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Time & Work

Chapter Outline & Learning Goals

By the end of this chapter, you will be able to:

1. Convert any "days to finish" statement into a work rate instantly.
2. Solve combined-work, alternate-day, and efficiency-ratio problems without equations.
3. Handle pipes & cisterns as a special case of Time & Work.
4. Tackle wages, men-days (M₁D₁H₁ = M₂D₂H₂), and "work left incomplete" traps.
5. Spot the examiner's favourite traps before they trap you.

The Hook: Two Men, One Wall, and a Question That Fools 80% of Aspirants

A can build a wall in 10 days. B can build it in 15 days. Working together, how long will they take?

If your instinct whispered "12.5 days" (the average), congratulations — you just fell into the most common trap in all of quantitative aptitude. The correct answer is 6 days. Not 12.5. Not even close.

Why? Because work doesn't add the way time does. Rates add. Times don't. That single sentence is the entire soul of this chapter. Master it, and 90% of Time & Work problems collapse into 10-second arithmetic.

Why This Chapter Feels Hard (And Why It Actually Isn't)

Let's be honest about the struggle. Time & Work frustrates students for three reasons:

• It feels backwards. "A finishes in 10 days" — so A's speed is... 1/10? Our brains resist fractions as speeds.
• Problems disguise themselves. Pipes, wages, alternate days, efficiency ratios — they look like five different chapters. They're one.
• The arithmetic gets ugly. LCMs of 12, 15, 20... unless you use the LCM method (coming soon), which makes them beautiful.

Every one of these problems dissolves once you stop thinking in *days* and start thinking in *units of work per day*.

Chunk 1: The Fundamental Conversion — Rate, Not Time

Core idea: If a person completes a job in *n* days, their one-day work (rate) is 1/n of the job.

• A finishes in 10 days → A does 1/10 of the work each day.
• B finishes in 15 days → B does 1/15 of the work each day.
• Together in one day: 1/10 + 1/15 = 3/30 + 2/30 = 5/30 = 1/6 of the work.

If 1/6 of the work is done per day, the whole work takes 6 days. Done. No averaging, no guessing.

The golden rule: When people work together, add their rates, never their times.

🧠 Active Recall: C finishes a job in 8 days, D in 24 days. What fraction of the job do they complete together in one day? How many days to finish together? *(Answer: 1/8 + 1/24 = 4/24 = 1/6 → 6 days.)*

Chunk 2: The LCM Method — Your New Best Friend

Fractions like 1/10 + 1/15 are fine, but examiners love 12, 15, and 18 together. Here's the trick that toppers use:

Assume the total work = LCM of the given days. Then every rate becomes a clean whole number.

Example: A does a job in 12 days, B in 18 days. Together?

• Total work = LCM(12, 18) = 36 units.
• A's rate = 36/12 = 3 units/day. B's rate = 36/18 = 2 units/day.
• Together = 5 units/day → Time = 36/5 = 7.2 days.

No fractions, no common denominators, no errors. This one method handles combined work, pipes, and alternate days alike.

🧠 Active Recall: Using the LCM method: A in 20 days, B in 30 days. Total work? Rates? Time together? *(60 units; 3 and 2 units/day; 12 days.)*

Chunk 3: Efficiency and Ratios

Sometimes the exam doesn't give days — it gives efficiency comparisons.

> "B is 50% more efficient than A."

This means if A's rate is 2 units/day, B's is 3 units/day (ratio 2:3). Now the key insight:

Efficiency and time are inversely proportional. If efficiency ratio is 2:3, the time ratio is 3:2.

So if A takes 15 days, B takes 15 × (2/3) = 10 days.

Chain rule for ratios: If A:B efficiency = 2:3 and B:C = 4:5, first make B common: A:B:C = 8:12:15. Times are then in ratio 1/8 : 1/12 : 1/15.

🧠 Active Recall: X is twice as efficient as Y. Y takes 24 days alone. How long does X take? How long together? *(12 days; together: LCM 24 → rates 2+1 = 3 → 8 days.)*

Chunk 4: The Work-Left and Joining/Leaving Traps

This is where Graduation-level papers (SSC CGL Tier-II, IBPS PO Mains) earn their reputation.

Type 1 — Someone leaves midway:
> A and B can finish a job in 12 and 18 days. They work together for 4 days, then A leaves. How long does B take to finish the rest?

• Total work = 36 units. Rates: A = 3, B = 2.
• In 4 days together: 4 × 5 = 20 units done. Remaining = 16.
• B alone: 16/2 = 8 more days.

Type 2 — Someone joins late: Same logic. Compute work done before the change, then recompute with the new team.

The trap: Examiners ask "in how many days was the work completed?" — meaning *total* time (4 + 8 = 12 days), not the remaining time (8 days). Read the question twice.

🧠 Active Recall: In the example above, what is the total time from start to finish? *(12 days — don't answer 8!)*

Chunk 5: Alternate Days

> A (12 days) and B (18 days) work on alternate days, A starting. When is the work done?

• Work = 36 units. A does 3, B does 2 per day.
• Every 2-day cycle: 5 units. After 7 cycles (14 days): 35 units. 1 unit left.
• Day 15 is A's turn; A does 3/day, so needs 1/3 of a day.
• Total = 14⅓ days.

Trap alert: Who starts matters. If B starts, the answer changes. Always track whose turn the leftover work falls on.

Chunk 6: Pipes & Cisterns — Time & Work in Disguise

A filling pipe is a worker. An emptying pipe is a negative worker — a saboteur on the team.

> Pipe A fills a tank in 6 hours, pipe B empties it in 12 hours. Both open: when is the tank full?

• Capacity = LCM(6, 12) = 12 units. A = +2/hr, B = −1/hr. Net = +1/hr → 12 hours.

Mnemonic 🧠: *"Fillers add, leakers subtract — treat the leak like a lazy worker who un-does work."* Think of the drain as an employee who spends all day deleting the team's files.

Chunk 7: Men, Days, Hours — The M₁D₁H₁ Formula

When the *amount of work* changes too (more men, more walls, more hours):

$$\frac{M_1 D_1 H_1}{W_1} = \frac{M_2 D_2 H_2}{W_2}$$

> 10 men build a wall in 8 days working 6 hours/day. How many days for 12 men working 5 hours/day to build the same wall?

• 10 × 8 × 6 = 12 × D₂ × 5 → D₂ = 480/60 = 8 days.

Include efficiency (E) if workers differ: M₁D₁H₁E₁/W₁ = M₂D₂H₂E₂/W₂.

🧠 Active Recall: If the second wall were *twice as big* in the example above, what would D₂ be? *(16 days — W₂ doubles, so days double.)*

Chunk 8: Wages — Work Done = Money Earned

Wages are divided in the ratio of work contributed, which for equal time means the ratio of efficiencies.

> A (10 days) and B (15 days) finish a job together and earn ₹600. Who gets what?

• Efficiency ratio = 1/10 : 1/15 = 3:2.
• A gets 3/5 × 600 = ₹360; B gets ₹240.

If people work for different durations, wages ∝ rate × days worked.

The Memory Hook: "RAT"

Remember the whole chapter with RAT:

• R — Rates add, times don't.
• A — Assume total work = LCM.
• T — Track who leaves, joins, or leaks.

If a problem feels hard, ask: *"Have I RAT-ted it yet?"*

Summary Table

📌 Review this chapter after 2 days. Redo the hook problem (10 & 15 days) and one leaving-midway problem from memory — if both take under 60 seconds each, this chapter is yours for life.