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.
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
- 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:
- Take the LCM of all given completion times as Total Work Units ($W$).
- Calculate individual 1-day or 1-hour work rates (units/day $= \frac{\text{Total Units}}{\text{Individual Days}}$).
- 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:
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:
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:
- Let B's daily efficiency $E_B = 100$ units/day.
- A is 60% more efficient, so A's daily efficiency $E_A = 100 + 60 = 160$ units/day.
- Ratio of efficiency $E_A : E_B = 160 : 100 = 8 : 5$.
- B takes 24 days, so Total Work $W = E_B \times 24 = 5 \times 24 = 120$ units.
- Combined daily efficiency of $(A + B) = 8 + 5 = 13$ units/day.
- 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:
- Take LCM of completion times (10 and 15) $= 30$ units (Total Work).
- A's daily rate $= \frac{30}{10} = 3$ units/day.
- B's daily rate $= \frac{30}{15} = 2$ units/day.
- Work cycle: Day 1 (A) $= 3$ units; Day 2 (B) $= 2$ units.
- 1 2-day cycle output $= 3 + 2 = 5$ units.
- Number of 2-day cycles required to reach 30 units $= \frac{30}{5} = 6$ cycles.
- 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:
- Take LCM of $(12, 15, 20) = 60$ units (Tank Capacity).
- Pipe A rate $= +\frac{60}{12} = +5$ units/hour.
- Pipe B rate $= +\frac{60}{15} = +4$ units/hour.
- Leak rate $= -\frac{60}{20} = -3$ units/hour.
- Net filling rate per hour $= +5 + 4 - 3 = +6$ units/hour.
- 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.
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?
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?
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?
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?