Question

two pipes running together can fill a cistern in 40/13 mins. if one pipe takes 3 mins more than the other to fill the cistern, find the time in which each pipe would fill the cistern.

Answer 1

Conditions

two pipes running together can fill a cistern in 40/13 mins. if one pipe takes 3 mins more than the other to fill the cistern, find the time in which each pipe would fill the cistern.

Solution

Let's the velocity of $1^{\text{st}}$ pipe is $x$, of $2^{\text{nd}}$ is $y$. Then:

$$\left\{ \begin{array}{l} \frac{1}{x + y} = \frac{40}{13} \\ \frac{1}{x} - \frac{1}{y} = 3 \end{array} \right.$$ $$\left\{ \begin{array}{l} 40(x + y) = 13 \\ y - x = 3xy \end{array} \right.$$ $$x = \frac{y}{3y + 1}$$ $$40\left(\frac{y}{3y + 1} + y\right) = 13$$ $$40\left(\frac{2y + 3y^2}{3y + 1}\right) = 13$$ $$120y^2 + 80y - 39y - 13 = 0$$ $$120y^2 + 41y - 13 = 0$$ $$D = 1681 + 6240 = 7921$$ $$y = \frac{-41 + 89}{240} = 0.2$$

The negative value of $y$ is rejected as velocity can't be negative.

$$x = \frac{y}{3y + 1} = \frac{0.2}{0.6 + 1} = 0.125$$ $$\frac{1}{x} = 8$$ $$\frac{1}{y} = 5$$

Answer: One pipe fills a cistern in 5 min, the other pipe – in 8 min