Let the time taken by first pipe to fill the tank = x hrsSo, the time taken by second pipe to fill tthe tank = x - 3 hrsNow, First pipe's 1 hour work = 1xNow, second pipe's 1 hour work = 1x-3Time taken by both the pipes together to fill the tank = 6 hr 40 min = 64060 hr = 623 hr = 203 hrsNow, amount of work done by both the pipes in 1 hr = 320Now, according to question : 1x + 1x - 3 = 320⇒x - 3 + xxx - 3 = 320⇒2x - 3x2 - 3x = 320⇒3x2 - 9x = 40x - 60⇒3x2 - 49x + 60 = 0⇒3x2 - 45x - 4x + 60 = 0⇒3xx - 15 - 4x - 15 = 0⇒3x - 4x - 15 = 0⇒x - 15 = 0 or 3x - 4 = 0⇒x = 15 or x = 43 rejectedSo, time taken by first pipe to fill the tank = 15 hrsNow, time taken by second pipe to fill the tank = 15 -3 = 12 hrs

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Let the time taken by first pipe to fill the tank = x hrs So, the time taken by second pipe to fill tthe tank = (x - 3) hrs Now, First pipe' s 1 hour work = \frac{1}{x} Now, second pipe' s 1 hour work = \frac{1}{x - 3} Time taken by both the pipes together to fill the tank = 6 hr 40 \min = 6 \frac{40}{60} hr = 6 \frac{2}{3} hr = \frac{20}{3} hrs Now, amount of work done by both the pipes in 1 hr = \frac{3}{20} Now, according to question : \frac{1}{x} + \frac{1}{x - 3} = \frac{3}{20} \Rightarrow \frac{x - 3 + x}{x (x - 3)} = \frac{3}{20} \Rightarrow \frac{2 x - 3}{x^{2} - 3 x} = \frac{3}{20} \Rightarrow 3 x^{2} - 9 x = 40 x - 60 \Rightarrow 3 x^{2} - 49 x + 60 = 0 \Rightarrow 3 x^{2} - 45 x - 4 x + 60 = 0 \Rightarrow 3 x (x - 15) - 4 (x - 15) = 0 \Rightarrow (3 x - 4) (x - 15) = 0 \Rightarrow x - 15 = 0 or 3 x - 4 = 0 \Rightarrow x = 15 or x = \frac{4}{3} (rejected) So, time taken by first pipe to fill the tank = 15 hrs Now, time taken by second pipe to fill the tank = 15 - 3 = 12 hrs

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