Time & work: Quantitative Aptitude Notes for Puducherry Police Sub-Inspector (SI)

Time & Work – Quantitative Aptitude Notes

Program: Puducherry Police Sub‑Inspector (SI)
Exam: Quantitative Aptitude – “Time & Work”

Goal: Understand the core concepts, solve the most common problem types, and be able to revise quickly before the exam.

1. Simple Explanation

Concept	What it Means	Example
Work	Total amount of job that needs to be completed. Usually measured in “man‑days” or “man‑hours”.	Building a wall = 120 man‑days
Rate (Speed)	How much work a person can finish in one unit of time. Denoted as R.	A worker can finish 1/10 of the wall in a day → R = 1/10 days⁻¹
Time	Duration needed to complete a task. Denoted as T.	5 days to finish the wall
Rate × Time = Work	Fundamental relation: `Work = Rate × Time`.	(1/10) × 5 = 1/2 wall → 0.5 of the wall done

2. Key Formulas & Relationships

Scenario	Formula	Intuition
Single worker	`T = W / R`	Time = Work ÷ Rate
Multiple workers together	`R_total = Σ R_i` `T = W / R_total`	Add individual rates (since they work simultaneously)
Work done in a given time	`Work_done = R × T`	Multiply rate by time
Work remaining	`Work_remaining = W – Work_done`	Subtract what’s already done
Work in terms of “man‑days”	`Man‑days = Number_of_workers × Time`	Equivalent to `W = Σ (1 / R_i) × T` when R_i = 1/Day for each worker
Faster worker	If worker A is `k` times as fast as B ⇒ `R_A = k × R_B`	Useful when converting rates

3. Common Problem Types

Problem	Key Idea	How to Approach
A & B together finish a job	Combined rate = `R_A + R_B`	`T = W / (R_A+R_B)`
A works alone for some time, then B joins	Separate phases → add work done in each	`Work_A = R_A × t1` `Work_B = R_B × t2` `Work_total = Work_A + Work_B`
One worker leaves	Subtract his contribution from total work	`Work_remaining = W – R_left × t_left`
Two workers work at different rates	Use ratio of speeds	`R_A : R_B = 1 : 2` → `R_A = 1/3`, `R_B = 2/3` (if total = 1/day)
Work done by one worker alone vs together	Compare time taken	`T_A = W / R_A` vs `T_total = W / (R_A+R_B)`
Work done in parts	Break job into segments (e.g., 1st part by A, 2nd by B)	Compute each part’s time separately

4. Worked Example

Problem

A and B can complete a piece of work in 12 days and 15 days respectively. If they work together for 4 days and then B leaves, how many more days will A need to finish the remaining work?

Step‑by‑Step Solution

Step	Explanation	Calculation
1. Find individual rates	Rate = 1 / days per job	`R_A = 1/12` `R_B = 1/15`
2. Combined rate for first 4 days	`R_total = R_A + R_B`	`R_total = 1/12 + 1/15 = 5/60 + 4/60 = 9/60 = 3/20`
3. Work done in first 4 days	`Work_done = R_total × 4`	`Work_done = (3/20) × 4 = 12/20 = 3/5`
4. Remaining work	`Work_remaining = 1 – Work_done`	`Work_remaining = 1 – 3/5 = 2/5`
5. Time for A alone	`T = Work_remaining / R_A`	`T = (2/5) / (1/12) = (2/5) × 12 = 24/5 = 4.8` days

Answer: A needs 4.8 days (≈ 4 days 19 hours) to finish the remaining work.

Tip: Always convert rates to a common denominator first, then simplify.

5. Quick Revision Summary

Work = Rate × Time – the universal rule.
Rate of a worker is the part of the job he can finish in one day (or hour).
Combined rate = sum of individual rates when workers work together.
Time required = Work / Combined rate.
Common trick: Convert all rates to a common denominator (e.g., days⁻¹) before adding/subtracting.
When a worker leaves or joins, treat his contribution as a separate time segment.
Check units: If work is in “man‑days”, rate is in “work per day”; if time is in hours, convert rates accordingly.
Always verify the answer by plugging back into the equation: Rate × Time should equal the work (or the remaining work).

6. Quick Practice Questions (Try before the exam)

**A can finish a job in 20