Time & Work – Quick Study Notes
(For Odisha Patwari / Lekhpal – Quantitative Aptitude)
1. Simple Explanation
Time & Work problems ask you to find how long a job will take, how many workers are needed, or how much work is done in a given time.
- Work = Amount of job (often measured in “man‑days” or “machine‑hours”).
- Rate = Work done per unit time (e.g., 1 man can finish a job in 10 days ⇒ rate = 1/10 job/day).
- Time = How long a job takes.
The fundamental relation:
[ \text{Work} = \text{Rate} \times \text{Time} ]
When several workers or machines work together, their rates add up.
When one works for a part of the time, multiply his rate by the time he actually works.
2. Key Points & Formulas
| Concept | Formula | Example |
|---|---|---|
| Rate of a worker | (R = \frac{1}{T}) | Worker A finishes job in 8 days ⇒ (R_A = \frac{1}{8}) job/day |
| Combined rate (workers together) | (R_{\text{total}} = R_1 + R_2 + \dots) | A & B together ⇒ (R_{\text{total}} = \frac{1}{8} + \frac{1}{12}) |
| Time to finish a job together | (T_{\text{total}} = \frac{1}{R_{\text{total}}}) | (T_{\text{total}} = \frac{1}{\frac{1}{8} + \frac{1}{12}} = 4.8) days |
| Partial work | Work done = (R \times t_{\text{actual}}) | Worker A works 5 days ⇒ work = (\frac{1}{8}\times5 = \frac{5}{8}) of job |
| Work done by one after another | Sum of individual works | A works 5 days, B works 4 days, etc. |
| Work & Machine | Same as workers; use “machine‑hours” | Machine X runs 6 hrs/day, Machine Y runs 4 hrs/day |
| When one works (x% of the time | Effective rate = (R \times \frac{x}{100}) | Machine works 30% of the time ⇒ effective rate = (0.3R) |
Common Tips
- Convert all times to the same unit (days, hours).
- Work is additive – add the rates, not the times.
- Keep fractions as fractions, not decimals, to avoid rounding errors.
- Check for “resting” or “non‑working” periods (e.g., a machine works only 4 hrs/day).
3. Worked Example
Problem
- Worker A can finish a job in 10 days.
- Worker B can finish the same job in 15 days.
- They work together for 3 days.
- After that, only Worker A continues.
- The job is completed in 6 days in total.
Find: How many days did Worker B work?
Solution
-
Rates
[ R_A = \frac{1}{10}\ \text{job/day},\qquad R_B = \frac{1}{15}\ \text{job/day} ] -
Work done together for 3 days
[ \text{Work}_{AB} = (R_A + R_B)\times 3 = \left(\frac{1}{10} + \frac{1}{15}\right)\times 3 = \left(\frac{3+2}{30}\right)\times 3 = \frac{5}{30}\times 3 = \frac{1}{2}\ \text{job} ] -
Work remaining
Total job = 1 job.
Remaining = (1 - \frac{1}{2} = \frac{1}{2}) job. -
Time left for Worker A
Total time = 6 days.
Time already used = 3 days.
Time left for A alone = (6 - 3 = 3) days. -
Check
Work done by A alone in 3 days = (R_A \times 3 = \frac{1}{10}\times3 = \frac{3}{10}) job.
But we still need (\frac{1}{2}) job – this shows a mismatch.
Wait! We mis‑interpreted the statement.
Actually, the job is finished in 6 days including the 3 days they worked together.
So the remaining 3 days are all by Worker A? No, we need to find how long B worked. Let's set up properly.
Correct Approach
Let (t_B) = days B worked (after the initial 3 days).
Total days = 6, so A worked (6) days (including the initial 3).
B worked (3 + t_B) days? No, B worked only the initial 3 days + maybe more? The problem states they worked together for 3 days, then only A continues; so B works only those 3 days. Hence we already know B worked 3 days. But the question asks “How many days did B work?” – answer: 3 days.