Pipes Leaks and Net Tank Flow

What you will learn

A tank-flow problem uses the same amount–rate–time reasoning as work, but water can leave as well as enter. You will track the water already present, use signs for changes in stored volume, and stop at the requested first-full or first-empty event.

You will combine inlets, subtract an outlet or leak, convert volume/time units, infer a constant leak from two fill experiments, and handle a valve change. You will also recognise zero or negative net accumulation and decide whether missing information matters before the target event.

These are idealised arithmetic models. Unless a question deliberately withholds a condition, each stated rate remains constant over the stated volume range; flows add independently; tank capacity is fixed; there are no unmentioned flows or delays. Actual pressure-dependent flow needs a different model. Stored water must stay between empty and capacity.

1 Separate calibration from the actual start

If an inlet alone fills an empty tank completely in 6 hours at constant rate, its rate is 1/6 tank per hour. If an outlet alone empties a completely full tank in 9 hours at constant rate, its outward rate magnitude is 1/9 tank per hour. These are full-capacity calibration experiments. They do not say that the actual problem starts empty or full.

For example, a tank that now starts one-third full has only two-thirds of its capacity left to fill, even though the inlet's full-tank calibration is unchanged. “It fills in 6 hours” without its starting and ending levels can be ambiguous. Identify both endpoints before taking a reciprocal.

Try these retrieval checks:

1. Add 1/4 and 1/6.
2. Subtract 1/9 from 1/6.
3. If a tank is one-third full, what fraction remains to fill?
4. Convert 0.5 L/min into L/h.

Feedback: 1/4 + 1/6 = 5/12; 1/6 − 1/9 = 1/18; the remaining fraction is 2/3; 0.5 × 60 = 30 L/h. Fractions of a tank and litres are two possible amount units, but do not mix them in one sum without conversion.

2 A sign tells you what happens to stored water

Use positive contributions for water entering and negative contributions for water leaving. The outlet's outward magnitude is positive; its contribution to the stored-water balance is negative. Thus:

net accumulation rate = total inflow − total outflow

During an interval where the rates remain valid, new stored volume = old stored volume + net rate × interval time. This expression must not be extended beyond empty or full without examining what changes there.

For two inlets, both signs are positive. As a quick transfer from work rates, suppose one inlet alone fills an empty tank completely in 4 hours and another in 6 hours. With both constant independent inlets active and no outflow, rate = 1/4 + 1/6 = 5/12 tank per hour. An empty tank fills in 12/5 hours = 2 hours 24 minutes. We have changed the context, not the rate-addition mechanism.

Worked case 1 Opposing flow and initial water

An inlet alone fills an empty tank completely in 6 hours. An outlet alone empties a full tank in 9 hours. Their constant rates remain valid throughout the specified filling range. Both operate together. Find the first-fill time from empty, then in a separate operation starting one-third full.

Panel 1 Inward and outward flow on one tank

Entering the tank	Water stored in the tank	Leaving the tank
Inlet → 1/6 tank per hour	Capacity 1 tank; initial amount 1/3 tank in the second operation	1/9 tank per hour → outlet
Adds +1/6 per hour	Net change = +1/6 − 1/9 = +1/18 per hour	Contributes −1/9 per hour

From empty, required accumulation is 1 tank, so time = 1/(1/18) = 18 hours. Starting one-third full, required accumulation is 1 − 1/3 = 2/3 tank, so time = (2/3)/(1/18) = 12 hours.

Check the second operation: 1/3 + 12×1/18 = 1. The same inlet alone would need (2/3)/(1/6) = 4 hours from that initial state, so the active outflow has correctly made filling slower.

An 18-unit scale explains the same subtraction: capacity is 18 units, inlet adds 3 units per hour and outlet removes 2. Net accumulation is 1 unit per hour. Starting with 6 units leaves 12 units to add. Subtracting the solo times 9 − 6 does not describe any of these flow amounts.

Pause and identify the required amount At a positive net rate of 1/10 tank per hour, a tank starts half full. Find the first-fill time. Would 1/(1/10) be the correct calculation for that starting state?

Feedback: only 1/2 tank must be added, so time = (1/2)/(1/10) = 5 hours. The full-tank reciprocal gives 10 hours and ignores the initial water.

Worked case 2 Litres per hour and litres per minute

A 720 L tank initially contains 180 L. A constant inlet supplies 90 L/h and a constant outlet removes 0.5 L/min. Both operate until the tank is first full. Find the time.

Convert before combining: 0.5 L/min × 60 min/h = 30 L/h outward. The minute units cancel in the conversion. Net accumulation is 90 − 30 = 60 L/h.

Panel 2 Volume balance in one consistent unit

Quantity	Value
Capacity	720 L
Initial water	180 L
Remaining capacity	720 − 180 = 540 L
Net accumulation	90 − 30 = 60 L/h
First-fill time	540/60 = 9 h

Check: 180 + 60×9 = 720 L. Over 9 hours, 810 L enters and 270 L leaves. Their difference is 540 L, the increase in stored water. Total incoming volume may exceed tank capacity when some water leaves; it is not the same as the amount held at one time.

You can also use tank fractions: initial 180/720 = 1/4; net rate 60/720 = 1/12 tank per hour; time = (3/4)/(1/12) = 9 hours. The result is unchanged. Subtracting 90 − 0.5 before converting would combine unlike time-based units.

Pause and keep the unit meaning A rate is 2 L/min. How much is that per hour? Is that conversion a change in the physical flow?

Feedback: 2×60 = 120 L/h. Only the unit description changes. Do not apply another factor of 60 after using the converted hourly rate with time already in hours.

3 A leak is an outward contribution

A leak can be inferred if an inlet's rate and the observed net fill rate are known under comparable conditions. The time difference is not itself the leak's emptying time.

Worked case 3 Infer the leak rate

With a leak closed, one inlet fills an empty tank completely in 6 hours. With the same leak open throughout, the same inlet fills that empty tank completely in 8 hours. Assume inlet and leak rates are constant over the full volume range, and the leak can drain the tank to empty. Find the leak rate and the time for the leak alone to empty a full tank.

Panel 3 One-hour accumulation with and without a leak

Operation	Inward rate	Outward leak magnitude	Net accumulation
Leak closed	1/6 tank per hour	0	1/6
Leak open	1/6 tank per hour	To find	1/8

Since net = inflow − leak, leak = inflow − net = 1/6 − 1/8 = 1/24 tank per hour outward. Its signed contribution to stored water is −1/24. A full tank would take 1/(1/24) = 24 hours to empty through that leak alone.

Check: 1/6 − 1/24 = 1/8, matching the observed fill. The 8 − 6 = 2 hour difference is an effect of the leak; it is not a rate and not the full-to-empty time.

This conclusion relies on the stated constant rate and ability to drain the entire tank. A hole above the bottom may stop draining before empty; a real leak may depend on pressure and level. Those are not supplied by the two fill times, so do not silently make a physical prediction about such cases.

4 Carry the actual level across a valve change

A valve event changes which flows are active. The initial water and all net water accumulated so far stay in the ledger. Use the new net rate only after the event, just as with joining and leaving workers.

Worked case 4 Close an outlet after three hours

An inlet alone fills an empty tank completely in 8 hours; an outlet alone empties a full tank in 12 hours. Assume constant rates over the stated range. In the actual operation, the tank starts one-quarter full. Both run for 3 hours, then the outlet closes while the inlet continues. Find additional time after closure and total time from the start.

Panel 4 Initial water plus two intervals

Stage	Active flows	Net rate in tanks per hour	Change in water	Stored amount after stage
Start	Not yet counted	—	Initial 1/4	1/4
First 3 hours	Inlet and outlet	1/8 − 1/12 = 1/24	3/24 = 1/8	1/4 + 1/8 = 3/8
After closure	Inlet only	1/8	Remaining 5/8	1 at first full

After 3 hours the level is 3/8, so neither full nor empty has occurred before the change. Remaining capacity = 5/8. Additional time = (5/8)/(1/8) = 5 hours. Total = 3 + 5 = 8 hours.

Check the whole operation: initial 1/4 + inlet contribution 8/8 − outlet contribution 3/12 = 1. The outlet contributes only during its three active hours. Applying its outflow for all eight hours would invent an operation after closure.

Pause and select the next row At the closure event, which pair starts the new interval: initial amount 1/4 with rate 1/24, or stored amount 3/8 with rate 1/8?

Feedback: use 3/8 with 1/8. The first pair describes the original start and earlier net rate. Carry the new state forward instead of restarting the tank.

5 The sign determines which boundary can be reached

For a tank below capacity, a positive net rate moves it toward full. A zero net rate preserves its stored amount. A negative net rate reduces it toward empty while water is available. A negative answer from a “time to fill” division is not a future filling event.

Worked case 5 Zero and negative net flow

Use two separate idealised constant-flow cases.

A: An inlet alone fills an empty tank completely in 6 hours and an outlet alone empties a full tank in 6 hours. In the actual operation, the tank starts one-quarter full with both running.

B: An inlet alone fills an empty tank completely in 12 hours and an outlet alone empties a full tank in 8 hours. In the actual operation, the tank starts one-half full with both running.

Does either case fill while its operation stays unchanged? What boundary is reached first in B, and when?

Panel 5 Direction and first boundary

Case	Initial amount	Net rate	Consequence before a change of operation
A	1/4 tank	1/6 − 1/6 = 0	Stays 1/4 full; no finite first-fill time
B	1/2 tank	1/12 − 1/8 = −1/24	Loses water; first reaches empty

For B, time to remove the initial half tank = (1/2)/(1/24) = 12 hours. Check: 1/2 − 12/24 = 0. Full is not reached while this unchanged operation continues.

Stop the negative-rate calculation at empty. Extrapolating to 24 hours would give 1/2 − 24/24 = −1/2 tank, which is impossible. At empty, actual outflow cannot exceed water available from storage and incoming supply. In the idealised pass-through situation where outflow capacity continues to meet or exceed inflow, there is no accumulation from an empty start; there is never negative stored water. Describing a different later valve action would require that action to be stated.

For A, do not divide by zero and do not say “zero hours to fill.” It starts below full and never gains water under the specified operation. This is different from a tank that starts already full: its first-full time is immediately 0, because the target state is already reached.

Pause and distinguish the questions

1. A half-full tank has zero net rate. Is a finite first-fill time determined?
2. A tank starts completely full. What is the first time it is full, measured from now?
3. A tank's constant positive net rate is known to be 1/12 tank per hour, but its starting amount is missing. Is one exact first-fill time determined?

Feedback: (1) No finite first-fill time exists while that zero rate continues. (2) Zero time: it is already at the target. (3) No unique time follows. Empty gives 12 hours; half full gives 6 hours. The net rate alone does not specify the amount still needed.

Worked case 6 Full before the planned outlet event

A tank starts three-quarters full. An inlet alone fills an empty tank completely in 4 hours at constant rate. An outlet is scheduled to open 2 hours after the inlet starts only if filling is still underway. Find the first-full time.

Panel 6 Stop at the first relevant event

Event	Time from start	Stored amount or role
Inlet starts	0	3/4 tank already present
First full	(1/4)/(1/4) = 1 h	Capacity reached; stop
Planned outlet opening	2 h, only if still filling	Occurs too late to affect first full

Remaining capacity = 1/4 tank. The inlet adds 1/4 tank per hour, so first full is at 1 hour, before the two-hour event. The outlet's rate is not needed for this question. Missing information is only an obstacle when it affects the requested event.

Do not run the first stage to two hours and retain 5/4 tank. Do not force the outlet to operate before the job it was intended to affect has finished. This is the same event-order check as a worker scheduled to leave after work is already complete.

6 A bounded-flow routine

1. Read empty-to-full and full-to-empty calibration endpoints; name capacity and actual initial water separately
2. Put all volumes and time-based rates in consistent units
3. Mark active inlet/outlet intervals and assign signs to stored-water contributions
4. Compute net rate and determine the direction before dividing
5. Compare time to the relevant boundary with time to the next valve change
6. At a change, carry the actual stored amount forward; at first full or first empty, stop if that is the target
7. Check the whole volume balance and that every intermediate amount stays between 0 and capacity

The review that follows will mix worker and tank contexts. Select the model from the stated conditions rather than a memorised keyword.

Independent practice

Try before reading the key. These are untimed formative checks: one mark for a correct answer, zero otherwise, no negative marking and no pass cutoff. These local learning settings are not the official RRB marking scheme.

1. Two inlets alone fill the same empty tank completely in 6 h and 10 h. Both now run together from empty at constant independent rates, with no outflow. Find the first-fill time.

A. 8 h

B. 3 h 45 min

C. 16 h

D. 4 h

2. An inlet alone fills an empty tank completely in 5 h; an outlet alone empties a full tank in 10 h. Both rates are constant over the full range. The tank now starts one-fifth full and both remain active. Find the first-fill time.

A. 2 h

B. 10 h

C. 8/3 h

D. 8 h

3. A 900 L tank contains 300 L initially. A constant inlet supplies 75 L/h and a constant outlet removes 0.25 L/min. Both operate together until first full. How long does it take?

A. 10 h

B. 15 h

C. 8 h

D. 40 h

4. With a leak closed, an inlet fills an empty tank completely in 4 h. With the same leak open throughout, it fills the same empty tank completely in 5 h. Assume unchanged constant inlet and leak rates over the full range and that the leak can drain to empty. How long would the leak alone take to empty a full tank?

A. 1 h

B. 5 h

C. 20 h

D. 20/9 h

5. An inlet alone fills an empty tank completely in 6 h; an outlet alone empties a full tank in 12 h. Both rates are constant. The tank now starts one-third full. Both run for 2 h, then the outlet closes and the inlet continues. Find total time to first full.

A. 3 h

B. 5 h

C. 7 h

D. 8 h

6. An inlet alone fills an empty tank completely in 7 h and an outlet alone empties a full tank in 7 h, at constant rates. In the actual operation, the tank starts two-fifths full and both remain active with no other flow. What happens while this operation stays unchanged?

A. It becomes full after 7 h

B. It is first full at time 0

C. It becomes empty after 7 h

D. It remains two-fifths full

7. An inlet alone fills an empty tank completely in 15 h; an outlet alone empties a full tank in 10 h. The actual tank starts three-fifths full. Both run at those constant rates until the first empty/full event. Which boundary occurs first and when?

A. Empty after 18 h

B. Empty after 30 h

C. Full after 12 h

D. Empty after 6 h

8. A tank starts four-fifths full. An inlet alone fills an empty tank completely in 10 h at constant rate. An outlet is scheduled to open after 3 h only if the tank is still filling; its rate is not supplied. Find the first-full time.

A. 3 h

B. 10 h

C. 2 h

D. Cannot be found without the outlet rate

9. An inlet alone fills an empty tank completely in 8 h; an outlet alone empties a full tank in 24 h. Both now run together at constant rates with no later changes. The actual initial water amount is not stated. Which claim about the first-fill time is justified?

A. It must be 12 h

B. It must be 6 h

C. It must be 8 h

D. No unique time is determined without the initial amount

Answer key and explanations

1. B; 2. D; 3. A; 4. C; 5. B; 6. D; 7. A; 8. C; 9. D

1. B is correct. Both are inlets, so rate = 1/6+1/10 = 4/15 tank per hour. Time = 15/4 h = 3 h 45 min. Check: (4/15)(15/4)=1, and time is below either solo time. A averages the solo times. C adds them. D subtracts 10−6, but subtracting times does not represent either adding incoming volumes or dividing the required amount by rate.

2. D is correct. Net rate = 1/5−1/10 = 1/10 tank per hour. Remaining capacity = 4/5, so time = (4/5)/(1/10)=8 h. Check: 1/5+8/10=1. A divides the initial amount 1/5 by the net rate instead of using the remaining capacity. B uses one full tank and ignores initial water. C adds the outlet as an inlet: (4/5)/(1/5+1/10)=8/3, a different operation.

3. A is correct. Outflow = 0.25×60 = 15 L/h. Net accumulation = 75−15 = 60 L/h; remaining volume = 900−300 = 600 L. Time = 600/60=10 h. Check: 300+10×60=900. B uses 900/60=15 h and ignores the starting 300 L. C uses 600/75=8 h, ignoring outflow. D uses 600/15=40 h, treating the outward rate magnitude as if it alone filled the remaining space. Convert minutes to hours before subtracting flows.

4. C is correct. Inflow is 1/4 and observed net rate is 1/5 tank per hour. Leak magnitude = 1/4−1/5=1/20, so leak-only emptying time is 20 h. Check: 1/4−1/20=1/5. A mistakes the fill-time difference 5−4 for emptying time. B takes the net filling time as a leak-only time. D uses the reciprocal of 1/4+1/5, wrongly treating the observed net rate as a second incoming flow.

5. B is correct. First net rate = 1/6−1/12=1/12; in 2 h it adds 1/6 tank. Stored amount becomes 1/3+1/6=1/2. Remaining half at inlet-only rate 1/6 needs 3 h; total is 5 h. Check: 1/3+5/6−2/12=1. A reports only time after closure. C omits initial water, leaving 5/6 after the first stage and incorrectly giving 2+5=7 h. D leaves the outlet active throughout: (2/3)/(1/12)=8 h is the result of a different schedule.

6. D is correct. Net rate = 1/7−1/7=0, so stored amount remains 2/5. A uses an inlet-only calibration without the active outlet or initial state. B confuses zero net rate with zero time to full; the tank starts below full. C ignores the balancing inlet and uses the full-to-empty calibration as though it applied to this actual operation. There is no finite first-fill time while this zero-net operation continues.

7. A is correct. Net rate = 1/15−1/10=−1/30 tank per hour, so water decreases. Time to empty = (3/5)/(1/30)=18 h; 3/5−18/30=0. B uses a full tank instead of the initial 3/5. C takes a magnitude from the invalid filling expression (2/5)/(−1/30)=−12 h; changing its sign cannot create a future filling event. D uses the outlet alone: (3/5)/(1/10)=6 h, omitting inflow. Stop at empty; negative stored water afterward is not allowed.

8. C is correct. Remaining capacity is 1/5 and inlet rate is 1/10 tank per hour, so time = (1/5)/(1/10)=2 h. Since 2<3, the outlet is not active before first full. A substitutes the planned event time for completion. B ignores the initial four-fifths. D demands a rate that is irrelevant to the requested earlier event. Check: 4/5+2/10=1.

9. D is correct. Net rate = 1/8−1/24=1/12 tank per hour. An empty start gives 12 h; a half-full start gives 6 h; a one-third-full start gives 8 h. Each is compatible with the unstated starting amount. A, B and C each assert one possible starting-state result as necessary. The full-tank calibration fixes the rate, not the actual initial stock or remaining capacity.

Let the error choose the revision

• Items 1–2: revisit inlet signs, remaining capacity and the initial-water distinction
• Item 3: revisit the litres/minute conversion and volume balance
• Item 4: revisit leak magnitude as the difference between inflow and observed net rate
• Item 5: rebuild the stage ledger at the outlet-closure event
• Items 6–7: determine the direction and first valid boundary before dividing
• Items 8–9: ask whether missing information changes the requested first event

Retry without the key and explain why all alternative choices fail.

Sources and boundary

RRB CEN 09/2025 §14.1, printed/physical p.28 names Time and Work and Pipes & Cistern. This is topic-scope evidence, not a frequency prediction. The list is illustrative and not necessarily exhaustive. Not all later amendments have been audited here.

NCERT Ganita Prakash Grade 8 Part II, Chapter 3 §3.6, printed pp.63–68 supplies fixed-work proportional reasoning and combined-rate foundations. Official prelims, PDF p.2 verify First Edition December 2025. NIOS Secondary Mathematics 211, Chapter 5 §§5.2–5.3, printed pp.142–146 supports forming and checking a simple balance equation; the accessible chapter does not establish its publication year. Our schedules, signed-flow applications and feasibility/boundary checks are explained extensions, not claims of named official question variants. All exposition, panels, cases and practice are original.

Alternating attendance/valve cycles, wage sharing, staffing-rounding rules, multiple-unknown systems and pressure-dependent flow physics remain outside this module.

Analogy

Think of a stockroom ledger. Deliveries add to the amount on the shelf and dispatches subtract from it. Counting deliveries alone would not tell you how much is currently stored; you also need the opening stock and the dispatches. A valve change resembles a change in which entries occur after a particular time.

This analogy explains the signs and why the old balance is carried into the next interval. It also gives a useful limit: you cannot dispatch stock that does not exist, and a capacity-limited store cannot keep an unlimited amount on the shelf.

Real stock arrives in batches and water flow may depend on level or pressure. The analogy does not establish constant rates or continuous divisibility. Our arithmetic cases state those assumptions and stop at the requested full/empty boundary; they are not engineering predictions about real pipe hardware.

Quick reference

• Keep full-capacity calibration separate from actual initial water: empty-to-full time gives an inlet rate; full-to-empty time gives an outlet magnitude
• Use one tank or litres consistently. Convert every rate to the same time unit before combining
• Two inlets add. Net accumulation = inflow − outflow, with outward magnitudes subtracted from stored water
• Within a valid constant-rate interval, final volume = initial volume + net rate × time
• For positive net rate, first-fill time = remaining capacity / net rate, provided no earlier schedule change occurs
• For negative net rate while water remains, first-empty time = initial stored amount / magnitude of net loss; a negative “fill time” is not a future filling event
• With zero net rate, a below-full tank stays below full; do not divide by zero. If the tank starts already at the target boundary, first-target time is 0
• Missing leak magnitude = inlet rate − observed net fill rate only under comparable full-fill and constant-rate conditions
• At a valve change, carry the actual water level forward and use only the now-active rates
• Compare the next boundary time with the next valve event; first-full/first-empty questions stop at that boundary
• Stored amount must remain between 0 and capacity. At empty, actual discharge is limited by available water; do not extend a negative stock calculation
• An unspecified initial amount can prevent a unique time even when the net rate is known; an outlet scheduled after first full may be irrelevant