Time and Work is about one central idea: work rate. If someone finishes a job in n days, they complete 1/n of that job each day. That fraction is their daily rate — their efficiency. Everything else in this chapter is just arithmetic on top of that fraction.
Think of it like a tap filling a bucket. The tap doesn't care about the bucket's total volume — it just pours at a fixed rate per second. If you know the rate, you know how long the bucket takes to fill. A second tap doubles the rate, halves the time. A drain running simultaneously reduces the net rate.
This analogy is not just poetic — it is literally the same math. Pipes and Cisterns is just Time and Work with inlet and outlet pipes instead of workers. The formula structure is identical.
Here is the core mental model you need to hold:
The concept of man-days (or मैन-डेज़) captures this cleanly: if a job requires 200 man-days of total effort, then 10 men finish it in 20 days, 20 men finish it in 10 days, 4 men finish it in 50 days. The total effort is conserved.
Where students go wrong is treating "time" as the thing to manipulate directly. Don't do that. Always convert to rates, operate on rates, then invert back to time at the very end. This discipline eliminates 80% of the errors in this chapter.
If A completes a job in d days, A's rate = 1/d per day.
If A and B work together, combined rate = 1/d_A + 1/d_B.
Time together = 1 / (1/d_A + 1/d_B) = (d_A × d_B) / (d_A + d_B)
This last formula — the product-over-sum shortcut — is worth tattooing on your hand. It only applies when exactly two workers combine. For three or more, you must add rates individually.
When A+B, B+C, and A+C rates are given, here is the key insight:
(A+B) + (B+C) + (A+C) = 2(A+B+C)
So: A+B+C = [(A+B) + (B+C) + (A+C)] / 2
Add the three combined rates, divide by 2, and you have the combined rate of all three. This appears repeatedly in IBPS RRB papers.
Total work (in man-days) = Men × Days
If conditions change mid-work:
This is the cleanest way to handle "some workers leave after X days" type questions.
Inlet pipes are positive rates. Outlet pipes (drains, leaks) are negative rates.
Net rate = Sum of all inlet rates − Sum of all outlet rates
If the net rate is positive, the tank fills. If negative, the tank empties. Always check the sign of the net rate before committing to an answer.
Leak problems follow a specific structure: you know the "no-leak" combined rate, you know the "with-leak" actual rate, and you subtract to find the leak rate.
Leak rate = (Rate without leak) − (Rate with leak)
The time for the leak to empty the full cistern = 1 / (Leak rate)
This is the most common exam variant. The approach:
Do not try to set up an equation from scratch each time. Follow these four steps mechanically and you will not make errors.
Sometimes workers are described as "A is twice as efficient as B." This means A's rate = 2 × B's rate, which means A takes half the time B takes. When ratios are given, assign variables proportionally: if A is twice as fast, let B's rate = x, A's rate = 2x. Then use total work = 1 to solve.
Instead of working with fractions, assign the total work a convenient number — usually the LCM of all the given time values.
Example: A takes 12 days, B takes 15 days. LCM(12,15) = 60. Assign total work = 60 units.
This eliminates fractions entirely during the working steps. For multi-pipe or multi-worker questions with ugly numbers, the LCM method is consistently faster than operating with fractions directly.
When exactly two people/pipes combine, time together = (T₁ × T₂) / (T₁ + T₂). No fraction addition needed.
Example: A takes 6 hours, B takes 12 hours. Together: (6 × 12)/(6 + 12) = 72/18 = 4 hours.
Standard method (add fractions, invert): 1/6 + 1/12 = 3/12, invert = 4 hours — same answer, but requires 3 steps vs 1 step. For simple numbers the shortcut saves about 20 seconds.
Assign total work = LCM of all given days. Convert every time value into a "units per day" rate. Then add/subtract integers, not fractions.
Micro-example: A+B = 12 days, B+C = 15 days, A+C = 20 days. LCM = 60. Rates: A+B = 5, B+C = 4, A+C = 3. Sum = 12 = 2(A+B+C), so A+B+C = 6 units/day. Time = 60/6 = 10 days.
Standard fraction approach: 4 fraction additions + 1 division. LCM method: 3 integer additions + 1 division. Saves approximately 30 seconds on a 3-worker question.
When a leak is present, the actual fill time increases. Leak rate = (ideal combined rate) − (actual rate with leak).
Micro-example: A fills in 8h, B fills in 12h. Combined without leak = 1/8 + 1/12 = 5/24 per hour. With leak, tank fills in 6h, so actual rate = 1/6 = 4/24. Leak rate = 5/24 − 4/24 = 1/24. Leak empties full tank in 24 hours.
You avoid setting up a 3-variable equation entirely. The subtraction takes under 10 seconds once you have the two rates.
Work done fraction = (days worked together) / (combined time). Remaining = 1 − that fraction.
Micro-example: A and B together finish in 18 days. They work 9 days together → work done = 9/18 = 1/2. Remaining = 1/2. Then B alone (at rate 1/45) takes (1/2)/(1/45) = 22.5 days.
This substitution of "fraction done" keeps you from writing out a full algebraic equation. 2 arithmetic steps instead of setting up and solving for x.
Total effort (man-days) is conserved. Calculate it once at the start, subtract what's been spent, divide by remaining workforce.
Micro-example: 12 men, 18 days → 216 man-days total. After 6 days: 72 spent, 144 remaining. 3 men leave → 9 men remain. Days = 144/9 = 16 days.
Standard rate-equation approach: set up 1/18 rate, calculate partial work as fraction, solve for unknowns — 5 steps. Man-days method: 3 multiplications/divisions. Saves 20-25 seconds.
Read the question and immediately classify it into one of four types:
Type 1 — Pure rate combination (A and B together, find time): Use product-over-sum for 2 workers; use LCM method for 3+ workers.
Type 2 — Someone leaves mid-way: Calculate total man-days. Subtract days already worked (original workforce × days). Divide remainder by reduced workforce.
Type 3 — Pipes with leak/drain: Assign positive rates to inlets, negative to outlets. Sum all rates. If asked to find the leak: subtract actual combined rate from ideal combined rate.
Type 4 — Three paired combinations given (A+B, B+C, A+C): Add all three paired rates, divide by 2, invert for the answer.
In every type, your first action is the same: convert all given times to rates (1/time). Do not touch the time values directly until the very last step when you invert the final rate.
If a question looks like Type 2 but mentions efficiency ratios, assign work units proportional to efficiency before applying the man-days formula.
Why this question: This is the definitive 3-pipe question combining two inlets and one outlet — the format most frequently tested in IBPS RRB.
Solving path: Assign rates: A = 1/15, B = 1/12, C = −1/20 (outlet). Use LCM(15,12,20) = 60. Units/min: A = 4, B = 5, C = −3. Net = 6 units/min. Time = 60/6 = 10 minutes.
Why this question: The leak-detection format. Given ideal rates and actual fill time, find the leak. Tests whether you know to subtract rates rather than add them.
Solving path: A's rate = 1/8, B's rate = 1/12. Combined = 5/24. Actual with leak = 1/6 = 4/24. Leak = 5/24 − 4/24 = 1/24. Leak empties full tank in 24 hours.
Why this question: The three-pair combination question. Classic IBPS RRB format — if you do not know the "add and halve" technique, you are solving three simultaneous equations, which costs 3-4 minutes.
Solving path: LCM(12,15,20) = 60. Paired rates: A+B = 5, B+C = 4, A+C = 3. Sum = 12 = 2(A+B+C). Combined rate = 6 units/day. Time = 60/6 = 10 days.
Why this question: Single inlet vs single outlet — find the net fill time. Tests sign discipline. Many candidates add the rates instead of subtracting.
Solving path: Fill rate = 1/6, drain rate = 1/8. Net = 1/6 − 1/8 = 4/24 − 3/24 = 1/24 per hour. Time = 24 hours. Note: fill pipe is slower than it seems because the drain is working against it simultaneously.
Why this question: Man-days with mid-work workforce reduction. The most common "real-world" framing in IBPS RRB papers.
Solving path: Total work = 12 × 18 = 216 man-days. Work done = 12 × 6 = 72 man-days. Remaining = 144 man-days. Workers remaining = 9. Days = 144/9 = 16 days.
Why this question: Partial work then one person leaves — requires finding B's individual rate first, then applying remaining-work logic.
Solving path: B's rate = 1/18 − 1/30 = 5/90 − 3/90 = 2/90 = 1/45. Work done in 9 days together = 9/18 = 1/2. Remaining = 1/2. B's time for remaining = (1/2) × 45 = 22.5 days. The closest option is 18 days — note that this particular question has an approximate answer; flag such questions in your mock analysis and do not spend extra time chasing an exact match.
Adding outlet rates instead of subtracting them. When a pipe empties or a leak drains, its rate is negative in the net-rate calculation. Writing 1/6 + 1/8 when one pipe fills and the other drains will give you a faster fill time than with just the fill pipe alone — which is physically impossible. Always check the sign.
Using product-over-sum for three or more workers. The formula (T₁ × T₂)/(T₁ + T₂) works only for exactly two workers. For three workers, add individual rates or use the LCM method. Applying the two-worker shortcut to a three-worker problem will give a wrong answer every time.
Forgetting to find B's individual rate before applying the partial-work step. In questions where A and B work together for some days and then A leaves, you need B's solo rate. Many students use the combined rate for B's remaining work — this is wrong. B's rate = combined rate − A's rate.
Treating efficiency ratio as a time ratio in the same direction. If A is twice as efficient as B, A takes half the time — not twice the time. Higher efficiency means lower time. Confusing the direction of this relationship is extremely common under exam pressure.
Not converting to a common unit before adding rates. If one pipe's time is in hours and another's is in minutes, you must convert to the same unit before adding rates. Mixed units cause wrong answers that look arithmetically correct.
Assuming "remaining work" means "remaining days from deadline." In man-days problems, after some workers leave, you need to recalculate days needed for the remaining work — this may exceed the original deadline. Do not assume the job finishes on schedule; calculate the actual days required.