To fill a s

Let the larger diameter pipe fill it in ‘a’ hours

The smaller diameter pipe fills it in ‘a + 10’ hours

In 1 hour, larger diameter pipe fills 1/a part of the pool.

In 1 hour, smaller diameter pipe fills 1/(a + 10) part of the pool.

Given, the pipe of larger diameter used for 4 hours and the pipe of smaller diameter for 9 hours, only half of the pool can be filled.

⇒ 4 × 1/a + 9 × 1/(a+10) = 1/2

⇒ 2(4a + 40 + 9a) = a² + 10a

⇒ 26a + 80 = a² + 10a

⇒ a² – 16a – 80 = 0

⇒ a² – 20a + 4a – 80 = 0

⇒ a(a – 20) + 4(a – 20) = 0

⇒ (a + 4)(a – 20) = 0

⇒ a = 20 hours

Time in which smaller diameter pipe fills the pool = 20 + 10 = 30 hours