Three pipes A, B, and C together fill a tank in 5 hours. Pipe C is twice as fast as B, and B is twice as fast as A. How long will A alone take to fill the tank?

Problem restatementGiven relative speeds of three pipes and their combined fill time, find the solo fill time of the slowest pipe A.

Given data

• Combined time = 5 h ⇒ combined rate = 1 ÷ 5 tank/hour.
• C is twice B; B is twice A.

Concept/ApproachLet A's rate be r tanks/hour. Then B = 2r and C = 4r. Sum rates, equate to 1/5, solve for r, then take reciprocal for A's time.

Step-by-step calculation Total rate = r + 2r + 4r = 7r 7r = 1/5 ⇒ r = 1/35 (tank/hour) Time for A alone = 1 ÷ r = 35 hours

Verification/AlternativeIndividual times: A = 35 h, B = 17.5 h, C = 8.75 h. Harmonic combination 1/35 + 1/17.5 + 1/8.75 = 1/5 tank/hour (checks).

Common pitfalls

• Assuming C is twice A directly (it is actually four times A since C = 2×B and B = 2×A).

Final Answer35 hours