One pipe can fill a cistern in 3 hours less than the other. The two pipes together can fill the cistern in 6 hours 40 minutes. Find the time that each pipe will take to fill be cistern.

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Solution

Let one pipe be A and other pipe be B .

Let the time taken by the pipe B to fill tank = x min

Given that the pipe A takes 3 hours (i.e 180 min) lessthan B = x - 180 min

Then time taken by A = x - 180 min

Also time taken by both A and B = 6 hr 40 min = 400 min

∴ Portion of tank filled in 1 min by pipe A = $fraction numerator 1 over denominator x minus 180 end fraction$

Portion of tank filled in 1 min by pipe B = $\frac{1}{x}$

∴ Portion of tank filled in 1 min by both pipes A and B = $\frac{1}{400}$

$1 over x plus fraction numerator 1 over denominator x minus 180 end fraction equals 1 over 400 fraction numerator x minus 180 plus x over denominator left parenthesis x minus 180 right parenthesis x end fraction equals 1 over 400$

$x^2 - 980 x -72000 = 0 $

$x^{2} - 900 x - 80 x - 72000 = 0$

$x (x - 900) - 80 (x - 900) = 0$

$(x - 80) (x - 900) = 0$

$x = 80, 900 m i n$

when x = 80 then time taken by A = x - 180 min = -100 it is not possible .

when x = 900 min then time taken by A = x - 180 min = 720 min = 12 hours

time taken by the pipe B to fill tank = x min = 900 min = 15 hours.

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