Time and Work — Study Notes

Overview

Time and Work is a staple topic in SOF IMO's Everyday Mathematics section. Problems test your ability to relate how much work a person or machine completes in a given time and how multiple workers combine their efforts. The key insight is treating work as a unitary quantity (usually "1 complete job") and expressing rates of work per unit time. Mastery requires understanding individual efficiency, joint work rates, and the special case of pipes filling or emptying tanks. Questions appear as single-worker problems, joint-work scenarios, or pipes-and-cisterns variants. Expect 2–3 questions in the exam, often with real-life contexts like painting walls, digging trenches, or filling swimming pools.

You must quickly convert "days to finish" into "work done per day," combine rates for multiple workers, and handle scenarios where some workers leave midway or pipes have opposite effects (filling vs. draining). The arithmetic is straightforward once you set up the equation correctly, but the exam rewards speed and accuracy under time pressure.

Key Concepts

• **Work is one unit**: Treat the entire task as 1 job. If A finishes in 10 days, A's one-day work = 1/10 of the job.
• **Rate = Work / Time**: A person's efficiency or work rate is the fraction of the job completed per unit time.
• **Combined work rate**: When A and B work together, their joint rate = (A's rate) + (B's rate). Time to finish together = 1 / (combined rate).
• **Man-days concept**: Total work = (number of workers) × (days worked). Use this to scale problems when the number of workers changes.
• **Pipes and cisterns**: Inlet pipes fill (positive rate), outlet pipes drain (negative rate). Net rate = sum of all pipe rates with correct signs.
• **Work left behind**: If workers complete part of a job and leave, calculate remaining work = 1 − (work done), then apply the rate of remaining workers to finish.
• **Negative time trap**: If you compute a negative time, check whether you've accidentally reversed filling/draining or miscounted the combined rate.
• **Efficiency comparison**: If A is twice as efficient as B, A's rate is double B's rate, so A takes half the time B takes.

Formulas / Key Facts

1. **Work done per day** = 1 / (days to complete alone). If A completes work in *n* days, A's rate = 1/*n* per day. 2. **Combined rate for two workers** = 1/*a* + 1/*b*, where *a* and *b* are their individual completion times. 3. **Time taken together** = 1 / (combined rate). For A and B: Time = 1 / (1/*a* + 1/*b*) = *ab* / (*a* + *b*) days. 4. **Man-days = (number of men) × (days)**. If *m* men complete work in *d* days, total work = *md* man-days. If *n* men work, days needed = *md* / *n*. 5. **Pipes formula**: Net rate = (sum of inlet rates) − (sum of outlet rates). Time to fill/empty = 1 / (net rate). 6. **Work and wage**: Wages are distributed in the ratio of work done (or man-days contributed). 7. **Fractional work**: Work completed in *t* days by a worker with rate *r* = *rt*. Remaining work = 1 − *rt*.

Worked Examples

**Example 1** — A can complete a job in 12 days, B in 15 days. Working together, how long to finish?

*Step 1*: A's rate = 1/12 per day, B's rate = 1/15 per day. *Step 2*: Combined rate = 1/12 + 1/15. Find LCM(12,15) = 60. So 1/12 = 5/60, 1/15 = 4/60. *Step 3*: Combined rate = 9/60 = 3/20 per day. *Step 4*: Time = 1 / (3/20) = 20/3 = 6⅔ days or 6 days 16 hours.

**Example 2** — A pipe fills a tank in 6 hours, another pipe empties it in 8 hours. If both are open, how long to fill the empty tank?

*Step 1*: Filling rate = 1/6 per hour (positive). Draining rate = 1/8 per hour (negative). *Step 2*: Net rate = 1/6 − 1/8. LCM(6,8) = 24. So 1/6 = 4/24, 1/8 = 3/24. *Step 3*: Net rate = 1/24 per hour. *Step 4*: Time to fill = 1 / (1/24) = 24 hours.

**Example 3** — A completes work in 10 days. A and B together finish in 6 days. How long does B alone take?

*Step 1*: A's rate = 1/10. Combined rate = 1/6. *Step 2*: B's rate = (combined rate) − (A's rate) = 1/6 − 1/10. *Step 3*: LCM(6,10) = 30. So 1/6 = 5/30, 1/10 = 3/30. B's rate = 2/30 = 1/15. *Step 4*: B alone takes 15 days.

**Example 4** — 12 men can build a wall in 8 days. How many men are needed to build it in 6 days?

*Step 1*: Total work = 12 men × 8 days = 96 man-days. *Step 2*: To finish in 6 days, number of men = 96 / 6 = 16 men.

Common Mistakes

1. **Adding times instead of rates** → Wrong: "A takes 10 days, B takes 15 days, together they take 10+15=25 days." Correct: Add rates (1/10 + 1/15), then invert to get time. 2. **Forgetting to subtract outlet rates** → When a drain is open, subtract its rate. Treating all pipes as positive leads to wrong net rate. 3. **Ignoring partial work done** → If A works for 3 days then leaves, compute work done = 3 × (A's rate), then apply B's rate to the remainder. Don't restart from "1 job." 4. **Mixing units** → If A finishes in 5 hours and B in 3 days, convert to the same unit (hours or days) before adding rates. 5. **Man-days confusion** → Doubling the number of workers halves the time, not doubles it. Inverse proportion: workers × days = constant.

Quick Reference

• **One worker's rate** = 1 / (time to complete alone). Invert to get time from rate.
• **Joint work time** = *ab* / (*a* + *b*) when A takes *a* days, B takes *b* days.
• **Pipes**: inlet (+), outlet (−). Net rate = sum with signs. Time = 1 / net rate.
• **Man-days product is constant**: *m*₁*d*₁ = *m*₂*d*₂. Use to scale workforce or duration.
• **Check units**: All rates in same time unit (days or hours). Convert before computing.
• **Partial work formula**: Remaining work = 1 − (rate × time already worked). Apply next worker's rate to remainder.