Answer: Option B. -> $$\frac{1}{r} = \frac{1}{p} - \frac{1}{q}$$
Net efficiency = q - p (∵ q > p)
Time required
$$\eqalign{
& {\text{r}} = \frac{{pq}}{{q - p}} \cr
& or\,\frac{1}{r} = \frac{1}{p} - \frac{1}{q} \cr} $$

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

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

Answer: Option B. -> $${\text{4}}\frac{2}{3}$$ hours
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

Answer: Option B. -> 24 minutes
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

Answer: Option D. -> 12:36 PM
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

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

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

Answer: Option A. -> 12 minutes
Part filled by A in 1 minute = $$\frac{1}{{20}}$$
Part filled by B in 1 minute = $$\frac{1}{{30}}$$
Part filled by (A + B) in 1 minute
$$\eqalign{
& = {\frac{1}{{20}} + \frac{1}{{30}}} = \frac{1}{{12}} \cr} $$
∴ Both pipes can fill the tank in 12 minutes

Answer: Option C. -> 35 hours
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.
$$\eqalign{
& \therefore \frac{1}{x} + \frac{2}{x} + \frac{4}{x} = \frac{1}{5} \cr
& \Rightarrow \frac{7}{x} = \frac{1}{5} \cr
& \Rightarrow x = 35\,{\text{hours}} \cr} $$