Question

A fish tank can be filled in 10 minutes using both pumps A and B simultaneously however pump B can pump water in OR out at the same rate if pump B is inadvertently run in reverse​ then the tank will be filled in 30 minutes. How long would it take each pump to fill the tank by itself? (Using cramer's rule)

Answer 1

Solution:

Let time taken by Pump A ,to fill the tank = x minutes

And, time taken by Pump B,to fill the tank =y minutes

Time taken  by both the pumps, if they work simultaneously together,to fill the tank = 10 minutes

$\frac{1}{x}+\frac{1}{y}=\frac{1}{10}\\\\\frac{1}{x}+\frac{1}{y}=\frac{60}{10}=6[\tex]</p><p>Also,pump B can pump water in OR out at the same rate if pump B is inadvertently run in reverse​ then the tank will be filled in 30 minutes. </p><p>[tex]\frac{1}{x}-\frac{1}{y}=\frac{1}{30}\\\\\frac{1}{x}-\frac{1}{y}=\frac{60}{30}=2\\\\A=\frac{1}{x},B=\frac{1}{y}\\\\A +B=6\\\\A-B=2\\\\D=\left[\begin{array}{ccc}1&1\\1&-1\end{array}\right] \\\\D=-1-1=-2\\\\D_{x}=\left[\begin{array}{ccc}6&2\\1&-1\end{array}\right] \\\\D_{x}=-6-2=-8\\\\D_{y}=\left[\begin{array}{ccc}1&1\\6&2\end{array}\right] \\\\D_{y}=-6+2= -4\\\\x=\frac{D_{x}}{D}=\frac{-8}{-2}=4\\\\y=\frac{D_{y}}{D}=\frac{-8}{-4}=2$

Time taken by Pump A ,alone to fill the tank completely $=\frac{1}{4} hours=\frac{60}{4}=15 minutes$

Time taken by Pump B ,alone to fill the tank completely$=\frac{1}{2} hours=\frac{60}{2}=30 minutes$