Unit 1: Efficiency and Inlet-Outlet Pipes

PEA308 — Advanced Analytical Skills-Ii 9 min read

I. Orientation

Efficiency and inlet-outlet problems are applications of the rate principle: work completed depends on the amount of work assigned and the rate at which a person, machine, pipe, or leak performs it. If a task is completed in time (T), its rate is the reciprocal (1/T) of the task per unit time.

Total work: Represent the complete task by (1), or by a convenient common multiple of individual work capacities.
Work rate: If a worker completes a task in (x) days, the worker’s one-day work is (1/x).
Combined rate: Independent rates are added for workers or inlet pipes and subtracted for leaks or outlet pipes.
Efficiency: Efficiency compares work rates, not merely completion times.
Time and rate: For a fixed amount of work, time varies inversely with rate.
Units: Keep time units consistent; convert hours to minutes or days before combining rates.
Assumption: Unless stated otherwise, each worker or pipe operates at a constant rate.

II. Efficiency Based Problems

A. Efficiency based problems

Efficiency based problems compare the rates of workers, machines, or groups and use the inverse relationship between efficiency and time. A more efficient worker completes the same work in less time.

Definition: If A completes work in (a) days, then A’s rate is (1/a) work per day.
Ratio of efficiencies: If A and B take (a) and (b) days respectively, then
TEXT
```
  Efficiency of A : Efficiency of B = b : a
```
The order reverses because efficiency is proportional to (1/\text{time}).
Percentage efficiency: If A is (p\%) more efficient than B:
TEXT
```
  A's efficiency = (100 + p)/100 × B's efficiency
```
Time relationship: If A is (p\%) more efficient than B, A’s time is
TEXT
```
  B's time × 100/(100 + p)
```
Worked example: B completes a job in 20 days. A is 25% more efficient than B. A’s time is (20 \times 100/125=16) days.
Important distinction: “A takes 25% less time” does not mean “A is 25% more efficient.” A time reduction from 20 to 15 days represents efficiency (20/15=4/3), or 33⅓% more efficiency.

B. Wages based problems

Wages based problems distribute payment according to each worker’s actual contribution, measured by work rate multiplied by time worked.

Basic rule: A worker’s wage share is proportional to:
TEXT
```
  Work contribution = Efficiency × Time worked
```
Equal work duration: If A and B work for the same period, their wages are in the ratio of their efficiencies.
Different durations: If A and B have efficiencies (e_A,e_B) and work for (t_A,t_B):
TEXT
```
  Wage ratio = e_A t_A : e_B t_B
```
Common work units: If rates are fractional, take the least common multiple of completion times. For 12-day and 18-day workers, total work may be (36) units; their daily rates become (3) and (2) units.
Worked example: A, B, and C receive ₹900. Their efficiencies are (2:3:4), and they work for (3,2,1) days. Contributions are (6:6:4), so wages are ₹300, ₹300, and ₹200.
Condition: The entire wage must be divided only among the stated contributors; unrelated attendance or seniority is not included unless specified.

III. Chain Rule

A. Chain rule

The chain rule solves problems in which work, workers, time, efficiency, and wages change successively. Each change is converted into a ratio and multiplied through.

Purpose: It connects quantities such as workers, days, hours per day, and efficiency when direct equations are inconvenient.

Standard work relation: For identical workers:

TEXT

  Work ∝ Number of workers × Days × Hours per day × Efficiency

General equation: For two situations:
TEXT
```
  W1/W2 = (n1 d1 h1 e1)/(n2 d2 h2 e2)
```
Here (W) is work, (n) workers, (d) days, (h) hours per day, and (e) relative efficiency.
Finding an unknown: Cancel equal factors, substitute known ratios, and solve for the missing variable.
Worked example: 12 workers complete a job in 15 days at 8 hours daily. At 10 hours daily, 18 workers of 80% efficiency complete the same job in (d) days:
TEXT
```
  12×15×8 = 18×d×10×0.8
  d = 10 days
```
Check: A larger workforce or longer daily schedule should reduce required days when work remains fixed.

IV. Alternate Work Problems

A. Alternate work problems

Alternate work problems involve workers or machines operating on different days or time intervals, so the rate must be accumulated in cycles rather than treated as continuously combined.

Daily contribution: Calculate each worker’s contribution separately for the day on which that worker operates.
Two-day cycle: If A works on day 1 and B on day 2, the cycle’s work is:
TEXT
```
  Cycle work = A's one-day work + B's one-day work
```
Cycle method: Divide total work by cycle work to find complete cycles, then calculate the remaining work after those cycles.
Different starting worker: Starting with A or B changes the final partial cycle, even though the complete two-day cycle is unchanged.
Worked example: A completes work in 6 days and B in 8 days, working on alternate days with A first. In two days they complete (1/6+1/8=7/24). Three cycles complete (21/24), leaving (3/24=1/8), which B completes on the seventh day. Total time is 7 days.
More than two workers: Form a complete cycle containing every scheduled worker, add all contributions, and then handle the remaining fraction in schedule order.
Restriction: Do not add rates as though both workers operate simultaneously when the statement specifies alternate operation.

V. Advanced Time and Work Problems

A. Advanced time and work problems

Advanced time and work problems combine partial completion, changing teams, efficiency differences, and workers joining or leaving. The safest method is to express every participant through a common work unit.

Common work: If A takes 12 days and B takes 18 days, choose total work (36) units; their rates are (3) and (2) units per day.
Partial work: Work completed in (t) days is:
TEXT
```
  Completed work = Daily rate × t
```
Joining or leaving: Divide the timeline into intervals. Use the correct combined rate in each interval.
Efficiency change: Replace a worker’s old rate by:
TEXT
```
  New rate = Old rate × efficiency multiplier
```

Remaining work: Always calculate:

TEXT

  Remaining work = Total work − Work already completed

Worked example: A completes a job in 12 days and B in 18 days. A works alone for 4 days, completing (4\times3=12) units out of 36. The remaining 24 units are completed by A and B together at (3+2=5) units daily, requiring (24/5=4.8) days. Total time is 8.8 days.
Verification: Add each interval’s work contribution and confirm that it equals the chosen total work exactly.

VI. Inlet-Outlet

A. Inlet-outlet

Inlet-outlet problems treat pipes as workers: an inlet adds water to a tank, while an outlet or leak removes water. Their rates are combined algebraically.

Inlet rate: A pipe filling a tank in (x) hours has rate (+1/x) tank per hour.
Outlet rate: A pipe emptying a tank in (y) hours has rate (-1/y) tank per hour.
Net rate: For inlets (I_1,I_2) and outlets (O_1,O_2):
TEXT
```
  Net rate = 1/I1 + 1/I2 − 1/O1 − 1/O2
```
Filling time: If the tank starts empty and net rate is positive:
TEXT
```
  Time = 1/(Net rate)
```
Emptying condition: If the total outlet rate is greater than or equal to the inlet rate, the tank cannot fill from empty.
Worked example: An inlet fills a tank in 6 hours and an outlet empties it in 9 hours. Net rate is (1/6-1/9=1/18), so the tank fills in 18 hours.
Common denominator: For rates (1/8), (1/12), and (1/24), use denominator 24 to obtain (3/24), (2/24), and (1/24).
Sign convention: Positive represents water entering; negative represents water leaving.

VII. Part of the Tank Filled

A. Part of the tank filled

Part of the tank filled problems ask for the time required to fill a fraction of the tank, or determine the fraction filled after a specified interval.

Fractional work: If net rate is (r) tank per hour, the fraction filled in (t) hours is:
TEXT
```
  Fraction filled = r × t
```
Time for a fraction: To fill fraction (f):
TEXT
```
  Time = f/r
```
Here (f) is between 0 and 1 and (r) is the net rate.
Initial water level: If the tank already contains fraction (f_0), the remaining fraction is (1-f_0).
Changing pipe conditions: Calculate the volume added during each interval separately; then add the fractions.
Worked example: Two inlets fill a tank in 10 and 15 hours. Their combined rate is (1/10+1/15=1/6). Time to fill (1/2) of the tank is ((1/2)/(1/6)=3) hours.
Partial closure: If a pipe is closed after (t_1) hours, first compute (r_1t_1), subtract it from the required fraction, and use the later rate for the remainder.
Limitation: A fraction refers to tank capacity, not necessarily to the time fraction; nonzero starting levels and leaks change the calculation.

VIII. Time-Based Problems

A. Time-based problems

Time-based problems determine how long a tank takes to fill, empty, or reach a specified level when pipe rates operate together or change over time.

Basic model: Water level is represented by volume (V), and the rate of change is:
TEXT
```
  Change in volume = Net rate × Time
```
Full tank target: If the tank begins with fraction (f_0) and must become full, use target volume (1-f_0).
Sequential operation: For pipes opening at different times, construct a timeline and assign a net rate to each interval.
Reversed question: If time (T) and net rate (r) are known, the filled portion is (rT); if the initial fraction is (f_0), final fraction is (f_0+rT).
Worked example: A pipe fills a tank in 8 hours. It operates alone for 2 hours, filling (2/8=1/4). A second inlet then joins, and together they fill the remaining (3/4) in 3 hours. Their combined rate is ((3/4)/3=1/4) tank per hour.
Overflow: Once the level reaches (1), later filling time is irrelevant unless overflow quantity is requested.
Consistency check: The net rate must have units of tank per hour, and multiplying it by time must produce a dimensionless fraction of the tank.