Time & Work – Study Notes
(Madhya Pradesh Police Constable – Quantitative Aptitude)
📌 Simple Explanation
Time & work problems ask you to find out how long it takes for workers (people, machines, etc.) to complete a job, or how much work is done in a given time.
- Rate of work = Work done per unit time.
- Work = Total amount of job needed to be finished (often measured in “man‑days” or “units”).
- Time = Duration needed to finish the job.
When two or more workers collaborate, their rates add up.
When a worker stops for a period (break, rest, etc.) the effective working time is reduced.
📚 Key Points & Formulas
| Concept | Symbol | Formula | Notes |
|---|---|---|---|
| Rate of work | ( R ) | ( R = \frac{W}{T} ) | (W)=total work, (T)=time taken |
| Work done | ( W ) | ( W = R \times T ) | |
| Time required | ( T ) | ( T = \frac{W}{R} ) | |
| Combined rate (two workers) | ( R_{\text{total}} ) | ( R_{\text{total}} = R_1 + R_2 ) | |
| Time with rest | ( T_{\text{effective}} ) | ( T_{\text{effective}} = \frac{T_{\text{actual}}}{\text{Working fraction}} ) | e.g., 80 % working → ( T_{\text{effective}} = \frac{T_{\text{actual}}}{0.8} ) |
| Proportion method | – | ( \frac{W_1}{R_1} = \frac{W_2}{R_2} ) | Use when work is split among workers working at different rates |
| Work‑Rate Table (for quick recall) | – | ( \text{Man‑days} = \text{Rate (man/day)} \times \text{Time (days)} ) |
Common Problem Types
- Single worker, given rate & work → Find time.
- Two workers, one slower → Find time for one to finish alone.
- Combined work → Find total time for multiple workers.
- Rest/Breaks → Adjust effective working time.
- Work split among workers → Use proportion or simultaneous work formulas.
✅ Worked Example
Problem
Three men, A, B and C, can complete a job in 12, 18 and 30 days respectively.
– A and B work together for 4 days.
– Then B and C work together for 6 days.
– Finally, only C works for the rest of the job.
In how many days will the whole job be finished?
Step‑by‑Step Solution
| Step | Description | Calculation | Result |
|---|---|---|---|
| 1 | Find individual rates (in “job per day”) | ( R_A = \frac{1}{12} ) ( R_B = \frac{1}{18} ) ( R_C = \frac{1}{30} ) |
– |
| 2 | Combined rate of A+B | ( R_{AB} = R_A + R_B = \frac{1}{12} + \frac{1}{18} = \frac{5}{36} ) | – |
| 3 | Work done by A+B in 4 days | ( W_{AB} = R_{AB} \times 4 = \frac{5}{36} \times 4 = \frac{5}{9} ) | 5/9 of job |
| 4 | Combined rate of B+C | ( R_{BC} = R_B + R_C = \frac{1}{18} + \frac{1}{30} = \frac{8}{90} = \frac{4}{45} ) | – |
| 5 | Work done by B+C in 6 days | ( W_{BC} = R_{BC} \times 6 = \frac{4}{45} \times 6 = \frac{24}{45} = \frac{8}{15} ) | 8/15 of job |
| 6 | Total work done so far | ( W_{\text{done}} = \frac{5}{9} + \frac{8}{15} ) Convert to common denominator 45: ( = \frac{25}{45} + \frac{24}{45} = \frac{49}{45} ) |
49/45 ≈ 1.0889 → >1? → Mistake! |
| 7 | Re‑check calculations | Actually, step 3 had 5/9 = 25/45. Step 5 had 8/15 = 24/45. Summation = 49/45 > 1 → indicates that the job would finish earlier than 4+6 days. | |
| So we need to find when the job completes: The job completes before the full 6 days of B+C. | |||
| 8 | Let (x) days be the time B+C actually work (after the first 4 days). | Work done after first 4 days: 5/9. Remaining work: (1 - \frac{5}{9} = \frac{4}{9}). | |
| 9 | Work done by B+C in (x) days: ( R_{BC} \times x = \frac{4}{45}x ). | Set equal to remaining work: (\frac{4}{45}x = \frac{4}{9}). | |
| 10 | Solve for (x): | ( x = \frac{4}{9} \times \frac{45}{4} = 5 ) days. | B+C work 5 days (not 6). |
| 11 | Total days spent: 4 (A+B) + 5 (B+C) = 9 days. | ||
| 12 | Any remaining work? | After |