Time and work is one of those topics where students accumulate five or six different formulas — one for two workers, another for three, a separate one for pipes, another for wages — and then freeze during exams trying to remember which formula applies to which situation.
There is a better way.
Every time and work problem — without exception — is a direct application of one relationship:
Work = Efficiency × Time
Everything else is algebra applied to this single equation. Once you see every problem through this lens, you stop memorizing separate formulas and start deriving answers from first principles. This guide builds that unified understanding, then shows how it applies to every major problem type with fully worked examples.
The One Formula Framework
Defining the Terms
Work (W): The total task to be completed. Always treated as 1 unit (one complete job) unless stated otherwise.
Efficiency (E): The fraction of work completed per unit time. If A completes a job in 12 days, A's efficiency = 1/12 per day.
Time (T): The time taken to complete the work.
The relationship:
W = E × T
Since W = 1 for a complete job:
1 = E × T → E = 1 ÷ T → T = 1 ÷ E
The LCM Method — Making Fractions Disappear
The fastest practical application of this framework uses LCM to convert fractional efficiencies into whole numbers — eliminating fractions entirely from the calculation.
How it works:
- Assign total work = LCM of all time values given
- Each worker's efficiency = Total work ÷ Their individual time
- Combined efficiency = Sum of individual efficiencies
- Time together = Total work ÷ Combined efficiency
Example: A completes work in 12 days, B in 8 days. Time together?
- Total work = LCM(12, 8) = 24 units
- A's efficiency = 24 ÷ 12 = 2 units/day
- B's efficiency = 24 ÷ 8 = 3 units/day
- Combined = 5 units/day
- Time = 24 ÷ 5 = 4.8 days
Compare this to the fraction method: 1/12 + 1/8 = 2/24 + 3/24 = 5/24 → time = 24/5 = 4.8 days. Same answer — but the LCM method uses only whole numbers throughout, eliminating fraction arithmetic errors.
Problem Type 1: Two or Three Workers Together
Two Workers — Direct Application
Formula: Combined time = (A × B) ÷ (A + B), where A and B are individual times
Worked Example 1:
A: 15 days, B: 10 days. Together?
- = (15 × 10) ÷ (15 + 10) = 150 ÷ 25 = 6 days
Worked Example 2 (LCM method):
A: 18 days, B: 12 days, C: 9 days. All together?
- Total work = LCM(18, 12, 9) = 36 units
- A = 36 ÷ 18 = 2 units/day
- B = 36 ÷ 12 = 3 units/day
- C = 36 ÷ 9 = 4 units/day
- Combined = 9 units/day
- Time = 36 ÷ 9 = 4 days
Worked Example 3:
A and B together complete work in 12 days. A alone takes 20 days. B alone?
- Combined efficiency = 1/12
- A's efficiency = 1/20
- B's efficiency = 1/12 − 1/20 = 5/60 − 3/60 = 2/60 = 1/30
- B alone = 30 days
Problem Type 2: Partial Work Problems
These problems involve one worker starting alone, then another joining partway through — or one worker leaving before completion.
Template: A works alone for x days, then B joins
Method (LCM approach):
- Find total work using LCM
- Calculate work done during solo period
- Remaining work = Total − Solo work
- Time for remaining = Remaining work ÷ Combined efficiency
Worked Example 1:
A: 20 days, B: 30 days. A works alone for 5 days, then B joins. Total time?
- Total work = LCM(20, 30) = 60 units
- A's efficiency = 3 units/day, B's = 2 units/day
- Work by A in 5 days = 5 × 3 = 15 units
- Remaining = 60 − 15 = 45 units
- Combined efficiency = 5 units/day
- Time for remaining = 45 ÷ 5 = 9 days
- Total time = 5 + 9 = 14 days
Worked Example 2:
A: 15 days, B: 20 days, C: 25 days. All start together. After 4 days, C leaves. How long does the work take total?
- Total work = LCM(15, 20, 25) = 300 units
- A = 20, B = 15, C = 12 units/day
- Work in first 4 days = (20 + 15 + 12) × 4 = 47 × 4 = 188 units
- Remaining = 300 − 188 = 112 units
- A + B combined = 35 units/day
- Time for remaining = 112 ÷ 35 = 3.2 days
- Total = 4 + 3.2 = 7.2 days
Template: A leaves early, B finishes alone
Worked Example:
A: 20 days, B: 30 days (Combined: 12 days). A leaves after 4 days. B finishes alone. Total time?
- Total work = LCM(20, 30) = 60 units
- A = 3, B = 2 units/day
- Work in 4 days together = 4 × 5 = 20 units
- Remaining = 60 − 20 = 40 units
- B alone at 2 units/day = 40 ÷ 2 = 20 days
- Total = 4 + 20 = 24 days
Problem Type 3: Pipes and Cisterns
Pipes and cisterns are identical to time and work — filling a tank is the "work," and filling pipes have positive efficiency while emptying pipes have negative efficiency.
The only rule to remember:
- Filling pipe efficiency = +1/T
- Emptying pipe efficiency = −1/T
- Net rate = algebraic sum of all rates
Worked Example 1:
Pipe A fills in 6 hours, Pipe B fills in 8 hours. Both open together. Time to fill?
- Net rate = 1/6 + 1/8 = 4/24 + 3/24 = 7/24
- Time = 24/7 ≈ 3.43 hours
LCM method:
- Total = LCM(6, 8) = 24 units
- A = 4 units/hr, B = 3 units/hr
- Combined = 7 units/hr
- Time = 24 ÷ 7 hours ✓
Worked Example 2:
Pipe A fills in 12 hours, Pipe B fills in 15 hours, Pipe C empties in 20 hours. All open together. Time to fill?
- Total = LCM(12, 15, 20) = 60 units
- A = +5, B = +4, C = −3 units/hr
- Net = 6 units/hr
- Time = 60 ÷ 6 = 10 hours
Worked Example 3:
A tank is 3/4 full. Pipe A fills in 12 hrs, Pipe B empties in 8 hrs. If both open, when does tank empty?
- Total = LCM(12, 8) = 24 units
- Current volume = 3/4 × 24 = 18 units
- A = +2, B = −3 units/hr → Net = −1 unit/hr (emptying)
- Time to empty 18 units = 18 hours
Worked Example 4 — Finding leak:
A pipe fills tank in 6 hours. Due to a leak, it takes 8 hours. Leak empties in?
- Filling rate = 1/6
- Net rate (with leak) = 1/8
- Leak rate = 1/6 − 1/8 = 4/24 − 3/24 = 1/24
- Leak empties tank in 24 hours
Problem Type 4: Efficiency Ratio Problems
When workers have different efficiency levels (not times), convert efficiency ratios to time ratios by inverting.
Rule: If efficiency ratio A:B = m:n, then time ratio A:B = n:m (inverse)
Worked Example 1:
A is twice as efficient as B. Together they finish in 12 days. A alone?
- Efficiency ratio A:B = 2:1
- Let A take x days → B takes 2x days
- 1/x + 1/2x = 1/12
- 3 ÷ 2x = 1/12
- 2x = 36 → x = 18 days (A alone)
- B alone = 36 days
Worked Example 2:
A is 3 times as efficient as B and 2 times as efficient as C. Together in 5 days. Individual times?
- Efficiency A:B:C = 6:2:3
- If A's efficiency = 6k, B = 2k, C = 3k → Combined = 11k
- Total work = 11k × 5 = 55k
- A alone = 55k ÷ 6k = 55/6 ≈ 9.17 days
- B alone = 55k ÷ 2k = 27.5 days
- C alone = 55k ÷ 3k ≈ 18.33 days
Worked Example 3:
A and B together: 10 days. B and C together: 15 days. A and C together: 12 days. All three together?
- 1/A + 1/B = 1/10
- 1/B + 1/C = 1/15
- 1/A + 1/C = 1/12
- Adding all three: 2 × (1/A + 1/B + 1/C) = 1/10 + 1/15 + 1/12
- = 6/60 + 4/60 + 5/60 = 15/60 = 1/4
- 1/A + 1/B + 1/C = 1/8
- All three together = 8 days
Problem Type 5: Work and Wages
When wages are involved, distribute payment in the ratio of work done — which equals the ratio of efficiency (for equal time) or efficiency × time (for unequal time).
Equal Time — Wages in Efficiency Ratio
Worked Example 1:
A and B work together for 5 days and earn ₹3,500. A's daily rate is twice B's. Find each person's share.
- Efficiency ratio A:B = 2:1
- A's share = 2/3 × 3500 = ₹2,333
- B's share = 1/3 × 3500 = ₹1,167
Unequal Time — Wages Proportional to Work Done
Worked Example 2:
A: 10 days to complete work. B: 15 days. A works 4 days, B works 5 days, then C finishes in 2 days. Total wages = ₹9,000. Find C's share.
- Total work = LCM(10, 15) = 30 units
- A's efficiency = 3, B's = 2 units/day
- Work by A = 4 × 3 = 12 units
- Work by B = 5 × 2 = 10 units
- Work by C = 30 − 12 − 10 = 8 units
- C's share = 8/30 × 9000 = ₹2,400
Worked Example 3:
A, B, C together complete work in 6 days for ₹4,800. A alone: 12 days, B alone: 18 days. Find C's daily wage.
- Total work = LCM(12, 18, 6) = 36 units
- A = 3, B = 2, Combined = 6 units/day
- C = 6 − 3 − 2 = 1 unit/day → C alone = 36 days
- Wage ratio A:B:C = 3:2:1
- C's total wage = 1/6 × 4800 = ₹800
- C's daily wage = 800 ÷ 6 = ₹133.33
Problem Type 6: Work Done in Alternate Days
Some problems involve workers working on alternate days rather than together.
Method: Find work done in one complete 2-day cycle, then find how many complete cycles fit, then handle the remainder.
Worked Example 1:
A: 10 days, B: 15 days. They work on alternate days, A starting first. Total time?
- Total work = LCM(10, 15) = 30 units
- A = 3 units/day, B = 2 units/day
- In every 2-day cycle: 3 + 2 = 5 units
- 30 ÷ 5 = 6 complete cycles = 12 days exactly
Worked Example 2:
A: 12 days, B: 18 days. Alternate days, B starts first. Total time?
- Total work = LCM(12, 18) = 36 units
- A = 3, B = 2 units/day
- Per 2-day cycle = 5 units
- 36 ÷ 5 = 7 cycles (35 units) with 1 unit remaining
- After 14 days: 35 units done. Day 15 = A's turn → A does 3 units but only 1 needed
- Time = 14 + 1/3 day = 14⅓ days
Problem Type 7: Men, Women, Children — Variable Efficiency
Worked Example:
4 men and 6 women complete work in 8 days. 3 men and 7 women complete in 10 days. In how many days will 10 women complete it?
- Let man's daily work = m, woman's = w
- 8(4m + 6w) = 1 → 32m + 48w = 1 … (1)
- 10(3m + 7w) = 1 → 30m + 70w = 1 … (2)
- From (1) and (2): 32m + 48w = 30m + 70w
- 2m = 22w → m = 11w
- Substitute in (1): 32(11w) + 48w = 1 → 352w + 48w = 1 → 400w = 1
- w = 1/400
- 10 women: rate = 10/400 = 1/40
- Time = 40 days
Quick Reference — All Problem Templates
| Problem Type | Method | Key Formula |
|---|---|---|
| Two workers together | LCM method | T = (A × B) ÷ (A + B) |
| Three workers together | LCM method | Total ÷ Sum of efficiencies |
| One joins later | Work done before + after | LCM to avoid fractions |
| Pipes filling | Same as workers, emptying = −rate | Net = Sum of signed rates |
| Leak problem | Filling rate − Net rate | Leak = 1/Fill − 1/Net |
| Efficiency ratio | Invert ratio for time | Wages ∝ work done |
| Alternate days | 2-day cycle method | Handle remainder carefully |
| A+B, B+C, A+C given | Add all three, halve | A+B+C = half the sum |