Pipes and Cisterns – Two inlets plus a constant waste pipe (capacity asked): Two inlet pipes can fill a tank in 20 minutes and 24 minutes, respectively. A waste pipe empties water at a constant 6 gallons per minute. With all three operating together, the tank becomes full in 15 minutes. What is the capacity of the tank in gallons?

Introduction / Context:
Here, two variable rates depend on tank capacity (inlet pipes), while the waste pipe has a fixed volumetric rate (6 gallons per minute). Form a linear equation in the unknown capacity and solve.

Given Data / Assumptions:

• Pipe 1: capacity/20 gallons per min.
• Pipe 2: capacity/24 gallons per min.
• Waste: 6 gallons per min (subtract).
• Net fill time with all pipes: 15 min ⇒ net rate = capacity/15 gallons per min.

Concept / Approach:
Set (V/20) + (V/24) − 6 = V/15 and solve for V, the tank capacity.

Step-by-Step Solution:
(V/20) + (V/24) − 6 = V/15.LCM 120 ⇒ 6V + 5V − 720 = 8V ⇒ 11V − 720 = 8V ⇒ 3V = 720 ⇒ V = 240.

Verification / Alternative check:
Check numerically: Inlets sum = 12 + 10 = 22 gpm; net must be V/15 = 16 gpm ⇒ waste = 22 − 16 = 6 gpm, consistent.

Why Other Options Are Wrong:
210, 150, 180, and 50 gallons do not satisfy the equation with a 6 gpm waste and 15-minute completion.

Common Pitfalls:
Forgetting to keep units consistent (minutes vs hours) or omitting the constant waste term.

Final Answer:
240 gallon