Pipes A and B together can fill a cistern in 15 hours. Working separately, pipe A takes 40 hours less than pipe B to fill the cistern. In how many hours will pipe A alone fill the cistern?

Introduction / Context:
This question is a classic two pipe time difference problem. The combined time to fill a tank is known, and a difference between the individual times is given. The task is to find the time taken by one particular pipe working alone. These questions are directly related to work and time problems where two persons can complete a job together in a known time and have a difference in their individual times.

Given Data / Assumptions:

Pipes A and B together fill the cistern in 15 hours.

Pipe A takes 40 hours less than pipe B to fill the cistern alone.

There are no leaks and flow rates are constant.

Concept / Approach:
The usual approach is to represent the individual times in terms of a single variable, using the given difference. From there we express individual rates as reciprocals of those times and then sum them to equal the combined rate 1/15. This generates a quadratic equation in the variable representing one of the times. Solving that quadratic and discarding any negative or unrealistic value gives the physically meaningful time for pipe A.

Step-by-Step Solution:
Let the time taken by pipe B to fill the cistern alone be t hours.Then the time taken by pipe A is t - 40 hours, because A is faster.The rate of pipe A is 1 / (t - 40) tank per hour.The rate of pipe B is 1 / t tank per hour.The combined rate is 1 / 15 tank per hour, so we have 1 / (t - 40) + 1 / t = 1 / 15.Combine the left-hand side using a common denominator: [t + (t - 40)] / [t(t - 40)] = (2t - 40) / [t(t - 40)].Thus (2t - 40) / [t(t - 40)] = 1 / 15.Cross multiply: 15(2t - 40) = t(t - 40).This gives 30t - 600 = t^2 - 40t.Rearrange: t^2 - 40t - 30t + 600 = 0, so t^2 - 70t + 600 = 0.Solve the quadratic equation t^2 - 70t + 600 = 0.The roots are t = 10 or t = 60 hours.If t = 10, then A's time is t - 40 = -30 hours, which is impossible; so discard t = 10.Thus t = 60 hours for B, and A's time is 60 - 40 = 20 hours.
Verification / Alternative check:
Check using A = 20 hours and B = 60 hours.Rate of A = 1/20, rate of B = 1/60.Combined rate = 1/20 + 1/60 = (3 + 1) / 60 = 4/60 = 1/15 tank per hour.Therefore, together they fill the cistern in 15 hours, confirming the solution.
Why Other Options Are Wrong:
30 hours, 40 hours, 60 hours and 15 hours do not satisfy both conditions at the same time. For example, if A took 30 hours, B would need to take 70 hours, and the combined rate would not be exactly 1/15 tank per hour. Only 20 hours gives a consistent pair of times (20 and 60) that match the given combined time and the 40 hour difference.

Common Pitfalls:
Some students mistakenly set the difference in rates equal to 40 instead of the difference in times. Others might incorrectly assume that the harmonic mean or arithmetic mean of the times is equal to the joint time. Algebraic handling of the quadratic can also cause issues if signs are not tracked carefully. The key is to build the equation 1/(t - 40) + 1/t = 1/15 correctly and then solve it systematically.

Final Answer:
Pipe A alone will fill the cistern in 20 hours.