a tap can empty a tank in an hour and half and another tap can empty the same tank in half an hour. if both the taps operate simultaneously. then how much time(in minutes) is requires to empty the rank.
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0
Answer:
The answer is 1hour 30min
Answered by
1
Answer:
20 mins is the answer
Step-by-step explanation:1 hour = 60 minutes
Rate of emptying the tank by the two taps are \frac{1}{60} and \frac{1}{30} of the tank per minute respectively.
Rate of emptying the tank when both operate simultaneously =
= \frac{1}{60} + \frac{1}{30} = \frac{1+2}{60} = \frac{3}{60} = \frac{1}{20}
of the tank per minute.
∴ Time taken by the two taps together to empty the tank = 20 minutes.
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