Time and Work is built on one idea so simple it's almost embarrassing: rate × time = work done. That's it. Everything — from A-and-B-together problems to pipes filling cisterns to contractors adding extra workers — reduces to this single equation.
Think of it like water flowing from taps. If one tap fills a bucket in 4 minutes, it delivers 1/4 of a bucket per minute — that's its rate. A second tap filling the same bucket in 6 minutes delivers 1/6 per minute. Open both together: 1/4 + 1/6 = 5/12 per minute. Bucket fills in 12/5 = 2.4 minutes. You never thought about buckets differently — you just added rates.
The analogy holds for human workers. A person who finishes a job in 20 days completes 1/20 of the job per day. Pair them with someone who finishes in 30 days (rate: 1/30 per day), and together they do 1/20 + 1/30 = 5/60 = 1/12 per day, finishing in 12 days.
Here's why this matters for AFCAT specifically: the exam tests not just your formula recall but your speed. A Time and Work problem that takes a textbook student 3 minutes should take you under 90 seconds. The difference is not effort — it's method. Most AFCAT Time and Work questions fall into four recognizable templates, and once you pattern-match correctly, you're filling in slots rather than solving from scratch.
The four templates are:
The underlying mathematics is identical across all four. What changes is only the surface language.
Total work is conventionally taken as 1 unit (one complete job). If a person completes the job in n days, their rate is 1/n per day.
For multiple workers: if rates are a, b, c per day, their combined rate is a + b + c, and time to finish is 1/(a + b + c).
This gives you the master formula for two-person problems:
If A takes p days and B takes q days, together they finish in:
Do not memorize this as a separate formula — it's just 1/(1/p + 1/q) simplified. Knowing the derivation means you can never get confused about when it applies.
For multi-person problems, fractions slow you down. Use LCM of all time values as your "total work units."
Example: A finishes in 12 days, B in 18 days. LCM(12, 18) = 36. Assign total work = 36 units.
No fractions at any step. This is the method you should default to in the exam hall.
When the workforce, hours per day, or number of days changes, total work stays constant:
where M = men, D = days, H = hours per day. This is the man-hours identity. Plug in the known values, solve for the unknown. It works because total work = man-hours of effort required, and that doesn't change just because you reorganize the workforce.
This template catches a lot of people. The key is to calculate work done in each phase separately, then track remaining work.
Phase approach:
Watch for problems that say "A and B work together for X days, then B leaves." That's two phases. The total adds to 1.
Structurally identical to human workers. Filling pipes have positive rates; emptying (outlet) pipes have negative rates.
Net rate = (sum of filling pipe rates) − (sum of emptying pipe rates)
If net rate is positive, the cistern fills. Time = 1 / (net rate). If net rate is negative, the cistern empties — and the question likely tells you the current fill level and asks when it empties.
When you're given A+B, B+C, and C+A times — a classic AFCAT pattern — don't try to find individual rates separately. Add all three pair equations:
Divide by 2 to get the all-three rate directly. Time = reciprocal of that rate. This saves you solving a three-variable system.
These problems give you a deadline, show that work is behind schedule, and ask how many extra workers are needed. The approach:
When you see any problem with 2–4 workers each given individual completion times, immediately take LCM of those times as total work. Assign each worker their slots-per-day (LCM ÷ their time). Add slots, divide into total. Zero fraction arithmetic.
Example: A=15 days, B=20 days, C=30 days. LCM=60. Slots: A=4, B=3, C=2. Together=9/day. Time=60/9=20/3 days.
Standard fraction method: 4 addition steps, common denominator hunting — roughly 60s. LCM method: 3 divisions, 1 addition, 1 division — under 20s.
When given three pair-times (A+B, B+C, C+A), add all three pair rates and halve. This gives the all-three rate directly without solving simultaneous equations.
Standard approach (solving 3 equations for a, b, c individually): 6–8 algebraic steps, ~90s. Pair-sum method: 3 reciprocals, 1 addition, 1 halving — under 30s.
For M×D×H problems, draw a 2×3 table: rows = Case 1 / Case 2, columns = Men / Days / Hours. Fill known values. The unknown column is what you solve for. Cross-multiply: product of Case 1 = product of Case 2. No formula to remember — the identity is just product conservation.
This eliminates mis-substitution errors. Saves roughly 2 mis-steps (and potentially the entire question if you'd confused which variable to solve for).
Fill in the number of filling pipes with a + sign and emptying pipes with a − sign directly in one expression. Never convert to a common denominator first — add fractions with a sign directly using the formula (ad ± bc)/bd for two-pipe problems.
Two pipes: fill in 3h, empty in 5h. Net = 1/3 − 1/5 = (5−3)/15 = 2/15. Time = 7.5h. Three steps total versus the standard "find LCD=15, convert each fraction" approach which has 5 steps.
In partial-work problems, after computing work done, sanity-check using percentages. If A+B work for 6 days at rate 7/72 per day, work done = 42/72 ≈ 58%. Remaining ≈ 42%. This quick percentage estimate catches arithmetic errors before you commit to a wrong answer — takes 5 seconds and eliminates the most common careless mistake in multi-phase problems.
Look at the problem and answer these questions in order:
Step 1 — What type?
pq/(p+q) or LCM (2-person)Step 2 — Set total work. Always use LCM of the given time values unless the problem involves hours and days together (then use man-hours directly).
Step 3 — Write rates, not times. Convert every "finishes in X days" to "rate = total work / X." Work in units, not fractions, wherever possible.
Step 4 — Set up one equation. Rate × Time = Work. Solve for the unknown.
Step 5 — Check sign. If it's a pipes problem, confirm net rate is positive (filling). If your answer is negative days or more men than exist, you have a sign error somewhere.
Total time in exam hall: 60–90 seconds for standard problems, 90–120 seconds for multi-phase or contractor problems.
Why this question: The foundational man-hours identity — every AFCAT has at least one of these. If you can't do this in under 60 seconds, you're spending too long on setup.
Solving path: Total work = 12 × 8 × 15 = 1440 man-hours. New scenario: men × 6 × 20 = 1440. Men = 1440/120 = 12. The answer is the same as the original workforce — notice that before solving. The reduced daily hours and increased days exactly cancel. That's the kind of observation that saves 20 seconds.
Why this question: Workforce-departure mid-problem. Tests whether you can track phases without mixing up which workers are active when.
Solving path: Total work = 15 × 10 = 150 man-days (LCM is trivial here, just the product). Work done in 4 days by 15 men = 60 man-days. Remaining = 90 man-days. Workers now = 10. Days = 90/10 = 9. Clean, no fractions needed.
Why this question: Classic pipes problem. Tests whether you know that an emptying pipe subtracts from the filling rate — and whether you correctly take the reciprocal at the end.
Solving path: Filling rate = 1/3, emptying rate = 1/5. Net = 1/3 − 1/5 = 2/15 per hour. Time = 15/2 = 7.5 hours. The trap here is adding instead of subtracting — if you got 15/8, you added the rates.
Why this question: Three-pair system. The most powerful shortcut in all of Time and Work. If you know the pair-sum method, this is a 30-second problem. Without it, it's 2 minutes.
Solving path: A+B = 1/30, B+C = 1/24, C+A = 1/20. Sum = 1/30 + 1/24 + 1/20 = (4+5+6)/120 = 15/120 = 1/8. So 2(a+b+c) = 1/8, meaning a+b+c = 1/16. All three together finish in 16 days.
Why this question: Contractor/deadline problem — the hardest template in this chapter. Tests whether you can re-derive the effective rate from past performance and apply it forward.
Solving path: 20 men in 32 days completed 4/5 of the work. Rate per man = (4/5)/(20×32) = 4/(5×640) = 1/800 per man per day. Remaining work = 1/5. Remaining time = 8 days. Men needed: n × 8 × (1/800) = 1/5. n = 800/(5×8) = 800/40 = 20... wait, that gives total men = 20. So additional men = 20 − 20 = 0? Let me reread the explanation — the explanation gives 40 total men needed, so additional = 20.
Re-check: rate per man = (4/5) ÷ (20 men × 32 days) = (4/5)/640 = 1/800. To do 1/5 in 8 days: n × 8 / 800 = 1/5. n = 800/(5×8) = 20. That gives 20 total men needed, additional = 0. But the answer is 20 additional. The correct derivation per the explanation: total work = 20 × 40 = 800 man-days. Work done = 4/5 × 800 = 640 man-days. Remaining = 160 man-days. In 8 days: men = 160/8 = 20. Current men = 20. Additional = 0.
Look — there's an inconsistency in the setup and explanation here. The explanation in the spec itself notes internal calculation issues. The reliable approach: use total work = M×D = 20×40 = 800 man-days. Done in 32 days by 20 men = 640 man-days = 4/5 of 800. Remaining = 160 man-days in 8 days → need 20 men. Additional = 20−20 = 0. The answer stated is 20 additional, implying the intended total needed = 40, which would mean the work is not proportional as given. Treat this as a problem where you set up the man-days equation correctly and verify against options.
Adding instead of subtracting for outlet pipes. An emptying pipe reduces the net fill rate. If you see "one pipe fills, another empties," the emptying pipe's rate carries a minus sign. Always confirm which direction each pipe works before adding rates.
Using pq/(p+q) when work phases differ. The combined-time formula only works when both people work for the entire duration. The moment someone leaves or joins mid-problem, switch to phase-by-phase calculation.
Forgetting to subtract work already done. In partial-work problems, students compute the rate for the second phase correctly but then divide it into the total work instead of the remaining work. Always track what fraction is left before computing the second-phase duration.
Not simplifying LCM before computing. Taking LCM of 18, 24, 36 is 72 — not 18×24×36. Computing with an unnecessarily large total-work number doesn't cause wrong answers, but it multiplies your arithmetic steps. Always simplify LCM properly.
Misreading "additional men needed" vs. "total men needed." The contractor problem explicitly asks for additional men. If your computation gives the total workforce required, remember to subtract the men already working. This is the most common final-step error in this template.
Assuming equal efficiency across all workers. If the problem says "12 men" without qualification, assume equal efficiency. If the problem gives different individual times, do not average them — work with each person's rate separately. Averaging completion times is always wrong.