Answer: Option C. -> $${\text{2}}\frac{1}{2}$$ hours
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
$$\eqalign{
& {\text{ = }}\frac{{65}}{{2 \times \left( {6 + 4 + 3} \right)}} \cr
& = \frac{{65}}{{2 \times 13}} \cr
& = \frac{5}{2} \cr
& = 2\frac{1}{2}\,{\text{hours}} \cr} $$

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

Answer: Option C. -> 14
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

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

Answer: Option C. -> 7 hours
$$\eqalign{
& \left( {{\text{A + B}}} \right){\text{'s 1 hour work}} \cr
& {\text{ = }} {\frac{1}{{12}} + \frac{1}{{15}}} = \frac{9}{{60}} = \frac{3}{{20}} \cr
& \left( {{\text{A + C}}} \right){\text{'s 1 hour work}} \cr
& {\text{ = }} {\frac{1}{{12}} + \frac{1}{{20}}} = \frac{8}{{60}} = \frac{2}{{15}} \cr
& {\text{Part filled in 2 hrs}} \cr
& {\text{ = }} {\frac{3}{{20}} + \frac{2}{{15}}} = \frac{{17}}{{60}} \cr
& {\text{Part filled in 6 hrs}} \cr
& {\text{ = }} {3 \times \frac{{17}}{{60}}} = \frac{{17}}{{20}} \cr
& {\text{Remaining part}} \cr
& {\text{ = }} {1 - \frac{{17}}{{20}}} = \frac{3}{{20}} \cr} $$
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

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

Answer: Option C. -> 120 gallons
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.

Answer: Option D. -> P, Q and S are open
$$\eqalign{
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{(Total Capacity)}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{120}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\overline { \downarrow \,\,\,\,\,\,\,\,\,\, \downarrow \,\,\,\,\,\,\,\,\,\,\, \downarrow \,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow } \cr
& {\text{efficiency 30}}\,\,\,\,\,\,15\,\,\,\,\,\,\,\,10\,\,\, - 12 \cr
& {\text{hours}} \to \,\,\mathop 4\limits_{\left( {\text{P}} \right)}^ \downarrow \,\,\,\,\,\,\mathop 8\limits_{\left( {\text{Q}} \right)}^ \downarrow \,\,\,\,\,\,\,\mathop {12}\limits_{\left( {\text{R}} \right)}^ \downarrow \,\,\,\,\,\,\,\,\,\,\,\mathop {10}\limits_{\left( {\text{S}} \right)}^ \downarrow \cr} $$
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

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

Answer: Option D. -> 11 hours
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.
$$\eqalign{
& {\text{= 11}}\frac{1}{{19}}{\text{ hours}} \cr
& \cong 11\,{\text{hours }} \cr} $$