Time & Work

A work rate says what fraction of a job is finished in one unit of time. Add rates when people work together without interfering.

Use the LCM of individual times as total work. For changing teams, track work completed in each interval.

A machine fills a tank in 4 hours. What fraction does it fill in 1 hour at a constant rate? 1. Treat one full tank as one job. 2. Divide the job by 4 hours: the rate is 1/4 tank per hour. 3. In 2 hours it fills 2 × 1/4 = 1/2 tank.

One box is a quarter of the job. The first hour fills one box, so the rate is ¼ tank per hour. Two shaded boxes show the amount filled after two hours. When machines work together, add their work rates; a draining pipe contributes a negative rate.

A completes a job in 12 days, B in 18. Take total work=36 units, so rates are 3 and 2 units/day. Together for 4 days they finish 20 units. A then leaves; B needs (36-20)/2=8 more days. Total elapsed time=12 days. If a leak removes 1 unit/day during the first phase, only 16 units are completed there and B later needs 10 days without the leak.

Add fractions of work per hour, not the hours needed to finish. Times do not add when workers cooperate; rates do. State whether a leak or interruption continues in later phases.

The result must make sense: with both people working at their usual rates, the job should finish faster than either person working alone.

An inlet has a positive rate and a leak a negative rate. Keep rates in tank-per-hour units.

An inlet fills a tank in 6 hours and a leak empties it in 9. Net rate is 1/6 - 1/9 = 1/18, so the open inlet and leak fill it in 18 hours.

Use the LCM of individual times as total work. For changing teams, track work completed in each interval.

A completes a job in 12 days, B in 18. Take total work=36 units, so rates are 3 and 2 units/day. Together for 4 days they finish 20 units. A then leaves; B needs (36-20)/2=8 more days. Total elapsed time=12 days. If a leak removes 1 unit/day during the first phase, only 16 units are completed there and B later needs 10 days without the leak.

Times do not add when workers cooperate; rates do. State whether a leak or interruption continues in later phases.

If one worker completes a job in a days at a constant rate, the rate is 1/a job per day. Add rates for workers contributing simultaneously; do not add their completion times. For the same job, efficiency and completion time are inversely proportional.

A needs 6 days and B needs 12. Together their rate is 1/6+1/12=1/4 job per day, so they finish in 4 days. If A works alone for two days first, 1/3 of the job is complete; the remaining 2/3 takes (2/3)/(1/4)=8/3 days together. Worker-hour model With equal constant productivity per worker-hour, work is proportional to workers × days × hours per day. Ten workers taking eight days at eight hours per day use 640 worker-hours. Twice that work needs 1280 worker-hours; five workers at four hours per day need 1280/20=64 days.

The worker-hour model assumes workers can contribute independently without changing productivity. Account for breaks, different efficiencies and staged joining or leaving when specified.

Treat an inlet as a positive filling rate and an outlet as a negative rate, under a constant-rate model. A positive net rate fills an initially empty tank; zero net rate never fills it, and a negative net rate cannot fill it from empty.

An inlet fills a tank in 6 hours and an outlet empties a full tank in 9. Net rate = 1/6−1/9=1/18 tank/hour, so filling takes 18 hours when both maintain the stated rates. Two inlets needing x and y hours fill together in xy/(x+y) hours; an inlet and outlet give xy/(y−x) only when y>x. Clock motion The minute hand moves 6° per minute; the hour hand moves 0.5° per minute. At H hours and M minutes on a correct 12-hour clock, reduce |30H−5.5M| modulo 360 and take the smaller of that angle and its complement to 360. At 3:30, the separation is |90−165|=75°. Consecutive coincidences are 720/11 minutes apart, giving 22 over a half-open 24-hour interval. Counting coincidences alone is not a reliable test of whether a real clock runs fast or slow. Calendars An ordinary year advances a weekday by one day and a leap year by two. In the Gregorian calendar, leap years are divisible by 4, except century years must also be divisible by 400. Thus 2000 was leap and 1900 was not.

An outlet cannot continue removing liquid below an empty tank. Clock-angle calculations must include the hour hand's movement during the minutes.