Work and Wages

Module 03 — Lesson 0003 · In Depth · 35 min

Work problems are rate problems: Work = Efficiency × Time. Efficiency is inversely proportional to time — faster means fewer days. The LCM method turns 1/N fractions into integer units so you can add rates like ordinary numbers.

1. The Core Concept: Time vs Efficiency

Efficiency ∝ 1 / Time — if A is twice as fast as B, A takes half the time.

Work = Efficiency × Time — same as distance = speed × time, with work as distance.

2. Two Methods to Solve

Method 1: Fraction (Unitary) — Traditional

Assume total work = 1 unit.

A in N days → 1-day work = 1/N
B in M days → 1-day work = 1/M
Together: 1/N + 1/M per day.

Method 2: LCM (Efficiency) — Recommended (Faster)

Assume total work = LCM of all days given (in units).
Efficiency = Total Work / Days — integer units/day.
Solve: add efficiencies, divide total by combined.

Worked: A 10 days, B 15 days, together?
Fraction: 1/10+1/15=1/6 → 6 days
LCM: Total LCM(10,15)=30 units → A=3 u/d, B=2 u/d → together 5 u/d → time=30/5=6 days. Same answer, but LCM keeps numbers integer — faster under time pressure.

3. The "Man-Days" Concept (Group Work)

For groups: "10 men do a job in 20 days" → total man-days is the work.

(M₁ × D₁ × H₁) / W₁ = (M₂ × D₂ × H₂) / W₂

M: men/workers, D: days, H: hours/day, W: work amount
If work same → M₁D₁H₁ = M₂D₂H₂

Worked: 12 men in 8 days vs how many men in 6 days (same work)? 12×8 = M₂×6 → M₂=16. More men → fewer days (inverse). If hours differ (8h vs 6h), include H.

4. Wages Distribution

Money is distributed by work done, not time alone.

Same duration (together start to finish): wages ratio = efficiency ratio.
Different duration (leaves early / joins late): wages ratio = (Efficiency × Days worked) ratio.

Worked: A 3 u/d, B 2 u/d. Both work 6 days → work done 18:12=3:2 → wages 3:2.
If A works 6 days, B works 4 days → A work 18, B work 8 → wages 9:4. Same efficiencies, different days → weight by days.

5. Advanced Scenarios

A. Leaving / Joining (Before Completion)

Calculate work done for known duration, subtract from total → remaining work → divide by current workers' efficiency.

Remaining = Total − (Efficiency × Days worked), Time for remainder = Remaining / Efficiency of current

Worked: A(3) + B(2) together =5 u/d, total 30. After 2 days work done 10, remaining 20. If A leaves, B alone takes 20/2=10 days more. If B leaves after 2 days and A continues, A takes 20/3≈6.67 days.

B. Alternate Days

A works day1, B day2, A day3…

Cycle work (2 days) = E_A + E_B
Full cycles = Total / Cycle, remainder handled day-by-day (don't assume fractional cycle averages).

Worked: Total 30, A 3, B 2, alternate starting A. Cycle 2d=5u. 30/5=6 cycles=12 days exactly. If total 31, 6 cycles=30, day13 A does 3 but only 1 needed → day13 partial → total 12+1/3 days (but in discrete-day problems answer is 13 days). Always check if last cycle overfills.

C. Negative Work (Pipes Logic)

Worker destroying work → negative efficiency. Treat like outlet pipe: net = +E₁ − E₂. See Module 04.

D. Efficiency Changes

If efficiency doubles/drops mid-work, calculate work per phase separately and subtract. E.g., A at 3 u/d for 2 days → 6u, then efficiency doubles to 6 u/d for remaining 24u → 4 days more.

6. Summary of Variables

Symbol	Meaning
`W`	Total work (LCM of days, in units)
`E`	Efficiency (units/day)
`T`	Time (days/hours)
`M`	Men/workers

Efficiency ∝ ?

A 10d, B 15d — LCM method total?

In that LCM, efficiencies A and B?

Together time for A 10d + B 15d?

Wages same duration → ratio is?

Wages different duration → ratio is?

12 men ×8 days = M×6 days → M?

Alternate A(3), B(2), total 30, cycle 5 per 2 days → full cycles?

Notes

Work :- Basics :-

Efficiency ∝ 1/Time ; Work ⇒ Eff×T

Unitary ⇒ 1/N per day ; LCM ⇒ Total=LCM(days), Eff=Total/days

(eg A10,B15→30 units→3,2 → together 5→6 days)

Man-Days :-

M₁D₁H₁/W₁ = M₂D₂H₂/W₂ ; same work → M₁D₁=M₂D₂

Wages :-

Same time ⇒ wages = Eff ratio ; diff time ⇒ Eff×Days

Advanced :-

a) Leaving/joining ⇒ remaining = Total − Eff×days → /current Eff

b) Alternate ⇒ cycle = E_A+E_B per 2d → full cycles + remainder

c) Negative ⇒ −Eff (like pipes) ; d) Variable ⇒ phase-wise

Primary source: none pinned yet — drop your preferred video/book resource into RESOURCES.md and it will be linked here. Ask me anything that's unclear.

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