Two pipes together can fill a tank in 12 hours. If the first pipe can fill the tank 10 hours faster than the second then the second pipe will take [[1]] to fill the tank.

Let the second pipe, which is the slower of the two pipes, fill the tank in x hours. Since the first pipe can fill the tank 10 hours faster than the second pipe, it will take (x - 10) hours for the first pipe to fill the tank alone.

The part of the tank filled by the second pipe in one hour is:

1 / x

The part of the tank filled by the first pipe in one hour is:

1 / (x - 10)

When both pipes are functioning together, the two pipes can fill the entire tank in 12 hours. Therefore, the part of the tank filled by both pipes in one hour is:

1 / 12

Setting up the equation for the two pipes working together:

(1 / x) + (1 / (x - 10)) = 1 / 12

To simplify the equation, take the least common denominator (LCD) of x and (x - 10) and equate it to 1/12:

(x - 10) + x = (x(x - 10)) / 12

Multiply through by 12x(x - 10) to remove the fractions:

12(x - 10) + 12x = x(x - 10)

Expand and simplify the equation:

12x - 120 + 12x = x² - 10x

Combine like terms:

x² - 34x + 120 = 0

Factorize the quadratic equation:

(x - 30)(x - 4) = 0

From this, we get two possible solutions for x: 30 or 4. However, x = 4 is not valid because (x - 10) would be negative, which is not possible for the time taken by the first pipe.

Therefore, the valid solution is x = 30.

Thus, the second pipe, which is the slower of the two pipes, will take 30 hours to fill the tank on its own.

The detailed steps above explain how the two pipes work together to fill the tank and how the time for the slower pipe is calculated.