Three pipes with diameters 2 cm, 3 cm and 4 cm are used together to fill a cistern. The rate of water flow in each pipe is proportional to the square of its diameter. If the largest pipe alone (4 cm diameter) can fill the cistern in 58 minutes, in how much time will all three pipes running together fill the cistern?

Introduction / Context:
This question combines the concept of work and time with proportional reasoning. The idea is that the volume of water flowing through a pipe per unit time is proportional to the cross sectional area of the pipe, which in turn is proportional to the square of its diameter. The problem then asks you to compute the combined filling time when different diameter pipes are used together, a very typical aptitude scenario.

Given Data / Assumptions:

Pipe diameters are 2 cm, 3 cm and 4 cm.

The rate of water flow in a pipe is proportional to the square of its diameter.

The 4 cm diameter pipe alone fills the cistern in 58 minutes.

All three pipes are assumed to deliver water at constant proportional rates.

Concept / Approach:
The key concept is proportionality. If flow is proportional to the square of the diameter, then the relative rates of the three pipes are in the ratio 2^2 : 3^2 : 4^2, that is 4 : 9 : 16. Once we know the time taken by the largest pipe, we can express the volume of the cistern in terms of that pipe's rate and time, then divide by the sum of all three rates to obtain the combined time. Treating the volume of the cistern as one unit simplifies all calculations.

Step-by-Step Solution:
Let the capacity of the cistern be 1 unit.Let k be the constant of proportionality for flow rate.Rate of the 2 cm pipe = k * (2^2) = 4k.Rate of the 3 cm pipe = k * (3^2) = 9k.Rate of the 4 cm pipe = k * (4^2) = 16k.Given that the 4 cm pipe alone fills the cistern in 58 minutes, we have: 16k * 58 = 1.So k = 1 / (16 * 58).The combined rate of all three pipes together is (4k + 9k + 16k) = 29k.Time taken when all three run together = 1 / (29k).Substitute k = 1 / (16 * 58) to get time = 1 / (29 * 1 / (16 * 58)) = (16 * 58) / 29 minutes.Compute (16 * 58) / 29. Since 58 / 29 = 2, the time becomes 16 * 2 = 32 minutes.
Verification / Alternative check:
Re check the ratio logic. The relative rates are 4 : 9 : 16, so the largest pipe contributes 16 out of a total of 29 parts of the flow.If 16 parts of flow fill the tank in 58 minutes, then 1 part corresponds to 58 / 16 minutes of tank volume.The full 29 parts of flow take (58 / 16) * (16 / 29) * 29, which again simplifies to 32 minutes.This confirms that the combined time is indeed 32 minutes.
Why Other Options Are Wrong:
Options 24 minutes and 28 minutes are too small and would imply a combined rate that is unrealistically high given the known rate of the largest pipe alone. They conflict with the proportionality condition. Options 36 minutes and 40 minutes are too large and would underestimate the combined contribution of the three pipes. The calculations clearly identify 32 minutes as the correct value.

Common Pitfalls:
Students often make the mistake of assuming that flow is directly proportional to diameter instead of the square of the diameter, which leads to incorrect rate ratios and wrong answers. Another frequent error is neglecting to use the given time of 58 minutes correctly to determine the proportionality constant. Some also mix up the use of ratios and actual values, especially when substituting back into the formula for time. Writing the ratio clearly and then using algebra with a symbolic constant k greatly reduces confusion.

Final Answer:
All three pipes running together will fill the cistern in 32 minutes.