Two taps running together can fill a tank in `3(1/13)` hours. If one tap takes 3 hours more than the other to fill the tank, then how much time will each tap take to fill the tank ?

Explanation

`{:(,"Tap I","Tap II"),("Time (in has)",x,x+3):}`
Let the time taken by Tap I to fill a tank =x hrs
`because` The time taken by Tap II to fill a tank =(x+3) hrs
and time taken by both to fill a tank `=3(1)/(13)=(40)/(13)` hrs
`because` Tap I's 1 hrs work `=(1)/(x)`
Tap II's 1 hr work `=(1)/(x+3)`
and (Tap I +Tap II)'s 1 hr work `=(13)/(40)`
According to the problem,
`(1)/(x)+(1)/(x+3)=(13)/(40)`
`implies(x+3+x)/(x(x+3))=(13)/(40)`
`implies40(2x+3)=13x(x+3)`
`implies80x+120=13x^(2)+39x`
`implies13x^(2)-41x-120=0`
`implies13x^(2)=65x+24x-120=0`
`implies13x(5-x)+24(x-5)=0`
`(x-5)(13x+24)=0` `becuase` Either `x-5=0 or 13x+24=0`
`impliesx=5orx=(-24)/(13)`
But time cannot be negative, so we reject `x=(-24)/(13)`
`becuase` x=5
`{:("Hence, time taken by Tap I=5hrs"), ("and time taken by Tap II"=(5+3)hrs=8hrs.):}}`