Question 45662
Let the two pipes have fill rates of R1 and R2 m^3/min
Let the volume of the tank be V m^3
The time for pipe1 to fill the tank is,
T1 = V/R1 min
similiarly,
T2 = V/R2
The difference in times is 10 min, so we can write,
T1 - T2 = 10
(at the moment it doesn't matter whether you say T1-T2 or T2-T1, since we haven't said which of R1 and R2 is the greater. It will all work out in the end)
substituting for T1 and T2,
V/R1 - V/R2 = 10
V(1/R1 - 1/R2) = 10
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If pipe1 fills the tank at R1 m^3/min and pipe2 fills the tank at R2 m^3/min, then together they both fill the tank at a rate of (R1+R2) m^3/min.
T3 = V/(R1+R2)
12 = V/(R1 + R2)
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You now have two equations in R1 and R2 from which you can solve for R1 and R2 separately, in terms of V.
You should end up with T1 = 30 mins, T2 = 20 mins.