Pipes A and B together fill a cistern in 4 hours. If opened separately, B would take 6 hours longer than A. How many hours does A alone need to fill the cistern?

Problem restatement
Two inlet pipes fill together in 4 hours. Separately, B needs 6 hours more than A. Find A's individual time.

Given data

• Together time = 4 hours.
• B's time = A's time + 6 hours.

Concept/Approach
Let A's time be a hours, B's time be a + 6 hours. Use rates: 1/a + 1/(a + 6) = 1/4 and solve the quadratic for a > 0.

Step-by-step calculation
1/a + 1/(a + 6) = 1/4(2a + 6) / [a(a + 6)] = 1/44(2a + 6) = a(a + 6) ⇒ 8a + 24 = a² + 6aa² − 2a − 24 = 0 ⇒ (a − 6)(a + 4) = 0a = 6 hours (reject negative root)

Verification/Alternative
A = 6 h ⇒ B = 12 h. Combined rate = 1/6 + 1/12 = 1/4 ⇒ 4 h, matches.

Common pitfalls
Adding times instead of rates; forgetting to discard the negative quadratic root.

Final Answer
6 hours