Pipes And Cistern — Part 2 MCQs with Explanations

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

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

Let A and B are pipe and leakage B = 1 efficiently B = $$\frac{4}{1}$$ = 4 hrs

Pipe A till 9 O'clock = 3 × 2 = 6 Pipe B till 9 O'clock = 2 × 1 = 2 Remaining work = 60 - 6 - 2 = 52 Time = $$\frac{{52}}{6}$$ = $$8\frac{4}{6}$$ = 8 hrs 40 min = 9:00 + 8 hrs 40 min = 05:40 pm

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 $$ = \frac{{9 \times 12}}{6} = 18\min $$

$$\eqalign{ & A = 12\frac{1}{2} = \frac{{25}}{2} \cr & B = 25 \cr} $$ $$\eqalign{ & A + B = \frac{{25}}{{2 + 1}} = \frac{{25}}{3} \cr & = 8\,{\text{hr }}20\,\min \cr & A + B - C \cr & = 8\,{\text{hr }}20\,\min + 1\,{\text{hr 4}}0\,\min \cr & = 10{\text{ hr}} \cr} $$ C = 8 + 4 - 10 = 2 70% of tank capacity $$ = 100 \times \frac{{70}}{{100}} = 70$$ Time taken in leakage $$ = \frac{{70}}{2} = 35\,{\text{hrs}}$$

\[\begin{array}{*{20}{c}} {\text{B}}&{}&{\text{A}} \\ {x + 5}&:&x \end{array}\] Now go through option, Let A = 10 hours B = 15 hours A + B = $$\frac{{30}}{5}$$ = 6 hours

$$\eqalign{ & t = \frac{{7200}}{{60}}\min \cr & = 1200\min \cr & = \frac{{1200}}{{60}}{\text{ hours}} \cr & = 20{\text{ hours}} \cr} $$

6 × 10 × 3 = 9 × x × 1 x = 20 Answer

Total work in a hours by (A + B + C) = 4 + 3 - 6 = 1 A + B + C = $$\frac{{24}}{1}$$ = 24 hours

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

A → 6 am → 10 am = 4 × 8 = 32 B → 8 am → 10 am = 2 × 4 = 8 120 - 40 = 80 A + B + C = $$\frac{{80}}{{15}}$$ = 5 hr 20 min 10 am + 5 hr 20 min 3 : 20 pm

A + B + C = 14 4 + B - 5 = 14 B = 15 unit B = $$\frac{{60}}{{15}}$$ = 4 hours

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 $$ = \frac{{50}}{{15 + 10}} = 2$$

C's work done in 2 hour = 4 × 2 = 8 B's work done in 4 hour = 5 × 4 = 20 A's share of work = 100 - 28 = 72

Till 10 am 3A + 2B + (A + B + C)t = 120 3 × 4 + 2 × 3 + (4 + 3 + 2)t = 120 9t = 120 - 18 t = $$\frac{{102}}{9}$$ t = $$11\frac{1}{3}$$ t = 11 hr 20 min 10 am + 11:20 = 21:20 = 9:20 pm

At 11 am, A is closed, now (B + C) will work together B + C = 5 - 8 = -3 The time was taken to fill 4 unit $$t = \frac{4}{3} = 1\frac{1}{3} = 1\,{\text{hr}}\,20\,\min $$

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.