Time and Work

1Work is a rate, and rates add

If a job takes n days, then one day's work is \frac{1}{n} of it. That reciprocal is the only conversion in the chapter, and everything else follows: two people working together do \frac{1}{a} + \frac{1}{b} in a day, so the job takes \frac{1}{\frac{1}{a}+\frac{1}{b}} = \frac{ab}{a+b} days.

Times do not add and they do not average. Twelve days and eighteen days together give 7.2, which is neither 30 nor 15 — and it must be less than the smaller of the two, because adding a second worker cannot make a job slower. That inequality is the fastest check there is on an answer of this kind.

Figure. Each line's steepness is a rate, and working together stacks the two steepnesses into one — the joint line's slope is exactly the sum of the other two. That is what "rates add" means graphically. What you cannot do is add or average the times, which are the horizontal distances at which each line would reach the finish. Lines are drawn through the plot origin at (0.08, 0.92) with day 18 at the right edge, so the three slopes stand in the ratio 3:2:5.

How it works

1. Turn every time into a raten days becomes \frac{1}{n} of the job per day. Do this before anything else in the question.
2. Add or subtract the ratesHelpers add; a leak or a dropout subtracts. The combined rate is a single number.
3. Invert at the very endTime is one divided by the combined rate — the last step, never an intermediate one.

2The LCM method

Fractions are what make work problems slow, and they are avoidable. Instead of calling the job 1, call it the LCM of every time given in the question. Each worker's daily output then becomes a whole number, and the rest of the problem is integer arithmetic.

The choice is free because the job's size never appears in the answer — you divide by it again at the end. What you gain is that combinations, leaks, dropouts and part-days all stay in whole units until the final division.

Figure. Call the job \mathrm{LCM}(20,30,60)=60 units. Daily outputs become whole numbers 3, 2 and 1; together they are 6 units a day, so time is 60\div 6=10 days — the same answer as the fraction route, without the fractions.

3Efficiency and time are inverse

For the same job, efficiency (or the number of identical workers) and time taken are inversely proportional: \frac{E_A}{E_B} = \frac{T_B}{T_A}. Double the efficiency and you halve the days; the product E \times T is the job, so it is the same for every worker.

That is why a ratio of efficiencies is enough to recover every individual time once one of them is known — and why mixing the ratio the wrong way round is the dominant trap. If A is more efficient than B, A's alone-time must be the shorter one.

Figure. The alone-times are drawn to scale at 24 and 40, which already stand in the inverse of the efficiency ratio 5:3. The longer bar belongs to the less efficient worker; swapping the labels would contradict the ratio the stem gave. Bars carry the times; the efficiency numbers sit in the annotation because they are the reciprocal story, not a second length on the same axis.

How it works

1. Write the efficiency ratio as ratesE_A : E_B = 5 : 3 means A does 5k units a day and B does 3k, for any convenient k.
2. Fit the known timeIf A alone takes 24 days, the job is 5k \times 24. That fixes the total.
3. Read every other time off the same totalB alone is total ÷ 3k. Together is total ÷ 8k. No new arithmetic.

4Men, days and hours

When the number of workers, the days and the hours per day all change at once, the invariant is the total labour: \frac{M_1 D_1 H_1}{W_1} = \frac{M_2 D_2 H_2}{W_2}. If the job is the same on both sides the W's cancel, and the equation says that men \times days \times hours is a fixed quantity.

Think of it as an area. Workers up the side, days along the bottom, and the rectangle's area is the job. Squeeze the rectangle narrower and it must grow taller to keep the same area — which is inverse variation, drawn.

Figure. This is the same wall with the daily hours left alone, so only men and days trade off. The two bars use `thickness` for the men and `value` for the days, and both enclose 36 \times 18 = 27 \times 24 = 648 man-days. Taking a quarter of the men off the thickness forces the length out by a third. Shortening the working day as well, as the worked example does, is a third dimension the picture cannot show, and it stretches the length further still, to 32.

How it works

1. Multiply everything on the known sideMen \times days \times hours gives the job in man-hours. One number, computed once.
2. Divide by the new daily capacityNew men \times new hours is how much gets done each day under the new arrangement.
3. Scale for a different jobIf the second job is twice as big, multiply the answer by two — that is what the W's are for.

5Pipes and cisterns

An inlet contributes a positive rate and a leak or outlet a negative rate, so the net fill rate is their algebraic sum. A pipe that fills a tank in a hours does \frac{1}{a} of the tank per hour; a drain that empties it in b hours does -\frac{1}{b}. Together they fill at \frac{1}{a} - \frac{1}{b}.

A positive net rate means the tank eventually fills; a negative one means it empties. The trap is treating every pipe as a filler — if the drain is faster than the inlet, the tank never fills, and no positive answer exists.

Figure. With the tank at 60 units the two inlets contribute 5 and 4 units an hour and the outlet removes 3. The net bar is not a fourth independent measurement — it is 5+4-3, drawn so the eye can check the subtraction. A figure that stacked all three as positive fillers would teach the wrong sign rule.

How it works

1. Assign a signed rate to every pipeInlets positive, outlets negative. Use the LCM of the given times as the tank size so every rate is an integer.
2. Add the signed ratesThe sum is the net fill (or empty) rate. Check its sign before inverting.
3. Invert only a positive netTime to fill is tank ÷ net rate. A negative net means the tank empties, and the question is usually asking something else.

6Alternate days and part cycles

When two people work on alternate days the pair is not working together — each day only one of them is on the job. Compute the work in one full cycle of two days, see how many whole cycles fit under the target, and then finish the remainder with whoever's turn it is.

That last phrase is the question. The answer depends on who starts, because the leftover work is cleared at the starting worker's rate on the odd day, or the other's on the even one. Two otherwise identical problems give different answers, and the examiner knows it.

Figure. The bar is one cycle, not the whole calendar: A contributes 3 units and B contributes 2, and that block of 5 repeats. Seven copies clear 35 of the 36 units in 14 days; the leftover unit falls on day 15, which is A's turn when A starts. Swap who sits in the first segment and the tail changes owner — that is the whole point the figure is for.

How it works

1. Size one cycleTwo days, one turn each: 3 + 2 = 5 units. This is the repeating block.
2. Fit whole cycles under the targetNever over it. 7 cycles give 35 of 36 units; an 8th would overshoot.
3. Finish the tail at one person's rateThe remaining 1 unit falls to whoever's day comes next, at that person's rate alone.