A tap can empty a tank in 1 h. A second tap can empty in 30 min. If both the taps operate simultaneously, how much time is needed to empty the tank? |
A) 20 min
B) 30 min
C) 40 min
D) 45 min
Correct Answer: A
Solution :
1 h = 60 min |
Rate of empty 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. |
\[\therefore \]Time taken by the two taps together to empty the tank = 20 min |
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