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 restatement
Given 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/Approach
Let 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/Alternative
Individual 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 Answer
35 hours