Pipes and Cisterns Quiz 1

Solution - Suppose pipe A alone takes x hours to fill the tank.
Then pipes B and C will take x/3 and x/6 hours respectively to fill the tank.
Therefore \frac{1}{x}+\frac{3}{x}+\frac{6}{x}=\frac{1}{40}
⇒ x = 400hrs

Solution - (P+Q)’s 1 hour’s work = \frac{1}{10}+\frac{1}{20}=\frac{3}{20}
(A+C)’s 1 hour’s work = \frac{1}{10}+\frac{1}{30}=\frac{2}{15}
Part filled in 2 hrs = \frac{3}{20}+\frac{2}{15}=\frac{17}{60}
Part filled in 6 hrs = 3\times \frac{17}{60}=\frac{17}{20}
Remaining Part = 1- \frac{17}{20}=\frac{3}{20}
Now, it is the turn of P and Q and 3/20 part is filled by P and Q in 1 hour.
Therefore, Total time taken to fill the tank = (6+1) hrs = 7 hrs

Solution - Let M was turned off after x min.
Then, cistern filled by M in x min + cistern filled by N in (x+5) min = 1
⇒ \frac{x}{60}+\frac{x+5}{40}=1
⇒ 5x+15=120
⇒ x = 21 min.

Solution – Let the required time be x hours, then
\frac{1}{32}\left ( \frac{1}{4}x \right ) + \frac{1}{32}\left (x- \frac{1}{4}x \right ) + \frac{1}{40}\left (x- \frac{1}{4}x \right ) =1
Solving we get x = 20 hrs

Solution - (A+B)’s 1 hour work = \left ( \frac{1}{12} +\frac{1}{15}\right )=\frac{3}{20}
(A+C)’s 1 hour work=\left ( \frac{1}{12} +\frac{1}{20}\right )=\frac{2}{15}
Part filled in 2 hrs = \left ( \frac{3}{20} +\frac{2}{15}\right )=\frac{17}{60}
Part filled in 6 hrs = 3 \times \frac{17}{60} = \frac{17}{20}
Remaining part = \frac{3}{20}
Now its turn of A and B and \frac{3}{20}part is filled by A and B in 1 hr
So now total time = (6+1)=7 hrs

Solution:
Let the slower pipe alone fill the tank in x minutes.
Then,faster pipe will fill it in x/3 minutes.
=>\frac{1}{x}+\frac{3}{x}=\frac{1}{72}
=>x=288 minutes

Solution: Let pipe A takes p min to fill
Then,pipe B takes 3p min to fill
=> 3p - p = 64
=> p = 32 min => 3p = 96 min
Required, both pipes to fill = (96 x 32)/(96 + 32) min = 24 min.

Solution - Let the capacity of the tank be LCM(40, 120) = 120 units.
=> Efficiency of pipe A = 120 / 40 = 3 units / minute
=> Efficiency of pipe B = 120 / 120 = 1 units / minute
=> Combined efficiency of pipe A and pipe B = 4 units / minute
Therefore, time required to empty the tank if both pipes work = 120 / 4 = 30 minutes

Solution -
Let A was turned off after x min.
Then, cistern filled by A in x min + cistern filled by B in (x+5) min = 1
⇒ \frac{x}{120}+\frac{x+5}{80}=1
⇒ 5x+15=240
⇒ x = 45 min.

Solution – Let us assume the units of work = LCM(40,80,160) = 160
Work done by Tap A/min = 4units
Work done by Tap B/min = 2 units
Work done by leakage/min = -1unit
Net work done/min = 5 units/min
Total time taken = \frac{160}{5} = 32 min