Answer: Option A. -> $8/3$ hrs
Answer: (a)Part of the tank filled in 4 hours by pipe A = $4/6 = 2/3$Remaining part = ${1 - 2}/3 = 1/3$Time taken by pipe B in filling $1/3$ part = $8/3$ hoursUsing Rule 8,Here, x = 6, y = 8, t = 4Required time = $[y(1 –{t/x})]$ hours= $[8(1 - 4/6)]$ hours = $8/3$ hours

Answer: Option A. -> 8$1/4$ minutes
Answer: (a)Using Rule 1,Two taps 'A' and 'B' can fill a tank in 'x' hours and 'y' hours respectively. If both the taps are opened together, then how much time it will take to fill the tank?Required time = $({xy}/{x + y})$ hrs

Answer: Option D. -> $1/r = 1/p - 1/q$
Answer: (d)Since, P < q,On opening pipe and sink together,Part of the tub filled in 1 hour= $1/P - 1/q$Clearly, $1/P - 1/q = 1/r$

Answer: Option A. -> 90 minutes
Answer: (a)Let the pipe B fill the tank in x minutes.Part of the tank filled by pipes A and B in 1 minute = $1/36$Part of the tank filled by pipe A in 1 minute= $1/36 - 1/x$According to the question,30 × $1/x + 40(1/36 - 1/x) = 1$$30/x + 10/9 - 40/x$ = 1$40/x - 30/x = 10/9 - 1$$10/x = 1/9 ⇒ x = 90$ minutes

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

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

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

Answer: Option D. -> 14 hours
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

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

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