Time and Work — SSC CGL Study Notes

Overview

Time and Work is a high-scoring topic in SSC CGL Quantitative Aptitude, typically yielding 2–3 questions per exam. The core principle is simple: work is inversely proportional to time when efficiency remains constant. Mastering this topic means understanding how to combine individual work rates, handle multiple workers, and solve pipe-and-cistern variations where filling and emptying happen simultaneously.

This topic connects closely with Ratio and Proportion, since work distribution often involves ratio calculations. Questions range from straightforward "A can finish in 10 days, B in 15 days—how long together?" to complex scenarios with workers joining/leaving mid-project or pipes with different rates operating alternately. The key skill is converting qualitative statements (faster/slower, more/less efficient) into quantitative work rates, then applying the fundamental formulas consistently.

Students who practice 40–50 problems across all sub-types typically score full marks on Time and Work questions. The topic rewards systematic thinking over shortcuts, though certain tricks (like using LCM for total work) dramatically speed up calculations.

Key Concepts

• **Work is constant**: If A completes work in *n* days, A's one-day work = 1/n. Total work can be assumed as 1 unit or calculated as LCM of given times for easier arithmetic.

• **Inverse relationship**: If efficiency doubles, time halves. If time is reduced by 1/3, efficiency must increase by 1/2 (not 1/3). This trips many students.

• **Combined work rate**: When A and B work together, (A+B)'s one-day work = A's one-day work + B's one-day work. Time taken together = Total work ÷ Combined rate.

• **Work from days**: If a group completes work in *t* days, their one-day work = 1/t. If they complete *fraction f* of work in *t* days, their one-day work = f/t.

• **Negative work (pipes and cisterns)**: Filling pipes add positive work; emptying pipes add negative work. Net work rate = Σ(inlet rates) − Σ(outlet rates).

• **Alternate day working**: Calculate work done in one complete cycle (e.g., 2 days if A and B alternate), then find how many full cycles fit, and handle remaining work separately.

• **Man-days concept**: Total work = Number of workers × Days × Hours per day × Efficiency. Useful when workers are added/removed or working hours change mid-project.

• **Efficiency and wages**: Wages are proportional to work done. If A and B's efficiencies are in ratio 3:2 for the same time, their wages split 3:2.

Formulas / Key Facts

1. **One-day work**: If A finishes in *n* days → A's 1-day work = 1/n 2. **Time for combined work**: If A takes *a* days, B takes *b* days → Together = (a×b)/(a+b) days 3. **Work done in *t* days**: Work = (1-day work) × t 4. **Total work (LCM method)**: Total work = LCM(time₁, time₂, ...). Makes calculations integer-based. 5. **Efficiency ratio**: If A:B efficiency = m:n, then time ratio A:B = n:m (inverse). 6. **Pipes formula**: If inlet fills in *a* hrs, outlet empties in *b* hrs → Net time = (a×b)/(b−a) hrs (when b > a, else tank never fills). 7. **Remaining work**: If *f* fraction done, remaining = 1 − f. Time for remaining = (1−f) ÷ (combined rate). 8. **Man-days constant**: M₁D₁ = M₂D₂ when same work. If work differs: M₁D₁W₂ = M₂D₂W₁.

Worked Examples

**Example 1: Basic combined work** A can do work in 12 days, B in 15 days. How long to finish together?

*Solution:* A's 1-day work = 1/12 B's 1-day work = 1/15 (A+B)'s 1-day work = 1/12 + 1/15 = (5+4)/60 = 9/60 = 3/20 Time together = 20/3 = 6⅔ days

*Shortcut:* Time = (12×15)/(12+15) = 180/27 = 6⅔ days

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**Example 2: Three workers with one negative** A completes work in 10 days, B in 15 days, C destroys completed work in 30 days. All work together—how long to finish?

*Solution:* Total work = LCM(10,15,30) = 30 units A's rate = 30/10 = 3 units/day B's rate = 30/15 = 2 units/day C's rate = −30/30 = −1 unit/day (negative) Net rate = 3 + 2 − 1 = 4 units/day Time = 30/4 = 7.5 days

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**Example 3: Pipes and cistern** Pipe A fills tank in 6 hrs, pipe B empties in 8 hrs. Both open—when will tank fill?

*Solution:* A's rate = 1/6 per hour B's rate = −1/8 per hour Net rate = 1/6 − 1/8 = (4−3)/24 = 1/24 per hour Time to fill = 24 hours

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**Example 4: Workers joining mid-project** A can finish work in 20 days. A works 5 days, then B joins. Together they finish in 6 more days. Find B's time alone.

*Solution:* Total work = 20 units (assume) A's rate = 1 unit/day Work by A in 5 days = 5 units Remaining work = 15 units A + B do 15 units in 6 days → (A+B) rate = 15/6 = 2.5 units/day B's rate = 2.5 − 1 = 1.5 units/day B alone time = 20/1.5 = 13⅓ days

Common Mistakes

1. **Adding times instead of rates** → Wrong: "A takes 10 days, B takes 15 days, together = 25 days." Correct: Add rates (1/10 + 1/15), then invert for time.

2. **Confusing efficiency ratio with time ratio** → If A is twice as efficient as B, A takes half the time, not twice. Efficiency and time are inversely proportional.

3. **Ignoring negative work** → When a pipe empties or a worker undoes work, subtract that rate. Students often add all rates as positive, getting absurdly fast completion.

4. **Mishandling alternate days** → If A and B work alternate days, don't just average their rates. Calculate work for a 2-day cycle, then multiply by number of cycles, plus any remaining fractional cycle.

5. **Unit mismatch** → Mixing days, hours, and minutes without conversion. Always convert to a single time unit before calculating rates.

Quick Reference

• **Time together formula**: (t₁ × t₂)/(t₁ + t₂) when two workers combine.
• **LCM trick**: Total work = LCM(all times) makes all rates integers—faster arithmetic.
• **Efficiency ∝ 1/Time**: Double efficiency = half time; triple efficiency = one-third time.
• **Pipes net rate**: Inlet rate − Outlet rate. If net is negative, tank never fills.
• **Man-days equivalence**: 12 men in 10 days = 8 men in 15 days (both = 120 man-days for same work).
• **Fraction done in *t* days**: (One-day work) × t. Remaining = 1 − fraction done.