If two pipes function simultaneously, a tank is filled in 12 hours. One pipe fills the tank 10 hours faster than the other. How many hours does the faster pipe alone take to fill the tank?

Let the time taken for the slower pipe to fill the tank = x hours
Then, Time taken for the faster pipe to fill the tank = (x-10) hours
1 hour work of slower pipe = $$\dfrac{1}{x}$$
1 hour work of faster pipe = $$\dfrac{1}{x-10}$$
Given, Total time taken for both the pipes to fill the tank = 12 hours

$$\dfrac{1}{x}+\dfrac{1}{x-10} = \dfrac{1}{12}$$

=> $$\dfrac{x-10+x}{x(x-10)} = \dfrac{1}{12}$$

=> $$12(2x-10) = x^2-10x$$
=> $$x^2 - 10x-24x+120 = 0$$
=> $$x^2-34x+120 = 0$$
=> $$x^2-30x-4x+120 = 0$$
=> $$x(x-30)-4(x-30)=0$$
=> $$(x-30)(x-4)=0$$
$$x = 30$$ or $$x = 4$$
Here, Faster pipe fills the tank in 10 hours lesser time than slower pipe. Hence, x cannot be 4.
Therefore, Slower pipe can fill the tank in 30 hours and faster pipe can fill the tank in 20 hours.