Practice MCQ Questions and Answer on Pipes and Cistern
31.
Two pipes A and B can fill a cistern in $$37\frac{1}{2}$$ minutes and 45 minutes respectively. Both pipes are opened. The cistern will be filled in just half an hour, if the B is turned off after:
(A) 5 min.
(B) 9 min.
(C) 10 min.
(D) 15 min.
Solution:
Let B be turned off after x minutes. Then, Part filled by (A + B) in x min. + Part filled by A in (30 - x) min. = 1 $$\eqalign{ & \therefore x\left( {\frac{2}{{75}} + \frac{1}{{45}}} \right) + \left( {30 - x} \right).\frac{2}{{75}} = 1 \cr & \Rightarrow \frac{{11x}}{{225}} + \frac{{ {60 - 2x} }}{{75}} = 1 \cr & \Rightarrow 11x + 180 - 6x = 225 \cr & \Rightarrow x = 9 \cr} $$
32.
A water tank can be filled by a tap in 30 minutes and another tap can fill it in 60 minutes. If both taps are kept open for 5 minutes and then the first tap is closed, how long will it take foe the tank to be filled ?
(A) 20 minutes
(B) 25 minutes
(C) 30 minutes
(D) 45 minutes
Solution:
(A + B)'s filling (2 + 1) = 3 units/min In 5 minutes they will fill 3 × 5 = 15 units Remaining capacity = 60 - 15 = 45 units Second pipe (B) fills it in $$\eqalign{ & = \frac{{{\text{Remaining capacity}}}}{{{\text{efficiency of B}}}} \cr & = \frac{{45}}{1} \cr & = \,45\,{\text{minutes}} \cr} $$
33.
Pipes A and B can fill a tank in 5 and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in:
12 buckets of water fill a tank when the capacity of each bucket is 13.5 litres. How many buckets will be needed to fill the same tank, if the capacity of each bucket is 9 litres?
(A) 8
(B) 15
(C) 16
(D) 18
Solution:
Capacity of the tank = (12 × 13.5) litres = 162 litres Capacity of each bucket = 9 litres Number of buckets needed $$\eqalign{ & {\text{= }}\left( {\frac{{162}}{9}} \right) \cr & = 18 \cr} $$
35.
A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is:
(A) 6 hours
(B) 10 hours
(C) 15 hours
(D) 30 hours
Solution:
Suppose, first pipe alone takes x hours to fill the tank. Then, Second and third pipes will take (x - 5) and (x - 9) hours respectively to fill the tank. $$\eqalign{ & \therefore \frac{1}{x} + \frac{1}{{ {x - 5} }} = \frac{1}{{ {x - 9} }} \cr & \Rightarrow \frac{{x - 5 + x}}{{x\left( {x - 5} \right)}} = \frac{1}{{ {x - 9} }} \cr & \Rightarrow \left( {2x - 5} \right)\left( {x - 9} \right) = x\left( {x - 5} \right) \cr & \Rightarrow {x^2} - 18x + 45 = 0 \cr & \Rightarrow \left( {x - 15} \right)\left( {x - 3} \right) = 0 \cr & \Rightarrow x = 15{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \left[ {{\text{neglecting}}\,x = 3} \right] \cr} $$
36.
Pipes A and B can fill a tank in 12 minutes and 15 minutes, respectively. The tank when fill can be emptied by pipe C in x minutes. When all the three pipes are opened simultaneously, the tank is full in 10 minutes. The value of x is:
(A) 18
(B) 15
(C) 20
(D) 24
Solution:
$$\eqalign{ & A + B + C = \frac{{60}}{{10}} \cr & 5 + 4 + C = 6 \cr & C = - 3 \cr & C{\text{ can empty the tank }}x \cr & = \frac{{60}}{3} = 20{\text{ minute}} \cr} $$
37.
Two pipes A and B can fill a cistern in $$12\frac{1}{2}$$ hours and 25 hours, respectively. The pipes are opened simultaneously and it is found that due to a leakage in the bottom, it took 1 hour 40 minutes more to fill the cistern. When the cistern is full, in how much time will the leak empty the cistern?
(A) 48 hours
(B) 45 hours
(C) 42 hours
(D) 50 hours
Solution:
Given: A can fill the cistern 12.5 hr B can fill the cistern 25 hr Formula Used: Total work = Efficiency × Time Calculation: Let the capacity of the cistern be 25 units (LCM of 12.5 and 25) ⇒ Efficiency of A = $$\frac{{25}}{{12.5}}$$ = 2 units/hr ⇒ Efficiency of B = $$\frac{{25}}{{25}}$$ = 1 units/hr ⇒ Combined efficiency of A and B = 2 + 1 = 3 units/hr Time taken by A and B to fill the cistern without leakage ⇒ $$\frac{{25}}{3}$$ = $$8\frac{1}{3}$$ hr = 8 hr 20 min Time taken by A and B to fill the cistern with leakage ⇒ 8 hr 20 min + 1 hr 40 min = 10 hr The combined efficiency of A and B with leakage ⇒ $$\frac{{25}}{{10}}$$ = 2.5 units/hr ⇒ Efficiency of leakage = 3 - 2.5 = 0.5 units/hr Time required by the leak to empty the full cistern ⇒ $$\frac{{25}}{{0.5}}$$ = 50 hr ∴ The leak can empty the full cistern in 50 hours.
38.
Two pipes A and B can fill an empty tank in 10 hours and 15 hours respectively. Pipe C alone can empty the completely filled tank in 12 hours. First both pipes A and B are opened and after 5 hours pipe C is also opened. What is the total time (in hours) in which then tank will be filled?
(A) 7
(B) 5
(C) 9
(D) 8
Solution:
According to question, Efficiency of (A + B) = 18 + 12 = 30 litre/hr In 5 hrs, filled capacity by (A + B) = 30 × 5 = 150 litres ∴ Remaining capacity = 180 - 150 = 30 litres Total efficiency of (A + B + C) = 18 + 12 - 15 = 15 litre/hr ∴ Time taken by (A + B + C) to fill the remaining capacity $$ \Rightarrow \frac{{30}}{{15}} = 2\,{\text{hrs}}$$ ∴ Required time, to fill the complete tank = 5 hrs + 2 hrs = 7 hrs
39.
Pipes A and B can fill a tank in 16 hours and 24 hours, respectively, whereas pipe C and empty the full tank is 40 hours. All three pipes are opened together, but pipe C is closed after 10 hours. After how many hours will the remaining part of the tank be filled?
(A) 2
(B) $$2\frac{1}{2}$$
(C) 5
(D) $$5\frac{1}{2}$$
Solution:
All three pipe opened together in 10 hours = (15 + 10 - 6) × 10 = 190 Remaining work = 240 - 190 = 50 Remaining work done by A and B $$ = \frac{{50}}{{15 + 10}} = 2$$
40.
A pipe can fill a tank with water in 3 hours. Due to leakage in bottom, it takes $$3\frac{1}{2}$$ hours to fill it. In what time the leak will empty the completely filled tank?
(A) 12 hours
(B) 21 hours
(C) $$6\frac{1}{2}$$ hours
(D) $$10\frac{1}{2}$$ hours
Solution:
Let the total capacity of tank is 21 units A's efficiency is 7 units/hr A's efficiency after leakage 6 units/hr leakage efficiency = 7 - 6 = 1 unit/hr leakage will empty the filled tank $$\frac{{{\text{Total Capacity}}}}{{{\text{Efficiency }}}} = \frac{{21}}{1}$$ = 21hours