Time and Work is one of the highest-return chapters in AFCAT Numerical Ability: 1 to 3 questions appear almost every shift, and they are pure formula plus arithmetic. The trap is that most students fight with messy fractions like 1/12 + 1/15. The Cavalier teaches you the LCM unit method so you work only with whole numbers and finish each question in under 40 seconds.
The one idea behind every question
Time and Work rests on a single relationship: Work = Rate × Time. If one person finishes a job in n days, then in one day they do 1/n of the job. That one-day fraction is called the person's rate or efficiency.
So if A finishes a wall in 10 days, A's one-day work is 1/10. If B finishes it in 15 days, B's one-day work is 1/15. Working together, their combined one-day work is 1/10 + 1/15. Every AFCAT question is just a rearrangement of this idea: you are given two of the three quantities (work, rate, time) and asked for the third.
The whole job is always treated as one complete unit of work, no matter whether it is a wall, a field, a report or a tank of water. That is why a person who takes n days does 1/n of it daily. Once you internalise that the job equals 1, every fraction in this chapter starts to make sense, and you will see that filling a tank, building a wall and typing a document are mathematically identical problems wearing different clothes.
One-day work and number of days are reciprocals. More days → slower → smaller daily fraction. More efficiency → fewer days.
The LCM unit method (your main weapon)
Fractions waste time. Instead, assume the total work equals the LCM of the given days. Then each worker's daily output becomes a clean whole number. This is the single most important habit to build for AFCAT, because the calculator-free, time-bound exam punishes slow fraction work heavily. The answer never depends on what number you choose for total work, so choosing the LCM is purely a convenience that keeps every intermediate value an integer.
- Take LCM of all the given day-figures → call it total work (in "units").
- Each worker's daily units = total work ÷ that worker's days.
- Add or subtract daily units as the question demands.
- Required days = total work ÷ combined daily units.
A does a job in 12 days, B in 18 days. Working together, in how many days do they finish?
A's rate = 36 ÷ 12 = 3 units/day
B's rate = 36 ÷ 18 = 2 units/day
Together = 3 + 2 = 5 units/day
Days = 36 ÷ 5 = 7.2 days = 7⅕ days
Never touch 1/12 + 1/18 in the exam. The LCM method turns it into 3 + 2 = 5. Whole numbers are faster and less error-prone under time pressure.
The two-worker shortcut formula
When exactly two workers act together, there is a direct formula you can apply mentally without even setting up the LCM. This is the fastest route when only two people are involved and the day-figures are small enough to multiply in your head.
If A takes x days and B takes y days, together they take (x × y) ÷ (x + y) days. The numerator is the product of the two times and the denominator is their sum, so the answer is always smaller than the faster person's time alone, which is a handy sanity check before you commit to an option.
A in 20 days, B in 30 days. Together?
= 600 ÷ 50 = 12 days
This product-over-sum shortcut works only for two workers. For three or more, fall back on the LCM unit method.
When one person leaves or undoes work
Sometimes one worker builds and another destroys, or someone leaves partway. Treat destroying as a negative rate and subtract. The mental model is a bank account: deposits are positive, withdrawals are negative, and the net daily change tells you how fast the balance, that is the completed work, grows.
A can build a wall in 8 days; B can demolish it in 12 days. If both work together, how long to build?
A builds = 24 ÷ 8 = +3 units/day
B demolishes = 24 ÷ 12 = −2 units/day
Net = 3 − 2 = 1 unit/day
Days = 24 ÷ 1 = 24 days
Students add B's rate instead of subtracting it. Read whether the second agent helps or opposes the job before choosing the sign.
Someone joins or leaves midway
For problems where a worker leaves after a few days, compute the work done so far in units, subtract from total, and let the remaining worker finish the rest.
A and B can finish a job in 10 and 15 days. They start together but A leaves after 2 days. In how many more days does B finish?
A = 3 units/day, B = 2 units/day
Day 1 to 2 together = (3 + 2) × 2 = 10 units done
Remaining = 30 − 10 = 20 units
B alone = 20 ÷ 2 = 10 more days
Working on alternate days
When workers take turns day by day, sum one full cycle (two days) of units, see how many complete cycles fit, then finish the leftover.
A finishes in 9 days, B in 18 days. They work on alternate days starting with A. How many days total?
A = 2 units/day, B = 1 unit/day
One cycle (A then B) = 2 + 1 = 3 units in 2 days
5 cycles = 15 units in 10 days
Remaining = 3 units; Day 11 is A's turn → A does 2, leaving 1
Day 12 is B's turn → B does 1 → done
Total = 12 days
Always note who starts the cycle. Starting with the faster worker can shave off the final partial day.
Efficiency ratios and 'twice as fast'
Efficiency is inversely proportional to time. If A is twice as efficient as B, A takes half the time of B. So an efficiency ratio of A:B = 2:1 means a time ratio of A:B = 1:2. Phrases like "50% more efficient", "twice as good" or "works at three-fourths the speed" are all just efficiency ratios in words; translate them into a clean ratio first, then flip to get the time ratio.
A is thrice as efficient as B and finishes a job 40 days sooner than B. Find the time each takes.
Let A = t, B = 3t; difference 3t − t = 2t = 40
t = 20 → A = 20 days, B = 60 days
Convert any efficiency ratio into a time ratio by simply flipping it. This single move solves most ratio-based questions.
Sharing of wages
Wages are divided in the ratio of work actually done, which equals the ratio of efficiencies (when everyone works the same number of days) or, more generally, efficiency × days each worked.
A, B and C finish a job in 6, 8 and 12 days respectively, working together. They earn ₹3,000. Find C's share.
Rates: A = 4, B = 3, C = 2 units/day
Ratio of work = 4 : 3 : 2 (total 9 parts)
C's share = (2 ÷ 9) × 3000 = ₹666.67
Do not split wages in the ratio of days taken. Split in the ratio of work done (efficiency × time). The slower worker earns less, not more.
The M×D×H work-equation
When the number of men, days, hours per day or amount of work changes between two situations, do not try to track individual rates. Use the master equation, which captures the fact that men, days and hours all multiply together to produce work:
(M₁ × D₁ × H₁) ÷ W₁ = (M₂ × D₂ × H₂) ÷ W₂
Here M = men, D = days, H = hours/day, W = work. Plug in the seven known values and solve for the eighth.
15 men working 8 hours a day finish a road in 12 days. How many days will 18 men working 10 hours a day take for the same road?
15 × 8 × 12 = 18 × D₂ × 10
1440 = 180 × D₂
D₂ = 1440 ÷ 180 = 8 days
If the second job is bigger or smaller, divide each side by its W. Same job both sides → drop W entirely.
Pipes and cisterns
Pipes and cisterns is Time and Work in disguise, and AFCAT loves to set at least one such question per paper. A filling pipe is a positive rate; a drain or leak is a negative rate. The tank capacity is your LCM unit total. The only new wrinkle is direction: water can be added or removed, so you must read each pipe carefully and tag it with the correct sign before you start adding rates.
Pipe A fills a tank in 6 hours, pipe B in 8 hours, and an outlet C empties it in 12 hours. All open together, how long to fill?
A = +4, B = +3, C = −2 units/hour
Net = 4 + 3 − 2 = 5 units/hour
Time = 24 ÷ 5 = 4.8 hours = 4 hr 48 min
If a question says a leak empties a full tank in t hours, treat the leak exactly like a worker who removes capacity-units per hour. Same maths, just a minus sign.
Speed shortcuts to memorise
- Two workers: together time = xy ÷ (x + y).
- Efficiency ↔ time: flip the ratio. 2:3 efficiency → 3:2 time.
- Work done in d days = rate × d (in LCM units).
- Wages ∝ efficiency × days worked, never days taken.
- n workers each of t days finish in t ÷ n days.
- If A does work in a days and A&B together in c days, then B alone = ac ÷ (a − c) days.
- Set total work = LCM of all given days.
- Rate = total ÷ each person's days; add to combine, subtract to oppose.
- Days = total work ÷ net rate.
- Two workers: xy ÷ (x + y). Efficiency ratio flips to time ratio.
- Wages split by work done; pipes use the same minus-sign trick for leaks.
Previous-year style practice
Q. A and B together can complete a piece of work in 12 days, while B and C together can do it in 16 days. A works alone for 5 days, then B works alone for 7 days, and C finishes the remaining work in 13 days. In how many days can C alone complete the whole work?
Answer: Let total work = LCM(12, 16) = 48 units. A+B = 4 units/day, B+C = 3 units/day. The phased work gives: 5·A + 7·B + 13·C = 48. Writing 5A + 5B = 5(A+B) = 20 and 2B + 2C = 2(B+C) = 6, regroup: 5(A+B) + 2(B+C) + 11C = 48 → 20 + 6 + 11C = 48 → 11C = 22 → C = 2 units/day. So C alone = 48 ÷ 2 = 24 days.
Phased problems crack open when you regroup terms into the combined rates you already know (A+B, B+C). Look for those pairings before brute-forcing.
Frequently asked questions
How many Time and Work questions come in AFCAT?
Typically 1 to 3 questions per shift in the Numerical Ability section, often clubbed with pipes and cisterns. It is a high-frequency, high-accuracy topic, so it deserves dedicated practice.
Should I use fractions or the LCM unit method?
Use the LCM unit method. By setting total work equal to the LCM of the given days, every rate becomes a whole number, so you avoid slow fraction addition and reduce careless slips under exam pressure.
How is the wages-sharing ratio decided?
Wages are shared in the ratio of work actually done, which equals efficiency multiplied by the number of days each person worked. Never divide wages in the ratio of days taken to finish alone.
How are pipes and cisterns different from Time and Work?
They are the same maths. A filling pipe is a positive rate, an outlet or leak is a negative rate, and the tank capacity is your LCM unit total. Add inflows, subtract outflows, then divide capacity by the net rate.
What is the fastest formula for two people working together?
If A alone takes x days and B alone takes y days, together they take (x times y) divided by (x plus y) days. Use this product-over-sum shortcut only for exactly two workers.
How do I convert an efficiency ratio into days?
Flip it. Efficiency and time are inversely proportional, so an efficiency ratio of A to B as 3 to 1 means a time ratio of 1 to 3. The more efficient worker takes proportionally fewer days.
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