Answer: Option C. -> 6 hours
Let the cistern be filled by pipe A alone in x hours.
Then, pipe B will fill it in (x + 6) hours
$$\eqalign{
& \therefore \frac{1}{x} + \frac{1}{{ {x + 6} }} = \frac{1}{4} \cr
& \Rightarrow \frac{{x + 6 + x}}{{x\left( {x + 6} \right)}} = \frac{1}{4} \cr
& \Rightarrow {x^2} - 2x - 24 = 0 \cr
& \Rightarrow \left( {x - 6} \right)\left( {x + 4} \right) = 0 \cr
& \Rightarrow x = 6\,{\kern 1pt} {\kern 1pt} \left[ {{\text{neglecting the negative value of }}x} \right] \cr} $$

Answer: Option C. -> 40
We have the following table:
Period
Supply
Demand
Calculation
Excess Qty. in tank
0 - 3 hrs
30000
10000
30000 - 10000
20000
3 - 6 hrs
30000
10000
20000 + 30000 - 10000
40000
6 - 9 hrs
30000
45000
40000 + 30000 - 45000
25000
9 - 12 hrs
30000
25000
25000 + 30000 - 25000
30000
12 - 15 hrs
30000
40000
30000 + 30000 - 40000
20000
15 - 18 hrs
30000
15000
20000 + 30000 - 15000
35000
18 - 21 hrs
30000
60000
35000 + 30000 - 60000
5000
21 - 24 hrs
30000
35000
5000 + 30000 - 35000
0
The excess quantity in tank at any time does not exceed 40000 litres,
Which is the required minimum capacity to avoid wastage.

Answer: Option D. -> 12 hours
Part of the cistern filled in 1 hour
$$ = \frac{1}{3} + \frac{1}{4} - \frac{1}{2}$$
(Cistern filled by 1st pipe + Cistern filled by 2nd pipe – Cistern emptied by 3rd pipe )
$$\eqalign{
& \frac{{4 + 3 - 6}}{{12}} \cr
& = \frac{1}{{12}} \cr} $$
Hence, the cistern will be filled in 12 hours.

Answer: Option D. -> $$\frac{{xy}}{{y - x}}$$ hours
Net part filled in 1 hour
$$\eqalign{
& {\text{= }}\left( {\frac{1}{x} - \frac{1}{y}} \right) \cr
& = \left( {\frac{{y - x}}{{xy}}} \right) \cr} $$
∴ The tank will be filled in
 $$ = \left( {\frac{{xy}}{{y - x}}} \right)\,\,{\text{hours}}$$

Answer: Option C. -> 40 liters
If tank has 4x liters of total capacity and its holds 3x liters of water and if 30 liters of water is taken out, then tank becomes empty.
It means 3x liters of water is taken out
3x = 30 liters
x = 10 liters
Capacity of tank
= 4x = 4 × 10 = 40 liters

Answer: Option B. -> 21 hours
Let the total capacity of tank is 21 units
A's efficiency is 7 units/hr
A's efficiency after leakage 6 units/hr
leakage efficiency = 7 - 6 = 1 unit/hr
leakage will empty the filled tank
$$\frac{{{\text{Total Capacity}}}}{{{\text{Efficiency }}}} = \frac{{21}}{1}$$
= 21hours

Answer: Option C. -> 16 hours
Part filled by (A + B) in 1 hour
$$\eqalign{
& {\text{= }}\left( {\frac{1}{{20}} + \frac{1}{{30}}} \right) \cr
& = \frac{1}{{12}} \cr} $$
So, A and B together can fill the tank in 12 hrs,
$$\frac{1}{3}$$ part is filled by (A + B) in
$$\left( {\frac{1}{3} \times 12} \right){\text{ = 4 hrs}}$$
Since the leak empties one - third water, so time taken to fill the tank
= Time taken by (A + B) to fill the whole tank + Time taken by (A + B) to fill one - third tank
= (12 + 4)
= 16 hours

Answer: Option C. -> 2 hours
The outlet pipe empties the one complete cistern in 3 hours
Time taken to empty $$\frac{2}{3}$$ Part of the cistern
$$\eqalign{
& {\text{= }}\frac{2}{3} \times 3 \cr
& = 2\,{\text{hours}} \cr} $$

Answer: Option D. -> 240 gallons
Work done by the waste pipe in 1 minute
$$\eqalign{
& {\text{ = }}\frac{1}{{30}} - \left( {\frac{1}{{40}} + \frac{1}{{48}}} \right) \cr
& = \left( {\frac{1}{{30}} - \frac{{11}}{{220}}} \right) \cr
& = - \frac{1}{{80}}\left[ { - \,{\text{Nagetive sign means emptying}}} \right] \cr
& \therefore {\text{Volume of }}\frac{1}{{80}}{\text{ part = 3 galons}} \cr
& {\text{Volume of whole tank}} \cr
& {\text{ = }}\left( {3 \times 80} \right){\text{gallons}} \cr
& {\text{ = 240 gallons}}{\text{}} \cr} $$

Answer: Option A. -> 10 km/hours
Rate of flow of water = x cm/minute
∴ Volume of water that flowed in the in 1 minutes
= (5 × 4 × x) = 20 x cu.cm.
∴ Volume of water that flowed in the tank in 6 hours 18 minutes.
i.e. (6 × 60 + 18) = 378 minutes
= 2x × 378 cu. cm.
According to question,
$$\eqalign{
& {\text{20}}x \times 378 = 700 \times 400 \times 450 \cr
& \Rightarrow x = \left( {\frac{{700 \times 400 \times 450}}{{20 \times 378}}} \right){\text{cm /minutes}} \cr
& \Rightarrow x = \left( {\frac{{700 \times 400 \times 450 \times 60}}{{100000 \times 20 \times 378}}} \right){\text{km/hours}} \cr
& \Rightarrow x{\text{ = 10 km/hours}} \cr} $$