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10.3 Time & Work, Efficiency Ratios & Pipes and Cisterns

Key Takeaways

  • The LCM Method establishes Total Work as the Least Common Multiple of individual completion times, converting fractional rates into integer daily efficiency units.

  • The Chain Rule equates productive capacity across varying manpower, durations, daily operating hours, and efficiency ratings: (M_1 * D_1 * H_1 * E_1) / W_1 = (M_2 * D_2 * H_2 * E_2) / W_2.

  • Individual efficiency is inversely proportional to completion time: if worker A is 50% more efficient than worker B, the efficiency ratio is 3 : 2 and the time ratio is 2 : 3.

  • In alternate-day scheduling, compute the work completed per complete multi-day cycle, determine total complete cycles, and allocate residual units sequentially to the scheduled starter.

  • In pipes and cisterns systems, inlet conduits represent positive efficiency (+E) while drainage leaks represent negative efficiency (-E); net filling requires sum(E_inlets) > sum(E_leaks).

Last updated: September 2026

10.3 Time & Work, Efficiency Ratios & Pipes and Cisterns

Time and Work and its engineering counterpart, Pipes and Cisterns, form one of the highest-weightage arithmetic modules in the SBI Clerk Preliminary and Main examinations. Standard textbook approaches rely on fractions (assigning total work as 1 and tracking fractional daily outputs 1A+1B\frac{1}{A} + \frac{1}{B}), which inevitably creates tedious common-denominator arithmetic under exam pressure. The LCM Efficiency Method replaces fractional algebra with integer arithmetic, enabling candidates to solve multi-worker schedules, alternating work shifts, mid-way departures, and drainage leak problems rapidly.


1. The LCM Efficiency Method Foundations

Instead of defining total work as 11, let:

Total Work=LCM(Individual Completion Times)\text{Total Work} = \text{LCM}(\text{Individual Completion Times}) Daily Efficiency (E)=Total Work (in units)Individual Completion Time (in days)\text{Daily Efficiency } (E) = \frac{\text{Total Work (in units)}}{\text{Individual Completion Time (in days)}} Time Taken to Complete Work=Remaining Work (in units)Combined Efficiency of Active Workers (∑Ei)\text{Time Taken to Complete Work} = \frac{\text{Remaining Work (in units)}}{\text{Combined Efficiency of Active Workers } (\sum E_i)}

LCM Efficiency Reference Architecture

Suppose Associate A completes an administrative audit in 12 days, Associate B completes it in 15 days, and Associate C completes it in 20 days.

  1. Compute LCM(12,15,20)=60 units\text{LCM}(12, 15, 20) = 60\text{ units} (Total Work).
  2. Establish individual daily efficiencies:
    • EA=6012=+5 units/dayE_A = \frac{60}{12} = +5\text{ units/day}
    • EB=6015=+4 units/dayE_B = \frac{60}{15} = +4\text{ units/day}
    • EC=6020=+3 units/dayE_C = \frac{60}{20} = +3\text{ units/day}
  3. Combined daily efficiency (EtotalE_{\text{total}}) = 5+4+3=12 units/day5 + 4 + 3 = 12\text{ units/day}.
  4. Time taken working concurrently = 6012=5 days\frac{60}{12} = \mathbf{5\text{ days}}.
AssociateIndividual Time (TiT_i)Total Work (LCM)Daily Efficiency (EiE_i)Output Share in Combined Work
Associate A12 days60 units60/12=5 units/day60 / 12 = \mathbf{5\text{ units/day}}5/12=41.67%5 / 12 = 41.67\%
Associate B15 days60 units60/15=4 units/day60 / 15 = \mathbf{4\text{ units/day}}4/12=33.33%4 / 12 = 33.33\%
Associate C20 days60 units60/20=3 units/day60 / 20 = \mathbf{3\text{ units/day}}3/12=25.00%3 / 12 = 25.00\%
Combined (A+B+C)—60 units12 units/day\mathbf{12\text{ units/day}}12/12=100.0%12 / 12 = 100.0\%

2. The Chain Rule & Workforce Equivalence

When workforce sizes, daily working hours, and physical output requirements vary simultaneously, the Chain Rule provides a single unifying equation:

M1×D1×H1×E1W1=M2×D2×H2×E2W2\frac{M_1 \times D_1 \times H_1 \times E_1}{W_1} = \frac{M_2 \times D_2 \times H_2 \times E_2}{W_2}

Where:

  • M=Number of workersM = \text{Number of workers}
  • D=Number of working daysD = \text{Number of working days}
  • H=Operating hours per dayH = \text{Operating hours per day}
  • E=Efficiency rating per workerE = \text{Efficiency rating per worker}
  • W=Quantity of work accomplished (pages verified, meters dug, loans processed)W = \text{Quantity of work accomplished (pages verified, meters dug, loans processed)}

Workforce Equivalence Mechanics (Men, Women, Children)

Examinations frequently feature composite teams (e.g., "6 men or 10 women can complete a task"). Convert heterogeneous worker groups into a single equivalent metric:

6M=10W  ⟹  3M=5W  ⟹  EMEW=536M = 10W \implies 3M = 5W \implies \frac{E_M}{E_W} = \frac{5}{3}

Assign integer efficiency ratings: each man possesses an efficiency of EM=5E_M = 5 units/day, while each woman possesses EW=3E_W = 3 units/day. Substitute these integer values directly into the Chain Rule.


3. Worker Departures and Arrivals Mid-Way

When team members join or leave mid-way through a project, two systematic techniques replace fractional timelines:

1. Direct Deduction Method (Worker Leaves After kk Days)

  • Calculate work completed by the initial team during the first kk days: Wdone=(∑Einitial)×kW_{\text{done}} = (\sum E_{\text{initial}}) \times k.
  • Determine remaining work: Wrem=Wtotal−WdoneW_{\text{rem}} = W_{\text{total}} - W_{\text{done}}.
  • Compute remaining days: Trem=Wrem∑EremainingT_{\text{rem}} = \frac{W_{\text{rem}}}{\sum E_{\text{remaining}}}.

2. Virtual Work Addition Method (Worker Leaves kk Days Before Completion)

  • When a question states that "Worker A leaves 3 days before the scheduled completion", do not work backwards with variable TT.
  • Shortcut: Force Worker A to stay! Add Worker A's phantom output for those 3 days to the Total Work: Wvirtual=Wtotal+(EA×3)W_{\text{virtual}} = W_{\text{total}} + (E_A \times 3)
  • Divide WvirtualW_{\text{virtual}} by the combined efficiency of the entire team (∑Eall)(\sum E_{\text{all}}). The result gives the total completion duration directly in a single step.

4. Alternate-Day Work Schedules

In alternating-day scenarios, workers operate in sequential rotation (e.g., Worker A works on Day 1, Worker B on Day 2, Worker A on Day 3, and so forth).

Step-by-Step Cycle Tracking Algorithm

  1. Define Cycle Length: For two alternating workers, 1 complete cycle =2 days= 2\text{ days}. For three workers, 1 complete cycle =3 days= 3\text{ days}.
  2. Compute Cycle Work: Sum the individual efficiencies across the full cycle: Wcycle=EA+EBW_{\text{cycle}} = E_A + E_B.
  3. Compute Complete Cycles: Number of Full Cycles=⌊WtotalWcycle⌋\text{Number of Full Cycles} = \left\lfloor \frac{W_{\text{total}}}{W_{\text{cycle}}} \right\rfloor
  4. Evaluate Cumulative Work & Time: Work Done in Full Cycles=Full Cycles×Wcycle\text{Work Done in Full Cycles} = \text{Full Cycles} \times W_{\text{cycle}} Time Elapsed=Full Cycles×(Days per Cycle)\text{Time Elapsed} = \text{Full Cycles} \times (\text{Days per Cycle})
  5. Allocate Remaining Units to Sequential Starters: Wremainder=Wtotal−Wdone in cyclesW_{\text{remainder}} = W_{\text{total}} - W_{\text{done in cycles}} Allocate WremainderW_{\text{remainder}} to the worker whose turn opens the next cycle. If the remainder is less than that worker's daily output, the fractional time added is WremainderEworker\frac{W_{\text{remainder}}}{E_{\text{worker}}}.

Warning

The Overshoot Trap: Never divide total work directly by cycle work if a fractional remainder exists, because the work concludes the exact moment the cumulative threshold is reached. A worker does not complete a full extra day if only a small fraction of work remains.


5. Pipes & Cisterns: Negative Efficiency Dynamics

Pipes and cisterns problems operate under identical mathematical principles as human labor, with one fundamental physical distinction:

Inlet Pipes contribute Positive Efficiency (+E)\text{Inlet Pipes contribute Positive Efficiency } (+E) Drainage Pipes / Structural Leaks contribute Negative Efficiency (−E)\text{Drainage Pipes / Structural Leaks contribute Negative Efficiency } (-E)

Worked Pipe and Structural Leak Problem

Problem: A main supply tap fills an overhead water tank at an SBI residential quarters in 8 hours. Due to an accidental leak in the base of the tank, it requires 10 hours to fill the tank completely. If the tank is completely full and the inlet tap is turned off, how long will the leak take to empty the entire tank?

  1. Establish Total Capacity: LCM(8,10)=40 units\text{LCM}(8, 10) = 40\text{ units}.
  2. Determine Inlet Efficiency (EinletE_{\text{inlet}}): Einlet=+408=+5 units/hourE_{\text{inlet}} = \frac{+40}{8} = +5\text{ units/hour}
  3. Determine Net Combined Efficiency (EnetE_{\text{net}}): Enet=Einlet+Eleak=+4010=+4 units/hourE_{\text{net}} = E_{\text{inlet}} + E_{\text{leak}} = \frac{+40}{10} = +4\text{ units/hour}
  4. Isolate Leak Efficiency (EleakE_{\text{leak}}): +5+Eleak=+4  ⟹  Eleak=4−5=−1 unit/hour+5 + E_{\text{leak}} = +4 \implies E_{\text{leak}} = 4 - 5 = -1\text{ unit/hour}
  5. Calculate Emptying Time: Time to Empty Full Tank=Total Capacity∣Eleak∣=401=40 hours\text{Time to Empty Full Tank} = \frac{\text{Total Capacity}}{|E_{\text{leak}}|} = \frac{40}{1} = \mathbf{40\text{ hours}}
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Time & Work / Pipes & Cisterns Solution Architecture
Test Your Knowledge

Associate A can complete a banking record digitisation assignment in 12 days, while Associate B can complete the exact same assignment in 18 days. If they work on alternate days with Associate A starting on the first day, in how many days will the entire assignment be completed?

A

14 1/3 days (approx 14.33 days)

B

14 1/2 days (14.50 days)

C

15 days

D

13 2/3 days (approx 13.67 days)

Test Your Knowledge

Senior Associate A is 50% more efficient than Junior Associate B. If Junior Associate B alone can complete a branch compliance review in 30 days, in how many days can Senior Associate A and Junior Associate B complete the same compliance review working together?

A

15 days

B

10 days

C

16 days

D

12 days

Test Your Knowledge

Two inlet pipes P and Q can fill an overhead water cistern at an SBI residential township in 20 minutes and 30 minutes, respectively. An outlet drainage pipe R can empty the entire cistern in 15 minutes. If all three pipes are opened simultaneously when the cistern is completely empty, how long will it take to fill the cistern to capacity?

A

45 minutes

B

60 minutes

C

50 minutes

D

75 minutes

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