Pipes & Cistern

1.Three pipes A, B and C can fill a tank from empty to full in 30 minutes, 20 minutes, and 10 minutes respectively. When the tank is empty, all the three pipes are opened. A, B and C discharge chemical solutions P,Q and R respectively. What is the proportion of the solution R in the liquid in the tank after 3 minutes?

1. ⁵/₁₁
2. ⁶/₁₁
3. ⁷/₁₁
4. ⁸/₁₁

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Part filled by (A + B + C) in 3 minutes \\ = 3 (\frac{1}{30} + \frac{1}{20} + \frac{1}{10}) \\ = (3 \times \frac{11}{60}) = \frac{11}{20} \\ Part filled by C in 3 minutes = \frac{3}{10} \\ ∴ Required ratio \\ = (\frac{3}{10} \times \frac{20}{11}) = \frac{6}{11} \end{aligned}

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

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2.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:

1. 1(¹³/₁₇) hours
2. 2(⁸/₁₁) hours
3. 3(⁹/₁₇) hours
4. 4(¹/₂) hours

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Net part filled in 1 hour \\ = (\frac{1}{5} + \frac{1}{6} - \frac{1}{12}) = \frac{17}{60} \\ ∴ The tank will be full in \frac{60}{17} hours \\ i . e ., 3 \frac{9}{17} hours \end{aligned}

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

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3.A pump can fill a tank with water in 2 hours. Because of a leak, it took 2(¹/₃) hours to fill the tank. The leak can drain all the water of the tank in:

1. 4(¹/₃) hours
2. 7 hours
3. 8 hours
4. 14 hours

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Work done by the leak in 1 hour \\ = (\frac{1}{2} - \frac{3}{7}) = \frac{1}{14} \\ ∴ Leak will empty the tank in 14 hrs . \end{aligned}

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

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4.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:

1. 6 hours
2. 10 hours
3. 15 hours
4. 30 hours

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Suppose, first pipe alone takes x hours to fll the tank . \\ Then, \\ Second and third pipes will take (x - 5) and (x - 9) hours respectively to fill the tank . \\ ∴ \frac{1}{x} + \frac{1}{(x - 5)} = \frac{1}{(x - 9)} \\ \Rightarrow \frac{x - 5 + x}{x (x - 5)} = \frac{1}{(x - 9)} \\ \Rightarrow (2 x - 5) (x - 9) = x (x - 5) \\ \Rightarrow x^{2} - 18 x + 45 = 0 \\ (x - 15) (x - 3) = 0 \\ \Rightarrow x = 15 [neglecting x = 3] \end{aligned}

$\eqalign{ & {\text{Suppose,}}\,{\text{first}}\,{\text{pipe}}\,{\text{alone}}\,{\text{takes}}\,x\,{\text{hours}}\,{\text{to}}\,{\text{fll}}\,{\text{the}}\,\,{\text{tank}}{\text{.}} \cr & {\text{Then,}} \cr & {\text{Second}}\,{\text{and}}\,{\text{third}}\,{\text{pipes}}\,{\text{will}}\,{\text{take}}\,\left( {x - 5} \right){\text{and}}\left( {x - 9} \right)\,{\text{hours}}\,{\text{respectively}}\,{\text{to}}\,{\text{fill}}\,{\text{the}}\,{\text{tank}}. \cr & \therefore \frac{1}{x} + \frac{1}{{\left( {x - 5} \right)}} = \frac{1}{{\left( {x - 9} \right)}} \cr & \Rightarrow \frac{{x - 5 + x}}{{x\left( {x - 5} \right)}} = \frac{1}{{\left( {x - 9} \right)}} \cr & \Rightarrow \left( {2x - 5} \right)\left( {x - 9} \right) = x\left( {x - 5} \right) \cr & \Rightarrow {x^2} - 18x + 45 = 0 \cr & \left( {x - 15} \right)\left( {x - 3} \right) = 0 \cr & \Rightarrow x = 15\,\,\,\,\,\,\left[ {{\text{neglecting}}\,x = 3} \right] \cr}$

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5.Two pipes can fill a tank in 20 and 24 minutes respectively and a waste pipe can empty 3 gallons per minute. All the three pipes working together can fill the tank in 15 minutes. The capacity of the tank is:

1. 60 gallons
2. 100 gallons
3. 120 gallons
4. 180 gallons

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Work done by the waste pipe in 1 minute \\ = \frac{1}{15} - (\frac{1}{20} + \frac{1}{24}) \\ = (\frac{1}{15} - \frac{11}{120}) \\ = - \frac{1}{40} [- v e sign means emptying] \\ ∴ Volume of \frac{1}{40} part = 3 gallons \\ Volume of whole \\ = (3 \times 40) gallons \\ = 120 gallons \end{aligned}

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

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6.A tank is filled in 5 hours by three pipes A, B and C. The pipe C is twice as fast as B and B is twice as fast as A. How much time will pipe A alone take to fill the tank?

1. 20 hours
2. 25 hours
3. 35 hours
4. Cannot be determined

Answer:Option 1

Explanation:

Solution:

\begin{aligned} 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 . \\ ∴ \frac{1}{x} + \frac{2}{x} + \frac{4}{x} = \frac{1}{5} \\ \Rightarrow \frac{7}{x} = \frac{1}{5} \\ \Rightarrow x = 35 hrs \end{aligned}

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

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7.Two pipes A and B together can fill a cistern in 4 hours. Had they been opened separately, then B would have taken 6 hours more than A to fill the cistern. How much time will be taken by A to fill the cistern separately?

1. 1 hour
2. 2 hours
3. 6 hours
4. 8 hours

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Let the cistern be filledby pipe A alone in x hours . \\ Then, pipe B will fill it in (x + 6) hours \\ ∴ \frac{1}{x} + \frac{1}{(x + 6)} = \frac{1}{4} \\ \Rightarrow \frac{x + 6 + x}{x (x + 6)} = \frac{1}{4} \\ \Rightarrow x^{2} - 2 x - 24 = 0 \\ \Rightarrow (x - 6) (x + 4) = 0 \\ \Rightarrow x = 6. [neglecting the negative value of x] \end{aligned}

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

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8.Two pipes A and B can fill a tank in 15 minutes and 20 minutes respectively. Both the pipes are opened together but after 4 minutes, pipe A is turned off. What is the total time required to fill the tank?

1. 10 min. 20 sec.
2. 11 min. 45 sec.
3. 12 min. 30 sec.
4. 14 min. 40 sec.

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Part filled in 4 minutes \\ = 4 (\frac{1}{15} + \frac{1}{20}) = \frac{7}{15} \\ Remaining part = (1 - \frac{7}{15}) = \frac{8}{15} \\ Part filled by B in 1 minute = \frac{1}{20} \\ ∴ \frac{1}{20} : \frac{8}{15} :: 1 : x \\ x = (\frac{8}{15} \times 1 \times 20) \\ = 10 \frac{2}{3} min \\ = 10 min . 40 \sec . \\ ∴ The tank will be full in \\ = (4 min . + 10 min . + 40 \sec .) \\ = 14 min . 40 \sec . \end{aligned}

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

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9.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:

1. 81 min.
2. 108 min.
3. 144 min.
4. 192 min.

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Let the slower pipe alone fill the tank in x minutes . \\ Then, faster pipe will fill it in \frac{x}{3} minutes . \\ ∴ \frac{1}{x} + \frac{3}{x} = \frac{1}{36} \\ \Rightarrow \frac{4}{x} = \frac{1}{36} \\ \Rightarrow x = 144 min \end{aligned}

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

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10.A large tanker can be filled by two pipes A and B in 60 minutes and 40 minutes respectively. How many minutes will it take to fill the tanker from empty state if B is used for half the time and A and B fill it together for the other half?

1. 15 min
2. 20 min
3. 27.5 min
4. 30 min

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Part filled by (A + B) in 1 minute \\ = (\frac{1}{60} + \frac{1}{40}) = \frac{1}{24} \\ Suppose the tank is filled in x minutes . \\ Then, \frac{x}{2} (\frac{1}{24} + \frac{1}{40}) = 1 \\ \Rightarrow \frac{x}{2} \times \frac{1}{15} = 1 \\ \Rightarrow x = 30 min . \end{aligned}

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

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