Time Work Master - Quick Revision
Time Work Master - Quick Revision
One-Liners
- Work done = 1 / Time taken (per day rate).
- If A does work in 8 days, daily work = 1/8.
- Combined 1-day work = sum of individual 1-day works.
- Total time = 1 / (combined 1-day work).
- Efficiency ∝ 1 / Time; higher efficiency → less time.
- M₁D₁T₁W₁ = M₂D₂T₂W₂ (men-days-time-work).
- If A is twice as efficient, A takes half the time of B.
- Chain rule: link workers, hours, days & output.
- Work remaining = 1 – work finished.
- Pipes: filling rate positive, emptying rate negative.
- Alternate days: add individual day-works sequentially.
- LCM method: treat total work = LCM of all times.
- Work ∝ (workers × days × hours).
- Fraction of work = (days worked) / (total days needed).
- If A & B together finish in 6 h, and A in 10 h, then B in 15 h.
- Break: subtract idle days before computing effective days.
- Double workforce halves the days (same hours).
- 1 / (A+B+C) = 1/A + 1/B + 1/C.
- Efficiency % = (base time – new time) / base time × 100.
- Always verify “work” units cancel to 1 (whole work).
| Formula |
Use |
| 1-day work = 1 / T |
Convert given days to daily rate |
| Together time = 1 / (1/T₁ + 1/T₂ + …) |
Combined finishing time |
| M₁D₁H₁/W₁ = M₂D₂H₂/W₂ |
Chain-rule for variable manpower/hours |
| Remaining work = 1 – (days×daily rate) |
After partial work |
| Efficiency ratio = Time₂ / Time₁ |
Compare two workers |
| Work = k × (men × days × hours) |
Constant k depends on job size |
| Alternate-day work = Σ (individual day rates) |
Day-wise summation |
| Net pipe rate = Σ fill – Σ empty |
Cistern in/out problem |
| LCM total work = LCM(T₁,T₂,…) |
Avoid fractions |
| % gain in efficiency = (T_old – T_new)/T_old × 100 |
Quick % change |
Memory Tricks
- “One-over-T” – daily work always 1/T (think “over” like flipping a fraction pancake).
- “Men-Days-Hands” – imagine many hands passing bricks → M-D-H-W constant.
- “Plus for Friends, Minus for Foes” – plus sign for helping pipes, minus for leaking pipes.
- “LCM = Laddoo Common Multiple” – sweet total work everyone can divide easily.
- “A+B = Fast Lane” – combined rate always bigger; denominator (time) therefore smaller.
Common Errors
| Error |
Correct |
| Adding times directly (3 d + 6 d = 9 d) |
Add rates: 1/3 + 1/6 = 1/2 → 2 d |
| Ignoring idle days |
Subtract idle period before plugging into formulas |
| Using wrong sign for outlet pipe |
Emptying rate is negative in net rate |
| Forgetting to invert final sum |
After adding 1-day works, remember 1/(sum) gives total days |
| Mixing hours & days without conversion |
Convert everything to same unit first |
5 Quick MCQs
Question 1: A does work in 10 d, B in 15 d. Together they finish in how many days?
> **Ans:** 6 days
Question 2: 8 men finish in 12 d. How many days for 6 men?
> **Ans:** 16 days
Question 3: Pipe A fills in 4 h, B empties in 6 h. Net time to fill if both open?
> **Ans:** 12 h
Question 4: A is 50 % more efficient than B. If B takes 18 d, A takes?
> **Ans:** 12 days
Question 5: Work completed in 7 d by 10 workers. For 5 workers to finish remaining half, how many extra days?
> **Ans:** 7 days
BETA
