Unit 1: Time and Work, Pipes and Cisterns

I. Orientation

Time and work problems apply the principle that completed work depends on the rate of working and the time for which that rate operates. Pipes and cisterns use the same model: an inlet contributes a positive rate, while an outlet contributes a negative rate. The standard unit of work is one complete job or one completely filled or emptied tank.

• Basic relationship:
  [
  \text{Work}=\text{Rate}\times\text{Time}
  ]
• Rate convention: If a person completes a job in (x) days, the one-day rate is (\frac{1}{x}) of the work.
• Common unit: Total work is usually taken as (1), or as the least common multiple of the given times when integer calculations are convenient.
• Combined action: When workers or pipes operate together, their rates are added for cooperating agents and subtracted for opposing agents.
• Constant conditions: Unless stated otherwise, each worker or pipe is assumed to work at a constant rate.
• Efficiency meaning: Efficiency is directly proportional to work rate. A person who is twice as efficient completes twice as much work in the same time.
• Time-rate inverse: For a fixed amount of work, time is inversely proportional to efficiency or rate.

II. Time and Work — Individual and Combined Work Rates

Time and work problems determine how long individuals or groups take to complete a fixed job. Their central method is to convert each completion time into a daily work rate and then combine those rates.

A. Problems based on time and work

Problems based on time and work connect the total job, the time taken, and the rate of completion through a single relationship.

• Work-rate model: If A completes a job in (a) days, A’s one-day work is:
  [
  \frac{1}{a}
  ]
  Here, (1) represents the entire job and (a) represents the number of days.
• Finding time: If a worker’s rate is (r) jobs per day and the total work is (W), then:
  [
  \text{Time}=\frac{W}{r}
  ]
• Partial work: If A works for (t) days and completes (\frac{t}{a}) of the job, the remaining work is:
  [
  1-\frac{t}{a}
  ]
• Sequential work: When workers act one after another, calculate each worker’s completed fraction separately and add the fractions.
• Worked example: A completes a job in 12 days and B in 18 days. Their individual rates are (\frac{1}{12}) and (\frac{1}{18}). In 3 days together they complete:
  [
  3\left(\frac{1}{12}+\frac{1}{18}\right)=3\left(\frac{5}{36}\right)=\frac{5}{12}
  ]
  Thus, (\frac{7}{12}) remains.

B. Formulae

Formulae provide quick relationships for individual work, combined work, efficiency, and wages.

• Individual rate: If time taken is (T), then:
  [
  R=\frac{1}{T}
  ]
  (R) is the fraction of work completed per unit of time.
• Total work: For rate (R) acting for time (T):
  [
  W=RT
  ]
• Two workers together: If A takes (a) days and B takes (b) days:
  [
  \text{Combined rate}=\frac{1}{a}+\frac{1}{b}
  =\frac{a+b}{ab}
  ]
  Therefore:
  [
  \text{Time together}=\frac{ab}{a+b}
  ]
• One worker joining later: If A works for (x) days and A+B complete the rest in (y) days:
  [
  xR_A+y(R_A+R_B)=1
  ]
• Efficiency-time relation: For the same work:
  [
  E_A:E_B=T_B:T_A
  ]
  (E) denotes efficiency and (T) denotes completion time.

C. Computation of work done together

Computing work done together requires adding the rates of all workers operating during the same interval.

• Rate addition: If A, B, and C complete (\frac{1}{a}), (\frac{1}{b}), and (\frac{1}{c}) of a job per day:
  [
  R_{\text{total}}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}
  ]
• LCM method: For times 12, 18, and 24 days, choose total work as:
  [
  \operatorname{LCM}(12,18,24)=72\text{ units}
  ]
  Their daily contributions are 6, 4, and 3 units, respectively.
• Group completion: The combined group rate is (6+4+3=13) units per day, so the group completes the job in:
  [
  \frac{72}{13}\text{ days}
  ]
• Changing groups: Divide the work into time intervals whenever a worker joins, leaves, or changes speed; calculate each interval independently.

D. Efficiency-based problems

Efficiency-based problems compare workers by their relative rates rather than by complete-job times.

• Direct proportion: If A is 50% more efficient than B, then:
  [
  E_A=1.5E_B
  ]
  In the same time, A does (1.5) times B’s work.
• Time comparison: If A and B have efficiencies (3:2), their times for the same work are:
  [
  T_A:T_B=2:3
  ]
• Combined efficiency: If B’s rate is (2) units per day and A is 50% more efficient, A’s rate is (3) units per day. Together they produce (5) units per day.
• Percentage caution: “A is 25% more efficient than B” means (E_A=1.25E_B), whereas “B is 25% less efficient than A” means (E_B=0.75E_A); these statements do not reverse symmetrically.

E. Men, women, and children-based problems

These problems express the work capacity of different categories of workers through an equivalent-unit relationship.

• Equivalent work units: If 2 men do the same work as 3 women, then:
  [
  1\text{ man}=\frac{3}{2}\text{ women}
  ]
• Conversion: In a workforce of 4 men and 6 women, using (1) man (=1.5) women:
  [
  4+6(1.5)=13\text{ woman-equivalent units}
  ]
• Rate interpretation: If one woman-equivalent unit completes 5 units of work per day, the workforce above completes:
  [
  13\times5=65\text{ units per day}
  ]
• Unknown capacity: If (m) men and (w) women complete a known job in (d) days, form an equation from:
  [
  \text{Total work}=(mE_m+wE_w)d
  ]
  (E_m) and (E_w) are the daily efficiencies of one man and one woman.
• Assumption: The stated equivalence applies only to the same type and quantity of work under the same conditions.

F. Wages-based work problems

Wages-based work problems distribute payment according to the amount of work performed, not merely the time spent.

• Proportional rule: If workers receive wages (W_A) and (W_B), then:
  [
  W_A:W_B=\text{Work done by A}:\text{Work done by B}
  ]
• Same duration: When A and B work for equal time, their wage ratio equals their efficiency ratio.
• Different durations: If A works (t_A) days at efficiency (E_A), and B works (t_B) days at efficiency (E_B):
  [
  W_A:W_B=t_AE_A:t_BE_B
  ]
• Worked example: A works 6 days at 4 units per day and B works 8 days at 3 units per day. Their work amounts are (24) and (24) units, so a wage of ₹960 is divided equally: ₹480 each.
• Shared payment: For total wages (P), a worker’s share is:
  [
  \frac{\text{Individual work}}{\text{Total work}}\times P
  ]

III. Pipes and Cisterns — Filling and Emptying Rates

Pipes and cisterns problems model the change in the quantity of water inside a tank. Inlets fill the tank, outlets drain it, and the net rate determines the required time.

A. Inlet-outlet

Inlet-outlet problems use signed rates: filling is positive and emptying is negative.

• Inlet rate: If an inlet fills a tank in (a) hours:
  [
  R_{\text{inlet}}=\frac{1}{a}\text{ tank per hour}
  ]
• Outlet rate: If an outlet empties it in (b) hours:
  [
  R_{\text{outlet}}=-\frac{1}{b}\text{ tank per hour}
  ]
• Net rate: With one inlet and one outlet open:
  [
  R_{\text{net}}=\frac{1}{a}-\frac{1}{b}
  ]
• Several pipes: Two inlets and one outlet give:
  [
  R_{\text{net}}=\frac{1}{a}+\frac{1}{b}-\frac{1}{c}
  ]
• Direction test: A positive net rate fills the tank; a negative net rate empties it. A zero net rate leaves the water level unchanged.
• Worked example: An inlet fills a tank in 6 hours and an outlet empties it in 9 hours. The net rate is:
  [
  \frac{1}{6}-\frac{1}{9}=\frac{1}{18}
  ]
  Hence, the tank fills in 18 hours when both operate.

B. Part of tank filled

Part of tank filled problems determine the time needed to reach a fraction of the tank or account for an initial water level.

• Fractional work: If the net filling rate is (R), time to fill fraction (f) is:
  [
  t=\frac{f}{R}
  ]
  Here, (f) is measured as a fraction of the full tank.
• Initial water: If a tank is already (\frac{2}{5}) full, the remaining fraction is:
  [
  1-\frac{2}{5}=\frac{3}{5}
  ]
• Changing rates: If one pipe operates first and another opens later, calculate the water fraction added in the first interval, then apply the new net rate to the remaining fraction.
• Tank capacity method: For a 600-litre tank, (\frac{3}{4}) full corresponds to:
  [
  600\times\frac{3}{4}=450\text{ litres}
  ]
  This allows rates stated in litres per minute to be used directly.
• Leak interpretation: A leak is an outlet; its rate must be subtracted even when the tank is being filled by an inlet.

C. Time-based problems

Time-based problems find the duration of filling, emptying, or reaching a specified level from known rates and starting conditions.

• Full-tank time: For net rate (R{\text{net}}), a tank initially empty takes:
  [
  T=\frac{1}{R{\text{net}}}
  ]
  provided (R_{\text{net}}>0).
• Known quantity rate: If a pipe supplies (q) litres per minute to a tank of capacity (C) litres:
  [
  T=\frac{C}{q}
  ]
• Partial starting level: If the tank starts with (p) of its capacity and must reach (q), the required fraction is (q-p), so:
  [
  T=\frac{q-p}{R_{\text{net}}}
  ]
• Alternating operation: When pipes are opened or closed at specified times, construct a time table with columns for interval, active pipes, net rate, and fraction added or removed.
• Emptying condition: A tank can be emptied only when the effective outlet rate exceeds the effective inlet rate:
  [
  R{\text{outlet}}>R{\text{inlet}}
  ]
• Units: Keep all rates in consistent units, such as litres per minute with minutes, or tank fractions per hour with hours. Converting (2) hours (30) minutes to (150) minutes avoids unit errors.