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.