Unit 1: Time, Work and Cisterns

I. Orientation — The Work–Rate–Time Principle

Time-and-work problems are governed by a proportional relationship: the amount of work completed equals the rate of work multiplied by the time spent. Pipes-and-cisterns problems apply the same principle to filling and emptying tanks, treating inlets as positive work and outlets as negative work.

• Defining relationship: Work, rate, and time satisfy (W=Rt), where (W) is work completed, (R) is work per unit time, and (t) is time.
• Standard convention: A complete job or full tank is usually represented by (1); fractions such as (1/5) indicate the part completed or filled.
• Uniform-rate assumption: Unless stated otherwise, each worker or pipe operates at a constant rate throughout the given period.
• Additive rates: Simultaneous workers or inlets have their rates added; an outlet’s emptying rate is subtracted.
• Common units: Rates may be measured as jobs per day, units per hour, or tanks per minute. All times must be converted to the same unit before combining rates.
• Efficiency connection: For the same work, efficiency is proportional to rate and inversely proportional to completion time.
• LCM alternative: Instead of taking total work as (1), it may be assigned a convenient number of units—often the least common multiple of the individual times.

II. Time and Work — Workers, Efficiency, and Payment

A. Time and work

Time and work analysis determines how long one or more agents require to complete a fixed job.

• Individual rate: If A completes one job in (a) days, A’s one-day work is (1/a).
• Work completed: If A works for (d) days, the completed fraction is (d/a).
• Remaining work: After completing fraction (x), the unfinished portion is (1-x).
• Inverse relation: For equal work, a faster worker takes less time; therefore (R\propto 1/t).
• Unit method: If total work is assigned (a) units and completed in (a) days, daily output is (1) unit.

W = R × t
R = W / t
t = W / R

Here, (W) denotes work, (R) denotes rate of work, and (t) denotes time.

B. Problems based on time and work

These problems translate statements about schedules, partial completion, or changing workers into work fractions or units.

• Step 1—Fix total work: Use (W=1) job or select an LCM-based total.
• Step 2—Find rates: Convert every completion time into daily or hourly work.
• Step 3—Construct phases: Calculate work separately when workers join, leave, or alternate.
• Step 4—Use the balance: Set total completed work equal to the complete job.
• Worked example: A completes a job in 12 days and works alone for 3 days; B then finishes it in 5 days.
  • A completes (3/12=1/4), leaving (3/4).
  • B’s rate is ((3/4)/5=3/20) job per day.
  • Hence B alone would require (1/(3/20)=20/3) days.

C. Formulae

Standard formulae allow work relationships to be represented consistently and solved without mixing time with rate.

• Single worker: A person finishing in (a) days has rate (1/a).
• Two workers together: If A and B require (a) and (b) days separately, their joint time is (ab/(a+b)).
• Three workers together: For individual times (a,b,c), joint time is (abc/(ab+bc+ca)).
• Efficiency ratio: If efficiencies are (E_A:E_B=m:n), corresponding times are (t_A:t_B=n:m).
• Workers and days: With equal efficiency and fixed work, (M_1D_1=M_2D_2).
• Workers, days, and hours: With equal efficiency, total work is proportional to (MDH).

R_A = 1/a
R_A+B = 1/a + 1/b
t_A+B = ab/(a + b)
W ∝ M × D × H × E

Here, (a,b) are individual times; (M) is the number of workers; (D) is days; (H) is hours per day; and (E) is efficiency per worker.

D. Computation of work done together

Joint work is computed by adding rates because each participant contributes simultaneously to the same total.

• Rate addition: If A and B have rates (R_A) and (R_B), their combined rate is (R_A+R_B).
• Time calculation: For one complete job, joint time equals (1/(R_A+R_B)).
• LCM computation: If A and B take 8 and 12 days, assign (24) units of work; their outputs are (3) and (2) units per day.
• Worked example: For the 8-day and 12-day workers, combined output is (3+2=5) units daily, so time is (24/5=4.8) days.
• Participation interval: If A works for (x) days and B for (y) days, their combined contribution is (xR_A+yR_B), not necessarily ((x+y)(R_A+R_B)).

E. Efficiency-based problems

Efficiency measures output per unit time and permits workers of unequal ability to be compared.

• Basic relation: Efficiency is equivalent to work rate, so (E=W/t).
• Equal-work comparison: If A is twice as efficient as B, (E_A:E_B=2:1), while (t_A:t_B=1:2).
• Percentage conversion: If A is 25% more efficient than B, then (E_A:E_B=125:100=5:4).
• Combined efficiency: Efficiencies expressed in compatible units may be added when workers operate together.
• Worked example: A takes 15 days. B is 50% more efficient, so (E_B:E_A=3:2); therefore (t_B:t_A=2:3), and B takes (15\times2/3=10) days.
• Caution: “20% less time” and “20% less efficient” are not equivalent because time and efficiency are inversely related.

F. Men, women, and children-based problems

These problems convert different worker categories into equivalent units by using their stated efficiency ratios.

• Equivalent labour: If (2) men do the same work as (3) women, then (2E_M=3E_W), giving (E_M:E_W=3:2).
• Mixed workforce rate: For (m) men, (w) women, and (c) children, total rate is (mE_M+wE_W+cE_C).
• Replacement rule: Workers can be substituted only after conversion through the efficiency ratio.
• Worked example: If one man equals two women and three men complete a job in 8 days, their daily rate is (1/8). Six women have the same total efficiency as three men and therefore also require 8 days.
• Fixed-work equation: When workforce composition changes, equate the products of total group efficiency and time.

W = (mE_M + wE_W + cE_C)t

Here, (m,w,c) are worker counts; (E_M,E_W,E_C) are their respective individual efficiencies; and (t) is working time.

G. Wages-based work problems

Wages are divided according to actual work contributed, not merely according to attendance or individual speed.

• Contribution principle: A worker’s share is proportional to rate multiplied by working time.
• Wage ratio: For A and B, the division ratio is (R_At_A:R_Bt_B).
• Equal duration: If both work for the same time, wages are divided in the ratio of efficiencies.
• Equal efficiency: If rates are equal, wages are divided according to time worked.
• Worked example: A is twice as efficient as B. A works 4 days and B works 6 days, so contributions are (2\times4:1\times6=8:6=4:3). From ₹2,100, A receives ₹1,200 and B ₹900.
• Group payment: First determine each person’s contribution units; then multiply total wages by that person’s units divided by all contribution units.

III. Pipes and Cisterns — Filling and Emptying Systems

A. Pipes and cisterns

A cistern problem treats a tank as the complete job and each pipe as a device performing positive or negative work.

• Tank representation: A full tank is (1), while half a tank is (1/2).
• Filling rate: A pipe filling the tank in (a) hours has rate (+1/a) tank per hour.
• Emptying rate: A pipe emptying it in (b) hours has rate (-1/b) tank per hour.
• Net rate: The algebraic sum of all active pipe rates determines whether the water level rises or falls.
• Capacity method: If capacity is (C) litres and inflow is (q) litres per minute, filling time is (C/q) minutes, provided there is no leakage or outlet flow.
• Assumption: Rates remain constant and pipes begin operating as stated unless delayed or intermittent operation is specified.

B. Inlet-outlet

Inlet–outlet problems combine positive filling rates with negative emptying rates to obtain the tank’s net change.

1. Inlets: Add water, so their rates carry positive signs.
2. Outlets: Remove water, so their rates carry negative signs.

• Net filling condition: The tank fills only when total inlet rate exceeds total outlet rate.
• Net emptying condition: The tank empties when outlet rate exceeds inlet rate.
• Worked example: An inlet fills a tank in 6 hours and an outlet empties it in 9 hours. Their net rate is
  [
  \frac{1}{6}-\frac{1}{9}=\frac{1}{18}.
  ]
  Thus, both operating together fill an empty tank in 18 hours.
• Equal rates: If inlet and outlet rates are equal, the water level remains unchanged while both operate.

C. Part of tank filled

A partially filled tank requires calculation only of the remaining fraction, using the active pipes’ net rate.

• Remaining capacity: If fraction (f) is already filled, the fraction still required is (1-f).
• Time required: With net filling rate (R), additional time is ((1-f)/R).
• Amount after time: Starting from fraction (f), the level after time (t) is (f+Rt), subject to the physical range (0\leq f+Rt\leq1).
• Worked example: A tank is (1/3) full, and a pipe fills (1/8) of the tank per hour. The remaining part is (2/3), so time required is
  [
  \frac{2/3}{1/8}=\frac{16}{3}\text{ hours}=5\text{ hours }20\text{ minutes}.
  ]
• Leakage effect: If a leak is active, use the reduced net rate rather than the inlet rate alone.

D. Time-based problems

Time-based cistern problems divide pipe operation into intervals whenever a pipe opens, closes, alternates, or starts late.

• Interval method: For each interval, multiply its duration by the net rate active during that interval.
• Delayed outlet: If an inlet works alone for (x) hours before an outlet opens, initial filling is (xR{\text{in}}); subsequent filling uses (R{\text{in}}-R_{\text{out}}).
• Alternating pipes: Compute the amount filled during one complete operating cycle, then determine the number of full cycles and the final partial cycle.
• Overflow boundary: Once accumulated work reaches (1), the tank is full; later inflow causes overflow rather than increasing the filled fraction.
• Worked example: An inlet fills a tank in 10 hours. It works alone for 2 hours, filling (2/10=1/5). An outlet that empties the tank in 15 hours then opens. The net rate becomes
  [
  \frac{1}{10}-\frac{1}{15}=\frac{1}{30}.
  ]
  The remaining (4/5) takes ((4/5)/(1/30)=24) hours, so total elapsed time is (2+24=26) hours.