3.5 Time, Work, Efficiency & Pipes & Cisterns

Key Takeaways

  • The LCM method assigns total work as the least common multiple of individual completion times, replacing fraction addition with integer work units.
  • Work done equals Efficiency multiplied by Time (W = E × T); efficiency and time are inversely proportional when work is held constant.
  • The chain rule formula (M1 × D1 × H1) / W1 = (M2 × D2 × H2) / W2 handles variable workforce, hours per day, and output scale.
  • Pipes and cisterns operate on identical principles to time and work, with inlet pipes performing positive work and outlet leaks performing negative work.
  • Alternate day working models require identifying complete multi-day work cycles before evaluating remaining fractional work.
Last updated: July 2026

Time, Work, Efficiency & Pipes & Cisterns

Time and Work problems test logical allocation of efficiency, combined work outputs, variable workforce chain rules, and reservoir filling/draining mechanics. Assigning arbitrary work units using the Least Common Multiple (LCM) method streamlines calculations and eliminates tedious fraction arithmetic.


1. Core Principles & LCM Method

Fundamental Equation

Work Done=Efficiency×Time    W=E×T\text{Work Done} = \text{Efficiency} \times \text{Time} \implies W = E \times T

  • Inverse Relationship: Efficiency ($E$) is inversely proportional to time ($T$) required to complete a fixed task. If A takes $D_A$ days, A's 1-day work efficiency is $\frac{1}{D_A}$.
  • LCM Unit Assignment Method:
    1. Take the LCM of all given completion times as Total Work Units ($W$).
    2. Calculate individual 1-day or 1-hour work rates (units/day $= \frac{\text{Total Units}}{\text{Individual Days}}$).
    3. Sum or subtract unit rates to solve combined work timelines.

2. Chain Rule & Workforce Equations

When workforce size, daily working hours, efficiency rates, and output quantities vary, use the generalized Chain Rule formula:

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 workers, $D$ = days, $H$ = hours per day, $E$ = efficiency factor, $W$ = units of work produced.


3. Pipes and Cisterns Mechanics

Pipes and Cisterns follow identical mathematical laws as Time and Work:

  • Inlet Pipe (Positive Work): Fills the tank. 1-hour rate $= +\frac{1}{\text{Fill Time}}$.
  • Outlet Pipe / Leak (Negative Work): Empties the tank. 1-hour rate $= -\frac{1}{\text{Empty Time}}$.
  • Net Filling Rate: Net 1-Hour Work=(Inlet Rates)(Outlet Rates)\text{Net 1-Hour Work} = \sum (\text{Inlet Rates}) - \sum (\text{Outlet Rates})

4. Step-by-Step Worked Math Examples

Worked Example 1: Efficiency Ratios & Combined Work

Problem: A is 60% more efficient than B. If B alone can complete a work in 24 days, in how many days can A and B working together finish the same work?

Step-by-Step Solution:

  1. Let B's daily efficiency $E_B = 100$ units/day.
  2. A is 60% more efficient, so A's daily efficiency $E_A = 100 + 60 = 160$ units/day.
  3. Ratio of efficiency $E_A : E_B = 160 : 100 = 8 : 5$.
  4. B takes 24 days, so Total Work $W = E_B \times 24 = 5 \times 24 = 120$ units.
  5. Combined daily efficiency of $(A + B) = 8 + 5 = 13$ units/day.
  6. Time taken together $= \frac{\text{Total Work}}{\text{Combined Efficiency}} = \frac{120}{13} = 9 \frac{3}{13} \text{ days}$.

Worked Example 2: Alternate Day Work Strategy

Problem: A can finish a work in 10 days and B in 15 days. If they work on alternate days starting with A, in how many days will the work be completed?

Step-by-Step Solution:

  1. Take LCM of completion times (10 and 15) $= 30$ units (Total Work).
  2. A's daily rate $= \frac{30}{10} = 3$ units/day.
  3. B's daily rate $= \frac{30}{15} = 2$ units/day.
  4. Work cycle: Day 1 (A) $= 3$ units; Day 2 (B) $= 2$ units.
  5. 1 2-day cycle output $= 3 + 2 = 5$ units.
  6. Number of 2-day cycles required to reach 30 units $= \frac{30}{5} = 6$ cycles.
  7. Total time $= 6 \text{ cycles} \times 2 \text{ days/cycle} = 12 \text{ days}$.

Worked Example 3: Cistern Inlet and Leak Calculations

Problem: Pipe A can fill a tank in 12 hours, and Pipe B can fill it in 15 hours. A leak at the bottom can empty the full tank in 20 hours. If all three are opened simultaneously, how long will it take to fill the tank?

Step-by-Step Solution:

  1. Take LCM of $(12, 15, 20) = 60$ units (Tank Capacity).
  2. Pipe A rate $= +\frac{60}{12} = +5$ units/hour.
  3. Pipe B rate $= +\frac{60}{15} = +4$ units/hour.
  4. Leak rate $= -\frac{60}{20} = -3$ units/hour.
  5. Net filling rate per hour $= +5 + 4 - 3 = +6$ units/hour.
  6. Total time to fill tank $= \frac{60}{6} = 10 \text{ hours}$.

5. Common Traps & Advanced Edge Cases

  • Incomplete Alternate Day Cycles: When work does not divide evenly into full 2-day cycles, evaluate remaining work units carefully. If 1 unit remains and it's A's turn (efficiency 3 units/day), add $\frac{1}{3}$ of a day to the full cycle count.
  • Negative Tank Filling: If outlet leak rate exceeds inlet rate, net work is negative, causing an already filled tank to empty rather than fill.
Test Your Knowledge

A tank can be filled by an inlet pipe in 8 hours, but takes 10 hours to fill due to a leak at the bottom. How long will the leak take to empty a full tank?

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Test Your Knowledge

A is twice as efficient as B. Working together, they can complete a task in 14 days. In how many days can A alone complete the task?

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Test Your Knowledge

A and B can do a piece of work in 12 days and 18 days respectively. They work on alternate days starting with A. In how many days will the work be completed?

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Test Your Knowledge

If 12 men working 8 hours a day can build a wall in 10 days, how many men working 6 hours a day can build the same wall in 16 days?

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