Pipes and Cisterns Formulas, Shortcut Tricks

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Pipes and Cisterns Formulas

BASIC CONDITION AND FORMULAS FOR PIPES And CISTERNS

Inlet : A pipe connected with a tank or a cistern or a reservoir, that fills it, is known as an inlet.
Outlet: A pipe connected with a tank or cistern or reservoir, emptying it, is known as an outlet.
If a pipe can fill a tank in x hours, then: part fill in 1 hour = $\frac{1}{x}$
If a pipe can empty a tank in y hours, then: part emptied in 1 hour = $\frac{1}{y}$
If a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours (where y > x), then on opening both the pipes, then the net part filled in 1 hour = $(\frac{1}{x}-\frac{1}{y})$
If a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours (where y > x), then on opening both the pipes, then the net part emptied in 1 hour = (1 - 1 )
Time for filling , (Filling pipe is bigger in size.) , F = $\frac{e\times f}{e-f}$
Time for emptying , (emptying pipe is bigger in size.) , E = $\frac{f\times e}{f-e}$
Pipes 'A' & 'B' can fill a tank in f1hrs & f2hrs respectively.Another pipe 'C' can empty the full tank in 'e'hrs.If the three pipes are opened simultaneously then the tank is filled in. F = $\left[\frac{L}{\frac{L}{F_{1}}+\frac{L}{F_{2}}-\frac{L}{e}}\right ]$
Two taps 'A' & 'B' can fill a tank in 't1' & 't2' hrs respectively.Another pipe 'C' can empty the full tank in 'e'hrs.If the tank is full & all the three pipes are opened simultaneously . Then the tank will be emptied in, E = $\left [ \frac{L}{\frac{L}{e}-\frac{L}{F_{1}}-\frac{L}{F_{2}}} \right ]$
Capacity of the tank is , F = $\left(\frac{f\times e}{e-f}\right)$

Pipes and Cisterns Shortcut Tricks

Question 1: Two pipes can fill a tank in 10 hours and 12 hours resp. while third pipe empties the full tank in 20 hours. If all the three pipes operate simultaneously, in how much time the tank will be filled?
solution : 1 minutes work of each of the three pipes
= $\frac{1}{10}+\frac{1}{12}-\frac{1}{20}$
= $\frac{6+5-3}{60}\;=\; \frac{8}{60}$
= $\frac{2}{15}\; =\; \frac{15}{2}\; =\; 7\frac{1}{2}$ = 7 hours 30 minutes.

question 2: A cistern can be filled in 9 hours but it takes 10 hours due to a leak in its bottom. If the cistern is full, then the time that the leak will take to empty it is:
solution : Time taken by A = 9 hours
Time taken by B = 10 hours
Time taken by A and B = $\frac{9\times 10}{10-9}$ = 90 hours.

question 3: Taps A and B can fill a bucket in 12 minutes and 15 minutes respectively. If both are opened and A is closed after 3 minutes. How much further time would it take for B to fill the bucket?
solution : For 3 minutes (I part) = $3\left (\frac{1}{12}+\frac{1}{15}\right)$
$3\left (\frac{15+12}{15\times 12}\right)\; =\; 3\left(\frac{3}{20}\right)\; =\; \frac{9}{20}$
Remaining part = $1-\frac{9}{20}\; =\: \frac{11}{20}$
Tap B fill the bucket in = $\frac{11}{20}-15\; =\; \frac{33}{4}\; =\; 8\frac{1}{4}$ = 8 hours 15 seconds.

question 4: If two pipes function simultaneously the reservoir will be filled in 12 hours, one pipe fills the reservoir 10 hours faster than the other. How many hours it takes the second pipe to fill the reservoir?
solution : Let the reservoir be filled by forst pipe in x hours, therefore the second pipe will fill it in (x+10) hours
$\therefore \frac{1}{x}+\frac{1}{x+10}\; =\; \frac{1}{12}$
$\frac{x+10+x}{x(x+10)}\; =\; \frac{1}{12}$
12 (10+2x) = $x^{2}+10x$
$x^{2}+10x-24x-120\; =\; 0$
$x^{2}-14x-120\; =\; 0$
= (x-20) (x+6) = 0
= x = 20 (x should nor –ve)
The second pipe can take (x+10) = 20 + 10 = 30 hours, to fill the reservoir.