<PRE><B>Question 396577</B>
Working together, two pipes can fill the tank in 2 hours and 6 minutes. Working alone, the larger pipe fills the tank in 4 hours less time than the smaller on. How long does the larger pipe take?

Let 1/x = hourly work rate of smaller pipe
1/(x-4) = hourly work rate of larger pipe
2 hrs and 6 min = 2.1 hours
1/2.1 = hourly work rate when both pipes working
sum of the individual work rates = work rate when working together

1/x + 1/(x-4) = 1/(2.1)
LCD =x(x-4)(2.1)
(x-4)(2.1)+x(2.1)=x(x-4)
2.1x-8.4+2.1x=x^2-4x
x^2-8.2x+8.4=0
solve with the following quadratic equation with a=1, b=-8.2, c=8.4

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 

x=(-(-8.2)+ - sqrt((8.2)^2-4*1*8.4)/2
 = (8.2 +-sqrt(67.24-33.6)/2
 = (8.2 +-sqrt(33.64)/2
 x=(8.2 +-5.8)/2
 x=14/2 or 2.4/2
 x=7 or x=1.2(reject because this would give a negative rate for the larger pipe.

ans: working alone the larger pipe would take 7-4 = 3 hours
     working alone the smaller pipe would take 7 hours</PRE>