Time & Work – Study Notes
(For West Bengal Patwari / Lekhpal – Quantitative Aptitude)
1. Simple Explanation
Time & Work deals with how long it takes or how fast a person (or a team) can complete a job, given their work rates.
- Work is measured in units (e.g., man‑days, tasks).
- Rate of a person = amount of work they can finish in one unit of time.
- Time = work ÷ rate.
When several people work together, their rates add up. When a person works for a part of the job and then stops, the remaining work is done by the rest.
2. Key Concepts & Formulas
| Concept | Formula | Explanation |
|---|---|---|
| Rate | ( R = \frac{W}{T} ) | Work (W) completed in Time (T). |
| Time to finish a job | ( T = \frac{W}{R} ) | Given work & rate. |
| Combined rate of (n) workers | ( R_{\text{total}} = R_1 + R_2 + \dots + R_n ) | Rates add because they work simultaneously. |
| Work done by a subset | ( W_{\text{done}} = R_{\text{subset}} \times T_{\text{subset}} ) | Use subset’s combined rate. |
| Remaining work | ( W_{\text{rem}} = W_{\text{total}} - W_{\text{done}} ) | Subtract already done portion. |
| Work‑rate equivalence | ( \frac{1}{R_{\text{person}}} = \text{time per unit work} ) | Useful when converting rates to “days per unit”. |
| Relative rates | e.g., “A is 2× faster than B” → ( R_A = 2 R_B ) | Helps set up equations. |
Tip: Always keep units consistent (e.g., days, hours).
Common trap: Mixing up “days per unit” with “units per day”.
3. Types of Problems
| Problem Type | Typical Question | Strategy |
|---|---|---|
| Single worker | “A can complete 100 units in 5 days.” | Compute rate → (R = 20) units/day; find time for other amounts. |
| Multiple workers together | “A & B together finish work in 8 days.” | (R_A + R_B = \frac{W}{8}). If one worker’s rate is known, find the other’s. |
| Work after part completion | “A works 3 days, B takes over, finish in total 10 days.” | Compute work by A, subtract from total, find B’s time. |
| Partial work with rate changes | “A works 2 days, then leaves; B & C work the rest.” | Break into phases, use combined rates. |
| Rate comparison | “A is 3 times faster than B.” | Set (R_A = 3 R_B), use total work. |
| Work & profit (occasionally) | “If work is done by X people, profit changes.” | Combine time, rate, and cost formulas. |
4. Worked Example
Problem
A and B can complete a piece of land in 6 days. B alone can finish it in 10 days. If A works for 2 days, then B works alone for 2 days, how many days will it take for B to finish the remaining work?
Step‑by‑Step Solution
-
Let total work (W = 1) unit (the whole land).
-
Rates
- (R_B = \frac{1}{10}) unit/day.
- Combined (R_{A+B} = \frac{1}{6}) unit/day.
Thus (R_A = R_{A+B} - R_B = \frac{1}{6} - \frac{1}{10} = \frac{5-3}{30} = \frac{2}{30} = \frac{1}{15}) unit/day.
-
Work done by A in 2 days
(W_A = R_A \times 2 = \frac{1}{15} \times 2 = \frac{2}{15}). -
Work done by B in 2 days
(W_B = R_B \times 2 = \frac{1}{10} \times 2 = \frac{1}{5}). -
Total work done so far
(W_{\text{done}} = \frac{2}{15} + \frac{1}{5} = \frac{2}{15} + \frac{3}{15} = \frac{5}{15} = \frac{1}{3}). -
Remaining work
(W_{\text{rem}} = 1 - \frac{1}{3} = \frac{2}{3}). -
Time for B to finish remaining work
(T_B = \frac{W_{\text{rem}}}{R_B} = \frac{\frac{2}{3}}{\frac{1}{10}} = \frac{2}{3} \times 10 = \frac{20}{3} = 6\frac{2}{3}) days.
Answer: B needs (6\frac{2}{3}) days (or 6.67 days) to finish the remaining work.
5. Quick Revision Summary
| Item | What to Remember |
|---|---|
| Rate of a worker | (R = \frac{1}{\text{days per unit}}) |
| Combined rate | Add individual rates |
| Work done | (W = R \times T) |
| Remaining work | Total – Done |
| Time to finish | (T = \frac{W_{\text{rem}}}{R_{\text{worker}}}) |
| Relative speeds | If A is 2× faster than B, then (R_A = 2R_B) |
| Common “words” | together, alone, after, then |