Pipes And Cistern - Aptitude MCQ Questions and Solutions with Explanations

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

$$\eqalign{ & \left[ {\frac{{{m_1} \times {h_1} \times {t_1}}}{{{w_1}}} = \frac{{{m_2} \times {h_2} \times {t_2}}}{{{w_2}}}} \right] \cr & {9_{taps}} \times {20_{\min }} = {t_{taps}} \times {15_{\min }} \cr & t = 12\,\,{\text{taps}} \cr} $$

A → 4 hours
B → 6 hours

According to question,
⇒ For the first hour tap A is opened and B for second hour
⇒ Work done by both in 2 hours
$$\eqalign{ & \to \left( {3\,{\text{lit/h}} + 2\,{\text{lit/h}}} \right) \times 2 = 10\,{\text{units}} \cr & \,\,\,\,\,\,\,\,\mathop {\,\,|\,\,\, \times 2}\limits_{{\text{4 hours}}}^{{\text{2 hours}}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mathop {\,\,|\,\,\, \times 2}\limits_{{\text{10 litres}}}^{{\text{5 liters}}\,} \cr} $$
⇒ Remaining part
= 12 - 10 = 2 liters
⇒ Again 5th hour A will be opened Tap A will fill the 2 liters water with its efficiency = $$\frac{2}{3}$$ hours
⇒ Therefore tank will be filled in
= $$\left( {4 + \frac{2}{3}} \right)$$  hours
= $${\text{4}}\frac{2}{3}$$ hours

A opened 2 hours early to B
In 2 hours A can do 3 × 2 = 6 unit work
Remaining work = 24 - 6 = 18
A + B can do it in
$$\eqalign{ & = \frac{{18}}{5}{\text{hours}} \cr & = 3\frac{3}{5}{\text{hours}} \cr & {\text{ = 3}}\,{\text{hours 36 minutes}} \cr} $$
∴ Tank will be full in 9 AM + 3 hours 36 minutes = 12.36 PM

A . . . . . (+) 30 minutes
B . . . . . (+) 45 minutes
C . . . . . (-) 36 minutes
⇒ Water filled by (A + B) in 12 min
= 12 × (6 + 4)
= 12 × 10 = 120 liters
⇒ Remaining capacity
= 180 - 120 = 60 liters
⇒ After 12 minutes emptied pipe C is also opened
⇒ Total capacity (A + B - C)
= (6 + 4 - 5) = 5 liters/minutes
⇒ Time taken by (A + B - C) with capacity 5 liters/minutes to fill the remaining part
$$ = \frac{{60\,\,{\text{liters}}}}{{5\,\,{\text{liters/minutes}}}}{\text{ = 12 minutes}}$$
⇒ Therefore, total time in which the tank will be filled up is
= 12 + 12
= 24 minutes

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
$$\eqalign{ & {\text{ = }}\frac{{65}}{{2 \times \left( {6 + 4 + 3} \right)}} \cr & = \frac{{65}}{{2 \times 13}} \cr & = \frac{5}{2} \cr & = 2\frac{1}{2}\,{\text{hours}} \cr} $$

$$\eqalign{ & {\text{1 second }} \to {\text{ 1 drop}} \cr & {\text{Number of second in 300 days}} \cr & \left( {{{24}_{{\text{hrs}}}} \times {{60}_{{\text{mins}}}} \times {{60}_{{\text{sec}}}}} \right) \times 300\,{\text{days}} \cr & {\text{Number of milli - liters wasted}} \cr & {\text{100}} \times \frac{{24 \times 60 \times 60 \times 300}}{{600}} \cr & = 43200 \times 100 \cr & = 4320000\,{\text{ml}} \cr & = \frac{{4320000}}{{1000}} \cr & = 4320\,{\text{liters}} \cr} $$

$$\eqalign{ & A + B + C = \frac{{60}}{{10}} \cr & 5 + 4 + C = 6 \cr & C = - 3 \cr & C{\text{ can empty the tank }}x \cr & = \frac{{60}}{3} = 20{\text{ minute}} \cr} $$

$$\eqalign{ & {\text{I fill the tank}} = 5\frac{1}{2} = \frac{{11}}{2}{\text{ in hour}} \cr & {\text{I - II fill the tank}} = 8\frac{1}{4} = \frac{{33}}{4}{\text{ in hour}} \cr & {\text{LCM }}\left( {11,\,33} \right) = 33 \cr} $$

$$\eqalign{ & {\text{Total work }}30\% = 33 \times \frac{{30}}{{100}} = \frac{{99}}{{10}} \cr & {\text{II}} = {\text{leak pipe efficiency}} = 6 - 4 = 2 \cr} $$
Time taken by pipe to empty the tank = $$\frac{{99}}{{\frac{{10}}{2}}}$$ = $$\frac{{99}}{{20}}$$ in hour

If pipe C continues for 2 more minutes, then = 2 × (-3) = -6 litre less fill
Then, 96 - 6 = 90 litre filled
A + B - C, 1 minute = 8 + 4 - 3 = 9
Time of A + B - C = $$\frac{{90}}{9}$$ = 10 minutes