Two taps running together can fill …

let the time taken by I tap to fill the cistern completely = x minutes.

thus, let the time taken by II tap to fill the cistern = (x + 5) minutes.

work done by tap I (in 1 min) = 1/x

work done by tap II (in 1 min) = 1/(x+5)

work don by tap I +work done by tap II in 1 minute = cistern filled in one minute = 1/6

thus the equation is 1/x + 1/(x+5) = 1/6.

6(2x+5) = x² + 5x

12x + 30= x² + 5x

x² -7x -30= 0.

x² +3x -10x -30 = 0.

x(x+3) -10(x+3) = 0.

time cannot be negative.

thus , x= 10.

thus, time taken by tap I = 10 minutes.

time taken by tap II = x+5 = 15 minutes.

Let the first tap take x minutes to fill the cistern, then second tap will take (x+5) minutes to fill it.

The first tap will fill the 1/x of the cistern in 1 minute and second tap will fill 1/(x+5).

Together, they will fill 1/6.

So, equation becomes

{tex}{1\over x}{/tex}+ {tex}{1\over x+5}{/tex}= {tex}{1\over 6}{/tex}

x²+5x= 12x+30

x²-7x-30 =0

x²-10x+3x-30=0

x(x-10)+3(x-10)=0

x= 10 

Therefore, first tap will fill in 10 minutes alone and the second tap will take 15 minutes to fill the cistern.