Three pipes fill a tank with uniform flow. The first two together take the same time as the third alone. The second pipe is 5 hours faster than the first and 4 hours slower than the third. How long does the first pipe take?

Problem restatementLet the individual times be t₁, t₂, t₃ for the first, second, third pipes. Given relational constraints and that the rate of the third equals the combined rate of first and second.

Given data

• t₂ = t₁ − 5
• t₂ = t₃ + 4 ⇒ t₃ = t₁ − 9
• 1/t₁ + 1/t₂ = 1/t₃

Concept/ApproachTranslate the time relations into rates, substitute t₂, t₃ in terms of t₁, and solve the resulting rational equation.

Step-by-step calculation1/t₁ + 1/(t₁ − 5) = 1/(t₁ − 9)[(t₁ − 5) + t₁] / [t₁(t₁ − 5)] = 1/(t₁ − 9)(2t₁ − 5)(t₁ − 9) = t₁(t₁ − 5)2t₁² − 23t₁ + 45 = t₁² − 5t₁t₁² − 18t₁ + 45 = 0Discriminant = 18² − 4×45 = 144 ⇒ t₁ = (18 ± 12)/2 ⇒ 15 or 3Valid time: 15 hours (3 h would make t₂ negative)

Verification/Alternativet₁=15 ⇒ t₂=10, t₃=6; rates 1/15 + 1/10 = 1/6 (true).

Common pitfallsEquating times instead of rates: the statement implies equal rates over the same time interval.

Final Answer15 hours