Time & Work: From Basics to Mains Level — A Complete Blog-Style Guide

I’ve been revising Time & Work from basic to advanced levels, and I thought I’d write a complete blog-style explanation of every concept. Instead of just repeating the same examples, I’ve changed the numbers and questions while keeping the underlying concepts exactly the same. By the end, you’ll have a solid mental map of how to handle almost any Time & Work problem.

1. The Foundation: Total Work, LCM, and Efficiency

The core idea is simple: Total Work = LCM of the times taken by individuals.
Once you have total work, each person’s efficiency is:

Efficiency is just “work per day” (or per hour). It’s the most useful unit because you can add efficiencies when people work together.

Example:
A can do a job in 12 days, B in 18 days.
LCM of 12 and 18 = 36 units (Total Work).
A’s efficiency = 36/12 = 3 units/day.
B’s efficiency = 36/18 = 2 units/day.
Together = 5 units/day.
Time together = 36/5 = 7.2 days (or 7⅕ days).

2. Finding a Missing Person’s Time

If A+B together take x days and B alone takes y days, then:

But the LCM method is faster.

Example:
A+B together = 15 days, B alone = 40 days. Find A alone.
LCM of 15 and 40 = 120 units.
A+B efficiency = 120/15 = 8.
B efficiency = 120/40 = 3.
A efficiency = 8 − 3 = 5.
A alone = 120/5 = 24 days.

3. Fraction of Work Left

Calculate work done, subtract from total, and express as a fraction.

Example:
A = 60 days, B = 90 days. They work together for 10 days. What fraction is left?
LCM = 180.
A = 3, B = 2 → combined = 5 units/day.
Work done in 10 days = 50 units.
Remaining = 130 units.
Fraction left = 130/180 = 13/18.

4. Efficiency and Time Are Inversely Proportional

If A : B time = 4 : 7, then A : B efficiency = 7 : 4.
More time → less efficiency; less time → more efficiency.

Example:
A takes 4 days, B takes 7 days for the same work.
Efficiency ratio A : B = 7 : 4.

5. Alternate Working (One Person Per Day)

Take LCM to get total work, then find the work per cycle.

• Cycle = the repeating pattern (e.g., A+B, B+A, or a 3-day rotation).
• Number of complete cycles = Total Work ÷ Cycle Work.
• Remaining work is done by the next person in sequence.
• To minimize time, start with the most efficient person.

Example 1 (Two people):
A = 12 days, B = 18 days. A starts, they alternate.
LCM = 36. A = 3, B = 2.
Cycle (A+B) = 5 units in 2 days.
7 cycles = 35 units in 14 days.
Remaining = 1 unit. A does 1 unit in 1/3 day.
Total = 14⅓ days.

If B starts: 7 cycles = 35 units in 14 days. Remaining 1 unit. B does 1/2 day.
Total = 14½ days.

Example 2 (Three people):
A = 6 days, B = 8 days, C = 12 days. Cycle A→B→C.
LCM = 24. A = 4, B = 3, C = 2.
Cycle = 9 units in 3 days.
2 cycles = 18 units in 6 days.
Remaining = 6 units.
Day 7: A does 4 → remaining 2.
Day 8: B does 3, but only 2 needed → 2/3 day.
Total = 7⅔ days.

6. Efficiency Comparisons and Percentage Changes

• “A is x% more efficient than B” →
• “A takes x% more time than B” →
• Always convert to efficiency ratio.

Example:
P is 25% more efficient than Q.
Q is 20% less efficient than R.
P and R together take 54 days. Find Q and R together.
P : Q = 5 : 4.
Q : R = 4 : 5 (since Q is 20% less than R).
So P : Q : R = 5 : 4 : 5.
P+R efficiency = 10. Total work = 10 × 54 = 540.
Q+R efficiency = 9.
Time = 540/9 = 60 days.

7. Efficiency Equations (A = B + C, etc.)

Example:
A is as efficient as B + C.
A+B together take 20 days, C alone takes 30 days. Find B alone.
Let .
, .
So and .
Substitute: → → .
B alone = 120 days.

8. Man-Day, Man-Hour, and Different Workers

Man-Day:

Example:
18 men can do a work in 20 days. How many men to do it in 15 days?
→ . Additional = 6 men.

Man-Hour:

Example:
6 men working 8 hours/day can do a work in 12 days. How many men working 9 hours/day to do it in 8 days?
→ → .

Different Workers:
If 2 men = 3 women = 6 boys, find time for 1 man + 1 woman + 1 boy.
Let 2M = 3W = 6B = k.
M = k/2, W = k/3, B = k/6.
Efficiency ratio M:W:B = 3:2:1.
If 2 men do in 60 days, total work = 2×60 = 120 man-days.
1M+1W+1B = 3+2+1 = 6 units. Time = 120/6 = 20 days.

9. Workers Leaving / Joining

Let total time = x. Express each person’s working days as (x − leave days).
Set sum of work = total work.

Example:
30 workers can finish in 40 days. After how many days should 6 workers leave so work completes in 45 days?
Let x days.

→ → days.

10. Pairwise Efficiency Problems

Example:
A+B = 24 days, B+C = 30 days, A+C = 40 days. Find A+B+C together.
LCM = 120.
A+B = 5, B+C = 4, A+C = 3.
Sum = 12 = 2(A+B+C) → A+B+C = 6.
Time = 120/6 = 20 days.

11. Wages Distribution

Wages ∝ Work done = Efficiency × Time.
Individual share = Total Wages × (Individual Work / Total Work).

Example:
A = 10 days, B = 15 days, C = 20 days. Together they get Rs 2600. Find C’s share.
LCM = 60. A = 6, B = 4, C = 3. Combined = 13.
Time together = 60/13 days.
C’s work = 3 × 60/13 = 180/13 units.
Total work = 60.
C’s share = Rs 600.

12. The Quadratic Root Formula

If A takes days, B takes days, and together they take days, then:

Example:
A takes x+8 days, B takes x+18 days, together x days.
.
A = 20 days, B = 30 days, together = 12 days. Check: 1/20 + 1/30 = 1/12.

13. Advanced Mains-Level Patterns

13.1 Partial Work with Pairwise Efficiencies

Example:
A+B = 15 days, B+C = 20 days.
A works 5 days, B works 10 days, C works 15 days and completes the work. Find time together.
LCM = 60. A+B = 4, B+C = 3.
Let C = x, B = 3−x, A = 1+x.
Work:
→ → .
A = 3.5, B = 0.5, C = 2.5. Combined = 6.5.
Time = 60/6.5 = 120/13 days.

13.2 Rotation with One Person Per Day

Example:
M = 15 days, K = 10 days, S = 30 days. They work one person per day in rotation. Who should start to minimize time?
LCM = 30. Eff: M=2, K=3, S=1.
Start with K (highest efficiency).
Cycle K,S,M = 6 units in 3 days. 5 cycles = 30 units in 15 days.
Total = 15 days.

13.3 Mixed Workforce and Wages

Example:
3 men = 4 women = 6 boys.
5 men, 6 women, 8 boys complete a work in 10 days. Find total work.
Efficiency ratio M:W:B = 4:3:2.
Combined = 5×4 + 6×3 + 8×2 = 20+18+16 = 54 units/day.
Total work = 54 × 10 = 540 units.
If total wages = Rs 5400, 1 boy’s daily wage?
Boy efficiency = 2 units/day. Work by boys = 8×2×10 = 160 units.
Boy share = .
Total boy-days = 8×10 = 80. Daily wage per boy = 1600/80 = Rs 20.

13.4 Changing Workforce Over Time

Example:
6 women complete a work in 30 days. Every 4 days, 1 woman leaves and 1 boy joins. Boy efficiency = 1/3 of woman. Find total time.
Let 1W = 3, 1B = 1. Total work = 6×3×30 = 540 units.
Segments of 4 days:
6W=18, 5W+1B=16, 4W+2B=14, 3W+3B=12, 2W+4B=10, 1W+5B=8.
Sum first 6 segments = 4×(18+16+14+12+10+8) = 4×78 = 312.
Remaining = 540−312 = 228.
6 boys = 6 units/day → 228/6 = 38 days.
Total = 24 + 38 = 62 days.

14. Quick Formula Sheet

Final Thoughts

Time & Work is not about memorizing dozens of formulas. It’s about converting everything into work units and efficiencies. Once you do that, even the nastiest Mains-level problem becomes a simple arithmetic puzzle. Practice the patterns above, change the numbers yourself, and you’ll be ready for any exam question.

Happy solving!