Question

Two water taps together can fill a tank in 9 3/8 hours. The tap of larger diameter takes 10 hours leas than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Answer 1

let the smaller tap fill the tank in X hours.


Then , the larger tap fill it in ( X - 10 ) hrs.



Time taken by both together to fill the tank = 75/8 hrs.



Part filled by the smaller tap in 1 hour = 1/X.


Part filled by the larger tap in 1 hour = 1/(X-10).


Part filled by both the taps in 1 hr = 8/75 .



Therefore,



1/X + 1/(X-10) = 8/7

=> ( X-10) + X / X(X-10) = 8/75



=> (2X-10)/X(X-10) = 8/75



=> 75 ( 2X - 10) = 8X ( X - 10)


=> 150X - 750 = 8x² - 80X



=> 8X² - 230X + 750 = 0



=> 4X² - 115X + 375 = 0



=> 4X² - 100X - 15X + 375 = 0


=> 4X ( X - 25) - 15 ( X - 25 ) = 0



=> ( X - 25 ) ( 4X - 15 ) = 0



=> X = 25 or X = 15/4





=> X = 25 [ Since X = 15/4 => ( X-10)<0 ].



Hence,


Time taken by the smaller tap to fill the tank = 25 hours.


And,


The time taken by the larger tap to fill the tank = (25-10) = 15 hrs.