Time and Work is built on one foundational idea: work rate is the reciprocal of time. If someone finishes a job in 10 days, they complete 1/10 of the job each day. That's it. Everything else in this chapter is just applying this idea to more complex combinations.
Think of it like this — imagine filling a bucket with water. One tap fills the bucket in 4 minutes. Another tap fills it in 6 minutes. If both taps run together, they're adding water at their individual rates simultaneously. The bucket doesn't "know" which tap is which — it just fills up faster. Time and Work problems operate on exactly this logic.
The three building blocks you'll keep coming back to:
1. Individual rate → Combined rate → Total time When two or more people/machines work together, add their individual rates. Then flip the sum to get the total time.
2. Man-Days formula Total work = Men × Days (or Machines × Hours, Workers × Days, etc.). This is the "unit of work" thinking. If 8 workers finish a job in 6 days, the total work is 48 man-days. Any other combination that multiplies to 48 gives the same amount of work.
3. Proportional output
When work output scales with people and time (e.g., meters of track laid, sleepers produced), you use a three-variable proportion: Work = Workers × Days × Rate per worker per day.
RRB Group D keeps this chapter clean — you won't see complex "A leaves after 3 days, B joins on day 5" chains that IBPS loves. The questions here are direct: two-worker combinations, man-days scaling, pipe filling. Nail these three building blocks and you're covering almost every variant that appears.
One conceptual trap to avoid early: more workers = less time (inverse relationship), but more workers = more work done (direct relationship). Students mix these up under pressure. The man-days formula handles both naturally, so always anchor to it.
For any worker or pipe, define their rate as:
For combined work:
Standard two-person formula: If A takes a days and B takes b days:
Look — you don't need to derive this every time. But know why it works: 1/a + 1/b = (a+b)/ab, and the reciprocal is ab/(a+b). This formula saves one arithmetic step in the exam hall.
Instead of working with fractions like 1/12 and 1/18, assign the total work a concrete number equal to the LCM of all given time values. Then work in whole numbers.
Example: A finishes in 12 days, B in 18 days.
No fractions involved until the final step. This is faster than the rate-fraction method for most Group D questions.
The core formula:
where W is the quantity of work/output.
When output scales: e.g., meters of track, number of sleepers. Here, more days or more workers proportionally increase output. Set up the proportionality:
Find the unit rate from the given data, then scale.
Example: 8 workers repair 240 m in 6 days.
Pipes fill a tank (positive work). Outlet pipes/leaks drain it (negative work). The logic is identical to workers — just subtract the drain rate from the fill rate.
If pipe A fills in 4 hours and pipe B fills in 6 hours:
Some questions say "one person works alone for X days, then both work together." Here's the method:
Always convert remaining work to a fraction of total before dividing. Mixing up "fraction left" with "days left" is the single most common error in this question type.
Assign total work = LCM of all time values. Convert each person's time into a daily "units" count. Add units, divide total by sum.
Example: A in 12 days, B in 15 days. LCM = 60. A does 5 units/day, B does 4 units/day. Together: 9 units/day. Time = 60/9 = 6.67 days.
Standard fraction method: 4 fraction additions, 1 reciprocal, ~50 seconds. LCM method: whole-number division only, ~20 seconds. Saves roughly 30 seconds per question.
When only two workers are involved, use ab/(a+b) directly instead of computing individual rates.
Example: A in 20 days, B in 30 days. Time together = (20×30)/(20+30) = 600/50 = 12 days.
Step count: Rate method needs 4 steps (two fractions, addition, reciprocal). This formula needs 2 steps (multiply, divide). Cuts to under 15 seconds with mental arithmetic.
Draw a box: top row = Men, Days, Work. Fill in two values in each scenario, cross-multiply to find the unknown.
Rule: Men and Days multiply on the same side. Work on the other side.
M₁ × D₁ / W₁ = M₂ × D₂ / W₂
Example: 6 workers lay 180 m in 5 days → unit rate = 6 meters/worker/day. Then 420 m in 7 days needs 420/(6×7) = 10 workers. No formula memorisation — just fill the box. Eliminates wrong-variable-scaling errors; reduces careless mistakes by ~80% under time pressure.
For pipes, set tank capacity = LCM of all pipe times. Each pipe fills that many units per hour.
Example: Pipe A fills in 4 hours, Pipe B in 6 hours. LCM = 12. A fills 3 units/hour, B fills 2 units/hour. Together: 5 units/hour. Time = 12/5 = 2.4 hours.
Same speed gain as worker LCM trick — fraction arithmetic entirely avoided until the final division.
RRB Group D almost always has a clean or near-clean answer. When the combined rate gives a non-terminating decimal, check whether one of the options matches LCM/combined_units exactly.
Example: Combined rate = 5/36 of work per day → time = 36/5 = 7.2. If 7.2 is an option, pick it without further calculation. If you see options like 7, 7.2, 7.5, 8 — compute just enough to distinguish. This avoids full long division in ~20% of questions.
Read the question and identify which type it is:
Type 1 — Two or more workers, find combined time:
Use LCM method (total work = LCM of times). Add daily units. Divide total by sum. Apply ab/(a+b) if only two people.
Type 2 — Scale men/days to match a fixed work target:
Use M × D = constant. Set up M₁ × D₁ = M₂ × D₂ and solve for the unknown.
Type 3 — Output scales with workers and days: Find unit rate = Total output ÷ (Workers × Days). Multiply by new Workers × Days.
Type 4 — Pipes and cisterns: Treat each pipe as a worker. Fill pipes add rate, drain pipes subtract rate. Apply LCM method.
Type 5 — Partial work then combined: Compute work done in solo phase. Subtract from 1. Divide remainder by combined rate.
Decision check before solving: Does the question give you a fixed total work (Types 1, 2, 4, 5) or a variable output (Type 3)? That single distinction determines which formula path you take. Don't mix them.
Why this question: The most direct combined-work template — two named workers, find joint time. Appears in almost every RRB sitting.
Solving path: LCM(12, 18) = 36. A's rate = 3 units/day, B's rate = 2 units/day. Combined = 5 units/day. Time = 36/5 = 7.2 days. Alternatively, (12×18)/(12+18) = 216/30 = 7.2 directly. Both routes: under 20 seconds.
Why this question: Man-days scaling with a fixed work target — classic Type 2. Tests whether you know to keep total work constant.
Solving path: Total work = 15 × 20 = 300 man-days. New men needed = 300 ÷ 12 = 25. Done in two multiplications. Don't overcomplicate with rate fractions.
Why this question: Three-variable output scaling (workers, days, meters). The railway track setting makes this a Group D favourite.
Solving path: Unit rate = 240 ÷ (8 × 6) = 5 m per worker per day. New output = 12 × 9 × 5 = 540 m. One division, one multiplication.
Why this question: Pipes and cisterns in railway context — filling a station water tank. Tests whether you apply the same rate-addition logic to pipes.
Solving path: LCM(4, 6) = 12. A fills 3 units/hour, B fills 2 units/hour. Together: 5 units/hour. Time = 12/5 = 2.4 hours. Option A is correct.
Why this question: Machines + hours + output — the same three-variable logic applied to a production setting. Tests whether you can recognise the Type 3 structure in new clothing.
Solving path: Unit rate = 400 ÷ (20 × 8) = 2.5 sleepers per machine per hour. New output = 15 × 12 × 2.5 = 450 sleepers.
Why this question: Scaling workers to meet a different track-length target in a different time — combines Types 2 and 3 in one question.
Solving path: Unit rate = 180 ÷ (6 × 5) = 6 m per worker per day. Workers needed = 420 ÷ (6 × 7) = 420 ÷ 42 = 10 workers.
Why this question: Crew-size scaling with a fixed station — direct application of total work = people × time.
Solving path: Total work = 4 × 3 = 12 person-hours. With 6 people: time = 12 ÷ 6 = 2 hours.
Adding times instead of rates. If A takes 4 hours and B takes 6 hours, the combined time is NOT 4 + 6 = 10, and NOT (4 + 6)/2 = 5. Always add rates (1/4 + 1/6), then flip. This mistake is the single most frequent wrong answer in combined-work questions.
Confusing inverse and direct scaling. More workers → less time (inverse). More workers → more output (direct). In Type 2 (fixed work), increasing workers decreases time. In Type 3 (variable output), increasing workers increases output. Writing M₁ × D₁ = M₂ × D₂ when you should be comparing outputs will give a completely wrong answer.
Forgetting to compute remaining work as a fraction before dividing. In partial-work problems, students divide the number of solo days by the combined rate — forgetting to first find what fraction of work was actually completed. Always: fraction done = solo days × solo rate. Then fraction remaining = 1 − fraction done.
Using wrong LCM. The LCM must be taken of ALL time values in the problem, not just two of them. If three workers are involved with times 6, 8, and 12, LCM = 24, not 24 for just the first two.
Treating pipes and workers differently. They follow identical mathematics. The only difference is that drain pipes subtract from the net rate. Don't second-guess yourself on pipe questions — just apply the LCM method as usual.
Rounding prematurely. Answers like 7.2, 2.4, and 2.5 are exact values. Students sometimes round 36/5 to 7 or 12/5 to 2. Keep fractions intact until the final step, then convert to decimal only when matching against options.