Question

Together, 2 pipes can fill a reservoir in 3/4 of an hour. Pipe 1 needs 1 hr ten minutes ( 1 1/6 hrs) to fill reservoir by itself. How long would pipe 2 need to fill the reservoir by itself?

Answer 1

Given: Together, 2 pipes can fill a reservoir in $\dfrac{3}{4}$ of an hour.

Pipe 1 needs 1 hour ten minutes (i.e. $1\dfrac{1}{6}$ hours ) to fill reservoir by itself.

Here, $1\dfrac{1}{6}=\dfrac{7}{6}\text{ hours}$

To find: Time taken by Pipe 2 to fill reservoir itself.

Let t be time taken by pipe 2 to fill the reservoir by itself.

then,

$\dfrac{1}{\text{Time taken together}}=\dfrac{1}{\text{Time taken by Pipe1}}+\dfrac{1}{\text{Time taken by Pipe2}}\\\\\Rightarrow\ \dfrac{1}{\dfrac{3}{4}}=\dfrac{1}{\dfrac{7}{6}}+\dfrac{1}{t}\\\\\Rightarrow\ \dfrac{4}{3}=\dfrac{6}{7}+\dfrac{1}{t}\\\\\Rightarrow\ \dfrac{1}{t}=\dfrac{4}{3}-\dfrac{6}{7}\\\\\Rightarrow\ \dfrac{1}{t}=\dfrac{28-18}{21}=\dfrac{10}{21}\\\\\Rightarrow\ t=\dfrac{21}{10}=2\dfrac{1}{10}$

Hence, Pipe 2 will take $2\dfrac{1}{10}$ hours to fill the reservoir by itself.