2 pipes connected fill a tank in 4 hours. the large pipe alone can fill the tank in 6 hours less time than it will take the small one. how long will it take the small pipe to fill the tank alone?

Answer 1

Let 1 designate the large pipe and 2 designate the small pipe.

Let R₁ = flow rate of large pipe defined as gallons/hour & R₂ = flow rate of small pipe as gallons/hour

Let T₁ = Time for the large pipe to fill the tank and T₂ = Time it takes to fill the tank

Let's assume that the tank holds 1000 gallons (you could assume another value but 1000 makes the math easier...... you can check the answers with another total volume assumed!)

We know

4*R₁ + 4*R₂ = 1000

T₂ = T₁ + 6

R₁*T₁ = 1000 = R₂*T₂ = R₂*(T₁+6) So, R₁ = 1000/T₁ & R₂ = 1000/(T₁ + 6)

Substituting into the first equation:

4*1000/T₁ + 4*1000/(T₁ + 6) = 1000

Then 4/T₁ + 4/(T₁ + 6) = 1 getting a common denominator & collecting terms, we obtain

T₁² - 2T₁ - 24 = 0 from which the only positive root is T₁ = 6

Therefore, T₂ (Rate for small pipe to fill the tank) = T₁ + 6 = 12 hours