Question

2. (2) One min's work of first ( operatorname{tap}=frac{1}{60} ) and one min's work of second tap ( =-frac{1}{30} ) One min's work of both taps ( =frac{1}{60}-frac{1}{30} )
[
=frac{1-2}{60}=frac{-1}{60}
]
Hence, It is clear that tank will be empited in ( 60 mathrm{min} )

# 2. A tap can fill a tank in one hour. A second tap can empty it in 30 min. If both the taps operate simulataneously, how much time is needed to empty the tank? (1) 20 min S (2) 60 min (3) 40 min 16 (4) 45 min

Solution