One pipe fills a tank three times as fast as another. Together they fill in 36 minutes. How long will the slower pipe take alone?

Problem restatement
Two pipes have rates in the ratio 1 : 3 (slower : faster). Together they fill in 36 minutes. Find the slower pipe's solo time.

Given data

• Let slower rate = r tank/min; faster rate = 3r tank/min.
• Together rate = 1/36 tank/min.

Concept/Approach
Add rates to match the combined fill rate, then invert to get time.

Step-by-step calculation
r + 3r = 4r = 1/36 ⇒ r = 1/144 tank/min Slower's time = 1 ÷ r = 144 minutes

Verification/Alternative
Faster time = 1 ÷ (3r) = 48 min; check: 1/144 + 1/48 = (1 + 3)/144 = 4/144 = 1/36 (ok).

Final Answer
144 minutes