A cistern is provided with two pipes A and B. A can fill a cistern in 20 minutes and B can empty it in 24 minutes. If A and B pipes are opened on alternate minutes, find in how many minutes, the cistern shall be full?
Solution
Correct Answer: Option B
According to the given information,
A can fill the cistern in 20 minutes
⇒ in one min the part of cistern filled by A = 1/20
Similarly, B can empty the cistern in 24 minutes
⇒ in one min the part of cistern emptied by B = 1/24
Now, the Part of the cistern filled in a span of 2 minutes \(= \;\frac{1}{{20}} - \frac{1}{{24}} = \frac{{6 - 5}}{{120}}\; = \;\frac{1}{{120}}\)
∴ Part of the cistern filled in (114 × 2) = 228 minutes \(= \;\frac{{114}}{{120}} = \frac{{19}}{{20}}\)
Part of cistern remaining to be filled = 1 – (19/20) = 1/20
After 228 minutes, it’s pipe A’s turn to fill the cistern and it has to fill 1/20th part of the cistern.
As stated earlier, pipe A will take 1 minute to fill the remaining part.
∴ total time = 228 + 1 = 229 minutes = 3 hours 49 minutes
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