Pipe A can fill a tank in 45 hours and pipe B can fill it in 36 hours. If both pipes are opened together into an empty tank, in how many hours will the tank be completely full?

Introduction / Context:Pipes-and-cistern problems rely on constant rates. When multiple inlets are opened together, their filling rates add to give a combined rate that determines the total time to fill the tank.

Given Data / Assumptions:

• Pipe A fills the tank in 45 hours (rate = 1/45 tank/hour).
• Pipe B fills the tank in 36 hours (rate = 1/36 tank/hour).
• The tank is initially empty; both pipes run simultaneously without interruptions or losses.

Concept / Approach:For constant rates, net rate = sum of individual rates for inlets. Time to complete one tank = 1 / (net rate).

Step-by-Step Solution:Rate(A) = 1/45 tank per hour.Rate(B) = 1/36 tank per hour.Net rate = 1/45 + 1/36.LCM(45, 36) = 180 ⇒ 1/45 = 4/180, 1/36 = 5/180.Net rate = (4 + 5)/180 = 9/180 = 1/20 tank per hour.Time to fill = 1 / (1/20) = 20 hours.

Verification / Alternative check:In 20 hours, A contributes 20/45 = 4/9 tank and B contributes 20/36 = 5/9 tank; total 1 tank.

Why Other Options Are Wrong:10 hr and 15 hr underestimate the time by treating the faster pipe as much quicker than it is; 28 hr ignores correct rate addition.

Common Pitfalls:Adding times instead of rates or using an incorrect LCM when summing fractions.

Final Answer:20 hr