Unit 1: Efficiency and Inlet-Outlet Pipes

1. Advanced Work, Rate, and Efficiency Dynamics

1.1 The Mathematical Formalization of Efficiency

In advanced quantitative aptitude, Efficiency ($E$) is defined as the quantum of physical work accomplished by an operational agent per unit time. Under the assumption of linear production kinetics:

$\text{Work } (W) = \text{Efficiency } (E) \times \text{Time } (T)$

From this governing axiom, the reciprocal relationship between efficiency and completion duration for an invariant task ($W = \text{Constant}$) is established:

$E \propto \frac{1}{T} \iff \frac{E_1}{E_2} = \frac{T_2}{T_1}$

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       +-------------------------------------------------------------+
       |               The Efficiency Invariance Theorem             |
       +-------------------------------------------------------------+
       | If Worker A is 'k' times as efficient as Worker B:          |
       |                   E_A = k * E_B                             |
       | Then Worker A requires only (1/k) of B's completion time:   |
       |                   T_A = (1/k) * T_B                         |
       +-------------------------------------------------------------+

Percentage Efficiency Adjustments

• If worker $A$ is $p\%$ more efficient than worker $B$:
  $E_A = \left( 1 + \frac{p}{100} \right) E_B$
  $T_A = \frac{T_B}{1 + \frac{p}{100}} = \left( \frac{100}{100 + p} \right) T_B$
  Example: If $A$ is $25\%$ more efficient than $B$ ($E_A = 1.25 E_B = \frac{5}{4} E_B$), then $A$ takes $\frac{4}{5}$ ($80\%$) of $B$'s time—a $20\%$ reduction in completion duration.

• If worker $A$ is $p\%$ less efficient than worker $B$:
  $E_A = \left( 1 - \frac{p}{100} \right) E_B \implies T_A = \left( \frac{100}{100 - p} \right) T_B$

1.2 The General Multi-Variable Chain Rule (MDHEW Formula)

When projects involve variable worker headcounts ($M$), calendar days ($D$), daily operating shifts in hours ($H$), individual efficiencies ($E$), and quantifiable physical outputs ($W$):

$\text{Total Work Output } (W) \propto M \times D \times H \times E$
$\frac{M \times D \times H \times E}{W} = \text{Constant}$

$\mathbf{\frac{M_1 \cdot D_1 \cdot H_1 \cdot E_1}{W_1} = \frac{M_2 \cdot D_2 \cdot H_2 \cdot E_2}{W_2}}$

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       +-------------------------------------------------------------+
       |                 Variables in the Chain Rule                 |
       +-------------------------------------------------------------+
       | M = Number of active workers / machines                     |
       | D = Number of operational days                              |
       | H = Working hours per day                                   |
       | E = Relative efficiency factor (default = 1)                |
       | W = Physical quantum of work (units, meters dug, wall area) |
       +-------------------------------------------------------------+

1.3 Equivalence Conversion Models (Men, Women, and Children)

Recruitment and competitive tests frequently pose problems involving heterogenous labor forces with distinct demographic efficiencies:
$m \text{ Men} = w \text{ Women} = c \text{ Children}$

Algorithmic Resolution Protocol

1. Determine Relative Individual Work Rates:
   Equate the daily outputs of the different cohorts to a common reference variable:
   $m \cdot R_M = w \cdot R_W = c \cdot R_C = k$
   $R_M = \frac{k}{m}, \qquad R_W = \frac{k}{w}, \qquad R_C = \frac{k}{c}$
2. Normalize the Combined Group into a Single Equivalent Cohort:
   Convert a composite team of $(x \text{ Men} + y \text{ Women} + z \text{ Children})$ into equivalent Men (or equivalent work units per day):
   $R_{\text{team}} = x \cdot R_M + y \cdot R_W + z \cdot R_C$
3. Compute Completion Duration:
   $T = \frac{W_{\text{total}}}{R_{\text{team}}}$

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       Example: 4 Men OR 6 Women can build a wall in 20 days.
       1. Rate Equivalence: 4 Men = 6 Women ==> 1 Man = 1.5 Women.
       2. Total Work (in Woman-Days): W = 6 Women * 20 Days = 120 Woman-Days.
       3. Target Group: 2 Men + 3 Women.
          Equivalent in Women = 2*(1.5) + 3 = 3 + 3 = 6 Women.
       4. Completion Time: 120 / 6 = 20 Days.

1.4 Alternate-Day Working Paradigms

In an alternate working schedule, agents do not work concurrently; rather, they perform sequentially in discrete rotational cycles (e.g., Worker $A$ on Day 1, Worker $B$ on Day 2, Worker $C$ on Day 3, and then repeating).

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   Cycle Sequence:  Day 1 (A) -> Day 2 (B) -> Day 3 (A) -> Day 4 (B) -> ...
   Full Cycle Duration = 2 Days
   Cycle Work = Rate_A + Rate_B

Analytical Solution Protocol

1. Calculate Total Work ($W_{\text{total}}$):
   Use the Least Common Multiple (LCM) of individual completion times:
   $W_{\text{total}} = \text{LCM}(T_A, T_B)$
2. Calculate Cycle Work ($W_{\text{cycle}}$):
   For a two-worker alternating schedule over a 2-day period:
   $W_{\text{cycle}} = R_A + R_B \text{ units}$
3. Determine Number of Completed Full Cycles ($k$):
   $k = \left\lfloor \frac{W_{\text{total}}}{W_{\text{cycle}}} \right\rfloor$
   $\text{Days elapsed from full cycles} = 2 \times k$
   $\text{Residual Work Remaining } (W_{\text{rem}}) = W_{\text{total}} - (k \times W_{\text{cycle}})$
4. Allocate Residual Work Step-by-Step:
   Evaluate who is scheduled next in the rotational sequence:
   • If $W_{\text{rem}} \le R_{\text{next}}$, the residual time is a fraction:
     $t_{\text{frac}} = \frac{W_{\text{rem}}}{R_{\text{next}}} \text{ days}$
   • If $W_{\text{rem}} > R_{\text{next}}$, the next worker works the full day, and the remaining deficit carries over to the subsequent worker.

1.5 Staggered Departures and Delayed Arrivals

When workers join after a delay or depart prematurely before project conclusion:

• Case A: Worker departs $x$ days BEFORE scheduled completion:
  Let total project completion time be $D$ days.

  • Worker who stayed until the end works for $D$ days.
  • Worker who departed $x$ days before completion works for $(D - x)$ days.
    $\sum R_i \cdot t_i = W_{\text{total}} \implies R_A \cdot D + R_B \cdot (D - x) = W_{\text{total}}$
    Solve directly for $D$.

• Case B: Worker leaves $y$ days AFTER project initiation:

  • Both workers work together for $y$ days, completing $(R_A + R_B) \times y$ work.
  • The remaining worker completes the residual work $W_{\text{rem}} = W_{\text{total}} - y(R_A + R_B)$ at individual rate $R_{\text{remaining}}$.

1.6 Principles of Commercial Wage Distribution

When multiple contributors complete a task for a designated contractual sum, fairness and legal economics dictate that wages must be apportioned strictly in proportion to actual work contributed, NOT simply the number of days spent on site.

$\text{Wage Proportion } W_i \propto (\text{Work Contributed by Agent } i) = R_i \times t_i$

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Case 1: All workers work concurrently for the exact identical duration (t_1 = t_2 = ... = t_k).
        Since time is invariant, Work Contributed is directly proportional to Work Rate:
        Wage_1 : Wage_2 : ... : Wage_k = R_1 : R_2 : ... : R_k = E_1 : E_2 : ... : E_k

Case 2: Workers work for differing numbers of days.
        Wages must be divided in the composite ratio of (Rate * Days):
        Wage_1 : Wage_2 = (R_1 * d_1) : (R_2 * d_2)

2. Advanced Fluid Mechanics: Pipes and Cisterns

Pipes and cisterns problems are mathematically isomorphic to Time and Work, with one fundamental thermodynamic divergence:

• Inlet Pipes (Filling Pipes): Function as positive work agents ($+R$). They increase the fluid volume inside the reservoir.
• Outlet Pipes (Drain / Waste / Leak Pipes): Function as negative work agents ($-R$). They deplete the accumulated fluid volume.

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       +-------------------------------------------------------------+
       |                Inlet-Outlet Hydraulic System                |
       +-------------------------------------------------------------+
                                 Inlet Pipe A (+R_A)
                                        |
                                        v
                            +-----------------------+
                            |                       |
                            |   Accumulated Volume  |
                            |         V(t)          |
                            |                       |
                            +-----------------------+
                                        |
                                        v
                                 Outlet Pipe B (-R_B)
                                 
       Net Rate of Fluid Accumulation: R_net = R_inlet - R_outlet

2.1 The Governing Net Accumulation Equation

Let a reservoir of capacity $V$ be serviced by $m$ inlet pipes with individual filling times $A_1, A_2, \dots, A_m$ and $n$ outlet pipes with individual emptying times $B_1, B_2, \dots, B_n$.

Normalizing total reservoir capacity to $V = 1$ (or using the LCM of all times):
$\text{Net Rate } (R_{\text{net}}) = \sum_{i=1}^m \frac{1}{A_i} - \sum_{j=1}^n \frac{1}{B_j}$

• Condition for Filling ($R_{\text{net}} > 0$):
  $\sum \frac{1}{A_i} > \sum \frac{1}{B_j} \implies \text{Time to Fill } T_{\text{fill}} = \frac{1}{R_{\text{net}}}$
• Condition for Emptying ($R_{\text{net}} < 0$):
  $\sum \frac{1}{A_i} < \sum \frac{1}{B_j} \implies \text{Time to Empty } T_{\text{empty}} = \frac{1}{|R_{\text{net}}|}$
• Equilibrium ($R_{\text{net}} = 0$): Fluid volume remains constant; neither filling nor emptying occurs.

2.2 The Bottom Leak Phenomenon

A standard archetype involves a cistern that is filled by an inlet pipe in $T_{\text{inlet}}$ hours, but due to an undetected leak at the bottom, takes $T_{\text{actual}}$ hours ($T_{\text{actual}} > T_{\text{inlet}}$) to achieve full capacity.

Let the filling rate of the inlet be $R_{\text{in}}$ and the emptying rate of the leak be $R_{\text{leak}}$.
$R_{\text{net}} = R_{\text{in}} - R_{\text{leak}}$
$\frac{1}{T_{\text{actual}}} = \frac{1}{T_{\text{inlet}}} - \frac{1}{T_{\text{leak}}}$

Isolating the leak's emptying rate:
$\frac{1}{T_{\text{leak}}} = \frac{1}{T_{\text{inlet}}} - \frac{1}{T_{\text{actual}}} = \frac{T_{\text{actual}} - T_{\text{inlet}}}{T_{\text{inlet}} \cdot T_{\text{actual}}}$

$\mathbf{T_{\text{leak}} = \frac{T_{\text{inlet}} \cdot T_{\text{actual}}}{T_{\text{actual}} - T_{\text{inlet}}}}$

Physical Meaning: $T_{\text{leak}}$ represents the time the leak would take to completely drain a full cistern if the inlet pipe were closed.

2.3 Alternating Inlet and Outlet Cycles

When an inlet pipe $A$ and an outlet pipe $B$ are operated alternately for 1 minute (or 1 hour) each:

• Minute 1: Inlet $A$ operates $\implies$ Volume increases by $+R_A$.
• Minute 2: Outlet $B$ operates $\implies$ Volume decreases by $-R_B$.
• 2-Minute Cycle Net Gain: $\Delta V_{\text{cycle}} = R_A - R_B$.

Critical Operational Rule (The Threshold Buffer Principle)

$\text{Buffer Volume} = V_{\text{total}} - R_A$

• The reservoir is declared full the very instant the water level reaches $100\%$ capacity.
• Once the tank is full, the outlet pipe is NOT opened!
• Therefore, we must apply cycles ONLY up to the threshold volume $(V_{\text{total}} - R_A)$.
• The final segment is finished by Pipe $A$ in its positive cycle, terminating the process immediately.

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       Example: Tank capacity = 60 units.
       Inlet A fills 5 units/min; Outlet B empties 3 units/min.
       Cycle Net Gain (2 mins) = 5 - 3 = 2 units.
       
       WRONG APPROACH: 60 / 2 = 30 cycles * 2 = 60 minutes.
       (In this flawed approach, the tank reaches 60 units during an inlet cycle,
        and then the solver unnecessarily subtracts water!).
        
       CORRECT THRESHOLD APPROACH:
       1. Reserve final inlet burst: Threshold = 60 - 5 = 55 units.
       2. Full cycles needed to reach or exceed 55 units:
          Cycles = ceil(55 / 2) = 28 cycles.
       3. Volume after 28 cycles (56 minutes) = 28 * 2 = 56 units.
       4. Deficit remaining = 60 - 56 = 4 units.
       5. Pipe A opens on Minute 57:
          Time needed = 4 / 5 min = 0.8 min = 48 seconds.
       6. Total Time = 56 minutes 48 seconds!

2.4 Volumetric Discharge Capacity Calculations

When problems introduce explicit physical discharge rates (e.g., liters per minute):

1. Compute the fractional emptying/filling rate per minute using the LCM method.
2. Determine the time $T$ required for that specific pipe or leak to fill/empty the entire cistern.
3. Multiply the total time (in minutes) by the volumetric flow rate ($Q$ in $\text{L/min}$):

$\mathbf{\text{Total Capacity } (C) = Q \times T_{\text{pipe}}}$

3. Comprehensive Analytical Heuristics and Master Summary Table

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+------------------------------------+---------------------------------------------------------------+
| Scenario                           | Analytical Formulation / Governing Rule                       |
+------------------------------------+---------------------------------------------------------------+
| Efficiency vs. Time Ratio          | E_1 / E_2 = T_2 / T_1                                         |
| Percentage Efficiency p% higher    | E_A = E_B * (1 + p/100) ; T_A = T_B / (1 + p/100)             |
| Multi-Variable Chain Rule          | (M_1 * D_1 * H_1 * E_1) / W_1 = (M_2 * D_2 * H_2 * E_2) / W_2 |
| Demographic Equivalence            | Convert composite team to unified equivalent baseline units   |
| Commercial Wage Division           | Wage_i proportional to (Work Done) = Rate_i * Days_Worked_i   |
| Inlet-Outlet Net Rate              | R_net = Sum(1/A_i) - Sum(1/B_j)                               |
| Bottom Leak Drain Time             | T_leak = (T_inlet * T_actual) / (T_actual - T_inlet)          |
| Alternating In/Out Threshold       | Target Cycle Capacity = Total - R_inlet                       |
| Total Volumetric Capacity          | Capacity = Flow Rate (L/min) * Total Time (mins)              |
+------------------------------------+---------------------------------------------------------------+

4. Comprehensive Worked Mathematical Exemplars

4.1 Exemplar 1: Heterogeneous Labor Force Equivalence

Problem: 3 men or 5 women can cultivate an agricultural tract of 12 hectares in 14 days, working 8 hours per day. How many days will 5 men and 7 women require to cultivate an agricultural tract of 20 hectares, working 7 hours per day, assuming all workers operate at constant efficiency?

Detailed Mathematical Derivation:

1. Determine Relative Individual Rates:
   $3 \text{ Men} = 5 \text{ Women} \implies 1 \text{ Man} = \frac{5}{3} \text{ Women}$

2. Convert the Second Group into Woman-Equivalents:
   $\text{Team} = 5 \text{ Men} + 7 \text{ Women} = 5 \left( \frac{5}{3} \right) + 7 = \frac{25}{3} + \frac{21}{3} = \frac{46}{3} \text{ Women}$

3. Apply the General Chain Rule (MDHW Formula):
   $\frac{M_1 \cdot D_1 \cdot H_1}{W_1} = \frac{M_2 \cdot D_2 \cdot H_2}{W_2}$
   Here:

   • $M_1 = 5 \text{ Women}, \quad D_1 = 14 \text{ days}, \quad H_1 = 8 \text{ hours/day}, \quad W_1 = 12 \text{ hectares}$
   • $M_2 = \frac{46}{3} \text{ Women}, \quad D_2 = ? \text{ days}, \quad H_2 = 7 \text{ hours/day}, \quad W_2 = 20 \text{ hectares}$

   $\frac{5 \times 14 \times 8}{12} = \frac{\left( \frac{46}{3} \right) \times D_2 \times 7}{20}$
   $\frac{560}{12} = \frac{46 \times 7 \times D_2}{60} \implies \frac{140}{3} = \frac{322 \times D_2}{60}$
   $D_2 = \frac{140}{3} \times \frac{60}{322} = 140 \times \frac{20}{322} = \frac{2800}{322} = \frac{1400}{161} \approx \mathbf{8.696 \text{ days}}$

4.2 Exemplar 2: Staggered Departure Before Completion

Problem: Worker A can construct a masonry wall in 20 days, and Worker B can construct the same wall in 30 days. Both start working together, but Worker A departs 5 days before the wall is completely finished. What is the total duration required to complete the wall?

Detailed Mathematical Derivation:

1. Total Work via LCM Method:
   $W_{\text{total}} = \text{LCM}(20, 30) = \mathbf{60 \text{ units}}$

2. Daily Work Rates:
   $R_A = \frac{60}{20} = 3 \text{ units/day}, \qquad R_B = \frac{60}{30} = 2 \text{ units/day}$

3. Time-Work Formulation:
   Let the total duration to complete the wall be $T$ days.

   • Worker B works for all $T$ days.
   • Worker A leaves 5 days early, working for $(T - 5)$ days.

   $R_A \cdot (T - 5) + R_B \cdot T = 60$
   $3(T - 5) + 2T = 60 \implies 3T - 15 + 2T = 60$
   $5T = 75 \implies \mathbf{T = 15 \text{ days}}$

Conclusion: The total construction duration is exactly 15 days (A worked for 10 days, B worked for 15 days).

4.3 Exemplar 3: Alternating Pipes with Threshold Buffer Principle

Problem: Inlet Pipe A can fill an empty reservoir in 12 hours, while Outlet Pipe B can drain a full reservoir in 15 hours. If both pipes are opened alternately for 1 hour each, starting with Pipe A, in how many hours will the reservoir be completely filled?

Detailed Mathematical Derivation:

1. Capacity and Individual Hourly Rates:
   $V_{\text{total}} = \text{LCM}(12, 15) = \mathbf{60 \text{ units}}$
   $\text{Inlet Rate } R_A = +\frac{60}{12} = +5 \text{ units/hour}$
   $\text{Outlet Rate } R_B = -\frac{60}{15} = -4 \text{ units/hour}$
2. Cycle Analysis (2-Hour Cycle):
   • Hour 1: Pipe A fills $+5$ units.
   • Hour 2: Pipe B empties $-4$ units.
   • Net gain per 2-hour cycle $= 5 - 4 = \mathbf{+1 \text{ unit}}$.
3. The Threshold Buffer Principle:
   The tank is declared full the instant the water level hits 60 units. Once full, Pipe B will not be opened.
   $\text{Threshold Volume} = V_{\text{total}} - R_A = 60 - 5 = \mathbf{55 \text{ units}}$
4. Number of Complete Cycles:
   To reach 55 units at $+1$ unit per cycle:
   $\text{Number of Cycles} = \frac{55}{1} = 55 \text{ cycles}$
   $\text{Time Elapsed} = 55 \times 2 = \mathbf{110 \text{ hours}}$
   $\text{Volume Accumulated} = 55 \times 1 = 55 \text{ units}$
5. Final Pipe A Burst:
   On Hour 111, Pipe A opens:
   • Volume required $= 60 - 55 = 5 \text{ units}$.
   • Time needed $= \frac{5 \text{ units}}{5 \text{ units/hour}} = \mathbf{1 \text{ hour}}$.
6. Total Duration:
   $\text{Total Time} = 110 + 1 = \mathbf{111 \text{ hours}}$
   (Note: A naive calculation dividing 60 by 1 gives 60 cycles = 120 hours, which is flawed because it ignores that the reservoir was already filled at 111 hours!).

4.4 Exemplar 4: Volumetric Capacity Determination from a Discharge Leak

Problem: An inlet pipe can fill a cistern in 8 hours. Due to an undetected leak in the bottom of the cistern, it actually requires 10 hours to fill. If the leak discharges water at the constant rate of 6 liters per minute, what is the total fluid capacity of the cistern in liters?

Detailed Mathematical Derivation:

1. Determining the Leak's Draining Rate:
   Let $T_{\text{in}} = 8 \text{ hours}$, and $T_{\text{net}} = 10 \text{ hours}$.
   $\frac{1}{T_{\text{net}}} = \frac{1}{T_{\text{in}}} - \frac{1}{T_{\text{leak}}}$
   $\frac{1}{T_{\text{leak}}} = \frac{1}{T_{\text{in}}} - \frac{1}{T_{\text{net}}} = \frac{1}{8} - \frac{1}{10} = \frac{5 - 4}{40} = \frac{1}{40}$
   $T_{\text{leak}} = \mathbf{40 \text{ hours}}$
2. Converting Draining Time into Minutes:
   $T_{\text{leak, mins}} = 40 \times 60 = \mathbf{2,400 \text{ minutes}}$
3. Calculating Cistern Volumetric Capacity:
   Given the volumetric discharge rate $Q = 6 \text{ liters/minute}$:
   $\text{Capacity} = Q \times T_{\text{leak, mins}} = 6 \text{ L/min} \times 2,400 \text{ mins} = \mathbf{14,400 \text{ liters}}$

The cistern has an exact fluid capacity of 14,400 liters.

5. Master Summary Table and Traps Reference

1. Never sum times directly: Always invert times to rates before executing addition or subtraction.
2. Wage Proportionality: If two workers complete a job together starting and ending at the same instant, wages divide strictly in the ratio of their efficiencies. If one worker leaves early, wages divide in the ratio of actual work units contributed.
3. Threshold Buffer in Alternating Pipes: Never divide total capacity blindly by cycle gain when an outlet pipe is involved; always withhold the final positive inlet burst.
4. Fractional Days Allocation: In alternate work cycles, verify whether the residual work is absorbed entirely by the next worker or overflows into the subsequent worker's turn.