Two pipes running together can fill a tank in \(11\frac{1}{9}\) minu

1.	Two pipes running together can fill a tank in \(11\frac{1}{9}\) minutes. If one pipe takes 5 minutes more than the other to fill the tank separately, find the time in which each p ipe would fill the tank separately.
Answer» Let the faster pipe fill the tank in ‘a’ min Slower pipe fills it in ‘a + 5’ min. Given, the pipes running together can fill a tank in \(11\frac{1}{9}\) = 100/9 minutes. In 1 min, part of tank filled = 9/100 \(\Rightarrow \frac{1}{a}+\frac{1}{a\,+\,5}=\frac{9}{100}\) ⇒ 100(a + a + 5) = 9(a² + 5a) ⇒ 200a + 500 = 9a² + 45a ⇒ 9a² – 155a - 500 = 0 ⇒ 9a² – 180a + 25a - 500 = 0 ⇒ 9a(a – 20) + 25(a – 20) = 0 ⇒ (9a + 25)(a – 20) = 0 ⇒ a = 20 mins Slower pipe will fill it in 25 min

Two pipes running together can fill a tank in \(11\frac{1}{9}\) minutes. If one pipe takes 5 minutes more than the other to fill the tank separately, find the time in which each p ipe would fill the tank separately.

Answer»

Let the faster pipe fill the tank in ‘a’ min

Slower pipe fills it in ‘a + 5’ min.

Given, the pipes running together can fill a tank in \(11\frac{1}{9}\) = 100/9 minutes.

In 1 min, part of tank filled = 9/100

\(\Rightarrow \frac{1}{a}+\frac{1}{a\,+\,5}=\frac{9}{100}\)

⇒ 100(a + a + 5) = 9(a² + 5a)

⇒ 200a + 500 = 9a² + 45a

⇒ 9a² – 155a - 500 = 0

⇒ 9a² – 180a + 25a - 500 = 0

⇒ 9a(a – 20) + 25(a – 20) = 0

⇒ (9a + 25)(a – 20) = 0

⇒ a = 20 mins

Slower pipe will fill it in 25 min