * x * (x + 5);

2x + 5 = 1 / 6x ^ 2 + 5/6 * x;

12x + 30 = x ^ 2 + 5x;

x ^ 2 – 7x – 30 = 0;

By the Vieta converse theorem, x1 = 10; x2 = – 3, the negative root can be disregarded since the number of hours cannot be negative.

So, the pool is filled with the first pipe in 10 hours, the second pipe in 10 + 5 = 15 hours.

Answer: 15 hours.