Math, asked by debanti, 1 year ago




A tank is fitted with 2 taps , first tap can completly fill the tank
in 45 minutes and second tank can empty the full tank in 1 hour. If both the taps are opened alternatively for one minute, then in how many hours empty tank will be filled completely ?









Answers

Answered by Anonymous
10
A tank is fitted with 2 taps , first tap can completely fill the tank
in 45 minutes and second tank can empty the full tank in 1 hour. If both the taps are opened alternatively for one minute, then in how many hours empty tank will be filled completely ?
solution :
         Let the capacity of the tank = X
         Time taken by tap first to fill the tank 45 minutes .
 So tape first takes 45 minutes to fill tank of X litres ,
   Volume of water filled in 1 minute = x / 45 L.
Tape Second  takes 60 minutes to empty the tank of X litres .
       Volume emptied in one Minute = x / 60

Therefore the volume of water filled in  opening tap one for a minute then closing it (tap first )and opening of tap  second for a minute  = x/ 45 - x/ 60

            Let the total time taken to fill the tank by alternatively opening and closing of tape first and second tap = t minutes  ,

therefore ,
       volume filled in t minutes = capacity of tank
         t (x /45 - x/60) = x
         t (60x - 45x ) / 60*45 = x
         t (15 x ) = 2700 x
           t = 2700 /15
         t= 180 minutes .
         t = 3 hours
Hope this helps you.
        
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