Time and Work — RRB NTPC Study Notes

Overview

Time and Work is a high-scoring arithmetic topic in RRB NTPC that typically yields 2–3 questions per paper. The underlying principle is simple: **Work = Rate × Time**. Once you grasp that work is constant and inversely proportional to time, all problems—whether about men digging canals, pipes filling tanks, or combined efforts—follow predictable patterns.

Mastery requires understanding three core scenarios: individual work rates, combined work (multiple workers together), and pipes-and-cisterns (inlets and outlets). The RRB exam favors straightforward numerical problems with direct application of formulas, though occasional twists—like workers leaving mid-project or alternating work patterns—do appear. Comfort with fractions, LCM, and unitary method is essential because most solutions involve equating fractional work done per day. Expect 1–2 minutes per question if your formula recall is sharp.

Practice is non-negotiable. The topic is mechanical once you internalize the work-rate relationship, but exam pressure causes silly errors—confusing "days" with "rate," or adding rates incorrectly in pipe problems. Drill 20–30 problems, focusing on combined work and pipes-and-cisterns, to build speed and accuracy.

Key Concepts

• **Work is always taken as 1 unit (or 100%).** If A completes a job in 10 days, A's one-day work = 1/10. If B does it in 15 days, B's one-day work = 1/15. Work done is cumulative and additive.

• **Rate of work = 1/Time.** If time decreases, rate increases proportionally. A faster worker has a higher rate. This inverse relationship is the backbone of all problems.

• **Combined work: add rates, not times.** If A works at 1/10 per day and B at 1/15 per day, together they work at (1/10 + 1/15) per day. Never add 10 + 15 to find combined time.

• **Work efficiency is the ratio of work rates.** If A is twice as efficient as B, A's rate is double B's rate. Efficiency ratios directly convert to rate ratios.

• **Man-days or worker-days measure total work.** If 8 men complete work in 12 days, total work = 8 × 12 = 96 man-days. This quantity stays constant; if men increase, days decrease proportionally (and vice versa).

• **Pipes and cisterns: inlets add, outlets subtract.** An inlet pipe filling a tank in 6 hours has rate +1/6 per hour. An outlet (drain) emptying in 8 hours has rate –1/8 per hour. Net rate = sum of all rates with proper signs.

• **LCM method simplifies fractions.** When A finishes in 12 days and B in 18 days, assume total work = LCM(12,18) = 36 units. Then A does 3 units/day, B does 2 units/day. Integer arithmetic replaces messy fractions.

• **Alternate-day work or work with breaks requires day-by-day tracking.** If A and B work on alternate days, calculate work done in a 2-day cycle, find how many full cycles fit, then add remaining work.

Formulas / Key Facts

1. **One-day work = 1 / Total days.** If X completes job in n days, X's rate = 1/n per day.

2. **Combined rate (A and B together) = 1/a + 1/b,** where a and b are individual times. Combined time T = 1 / (1/a + 1/b) = ab/(a+b).

3. **Work = Rate × Time.** Rearrange to Time = Work / Rate or Rate = Work / Time.

4. **M₁D₁W₁/T₁ = M₂D₂W₂/T₂** (Man-Day-Work-Time formula). M=men, D=days, W=work, T=hours/day. If any variable changes, adjust others proportionally.

5. **Efficiency ratio = inverse ratio of time.** If A takes 6 days and B takes 9 days, efficiency ratio A:B = 9:6 = 3:2.

6. **Net rate in pipes = Σ(inlet rates) – Σ(outlet rates).** Time to fill or empty = 1 / |net rate|.

7. **Remaining work = 1 – (work done so far).** Always express work as fractions of the whole job.

8. **If n workers take d days for work W, one worker takes nd days for same work.** Or equivalently, work W = n × d man-days.

Worked Examples

**Example 1: Basic Combined Work** A can complete a task in 12 days, B in 15 days. How long to finish together?

*Solution:* A's rate = 1/12 per day. B's rate = 1/15 per day. Combined rate = 1/12 + 1/15 = (5 + 4)/60 = 9/60 = 3/20 per day. Time = 1 ÷ (3/20) = 20/3 = 6⅔ days or 6 days 16 hours.

**Example 2: LCM Method** A does work in 10 days, B in 15 days. Working together, they finish in how many days?

*Solution (LCM approach):* Total work = LCM(10,15) = 30 units. A's rate = 30/10 = 3 units/day. B's rate = 30/15 = 2 units/day. Combined rate = 3 + 2 = 5 units/day. Time = 30/5 = 6 days.

**Example 3: Pipes and Cisterns** Pipe A fills tank in 4 hours, pipe B empties in 6 hours. Both open together; time to fill?

*Solution:* A's rate = +1/4 per hour (filling). B's rate = –1/6 per hour (emptying). Net rate = 1/4 – 1/6 = (3 – 2)/12 = 1/12 per hour. Time to fill = 1 ÷ (1/12) = 12 hours.

**Example 4: Man-Days Problem** 8 men complete work in 10 days. How many men needed to finish in 5 days?

*Solution:* Total work = 8 × 10 = 80 man-days. Men needed = 80 / 5 = 16 men.

Common Mistakes

• **Adding times instead of rates.** Students write "A takes 10 days, B takes 15 days, together 25 days" — wrong. Correct: add rates 1/10 + 1/15, then invert.

• **Forgetting negative sign for outlet pipes.** In pipe problems, always subtract the emptying rate. Writing 1/4 + 1/6 instead of 1/4 – 1/6 flips the answer entirely.

• **Confusing work done with work remaining.** If A does 1/3 of job, remaining is 2/3, not 1/3. Track cumulative work carefully.

• **Misapplying efficiency ratios.** Saying "A is twice as efficient" means A's time is half (not double) B's time. Efficiency ratio 2:1 → time ratio 1:2.

• **Ignoring units in man-day formula.** If question mixes days and hours/day, include T (hours per day) in M₁D₁T₁ = M₂D₂T₂. Skipping T gives wrong answer.

Quick Reference

• **Combined time for A (a days) and B (b days) = ab/(a+b) days.**
• **If A does job in x days, one day work = 1/x; x days work = x × (1/x) = 1 (full job).**
• **Pipe filling in t₁, emptying in t₂, both open: net time = t₁t₂/|t₁–t₂|** (sign determines fill or empty).
• **Man-days constant: M₁D₁ = M₂D₂ when work and hours/day same.**
• **Efficiency ∝ 1/Time. Higher efficiency = lower time.**
• **LCM trick: assume work = LCM of all times, convert to integer rates, simplify arithmetic.**