Time & Work – Quantitative Ability
Program: ATMA
Subject: Quantitative Ability
Topic: Time & Work
📌 Simple Explanation
In “time & work” problems we treat work as a quantity that can be split into pieces and measured by how many people (or machines, etc.) can finish it in a given time.
- Rate of work = amount of work done per unit time.
- Work is usually expressed in units (e.g., “one job”, “100 units of work”).
- If a person/thing can finish a job in
Thours, their rate is1/Tunits per hour.
When multiple workers cooperate or work in stages, we add or subtract their rates, or use the combined rate formula.
📚 Key Points & Formulas
| Concept | Formula | Meaning |
|---|---|---|
| Rate of one worker | r = 1 / T |
Units of work per hour |
| Combined rate (workers working together) | r_total = r1 + r2 + … + rn |
Sum of individual rates |
| Time to finish a job together | T_total = 1 / r_total |
Hours needed for all workers |
Work done by a worker in t hours |
W = r * t |
Units of work completed |
Work left after t hours |
W_left = Total - W_done |
|
| Work done by different workers in stages | Sum of each worker’s contribution | |
| One worker’s time when another works partially | T1 = T2 * (r2 / r1) |
Derived from rate × time relationship |
| If one worker works 1/3 of the time | T_total = T1 + (1/3)T1 + (2/3)T2 |
Example of partial work |
| Work rate of machines/people | rate = 1 / time |
Often used for machines, engines, etc. |
Important Rules
- Always convert rates to the same unit (e.g., hours).
- Add rates when workers work simultaneously.
- Subtract rates when a worker works against another (rarely in standard tests).
- Work left after a stage = Total work – Work done.
- Time = Work / Rate – never mix units.
🧮 Worked Example
Problem
A and B can complete a job in 20 hours and 30 hours respectively.
A works alone for 4 hours, then leaves.
B then works alone until the job is finished.
Find the total time taken to finish the job.
Step‑by‑Step
| Step | Calculation | Result |
|---|---|---|
| 1. Compute individual rates | (r_A = \frac{1}{20}) units/hr, (r_B = \frac{1}{30}) units/hr | (r_A = 0.05), (r_B = 0.0333...) |
| 2. Work done by A in 4 hrs | (W_A = r_A \times 4 = 0.05 \times 4 = 0.20) units | 20% of the job |
| 3. Work left after A’s part | (W_{\text{left}} = 1 - 0.20 = 0.80) units | 80% of the job |
| 4. Time for B to finish remaining work | (t_B = \frac{W_{\text{left}}}{r_B} = \frac{0.80}{0.0333...} \approx 24) hrs | |
| 5. Total time | (T_{\text{total}} = 4 + 24 = 28) hrs |
Answer: The job is finished in 28 hours.
🔍 Quick Revision Summary
| Topic | Recall |
|---|---|
| Rate of work | rate = 1 / time_to_finish |
| Combined rate | Sum of individual rates when working together |
| Time to finish | time = 1 / combined_rate |
| Partial work | Compute work done in a phase, subtract from total, then find time with remaining worker’s rate |
| Key formula | ( \text{Work} = \text{Rate} \times \text{Time} ) |
| Common pattern | “X works Y hours, then Z works until job finishes” → split into phases |
| Units consistency | Always use hours for time, units of work as whole job = 1 |
| Check | After solving, verify by multiplying the final time by combined rate to get 1 (whole job) |
Quick Practice Question:
A can finish a job in 15 hrs, B in 25 hrs. A works alone for 3 hrs, then leaves. How long will B take to finish the remaining work?
Hint: Compute A’s rate, work done, remaining work, then divide by B’s rate.
Happy studying! 🎓✨