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Three pipes A,B and C attached to a cistern. A can fill it in 10 min, B in 15 min, C is a waste pipe for emptying it. After opening both the pipes A and B, a man leaves the cistern and returns when the cistern should have been just full. Finding, however, that the waste pipe had left open, he closes it and the cistern now gets filled in 2 min. In how much time the pipe C, if opened alone, empty the full cistern?
Answer & Solution
Correct Answer:
Option
C
Let pipe C alone can empty the cistern in x min.
A fills cistern in 1 min = $$\frac{1}{{10}}$$
B fills cistern in 1 min = $$\frac{1}{{15}}$$
A and B together fill in 1 min $$ = \frac{{10 \times 15}}{{10 + 15}} = \frac{{150}}{{25}} = 6\,{\text{min}}$$
Since, waste pipe was left open for 6 min then,
6 min, $$\frac{6}{x}$$ part of cistern will be emptied by C
Now,
$$\frac{6}{x}$$ part of the cistern would be filled by A and B in 2 min.
Hence,
cistern will be filled in $$\frac{3}{x}$$ min. And
$$\frac{x}{3}$$ = 6
x = 18 min.
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LoginLast 2 mins work done by Pipe A& B=1st 6 mins work done by Pipe C
so, 2/10+2/15=6/C... C=18 mins (Ans.)
i.e. in last 2 min. are extra caused due to opened C so water filled by A&B in last 2 min is equal to the water emptied by C in whole 6 min.(would have taken to Full the tank)
As A&B filling 10 litres in 2 min so C empties 10 litres in 6 min.
If C empties 10 litres in 6 min then empties whole (30litres) in 6×3=18 min.
Three pipes A, B and C attached to a cistern. A can fill it in 10 min, B in 15 min, C is a waste pipe for emptying it. After opening both the pipes A and B, a man leaves the cistern and returns when the cistern is supposed to be filled. Finding, however, that the waste pipe had left open, he closes it and now the cistern takes 2 minutes more to fill. In how much time the pipe C, if opened alone, empty the full cistern?
Time to fill (mins) 10 15
LCM (Total units to fill) 60
Efficiency = LCM/time (units/min) 60/10 = 6 60/15 = 4
Total efficiency = 10 units/min
Thus, time taken when they work together = LCM/Efficiency = 60/10 = 6 mins.
Or, A and B working together can fill the tank completely in 6 mins.
Thus, in 2 mins, they will fill = 2/6 = 1/3rd of the tank; which was filled when the man plugs off the emptying pipe "C".
i.e. C would have emptied 1/3rd of the tank in 6 mins.
Thus, C can empty the full tank in 6 x 3 = 18mins.