Pipes and cisterns

1. Basic model: a tank as one unit of work

Pipes and cisterns belong to the same mathematical family as time-and-work problems. A cistern is simply a storage container. An inlet adds water, while an outlet or leak removes it. Instead of counting workers and completed work, count pipes and the fraction of a tank filled. The model is especially useful in CSAT because a verbal description can usually be reduced to a small rate equation.

Take the tank’s total capacity as one unit. A pipe that fills an empty tank in 6 hours supplies 1/6 tank each hour. An outlet that empties a full tank in 9 hours removes 1/9 tank each hour. With both operating, the net rate is 1/6 − 1/9 = 1/18 tank per hour. An initially empty tank therefore takes 18 hours to fill.

Always distinguish quantity, rate and time. Quantity may be expressed in litres or fractions of capacity; rate is quantity per unit time; time must use a consistent unit. The governing relation is quantity change = net rate × elapsed time. This relation applies separately to each interval during which the operating pipes remain unchanged.

• Use positive signs for filling and negative signs for emptying.
• Treat stated filling or emptying times as referring to the whole tank unless the question specifies otherwise.

2. Combining rates and choosing a convenient capacity

For two filling pipes taking a and b hours separately, their combined rate is 1/a + 1/b. Their combined filling time is ab/(a + b). Thus, pipes taking 12 and 18 hours fill a tank together in 216/30 = 7.2 hours, or 7 hours 12 minutes. Adding the completion times or taking their arithmetic average does not produce the combined time.

For a filling pipe taking a hours and an outlet taking b hours, the net rate is 1/a − 1/b. If b is greater than a, the inlet is faster and the filling time from empty is ab/(b − a). If a equals b, the water level remains unchanged under the constant-rate model. If b is smaller than a, the tank cannot fill from empty with both operating.

The least common multiple method replaces fractions with convenient integers. If two inlets fill a tank in 12 and 18 minutes and an outlet empties it in 36 minutes, assume a capacity of 36 units. Their rates become 3, 2 and −1 units per minute. The net rate is 4 units per minute, so filling takes 9 minutes. This assumed capacity is a calculation aid, not a claim about actual tank size.

• Prefer the LCM method when several whole-number completion times occur.
• Prefer direct rates when capacities and flows are already given in litres and litres per minute.

3. Partial tanks, delayed operation and hidden leaks

An initially filled portion changes the quantity remaining, not the pipe’s rate. If a tank is already one-third full and a pipe can fill the whole tank in 8 hours, the remaining two-thirds takes 8 × 2/3 = 16/3 hours. When several pipes operate, divide the remaining fraction by their net rate rather than multiplying by any individual completion time.

For delayed openings or closures, split the operation into stages. Suppose A fills a tank in 12 hours and B in 18 hours. A works alone for 3 hours, filling 1/4 of the tank. Both then work at 5/36 tank per hour. The remaining 3/4 takes (3/4)/(5/36) = 27/5 hours. Total elapsed time is 3 + 5.4 = 8.4 hours. Clearly distinguish total time from the time measured after B opens.

A leak’s rate can be inferred from the difference between normal and observed filling rates. If an inlet normally fills a tank in 6 hours but takes 8 hours with a leak, the leak removes 1/6 − 1/8 = 1/24 tank per hour. It would empty a full tank in 24 hours under the same constant-rate assumption. Subtracting the two filling times directly cannot determine the leak’s emptying time.

• At every change in operation, record elapsed time, water already present and the new net rate.
• Check whether the tank fills before a scheduled pipe opening or closure; later stages may never occur.

4. Alternating pipes and conditional outlets

Alternating-pipe questions require attention to order. Suppose A fills a tank in 4 hours and B in 6 hours, and they operate for one hour each alternately, beginning with A. Each two-hour cycle fills 1/4 + 1/6 = 5/12. After two cycles, 5/6 is filled. A then fills the remaining 1/6 in (1/6)/(1/4) = 2/3 hour. Total time is 4 hours 40 minutes.

Do not automatically use the average cycle rate for the entire operation. The tank may become full partway through a filling interval, before the next outlet or slower pipe begins. Count complete cycles only while checking whether an earlier filling interval reaches capacity. In alternation involving an outlet, also check whether the tank becomes empty before that outlet’s scheduled operating period ends.

Some problems place a leak at a specified height. Such a leak acts only when water is above its opening. For a tank with uniform horizontal cross-section, a leak halfway up corresponds to half the capacity; this does not hold for every tank shape. Divide the calculation at the threshold and apply only the rates active in each stage.

• A fractional final interval is legitimate: stop the calculation as soon as the tank becomes full.
• Starting with a different pipe can change the answer even when the net quantity per complete cycle is unchanged.

5. CSAT shortcuts, checks and model limitations

Read the question first for the starting level, target level, operating sequence and required time unit. Then choose either unit capacity or a convenient LCM capacity. Write a signed rate expression before calculating. For a fixed tank, rates are inversely proportional to completion times: pipes taking 10 and 15 minutes have rates in the ratio 3:2, not 2:3.

Use bounds to detect mistakes. Two inlets working together must fill faster than either alone. Adding an outlet must increase the filling time, provided filling remains possible. Convert decimal time carefully: 0.4 hour is 24 minutes, not 40 minutes. If capacity is 900 litres and net inflow is 30 litres per minute, the answer must be 30 minutes.

Real fluid systems may depart from the examination model. Flow through an opening can vary with water pressure and water level; pumping rates may also change. Do not introduce these complications into an ordinary arithmetic question unless stated. Conversely, when the question explicitly makes flow variable or an outlet conditional, a single constant net rate is insufficient.

• Estimate the answer’s magnitude before matching it with an option.
• Practise representing changing operations in a small stage-wise quantity ledger.

Previous year questions

No UPSC question has been asked directly on this micro-topic yet. Use the practice questions below.

Practice questions