Question

Two taps can fill a tank in 58/7 hours. The tap of larger diameter takes 4 hours less than the smaller one to fill the tank separately. Find the time in which each top can separately fill the tank

Answer 1

Let the tap of the larger diameter fills the tank alone in (x – 10) hours.   In 1 hour, the tap of the  smaller diameter can fill 1/x part of the tank.   In 1 hour, the tap of the  larger diameter can fill 1/(x – 10) part of the tank.   Two water taps together can fii a tank in 75 / 8 hours.   But  in 1 hour the taps fill 8/75 part of the tank.   1 / x  +  1 / (x – 10) = 8 / 75.   ( x – 10 + x ) / x ( x – 10) =  8 / 75.   2( x – 5) / ( x2 – 10 x) = 8 / 75.   4x2 – 40x = 75x – 375.   4x2 – 115x + 375 = 0   4x2 – 100x – 15x + 375 = 0   4x ( x – 25) – 15( x – 25) = 0   ( 4x -15)( x – 25) = 0.   x = 25, 15/ 4.   But x = 15 / 4 then x – 10 = -25 /4 which is not possible since time   But x = 25 then x – 10 = 15.Larger diameter of the tap can the tank 15 hours and smaller diameter of the tank can fill the tank in 25 hours.  

Answer 2

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