Time and Work
Overview
Time and Work is a foundational quantitative aptitude topic that appears consistently in TNPSC Group IV examinations. This topic tests your ability to calculate how long workers take to complete jobs individually or together, and how pipes fill or empty tanks.
The core principle is simple: work done equals rate multiplied by time. Once you master the concept of "work per day" or "work per hour," most problems become straightforward fraction arithmetic. Students who struggle here usually have weak fundamentals in LCM and fraction operations, so ensure those basics are solid before tackling complex problems.
The three main problem types are: individual work problems, combined work problems, and pipes and cisterns. All three use identical mathematical logic—only the context changes.
Key Concepts
- Work as a unit: Assume total work = 1 unit (or LCM of given days for easier calculation). If A completes work in 10 days, A's 1 day work = 1/10.
- Combined work: When A and B work together, their combined 1 day work = 1/A + 1/B. Time to complete = 1/(combined rate).
- Inverse relationship: More workers = less time (assuming equal efficiency). If 5 men finish in 10 days, 10 men finish in 5 days.
- Efficiency concept: If A is twice as efficient as B, A does in 1 day what B does in 2 days. Efficiency and time are inversely proportional.
- Pipes filling tanks: Inlet pipe fills, outlet pipe empties. Net rate = filling rate − emptying rate.
- Negative work: When someone destroys work or a pipe empties a tank, treat their contribution as negative in calculations.
- LCM method: Taking total work as LCM of given times converts fractions to whole numbers, making arithmetic faster.
Formulas / Key Facts
Basic formulas:
- Work = Rate × Time
- If A completes work in 'n' days, A's 1 day work = 1/n
- If A's 1 day work = 1/n, A completes work in 'n' days
Combined work:
- A and B together: Time = (A × B)/(A + B) days, where A and B are individual completion times
- A, B, C together: 1 day work = 1/A + 1/B + 1/C
Man-days concept:
- M₁ × D₁ × H₁ / W₁ = M₂ × D₂ × H₂ / W₂
- (Men × Days × Hours per day) remains constant for same work
Pipes and cisterns:
- Inlet fills in 'a' hours: 1 hour work = 1/a (positive)
- Outlet empties in 'b' hours: 1 hour work = 1/b (negative)
- Net rate when both open = 1/a − 1/b
Alternating work:
- Find work done in one complete cycle, then calculate total cycles needed
Worked Examples
Example 1: Basic combined work A can do a work in 12 days, B can do it in 18 days. In how many days can they complete it together?
Solution:
- A's 1 day work = 1/12
- B's 1 day work = 1/18
- Combined 1 day work = 1/12 + 1/18 = (3 + 2)/36 = 5/36
- Time to complete = 36/5 = 7.2 days = 7 days and 4.8 hours
Alternative (formula): Time = (12 × 18)/(12 + 18) = 216/30 = 7.2 days
Example 2: Pipes and cisterns Pipe A fills a tank in 20 minutes, Pipe B fills it in 30 minutes, and Pipe C empties it in 15 minutes. If all three are opened together, how long to fill the tank?
Solution:
- A's 1 min work = 1/20 (fills)
- B's 1 min work = 1/30 (fills)
- C's 1 min work = 1/15 (empties, so negative)
- Net 1 min work = 1/20 + 1/30 − 1/15
- LCM of 20, 30, 15 = 60
- = 3/60 + 2/60 − 4/60 = 1/60
- Time to fill = 60 minutes
Example 3: Work with efficiency A is twice as efficient as B. Together they finish work in 12 days. How long would B alone take?
Solution:
- Let B's 1 day work = x, then A's 1 day work = 2x
- Combined: x + 2x = 3x
- 3x = 1/12 (they finish in 12 days)
- x = 1/36
- B alone takes 36 days
- (A alone takes 18 days)
Example 4: LCM method A completes work in 15 days, B in 10 days. They work together for 4 days, then A leaves. How many more days does B need?
Solution:
- Total work = LCM(15, 10) = 30 units
- A's 1 day = 30/15 = 2 units
- B's 1 day = 30/10 = 3 units
- Together in 1 day = 5 units
- Work in 4 days = 20 units
- Remaining = 30 − 20 = 10 units
- B alone needs = 10/3 = 3⅓ days
Common Mistakes
- Adding times instead of rates: Wrong: "A takes 10 days, B takes 15 days, together = 25 days." Correct: Add rates (1/10 + 1/15), not days.
- Forgetting negative sign for outlet pipes: Wrong: Adding emptying rate to filling rate. Correct: Subtract the outlet pipe's rate from inlet pipe's rate.
- Confusing efficiency with time: Wrong: "A is twice as efficient, so takes twice the time." Correct: Higher efficiency means less time (inverse relationship).
- Ignoring partial work in stages: Wrong: Assuming combined time formula works when workers join or leave midway. Correct: Calculate work done in each stage separately.
- LCM errors: Taking LCM incorrectly leads to wrong unit calculations. Double-check LCM before proceeding.
Quick Reference
- A alone in 'a' days, B alone in 'b' days → Together = ab/(a+b) days
- More men, less days: M₁D₁ = M₂D₂
- Inlet fills, outlet empties → Net = 1/inlet − 1/outlet
- Efficiency doubled = Time halved
- LCM method: Total work = LCM, individual rate = LCM/days
- 1 day work = 1/(total days to complete)