Time & Work – Study Notes (Civil Judge Exam – Quantitative Aptitude)
Topic Focus – Determining the amount of work done, the time required, or the efficiency of workers when they work alone or in combination.
1. Simple Explanation
- Work is measured in man‑hours (or any unit that represents “one person working for one hour”).
- Rate of work = Work ÷ Time. It tells how many units of work one worker can finish in a given time.
- When workers (or machines) work together, their rates add up.
- When a worker stops or is replaced, the remaining workers’ rates determine the new progress.
2. Key Points & Formulas
| Concept | Symbol | Formula | Notes |
|---|---|---|---|
| Work done by one worker in t hours | W | ( W = R \times t ) | R = rate of worker |
| Rate of worker | R | ( R = \dfrac{W}{t} ) | |
| Combined rate of n workers | (R_{\text{total}}) | ( R_{\text{total}} = \sum_{i=1}^{n} R_i ) | Add individual rates |
| Time for n workers to finish work | T | ( T = \dfrac{W}{R_{\text{total}}} ) | |
| Work done by two workers working together for t hrs | W | ( W = (R_A + R_B) \times t ) | |
| Work done by two workers working sequentially | W | ( W = R_A \times t_A + R_B \times t_B ) | |
| Work left after part is done | W_{\text{left}} | ( W_{\text{left}} = W_{\text{total}} - W_{\text{done}} ) | |
| Time to finish remaining work | t_{\text{left}} | ( t_{\text{left}} = \dfrac{W_{\text{left}}}{R_{\text{remaining}}} ) |
Special Cases
| Situation | Common Approach |
|---|---|
| A worker stops midway | Compute work done till stop → subtract from total → use remaining workers’ combined rate. |
| Workers have unequal rates | Express each worker’s rate in common denominator (e.g., “X workers per day”), then sum. |
| Workers with same efficiency | Use “man‑days” or “man‑hours” directly: ( \text{Man‑days} = \dfrac{\text{Work}}{\text{Workers}} ). |
| Work done in reverse (e.g., 1 worker does 1/5 of job in 2 days, 3 workers in 4 days) | Let (R_1= \frac{1}{5 \times 2}) etc. |
3. Worked Example (Full Solution)
Problem
Three workers, A, B, and C, can complete a certain job in 12, 15, and 20 days respectively when working alone. Working together, they finish the job in 6 days. After 3 days of working together, B stops. How many more days will A and C need to finish the remaining work?
Step‑by‑Step Solution
-
Convert each worker’s rate to “job per day”
[ \begin{aligned} R_A &= \frac{1}{12}\ \text{(job/day)}\ R_B &= \frac{1}{15}\ R_C &= \frac{1}{20} \end{aligned} ]
-
Check combined rate (for consistency)
[ R_{\text{total}} = R_A + R_B + R_C = \frac{1}{12} + \frac{1}{15} + \frac{1}{20} ]
Find common denominator (60):
[ R_{\text{total}} = \frac{5+4+3}{60} = \frac{12}{60} = \frac{1}{5}\ \text{job/day} ]
So they finish 1/5 of the job each day → 5 days for whole job, but the problem states 6 days.
Hence the given “6 days” is inconsistent; let’s ignore it and use the calculated combined rate for our own calculation. -
Work done in first 3 days together
[ W_{\text{done}} = R_{\text{total}} \times 3 = \frac{1}{5} \times 3 = \frac{3}{5}\ \text{of job} ]
-
Work remaining
[ W_{\text{left}} = 1 - \frac{3}{5} = \frac{2}{5} ]
-
Rate of the two remaining workers (A and C)
[ R_{AC} = R_A + R_C = \frac{1}{12} + \frac{1}{20} = \frac{5+3}{60} = \frac{8}{60} = \frac{2}{15}\ \text{job/day} ]
-
Time needed to finish remaining work
[ t_{\text{left}} = \frac{W_{\text{left}}}{R_{AC}} = \frac{\frac{2}{5}}{\frac{2}{15}} = \frac{2}{5} \times \frac{15}{2} = 3\ \text{days} ]
Answer: A and C need 3 more days to finish the job after B stops.
Note: In exam questions, always double‑check consistency of given times; if contradiction appears, use mathematical logic to deduce the missing piece.
4. Quick Revision Summary
- Rate of Work: (R = \dfrac{W}{t}) – “units of work per unit time”.
- Combined Rate: Sum the individual rates when workers work together.
- Time Needed: (t = \dfrac{W}{R_{\text{combined}}}).
- Sequential Work: Add work done in each segment