If two pipes function simultaneously, the reservoir will be filled in 12 hours. One pipe fills the empty reservoir 10 hours faster than the other. How many hours does it take the second pipe to fill the reservoir ?

Let first pipe fill the reservoir faster and it takes x hours.

So, in 1 hour it will fill $\dfrac{1}{x}$ of the reservoir.

Since, second pipe takes 10 more hours than first pipe, so it will take (x + 10) hours.

So, in 1 hour it will fill $\dfrac{1}{x + 10}$ of the reservoir.

Given,

Both pipes fill reservoir in 12 hours.

So, in hour they will fill $\dfrac{1}{12}$ of the reservoir together.

$\therefore \dfrac{1}{x} + \dfrac{1}{x + 10} = \dfrac{1}{12} \\[1em] \Rightarrow \dfrac{x + 10 + x}{x(x + 10)} = \dfrac{1}{12} \\[1em] \Rightarrow \dfrac{2x + 10}{x^2 + 10x} = \dfrac{1}{12} \\[1em] \Rightarrow x^2 + 10x = 12(2x + 10) \\[1em] \Rightarrow x^2 + 10x = 24x + 120 \\[1em] \Rightarrow x^2 + 10x - 24x - 120 = 0 \\[1em] \Rightarrow x^2 - 14x - 120 = 0 \\[1em] \Rightarrow x^2 - 20x + 6x - 120 = 0 \\[1em] \Rightarrow x(x - 20) + 6(x - 20) = 0 \\[1em] \Rightarrow (x + 6)(x - 20) = 0 \\[1em] \Rightarrow x + 6 = 0 \text{ or } x - 20 = 0 \\[1em] \Rightarrow x = -6 \text{ or } x = 20.$

x + 10 = 20 + 10 = 30.

Hence, second pipe will take 30 hours to fill the reservoir.