Time and Work

Overview

Time and Work appears about 1.8 times per paper across the last four AFCAT solved papers, placing it in the high yield band of Numerical Ability.

Time and work delivers roughly one to two AFCAT questions every paper — about 1.75 on the four-paper average. That is between three and six raw marks under the +3 / −1 scheme. The chapter sits in the high-yield Numerical Ability tier alongside time-speed-distance and ratio-proportion-and-mixtures, and like both of those it rewards method discipline far more than algebra.

Every AFCAT time-and-work item collapses to a single skill: converting the wording of the problem into a rate, in units of work per day or per hour. Once each agent has a rate, the rest of the question is one addition, one subtraction, or one division. The challenge is not the arithmetic; it is keeping the rates clean. The LCM-units method does exactly that — it replaces fractions with whole numbers and turns every problem into elementary integer manipulation.

This page builds the full toolkit. You get the side-by-side comparison of the fraction method versus the LCM-units method, the step-by-step LCM protocol, the two-agent and three-agent combined formulas, the efficiency-to-time-ratio conversion, a phase-by-phase template for staggered joining and leaving, the alternating-day cycle method, the wage-division rule, the group-size inverse-proportion shortcut, the integrated pipes-and-cisterns extension, and twelve worked examples that cover every AFCAT trap pattern. The aim is to walk into the exam treating every time-and-work item as a thirty-to-sixty-second pickup mark.

Why time-and-work is a reliable scorer

Three features make time-and-work one of the most bankable items on the AFCAT Numerical Ability paper.

• The wording is repetitive. Across the last four solved papers, every time-and-work question fell into one of five patterns: simple combined work, staggered joining, efficiency comparison, group-size change, or alternating-day work. AFCAT does not invent new structures here — it recycles the same five with new numbers.
• The arithmetic is small. Agent times are almost always single-digit or two-digit integers chosen so that the LCM is friendly — 12, 18, 20, 24, 30, 36, 60 and 72 appear repeatedly. Once the LCM is set, every rate is a clean whole number.
• The traps are spottable. Each of the five patterns has a single sentence in the question stem that flags which method to use. Train your eye to spot the flag and the rest of the work is mechanical.

Against the +3 / −1 scoring scheme, two correct time-and-work items add six marks before negatives. A candidate who has drilled the LCM protocol picks up both in under two minutes of paper time. A candidate who guesses on the staggered-joining trap walks away with one mark of damage and one mark of forgone gain — a four-mark swing on the same problem.

The two methods — fraction method vs LCM-units method

Every time-and-work problem can be solved two ways. The fraction method works in terms of ‘portion of total job completed per day’. The LCM-units method works in terms of ‘whole units of work completed per day’. Both are correct. One is faster.

The two methods deliver the same answer. The LCM method does it without writing a single fraction on the paper. For a single two-agent item the saving is small, perhaps ten seconds. For a three-agent item with pipes and emptying, the saving is forty to sixty seconds — enough to attempt one extra question elsewhere in the section.

Use the fraction method only when LCM computation itself is a hassle (rare in AFCAT, where the agent times are pre-chosen for cleanness). In every other situation, default to LCM.

The LCM-units protocol — step by step

Run this protocol every time. After ten drilled problems it becomes automatic.

1. List the agent times. Write each agent's solo time in a row at the top of your scratch space.
2. Compute the LCM. For two agents with times a and b, LCM = (a × b) ÷ HCF(a, b). For three agents, take the LCM of the first two, then LCM that with the third.
3. Set total work = LCM units. This is the key step. The total job is now a whole number of work-units, not a fraction.
4. Compute each rate. Agent's rate = total work ÷ agent's solo time. Because the LCM is divisible by every agent time, every rate is a clean integer.
5. Combine rates. Add the rates of agents working together. Subtract the rates of agents who oppose the work (emptying pipes, leak rates).
6. Solve. Time to finish = total work ÷ combined rate. If the question asks for work done in a given period, multiply the combined rate by that period.

Worked rationale. Take A alone in 20 days, B alone in 30 days, C alone in 60 days. LCM(20, 30, 60) = 60. Total work = 60 units. A's rate = 60/20 = 3 units/day; B = 60/30 = 2; C = 60/60 = 1. Combined = 6 units/day. Time together = 60 / 6 = 10 days.

Notice what did not appear: no common-denominator additions, no decimals, no carrying fractions through the calculation. Every number on the page was an integer.

Two-agent combined-work formula

When two agents work together, the shortcut formula avoids the LCM step entirely.

If A finishes a job alone in a days and B finishes alone in b days, together they finish in T = ab / (a + b) days.

This is a direct consequence of rate addition. A's rate is 1/a; B's rate is 1/b; combined rate is (a + b) / (ab); time is the reciprocal.

Worked check. A alone in 15 days, B alone in 10 days. T = (15 × 10) / (15 + 10) = 150 / 25 = 6 days. LCM verification: LCM(15, 10) = 30. A = 2 units/day; B = 3; combined = 5; time = 30 / 5 = 6 days. Both routes agree.

Use the ab / (a + b) shortcut whenever the problem has exactly two agents both working through the entire job. As soon as a third agent appears, or one agent joins or leaves part way, drop the shortcut and run the full LCM protocol instead.

Three-agent combined-work formula

For three agents, the rate-addition logic still holds but the algebraic shortcut is uglier.

If A, B, C finish alone in a, b, c days, together they finish in T = abc / (ab + bc + ca) days.

Most candidates compute this faster via the LCM method than by plugging into the formula. The formula is worth knowing for the rare case when AFCAT gives three odd agent times that share no friendly LCM — but in practice the LCM method wins on speed nine times out of ten.

Worked check. A alone in 6 days, B alone in 8 days, C alone in 12 days. By formula: T = (6 × 8 × 12) / (48 + 96 + 72) = 576 / 216 = 8/3 days ≈ 2.67 days. By LCM: LCM(6, 8, 12) = 24. Rates 4, 3, 2 units/day. Combined = 9. Time = 24/9 = 8/3 days. Match.

Efficiency comparison — k% more efficient

Efficiency is the inverse of time. If A is more efficient than B, A finishes the same job in less time. The exact conversion follows a single rule.

Ratio of times is the inverse of the ratio of efficiencies. If A's efficiency : B's efficiency = e₁ : e₂, then A's time : B's time = e₂ : e₁.

The conversion formula. If A is k% more efficient than B, then A's time / B's time = 100 / (100 + k). If A is k% less efficient than B, then A's time / B's time = 100 / (100 − k).

Staggered joining and leaving — phase-by-phase computation

When agents join or leave at different points in the job, the job has to be split into phases. Each phase has its own combined rate. Inside each phase the standard LCM logic applies.

Protocol.

1. Draw a simple timeline. Mark every event where an agent joins or leaves.
2. For each phase between two events, identify which agents are active.
3. Compute work done in that phase = combined rate of active agents × phase duration.
4. Sum the phase contributions. What remains is unfinished work, to be handled in the final phase.
5. If the final phase has a known set of active agents and an unknown duration, set its work contribution equal to the remaining units and solve.

Worked walk-through. A and B can together do a job in 12 days. A alone takes 20 days. They work together for 4 days, then A leaves and B finishes the rest alone. How long does the whole job take?

LCM(12, 20) = 60. Combined rate = 60/12 = 5 units/day. A's rate = 60/20 = 3 units/day. B's rate = 5 − 3 = 2 units/day. So B alone would take 60/2 = 30 days.

Phase 1 (A and B together, 4 days): work done = 5 × 4 = 20 units. Remaining = 60 − 20 = 40 units. Phase 2 (B alone): time = 40 / 2 = 20 days. Total project time = 4 + 20 = 24 days.

The phase diagram keeps the computation honest. Never try to do staggered-joining problems in your head — the working-out is short, but skipping it is the single biggest source of error in this chapter.

Alternating-day work — A on odd days, B on even days

Sometimes the question specifies that A and B do not work together but on alternate days — A on day 1, B on day 2, A on day 3, and so on. The two-day cycle, not the individual day, is the unit of computation.

Protocol.

1. Compute work done in one full cycle (two days for two alternating agents, three days for three).
2. Divide total work by cycle-work to find how many complete cycles fit before the job nears completion.
3. Handle the residual — the last fraction of a cycle — separately.
4. Total time = (full cycles × cycle length) + (residual time in the broken cycle).

Worked walk-through. A alone finishes a job in 18 days, B alone in 24 days. A starts on day 1, B on day 2, alternating. When is the job finished?

LCM(18, 24) = 72 units total work. A's rate = 4 units/day; B's = 3 units/day. One full two-day cycle delivers 4 + 3 = 7 units.

How many complete cycles fit inside 72 units? 72 ÷ 7 = 10 cycles with remainder 2 units. After 10 cycles (20 days) the work done is 70 units; 2 units remain. Day 21 is A's turn — A delivers 4 units in a full day but only 2 are needed, so A works for 2/4 = 1/2 day. Total time = 20 + 1/2 = 20.5 days.

Wage division — proportional to work done

When two or more agents finish a job and share a total wage, the share each agent receives is proportional to the work that agent contributed — not to the time each agent spent and not to the number of days each agent was present.

Wage ratio = ratio of work done = ratio of (rate × time present) for each agent.

For agents who worked through the entire job at their own rates, the work-done ratio equals the rate ratio, which equals the inverse of the solo-time ratio.

Worked walk-through. A alone in 10 days, B alone in 15 days. They work together and finish. Total wage ₹600. How is it split?

LCM(10, 15) = 30. A's rate = 3 units/day; B's = 2 units/day. Together they take 30 / 5 = 6 days. A's contribution = 3 × 6 = 18 units; B's = 2 × 6 = 12 units. Ratio = 18 : 12 = 3 : 2. A's share = 600 × 3/5 = ₹360; B's = 600 × 2/5 = ₹240.

Shortcut: when both agents work the entire time, the wage ratio is simply the ratio of their rates, which is the inverse of their solo-time ratio. Here solo times are 10 : 15 = 2 : 3, so wages are in the inverse ratio 3 : 2. No need to compute the actual project time.

If a third agent joins for only part of the project, however, you must use rate × time-present for each — the inverse-of-solo-time shortcut breaks the moment one agent's time on site differs from another's.

Group-size changes — inverse proportion

‘If 8 men finish a piece of work in 12 days, how many days will 12 men take?’ This family of problem rests on the work-equation M × D = constant, where M is the number of equally-skilled workers and D is the number of days.

M₁ × D₁ × H₁ = M₂ × D₂ × H₂, where M, D, H are workers, days and hours per day. Total man-hours are conserved.

If hours per day are constant, drop H and use just M × D = constant.

Worked walk-through one. 8 men finish work in 12 days. 12 men finish in? 8 × 12 = 12 × D → D = 96 / 12 = 8 days. More workers, fewer days, inverse proportion.

Worked walk-through two. 15 men working 8 hours a day finish a job in 20 days. How many days will 12 men working 10 hours a day take? 15 × 8 × 20 = 12 × 10 × D → 2400 = 120D → D = 20 days. Same number of days, because the reduced workforce is compensated by longer working hours.

Mixed group worked walk-through. AFCAT occasionally mixes men and women with different efficiencies. ‘1 man does the work of 2 women. If 6 men and 8 women finish a job in 10 days, how long will 4 men and 6 women take?’ Convert women to men-equivalents: 8 women = 4 men, so 6 men + 8 women = 10 men. Original work = 10 × 10 = 100 man-days. New crew = 4 men + 6 women = 4 + 3 = 7 men. New time = 100 / 7 ≈ 14.29 days.

Pipes and cisterns — the light version

Pipes-and-cisterns problems are time-and-work problems with a sign convention. A pipe that fills a tank in T hours has a rate of +1/T tank per hour. A pipe that empties a tank in T hours has a rate of −1/T per hour. The combined rate is the algebraic sum.

The LCM method extends cleanly. Set total tank capacity equal to the LCM of all pipe times. Each fill pipe contributes +(LCM / T) units per hour; each empty pipe contributes −(LCM / T) units per hour. Net rate gives net fill time.

Trap to watch. If the question says ‘pipe A fills the tank and pipe C empties it, and both are opened together, the tank fills in X hours’, the empty pipe must be slower than the fill pipe — otherwise the tank would never fill. AFCAT occasionally inverts the rates to test whether you catch the impossibility. If the empty rate exceeds the fill rate, the answer is ‘the tank will never fill’ or ‘the tank empties in N hours’ — read carefully before doing arithmetic.

Deep pipes-and-cisterns material (multiple inflow phases, tank initially half-full, leaks discovered after some hours) sits in the boats-streams-pipes-and-cisterns chapter. The light version covered here is all that AFCAT requires inside the time-and-work cluster.

Trap patterns AFCAT uses

Four wording patterns account for almost every wrong answer in this chapter. Memorise the flag word in each.

• Staggered joining, single combined number. ‘A and B together can do a job in 12 days. They work together for 5 days, then B leaves. A finishes the rest in 14 days. How long would A take alone?’ The flag is ‘then X leaves’ or ‘then X joins’. Compute work in phase one, subtract, set up the phase-two equation, solve.
• Efficiency phrased as a ratio of work, not time. ‘A is twice as good a workman as B’ means A is twice as efficient, so A takes half the time. Many candidates invert this — they read ‘twice as good’ and assume A takes twice as long.
• Group-size change with extra hours. ‘If 10 men can finish a job in 15 days working 8 hours a day, how long will 12 men take working 6 hours a day?’ The flag is the second hours-per-day figure. Use M × D × H = constant; never drop the H term when both rows specify hours.
• Wage division with unequal time on site. ‘A worked for 6 days, B for 4 days, C for 2 days, and they earned ₹X between them.’ The wage ratio is rate × time-present for each — never the rate alone, never the inverse-of-solo-time shortcut. The flag is any phrasing where the agents are present on site for different numbers of days.

If you can identify which of the four flags is in front of you within ten seconds of reading the stem, the rest of the problem is mechanical. The trap is in the misread, not in the maths.

Time budget and pacing

Across an AFCAT paper of 100 questions in 120 minutes, the average time budget is 72 seconds per item. Numerical Ability items run slightly hotter than average because they need scratch work; the practical target inside this section is 60 to 90 seconds per question.

• Simple combined work (two agents, both throughout): 30 to 45 seconds. The ab / (a + b) shortcut should land the answer in under thirty seconds for friendly numbers.
• Three-agent combined work or pipes: 60 to 75 seconds. Set LCM, compute three rates, sum, divide.
• Efficiency comparison: 45 seconds. The 100 / (100 ± k) ratio should be in your head within five seconds of reading the percentage.
• Staggered joining or wage division: 75 to 90 seconds. The phase diagram costs time but eliminates errors. Do not skip it.
• Alternating-day cycle: 60 to 75 seconds. The cycle-work computation is the slow step.

If a time-and-work item is still unsolved after 90 seconds, mark it for review and move on. The opportunity cost of one wrong attempt is four marks in the +3 / −1 scheme — a guess on this chapter rarely pays. Skipping a single hard item to attempt two easy items elsewhere is the correct trade.

Worked AFCAT-style examples

Exam-day strategy

1. Default to the LCM-units protocol on every problem. Treat the fraction method as a fallback for the rare item where the LCM is awkward.
2. For two-agent items where both work throughout, deploy the ab / (a + b) shortcut. It saves ten seconds per question and removes the LCM step entirely.
3. Translate efficiency wording into a time ratio inside five seconds of reading the stem. ‘k% more efficient’ means time ratio 100 : (100 + k); ‘k% less efficient’ means 100 : (100 − k); ‘x times as efficient’ means time ratio 1 : x. Get this conversion automatic.
4. Draw a phase diagram for every staggered-joining problem. It costs ten seconds but eliminates the single largest source of errors in this chapter.
5. Spot wage-division traps by the phrasing. If the agents are present on site for different numbers of days, the wage ratio is rate × time-present — never the rate alone.
6. On group-size problems, write the M × D × H equation in full before substituting. Dropping the H term when both rows specify hours is the most common slip.
7. On pipes-and-cisterns problems, write fill rates with a plus sign and empty rates with a minus sign in your scratch work. This single notational habit prevents sign errors.
8. Target 60 to 90 seconds per item. If you are still unsolved after 90 seconds, mark for review and move on — the +3 / −1 maths does not reward stubbornness on a single problem.

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