In order to fill an empty swimming pool completely, a pipe of larger diameter alone takes 10 hour less than the time taken by the pipe of the smaller diameter alone. If the pipe of the larger diameter is used for 4 hours and the pipe of the smaller diameter is used for 9 hours, half of the pool is filled. In how many hours will the pipe of the larger diameter alone fill the pool.

Let the time taken by the larger diameter pipe to fill the pool alone be L hours and the time taken by the smaller diameter pipe to fill the pool alone be S hours.

From the problem, we know that the larger diameter pipe takes 10 hours less than the smaller diameter pipe. Hence, we can write the relationship as:

⇒ L = S - 10

The rate of the larger pipe is $\dfrac{1}{\text{L}}$ of the pool per hour.

The rate of the smaller pipe is $\dfrac{1}{\text{S}}$ of the pool per hour.

It is given that when the larger pipe is used for 4 hours and the smaller pipe is used for 9 hours, half of the pool is filled. So, the total work done in this scenario can be expressed as:

⇒ $4 \times \dfrac{1}{\text{L}} + 9 \times \dfrac{1}{\text{S}} = \dfrac{1}{2}$

Substituting the value of L = S - 10, we get

$\Rightarrow 4 \times \dfrac{1}{S - 10} + 9 \times \dfrac{1}{\text{S}} = \dfrac{1}{2}\\[1em] \Rightarrow \dfrac{4}{S - 10} + \dfrac{9}{\text{S}} = \dfrac{1}{2}\\[1em] \Rightarrow \dfrac{4S}{S(S - 10)} + \dfrac{9(S - 10)}{S(S - 10)} = \dfrac{1}{2}\\[1em] \Rightarrow \dfrac{4S}{S(S - 10)} + \dfrac{9S - 90}{S(S - 10)} = \dfrac{1}{2}\\[1em] \Rightarrow \dfrac{4S + 9S - 90}{S(S - 10)} = \dfrac{1}{2}\\[1em] \Rightarrow \dfrac{13S - 90}{S^2 - 10S} = \dfrac{1}{2}\\[1em] \Rightarrow 2(13S - 90) = S^2 - 10S\\[1em] \Rightarrow 26S - 180 = S^2 - 10S\\[1em] \Rightarrow S^2 - 10S - 26S + 180 = 0\\[1em] \Rightarrow S^2 - 36S + 180 = 0\\[1em] \Rightarrow S^2 - 30S - 6S + 180 = 0\\[1em] \Rightarrow S(S - 30) - 6(S - 30) = 0\\[1em] \Rightarrow (S - 30)(S - 6) = 0\\[1em] \Rightarrow (S - 30) = 0 \text{ or } (S - 6) = 0\\[1em] \Rightarrow S = 30 \text{ or } S = 6\\[1em]$

Substituting the value of S in equation, L = S - 10

⇒ L = 30 - 10 = 20 or 6 - 10 = -4

As time cannot be negative.

Hence, pipe of the larger diameter alone will take 20 hours to fill the pool.