Answer: Option C. -> $$\frac{{\text{T}}}{{{\text{x}} - {\text{y}}}}$$  minutes
Net volume filled in 1 minute
= (x - y) liters
∴ The tank will be filled in
= $$\frac{{\text{T}}}{{\left( {x - y} \right)}}$$  minutes

Answer: Option D. -> 7200 litters
Let the total capacity of the tank is 30 units.
The efficiency of Leakage(Pipe A) will be $$\frac{30}{10}$$  = 3
And the efficiency of the leakage (Pipe A) and another Pipe (B) which is filling the tank will be $$\frac{30}{15}$$  = 2
Pipe A is emptying at 3 units/hr and when filling pipe B started then the emptying rate will come down to 2 units/hr.
∴ Filling Pipe B efficiency is 3 - 2 = 1unit/hr
Pipe B will be fill the tank in $$\frac{30}{1}$$  = 30 hrs
Filling rate of Pipe B per minute is 4 litter
∴ Total Capacity of tank will be = (4 × 60) × 30 = 7200 litters

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

Answer: Option C. -> 15 hours
Suppose first pipe alone takes x hours to fill the tank.
Then second and third pipes will takes (x - 5) and (x - 9) hours respectively to fill the tank.
$$\eqalign{
& \therefore \frac{1}{x} + \frac{1}{{\left( {x - 5} \right)}} = \frac{1}{{\left( {x - 9} \right)}} \cr
& \Rightarrow \frac{{x - 5 + x}}{{x\left( {x - 5} \right)}} = \frac{1}{{\left( {x - 9} \right)}} \cr
& \Rightarrow \left( {2x - 5} \right)\left( {x - 9} \right) = x\left( {x - 5} \right) \cr
& \Rightarrow {x^2} - 18x + 45 = 0 \cr
& \Rightarrow \left( {x - 15} \right)\left( {x - 3} \right) = 0 \cr
& \Rightarrow x = 15\left[ {{\text{neglecting }}x\,{\text{ = 3}}} \right] \cr} $$
So, first pipe alone takes 15 hrs to fill the tank.

Answer: Option D. -> 14 hours
Work done by the leak in 1 hour
$$\eqalign{
& {\text{ = }}\left( {\frac{1}{2} - \frac{3}{7}} \right) = \frac{1}{{14}} \cr} $$
∴ Leak will empty the tank in 14 hours

Answer: Option D. -> 18
Capacity of the tank
= (12 × 13.5) litres
= 162 litres
Capacity of each bucket = 9 litres
Number of buckets needed
$$\eqalign{
& {\text{= }}\left( {\frac{{162}}{9}} \right) \cr
& = 18 \cr} $$

Answer: Option C. -> 24 hours
Net part filled in 1 hour
$$\eqalign{
& {\text{ = }}\left( {\frac{1}{8} - \frac{1}{{12}}} \right) = \frac{1}{{24}} \cr} $$
∴ The tank will be filled in 24 hours

Answer: Option D. -> $$\left( {\frac{{xy}}{{y - x}}} \right){\text{hours}}$$
Time will be taken by both of them to fill the tank
$${\text{ = }}\frac{{xy}}{{y - x}}$$

Answer: Option D. -> 45 minutes
(A + B)'s filling (2 + 1) = 3 units/min
In 5 minutes they will fill 3 × 5 = 15 units
Remaining capacity = 60 - 15 = 45 units
Second pipe (B) fills it in
$$\eqalign{
& = \frac{{{\text{Remaining capacity}}}}{{{\text{efficiency of B}}}} \cr
& = \frac{{45}}{1} \cr
& = \,45\,{\text{minutes}} \cr} $$

Answer: Option C. -> 11.45 A.M.
Pipe A will fill 3 units till 11 A.M. Remaining capacity
= 6 - 3
= 3 units
Now both pipes will fill the tank in
$$\frac{{{\text{Total Capacity}}}}{{{\text{Efficiency }}}} = \frac{3}{{\left( {3 + 1} \right)}} = \frac{3}{4}{\text{ hours}}$$
So, $$\left( {11 + \frac{3}{4}} \right)$$  A.M., tank will be filled = 11.45 A.M.