Practice MCQ Questions and Answer on Pipes and Cistern
1.
When operated separately pipe A takes 5 hours less than pipe B to fill a cistern, and when operated together, the cistern gets filled in 6 hours. In how much time (in hours) will pipe A fill the cistern, if operated separately?
(A) 10
(B) 18
(C) 9
(D) 15
Solution:
\[\begin{array}{*{20}{c}} {\text{B}}&{}&{\text{A}} \\ {x + 5}&:&x \end{array}\] Now go through option, Let A = 10 hours B = 15 hours A + B = $$\frac{{30}}{5}$$ = 6 hours
2.
One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the slower pipe alone will be able to fill the tank in-
(A) 81 min
(B) 108 min
(C) 144 min
(D) 192 min
Solution:
Let the slower pipe alone fill the tank in x minutes Then, Faster pipe alone will fill it in $$\frac{x}{3}$$ minutes $$\eqalign{ & \therefore \frac{1}{x} + \frac{3}{x} = \frac{1}{{36}} \cr & \Rightarrow \frac{4}{x} = \frac{1}{{36}} \cr & \Rightarrow x = 144 \cr} $$ So slower pipe alone will fill the tank in 144 min.
3.
Two pipes can fill a tank in 12 hours and 16 hours respectively. A third pipe can empty the tank in 30 hours. If all three pipes are opened and functions simultaneously, how much time will the tank take to be full?( in hours )
(A) $$10\frac{4}{9}$$
(B) $$9\frac{1}{2}$$
(C) $$8\frac{8}{9}$$
(D) $$7\frac{2}{9}$$
Solution:
First pipe fill the tank in 1 hour = $$\frac{1}{{12}}$$ part of tank Second pipe fill the tank in 1 hour = $$\frac{1}{{16}}$$ part of tank Third pipe empty the tank in 1 hour = $$\frac{1}{{30}}$$ part of tank When all three pipes are opened simultaneously, part of the tank filled in 1 hour $$ = \frac{1}{{12}} + \frac{1}{{16}} - \frac{1}{{30}}$$ LCM of 12, 16 and 30 = 240 $$\eqalign{ & {\text{ = }}\frac{{20 + 15 - 8}}{{240}} \cr & = \frac{{27}}{{240}} \cr} $$ ∴ Required time taken by all the three pipes $${\text{ = }}\frac{{240}}{{27}} = \frac{{80}}{9} = 8\frac{8}{9}\,{\text{Hours}}$$
4.
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 $$\eqalign{ & \therefore x\left( {\frac{2}{{75}} + \frac{1}{{45}}} \right) + \left( {30 - x} \right).\frac{2}{{75}} = 1 \cr & \Rightarrow \frac{{11x}}{{225}} + \frac{{ {60 - 2x} }}{{75}} = 1 \cr & \Rightarrow 11x + 180 - 6x = 225 \cr & \Rightarrow x = 9 \cr} $$
5.
Three pipes A, B and C can fill a tank in 6 hours, 9 hours and 12 hours respectively. B and C are opened for half an hour, then A is also opened. The time taken by the three pipes together to fill the remaining part of the tank is -
(A) 3 hours
(B) 2 hours
(C) $${\text{2}}\frac{1}{2}$$ hours
(D) $${\text{3}}\frac{1}{2}$$ hours
Solution:
In half an hour (B + C) must have filled $$ = \frac{4}{2} + \frac{3}{2} = \frac{7}{2}\,{\text{units}}$$ Capacity left $${\text{ = 36}} - \frac{7}{2} = \frac{{65}}{2}\,{\text{units}}$$ Now all pipes will fill the remaining tank $$\eqalign{ & {\text{ = }}\frac{{65}}{{2 \times \left( {6 + 4 + 3} \right)}} \cr & = \frac{{65}}{{2 \times 13}} \cr & = \frac{5}{2} \cr & = 2\frac{1}{2}\,{\text{hours}} \cr} $$
6.
Having the same capacity 9 taps fill up a water tank in 20 minutes. How many taps of the same capacity are required to fill up the same water tank in 15 minutes ?
A swimming pool is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the pool in the same time during which the pool is filled by the third pipe alone. The second pipe fills the pool 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 takes (x - 5) and (x - 9) hours respectively to fill the tank. $$\eqalign{ & \therefore \frac{1}{x} + \frac{1}{{\left( {x - 5} \right)}} = \frac{1}{{\left( {x - 9} \right)}} \cr & \Rightarrow \frac{{x - 5 + x}}{{x\left( {x - 5} \right)}} = \frac{1}{{\left( {x - 9} \right)}} \cr & \Rightarrow \left( {2x - 5} \right)\left( {x - 9} \right) = x\left( {x - 5} \right) \cr & \Rightarrow {x^2} - 18x + 45 = 0 \cr & \Rightarrow \left( {x - 15} \right)\left( {x - 3} \right) = 0 \cr & \Rightarrow x = 15\left[ {{\text{neglecting }}x\,{\text{ = 3}}} \right] \cr} $$ So, first pipe alone takes 15 hrs to fill the tank.
8.
A tank with capacity T liters is empty. If water flows into the tank from pipe X at the rate of x liters per minute and water is pumped out by Y at the rate of y liters per minute and x > y, then how many minutes will the tank be filled?
Net volume filled in 1 minute = (x - y) liters ∴ The tank will be filled in = $$\frac{{\text{T}}}{{\left( {x - y} \right)}}$$ minutes
9.
Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is:
(A) 10
(B) 12
(C) 14
(D) 16
Solution:
Part filled in 2 hours = $$\frac{2}{6}$$ = $$\frac{1}{3}$$ Remaining part = $$ {1 - \frac{1}{3}} $$ = $$\frac{2}{3}$$ ∴ (A + B)'s 7 hour's work = $$\frac{2}{3}$$ (A + B)'s 1 hour's work = $$\frac{2}{{21}}$$ ∴ C's 1 hour's work = {(A + B + C)'s 1 hour's work} - {(A + B's 1 hour's work} $$\eqalign{ & = {\frac{1}{6} - \frac{2}{{21}}} \cr & = \frac{1}{{14}} \cr} $$ ∴ C alone can fill the tank in 14 hours
10.
Pipes A and B are filling pipes while pipe C is an emptying pipe. A and B can fill a tank in 72 and 90 minutes respectively. When all the three pipes are opened together, the tank gets filled in 2 hours. A and B are opened together for 12 minutes, then closed and C is opened. The tank will be empty after:
(A) 12 minute
(B) 15 minute
(C) 18 minute
(D) 16 minute
Solution:
A + B - C = 3 5 + 4 - C = 3 C = 6 In 12 min, (A + B) can fill = (5 + 4) × 12 = 9 × 12 The time was taken to empty (9 × 12) by 'C' Time $$ = \frac{{9 \times 12}}{6} = 18\min $$