Pipes And Cistern — Part 1 MCQs with Explanations

Part filled by (A + B + C) in 3 minutes $$\eqalign{ & = 3\left( {\frac{1}{{30}} + \frac{1}{{20}} + \frac{1}{{10}}} \right) \cr & = {3 \times \frac{{11}}{{60}}} = \frac{{11}}{{20}} \cr & {\text{Part}}\,{\text{filled}}\,{\text{by}}\,{\text{C}}\,{\text{in}}\,{\text{3}}\,{\text{minutes}} = \frac{3}{{10}} \cr & \therefore {\text{Required}}\,{\text{ratio}} \cr & = {\frac{3}{{10}} \times \frac{{20}}{{11}}} = \frac{6}{{11}} \cr} $$

$$\eqalign{ & {\text{Net}}\,{\text{part}}\,{\text{filled}}\,{\text{in}}\,{\text{1}}\,{\text{hour}} \cr & = {\frac{1}{5} + \frac{1}{6} - \frac{1}{{12}}} = \frac{{17}}{{60}} \cr & \therefore {\text{The}}\,{\text{tank}}\,{\text{will}}\,{\text{be}}\,{\text{full}}\,{\text{in}}\,\frac{{60}}{{17}}\,{\text{hours}} \cr & i.e.,\,3\frac{9}{{17}}\,{\text{hours}} \cr} $$

Work done by the leak in 1 hour $$\eqalign{ & = {\frac{1}{2} - \frac{3}{7}} = \frac{1}{{14}} \cr} $$ ∴ Leak will empty the tank in 14 hours

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 $$\eqalign{ & \therefore x\left( {\frac{2}{{75}} + \frac{1}{{45}}} \right) + \left( {30 - x} \right).\frac{2}{{75}} = 1 \cr & \Rightarrow \frac{{11x}}{{225}} + \frac{{ {60 - 2x} }}{{75}} = 1 \cr & \Rightarrow 11x + 180 - 6x = 225 \cr & \Rightarrow x = 9 \cr} $$

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. $$\eqalign{ & \therefore \frac{1}{x} + \frac{1}{{ {x - 5} }} = \frac{1}{{ {x - 9} }} \cr & \Rightarrow \frac{{x - 5 + x}}{{x\left( {x - 5} \right)}} = \frac{1}{{ {x - 9} }} \cr & \Rightarrow \left( {2x - 5} \right)\left( {x - 9} \right) = x\left( {x - 5} \right) \cr & \Rightarrow {x^2} - 18x + 45 = 0 \cr & \Rightarrow \left( {x - 15} \right)\left( {x - 3} \right) = 0 \cr & \Rightarrow x = 15{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \left[ {{\text{neglecting}}\,x = 3} \right] \cr} $$

Work done by the waste pipe in 1 minute $$\eqalign{ & = \frac{1}{{15}} - \left( {\frac{1}{{20}} + \frac{1}{{24}}} \right) \cr & = {\frac{1}{{15}} - \frac{{11}}{{120}}} \cr & = - \frac{1}{{40}}\,\,\,\,\,\left[ { - ve\,{\text{sign}}\,{\text{means}}\,{\text{emptying}}} \right] \cr & \therefore {\text{Volume}}\,{\text{of}}\,\frac{1}{{40}}{\text{part}} = 3\,{\text{gallons}} \cr & {\text{Volume}}\,{\text{of}}\,{\text{whole}} \cr & = \left( {3 \times 40} \right){\text{gallons}} \cr & = 120\,{\text{gallons}} \cr} $$

Suppose pipe A alone takes x hours to fill the tank. Then, pipes B and C will take $$\frac{x}{2}$$ and $$\frac{x}{4}$$ hours respectively to fill the tank. $$\eqalign{ & \therefore \frac{1}{x} + \frac{2}{x} + \frac{4}{x} = \frac{1}{5} \cr & \Rightarrow \frac{7}{x} = \frac{1}{5} \cr & \Rightarrow x = 35\,{\text{hours}} \cr} $$

Let the cistern be filled by pipe A alone in x hours. Then, pipe B will fill it in (x + 6) hours $$\eqalign{ & \therefore \frac{1}{x} + \frac{1}{{ {x + 6} }} = \frac{1}{4} \cr & \Rightarrow \frac{{x + 6 + x}}{{x\left( {x + 6} \right)}} = \frac{1}{4} \cr & \Rightarrow {x^2} - 2x - 24 = 0 \cr & \Rightarrow \left( {x - 6} \right)\left( {x + 4} \right) = 0 \cr & \Rightarrow x = 6\,{\kern 1pt} {\kern 1pt} \left[ {{\text{neglecting the negative value of }}x} \right] \cr} $$

Part filled by A in 1 minute = $$\frac{1}{{20}}$$ Part filled by B in 1 minute = $$\frac{1}{{30}}$$ Part filled by (A + B) in 1 minute $$\eqalign{ & = {\frac{1}{{20}} + \frac{1}{{30}}} = \frac{1}{{12}} \cr} $$ ∴ Both pipes can fill the tank in 12 minutes

$$\eqalign{ & {\text{Part}}\,{\text{filled}}\,{\text{in}}\,{\text{4}}\,{\text{minutes}} \cr & = 4\left( {\frac{1}{{15}} + \frac{1}{{20}}} \right) = \frac{7}{{15}} \cr & {\text{Remaining}}\,{\text{part}} = {1 - \frac{7}{{15}}} = \frac{8}{{15}} \cr & {\text{Part}}\,{\text{filled}}\,{\text{by}}\,B\,{\text{in}}\,{\text{1}}\,{\text{minute}} = \frac{1}{{20}} \cr & \therefore \frac{1}{{20}}:\frac{8}{{15}}::1:x \cr & x = {\frac{8}{{15}} \times 1 \times 20} \cr & \,\,\,\,\,\, = 10\frac{2}{3}\,\min \cr & \,\,\,\,\,\, = 10\min .\,40\,\sec . \cr & \therefore {\text{The}}\,{\text{tank}}\,{\text{will}}\,{\text{be}}\,{\text{full}}\,{\text{in}}\, \cr & = {4\min . + 10\min . +\, 40\sec .} \cr & = 14\min .\,40\sec . \cr} $$

Let the slower pipe alone fill the tank in x minutes. Then, faster pipe will fill it in $$\frac{x}{3}$$ minutes. $$\eqalign{ & \therefore \frac{1}{x} + \frac{3}{x} = \frac{1}{{36}} \cr & \Rightarrow \frac{4}{x} = \frac{1}{{36}} \cr & \Rightarrow x = 144\,\text{minutes} \cr} $$

$$\eqalign{ & {\text{Part}}\,{\text{filled}}\,{\text{by}}\,(A + B)\,{\text{in}}\,{\text{1}}\,{\text{minute}} \cr & = {\frac{1}{{60}} + \frac{1}{{40}}} = \frac{1}{{24}} \cr & {\text{Suppose}}\,{\text{the}}\,{\text{tank}}\,{\text{is}}\,{\text{filled}}\,{\text{in}}\,x\,{\text{minutes}} \cr & {\text{Then}},\,\frac{x}{2}\left( {\frac{1}{{24}} + \frac{1}{{40}}} \right) = 1 \cr & \Rightarrow \frac{x}{2} \times \frac{1}{{15}} = 1 \cr & \Rightarrow x = 30\,\text{minutes} \cr} $$

Time taken by one tap to fill half of the the tank = 3 hours Part filled by the four taps in 1 hour $$\eqalign{ & = {4 \times \frac{1}{6}} = \frac{2}{3} \cr & {\text{Remaining}}\,{\text{part}} = {1 - \frac{1}{2}} = \frac{1}{2} \cr & \therefore \frac{2}{3}:\frac{1}{2}::1:x \cr & \Rightarrow x = {\frac{1}{2} \times 1 \times \frac{3}{2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{4}\,hrs.\,\,i.e.,\,45\,\operatorname{mins} . \cr & {\text{So,}}\,{\text{total}}\,{\text{time}}\,{\text{taken}} = 3\,hrs.\,45\,mins. \cr} $$

$$\eqalign{ & \left( {{\text{A + B}}} \right){\text{'s 1 hour work}} \cr & {\text{ = }} {\frac{1}{{12}} + \frac{1}{{15}}} = \frac{9}{{60}} = \frac{3}{{20}} \cr & \left( {{\text{A + C}}} \right){\text{'s 1 hour work}} \cr & {\text{ = }} {\frac{1}{{12}} + \frac{1}{{20}}} = \frac{8}{{60}} = \frac{2}{{15}} \cr & {\text{Part filled in 2 hrs}} \cr & {\text{ = }} {\frac{3}{{20}} + \frac{2}{{15}}} = \frac{{17}}{{60}} \cr & {\text{Part filled in 6 hrs}} \cr & {\text{ = }} {3 \times \frac{{17}}{{60}}} = \frac{{17}}{{20}} \cr & {\text{Remaining part}} \cr & {\text{ = }} {1 - \frac{{17}}{{20}}} = \frac{3}{{20}} \cr} $$ Now it is the turn of A and B and $$\frac{3}{{20}}$$ part is filled by A and B in 1 hour ∴ Total time taken to fill tank = (6 + 1) hrs = 7 hrs

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

We have the following table: Period Supply Demand Calculation Excess Qty. in tank 0 - 3 hrs 30000 10000 30000 - 10000 20000 3 - 6 hrs 30000 10000 20000 + 30000 - 10000 40000 6 - 9 hrs 30000 45000 40000 + 30000 - 45000 25000 9 - 12 hrs 30000 25000 25000 + 30000 - 25000 30000 12 - 15 hrs 30000 40000 30000 + 30000 - 40000 20000 15 - 18 hrs 30000 15000 20000 + 30000 - 15000 35000 18 - 21 hrs 30000 60000 35000 + 30000 - 60000 5000 21 - 24 hrs 30000 35000 5000 + 30000 - 35000 0 The excess quantity in tank at any time does not exceed 40000 litres, Which is the required minimum capacity to avoid wastage.

Net part filled in 1 hour $$\eqalign{ & {\text{= }}\left( {\frac{1}{x} - \frac{1}{y}} \right) \cr & = \left( {\frac{{y - x}}{{xy}}} \right) \cr} $$ ∴ The tank will be filled in $$ = \left( {\frac{{xy}}{{y - x}}} \right)\,\,{\text{hours}}$$

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

Part of the cistern filled in 1 hour $$ = \frac{1}{3} + \frac{1}{4} - \frac{1}{2}$$ (Cistern filled by 1 st pipe + Cistern filled by 2 nd pipe – Cistern emptied by 3 rd pipe ) $$\eqalign{ & \frac{{4 + 3 - 6}}{{12}} \cr & = \frac{1}{{12}} \cr} $$ Hence, the cistern will be filled in 12 hours.

If tank has 4x liters of total capacity and its holds 3x liters of water and if 30 liters of water is taken out, then tank becomes empty. It means 3x liters of water is taken out 3x = 30 liters x = 10 liters Capacity of tank = 4x = 4 × 10 = 40 liters

Net volume filled in 1 minute = (x - y) liters ∴ The tank will be filled in = $$\frac{{\text{T}}}{{\left( {x - y} \right)}}$$ minutes

Net part filled in 1 hour $$\eqalign{ & {\text{ = }}\frac{1}{3} - \left( {\frac{1}{7} + \frac{1}{{10}}} \right) \cr & = \frac{1}{3} - \frac{{17}}{{70}} \cr & = \frac{{19}}{{210}} \cr} $$ ∴ The tank will be filled in $$\frac{{210}}{{19}}$$ hours i.e. $$\eqalign{ & {\text{= 11}}\frac{1}{{19}}{\text{ hours}} \cr & \cong 11\,{\text{hours }} \cr} $$

$$\eqalign{ & {\text{Net part filled in 1 hour}} \cr & = \frac{2}{3} - \left( {\frac{2}{9} + \frac{1}{3}} \right) \cr & = \left( {\frac{2}{3} - \frac{5}{9}} \right) \cr & = \frac{1}{9} \cr & \therefore \,\frac{1}{9}\,:\,\frac{1}{2}\,::\,1\,:\,x \cr & {\text{or}}\,\,\,x = \left( {\frac{1}{2} \times 9} \right) = 4\frac{1}{2}{\text{ hours}} \cr & {\text{So, the tank will be emptied in}} \cr & {\text{ = }}4\frac{1}{2}{\text{ hours}} \cr} $$

Work done by the C pipe in 1 minute $$\eqalign{ & = \frac{1}{{15}} - \left( {\frac{1}{{20}} + \frac{1}{{24}}} \right) \cr & = \left( {\frac{1}{{15}} - \frac{{11}}{{120}}} \right) \cr & = - \frac{1}{{40}}\,\left[ { - {\text{ve}}\,{\text{means}}\,{\text{emptying}}} \right] \cr} $$ ∴ Volume of $$\frac{1}{{40}}$$ part = 3 gallons. Volume of whole = (3 × 40) gallons = 120 gallons.

$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{(Total Capacity)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{120}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\overline { \downarrow \,\,\,\,\,\,\,\,\,\, \downarrow \,\,\,\,\,\,\,\,\,\,\, \downarrow \,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow } \cr & {\text{efficiency 30}}\,\,\,\,\,\,15\,\,\,\,\,\,\,\,10\,\,\, - 12 \cr & {\text{hours}} \to \,\,\mathop 4\limits_{\left( {\text{P}} \right)}^ \downarrow \,\,\,\,\,\,\mathop 8\limits_{\left( {\text{Q}} \right)}^ \downarrow \,\,\,\,\,\,\,\mathop {12}\limits_{\left( {\text{R}} \right)}^ \downarrow \,\,\,\,\,\,\,\,\,\,\,\mathop {10}\limits_{\left( {\text{S}} \right)}^ \downarrow \cr} $$ In order to fill the cistern in less time. So, efficiency of filling should be more now, check all options (A) → Q efficiency 15 units/hr (B) → (P + R - S) efficiency = 30 + 10 - 12 = 28 units/hr (C) → (P - S) efficiency = 30 - 12 = 18 units/hr (D) → (P + Q - S) efficiency = 30 + 15 - 12 = 33 units/hr Option (D) is answer Since efficiency of option (D) is highest

Let the total capacity of the tank is 30 units. The efficiency of Leakage(Pipe A) will be $$\frac{30}{10}$$ = 3 And the efficiency of the leakage (Pipe A) and another Pipe (B) which is filling the tank will be $$\frac{30}{15}$$ = 2 Pipe A is emptying at 3 units/hr and when filling pipe B started then the emptying rate will come down to 2 units/hr. ∴ Filling Pipe B efficiency is 3 - 2 = 1unit/hr Pipe B will be fill the tank in $$\frac{30}{1}$$ = 30 hrs Filling rate of Pipe B per minute is 4 litter ∴ Total Capacity of tank will be = (4 × 60) × 30 = 7200 litters

Work done by the leak in 1 hour $$\eqalign{ & {\text{ = }}\left( {\frac{1}{2} - \frac{3}{7}} \right) = \frac{1}{{14}} \cr} $$ ∴ Leak will empty the tank in 14 hours

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.

Capacity of the tank = (12 × 13.5) litres = 162 litres Capacity of each bucket = 9 litres Number of buckets needed $$\eqalign{ & {\text{= }}\left( {\frac{{162}}{9}} \right) \cr & = 18 \cr} $$

Part filled by (A + B) in 1 hour $$\eqalign{ & {\text{= }}\left( {\frac{1}{{20}} + \frac{1}{{30}}} \right) \cr & = \frac{1}{{12}} \cr} $$ So, A and B together can fill the tank in 12 hrs, $$\frac{1}{3}$$ part is filled by (A + B) in $$\left( {\frac{1}{3} \times 12} \right){\text{ = 4 hrs}}$$ Since the leak empties one - third water, so time taken to fill the tank = Time taken by (A + B) to fill the whole tank + Time taken by (A + B) to fill one - third tank = (12 + 4) = 16 hours

Work done by the waste pipe in 1 minute $$\eqalign{ & {\text{ = }}\frac{1}{{30}} - \left( {\frac{1}{{40}} + \frac{1}{{48}}} \right) \cr & = \left( {\frac{1}{{30}} - \frac{{11}}{{220}}} \right) \cr & = - \frac{1}{{80}}\left[ { - \,{\text{Nagetive sign means emptying}}} \right] \cr & \therefore {\text{Volume of }}\frac{1}{{80}}{\text{ part = 3 galons}} \cr & {\text{Volume of whole tank}} \cr & {\text{ = }}\left( {3 \times 80} \right){\text{gallons}} \cr & {\text{ = 240 gallons}}{\text{}} \cr} $$

The outlet pipe empties the one complete cistern in 3 hours Time taken to empty $$\frac{2}{3}$$ Part of the cistern $$\eqalign{ & {\text{= }}\frac{2}{3} \times 3 \cr & = 2\,{\text{hours}} \cr} $$

Rate of flow of water = x cm/minute ∴ Volume of water that flowed in the in 1 minutes = (5 × 4 × x) = 20 x cu.cm. ∴ Volume of water that flowed in the tank in 6 hours 18 minutes. i.e. (6 × 60 + 18) = 378 minutes = 2x × 378 cu. cm. According to question, $$\eqalign{ & {\text{20}}x \times 378 = 700 \times 400 \times 450 \cr & \Rightarrow x = \left( {\frac{{700 \times 400 \times 450}}{{20 \times 378}}} \right){\text{cm /minutes}} \cr & \Rightarrow x = \left( {\frac{{700 \times 400 \times 450 \times 60}}{{100000 \times 20 \times 378}}} \right){\text{km/hours}} \cr & \Rightarrow x{\text{ = 10 km/hours}} \cr} $$

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 $${\text{ = }}\frac{{240}}{{27}} = \frac{{80}}{9} = 8\frac{8}{9}\,{\text{Hours}}$$

Net part filled in 1 hour $$\eqalign{ & {\text{ = }}\left( {\frac{1}{8} - \frac{1}{{12}}} \right) = \frac{1}{{24}} \cr} $$ ∴ The tank will be filled in 24 hours

Pipe A will fill 3 units till 11 A.M. Remaining capacity = 6 - 3 = 3 units Now both pipes will fill the tank in $$\frac{{{\text{Total Capacity}}}}{{{\text{Efficiency }}}} = \frac{3}{{\left( {3 + 1} \right)}} = \frac{3}{4}{\text{ hours}}$$ So, $$\left( {11 + \frac{3}{4}} \right)$$ A.M., tank will be filled = 11.45 A.M.

(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 $$\eqalign{ & = \frac{{{\text{Remaining capacity}}}}{{{\text{efficiency of B}}}} \cr & = \frac{{45}}{1} \cr & = \,45\,{\text{minutes}} \cr} $$

Time will be taken by both of them to fill the tank $${\text{ = }}\frac{{xy}}{{y - x}}$$

Net efficiency = q - p (∵ q > p) Time required $$\eqalign{ & {\text{r}} = \frac{{pq}}{{q - p}} \cr & or\,\frac{1}{r} = \frac{1}{p} - \frac{1}{q} \cr} $$

(A + B + C)'s efficiency = 3 + 2 + 1 = 6 units/hr (A + B + C) can fill the tank in $$\eqalign{ & {\text{ = }}\frac{{{\text{Total Capacity}}}}{{{\text{Efficiency of}}\left( {{\text{A + B + C}}} \right){\text{ }}}} \cr & = \frac{{12}}{6} \cr & = 2{\text{ hours}} \cr} $$