Pipes And Cistern - Aptitude MCQ Questions and Solutions with Explanations

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

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

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

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,
$$\eqalign{ & {\text{20}}x \times 378 = 700 \times 400 \times 450 \cr & \Rightarrow x = \left( {\frac{{700 \times 400 \times 450}}{{20 \times 378}}} \right){\text{cm /minutes}} \cr & \Rightarrow x = \left( {\frac{{700 \times 400 \times 450 \times 60}}{{100000 \times 20 \times 378}}} \right){\text{km/hours}} \cr & \Rightarrow x{\text{ = 10 km/hours}} \cr} $$

First pipe fill the tank in 1 hour = $$\frac{1}{{12}}$$ part of tank
Second pipe fill the tank in 1 hour = $$\frac{1}{{16}}$$ part of tank
Third pipe empty the tank in 1 hour = $$\frac{1}{{30}}$$ part of tank
When all three pipes are opened simultaneously, part of the tank filled in 1 hour
$$ = \frac{1}{{12}} + \frac{1}{{16}} - \frac{1}{{30}}$$
LCM of 12, 16 and 30 = 240
$$\eqalign{ & {\text{ = }}\frac{{20 + 15 - 8}}{{240}} \cr & = \frac{{27}}{{240}} \cr} $$
∴ Required time taken by all the three pipes
$${\text{ = }}\frac{{240}}{{27}} = \frac{{80}}{9} = 8\frac{8}{9}\,{\text{Hours}}$$

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

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.

(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
$$\eqalign{ & = \frac{{{\text{Remaining capacity}}}}{{{\text{efficiency of B}}}} \cr & = \frac{{45}}{1} \cr & = \,45\,{\text{minutes}} \cr} $$

Time will be taken by both of them to fill the tank
$${\text{ = }}\frac{{xy}}{{y - x}}$$

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