Two water taps together can fill a tank in 9(3/8) hours. Find the time in which each tap can separately fill the tank.

Question

Two water taps together can fill a tank in \(9\frac{3}{8}\) hours.The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Answer 1

Let the smaller diameter tap fill the reservoir in ‘a’ hours

Larger diameter tap fills it in ‘a – 10’ hours.

Given, two water taps together can fill a tank in \(9\frac{3}{8}=\) 75/8 hours.

In 1 hour, part of tank filled = 8/75

\(\Rightarrow \frac{1}{a}+\frac{1}{a\,-\,10}=\frac{8}{75}\)

⇒ 75(a + a – 10) = 8a² – 80a 

⇒ 150a – 750 = 8a² – 80a 

⇒ 8a² – 230a + 750 = 0 

⇒ 4a² – 115a + 375 = 0 

⇒ 4a² – 100a – 15a + 375 = 0 

⇒ 4a(a – 25) – 15(a – 25) = 0 

⇒ (4a – 15)(a – 25) = 0 

Value of a can’t be 15/4 as (a – 10) will be negative 

Thus a = 25 

Time taken by faster tap = 25 – 10 = 15 hours