% Cistern is filled by 1^st pipe in one minute = ¹⁰⁰/₂₀ = 5%;

% Cistern is filled by 2^nd pipe in one minute = ¹⁰⁰/₃₀ = 3.33%;

% cistern filled by 1^st and 2^nd pipes in one minute = 8.33%;

According to question,

Cistern is totally filled by 2^nd pipe in last 10 minute.

That means 2^nd pipe filled 33.3% of the cistern in last 10 minute and 66.66% of cistern is filled by 1^st and 2^nd pipe together in = ^66.66/_8.33 = 8 minutes;

Thus, after 8 minute, 1^st pipe must be turned off.

1^st Method:

\begin{align} & \left( {B + C} \right)'s\, 2\, {\text{days}}\, {\text{work}} \cr & = 2 \times \left( {\frac{1}{{20}} + \frac{1}{{30}}} \right) \cr & = 2 \times \left( {\frac{{3 + 2}}{{60}}} \right) \cr & = \frac{1}{6}{\text{part}} \cr & {\text{Remaining}}\, {\text{work}} \cr & = 1 - \frac{1}{6} \cr & = \frac{5}{6}part \cr & {\text{A's}}\, {\text{one}}\, {\text{day's}}\, {\text{work}} \cr & = \frac{1}{{18}}{\text{part}} \cr & {\text{Time taken to complete the work}} \cr & = \frac{{\left( {\frac{5}{6}} \right)}}{{\left( {\frac{1}{{18}}} \right)}}\, {\text{days}} \cr & {\text{Hence, }} \cr & {\text{Time taken to complete the work}} \cr & = \left( {\frac{5}{6}} \right) \times 18 \cr & = 15\, {\text{days}} \cr\end{align} 2^nd Method:

% of work B completes in one day = ¹⁰⁰/₂₀ = 5%;

% of work C completes in one day = ¹⁰⁰/₃₀ = 3.33%;

% of work (A+B) completes together in one day = 5+3.33 = 8.66%;

% work (A+B) completes together in 2 days = 8.66*2 = 17.32%;

Remaining work = 100-17.32 = 82.68%;

% of work A completes in 1 day = ¹⁰⁰/₁₈ = 5.55%

Time taken to complete the remaining work by A

= ^82.68/_5.55 = 15 days.

Pipe A can fill empty tank in 30 min.

Pipe A can fill the tank = ¹⁰⁰/₃₆ = 2.77% per minute.

Pipe B can fill empty tank in 45 min.

Pipe B can fill the tank = ¹⁰⁰/₄₅ = 2.22% per min.

A and B can together fill the tank,

= (2.77 + 2.22)= 5% per minute.

So, A and B can fill the tank in 7 min.,

= 7*5 = 35% of the tank.

Rest tank to be filled = 100 - 35 = 65%.

C can empty the full tank in 30 min.

C can empty the tank = ¹⁰⁰/₃₀ = 3.33% per min.

C is doing negative work i.e. emptying the tank.

A, B and C can together fill the tank,

= 2.77% + 2.22% - 3.33% = 1.67% tank per minute.

So, A, B and C will take time to fill 65% empty tank,

= ⁶⁵/_1.67 = 39 min. (Approx.)

\begin{align} & \left( {{\text{A + B + C}}} \right){\text{'s}}\, {\text{1}}\, {\text{day's}}\, {\text{work}} = \frac{1}{4} \cr & {\text{A's}}\, {\text{1}}\, {\text{day's}}\, {\text{work}} = \frac{1}{{16}} \cr & {\text{B's}}\, {\text{1}}\, {\text{day's}}\, {\text{work}} = \frac{1}{{12}} \cr & \therefore {\text{C's}}\, {\text{1}}\, {\text{day's}}\, {\text{work}} \cr & = \frac{1}{4} - \left( {\frac{1}{{16}} + \frac{1}{{12}}} \right) = \left( {\frac{1}{4} - \frac{7}{{48}}} \right) = \frac{5}{{48}} \cr & {\text{So, }}\, {\text{C}}\, \, {\text{alone}}\, {\text{can}}\, {\text{do}}\, {\text{the}}\, {\text{work}}\, {\text{in}} \cr & \frac{{48}}{5} = 9\frac{3}{5}\, {\text{days}} \cr\end{align}

Vimal's 1 day work = ¹/₂₀.

Since, Vimal and Kamal can together complete in 12 days.

i.e. (Vimal + Kamal)'s 1 day work = ¹/₁₂

Then,

Kamal's 1 day work,

= ¹/₁₂ - ¹/₂₀ = ²/₆₀ = ¹/₃₀

If Kamal Works only for half a day daily, then his 1 day work becomes (¹/₂)(¹/₃₀) = ¹/₆₀

Therefore, 1 day work of both Vimal and Kamal,

= ¹/₂₀ + ¹/₆₀ = ⁴/₆₀ = ¹/₁₅

Hence, the work will be completed in 15 days.