Quick Revision

One Day's Work and the LCM Method

A job done in n days means $\frac{1}{n}$ of it a day, and $\frac{d}{n}$ in d days.
LCM method: total work = LCM of the times; each person's units a day = LCM ÷ own time.
A in 25 days, B in 100: work 100 units, A does 4 a day and B 1, so 5 together and 20 days.
Part of a job: $\frac{2}{7}$ in 4 days means the whole job in 14.
Trap: never add or average times. Add rates.

Two people: $T = \frac{ab}{a + b}$ .
One person from the pair: $\frac{1}{b} = \frac{1}{T} - \frac{1}{a}$ . A and B in 24 days, A alone in 40: $\frac{1}{24} - \frac{1}{40} = \frac{1}{60}$ , so B takes 60 days.
Three people: add three rates. 30, 45 and 90 days: work 90 units, 3 + 2 + 1 = 6 a day, so 15 days.
$\frac{ab}{a + b}$ is for two people only; for three, use units.
Together is always quicker than the fastest person alone. Use that to strike out options.

Add the three pair rates and halve: that is all three together.
One person = all three − the other two.
A + B in 20 days, B + C in 28, C + A in 35: work 140, pairs 7, 5 and 4 a day. All three = 16 ÷ 2 = 8 a day, so $17\frac{1}{2}$ days. C = 8 − 7 = 1 a day, 140 days.
Trap: forgetting to halve gives half the true time.

Time runs the other way to efficiency: $T \propto \frac{1}{E}$ .
A is x% more efficient than B: A : B = (100 + x) : 100, times the other way. 40% more: 7 : 5, so times 5 : 7; B 35 days, A 25.
Half as much work a day means twice the time.
Ratio with joint time: work = ratio sum × days. 2 : 5 together in 10 days: work 70, A alone 35.
Trap: 25% more efficient cuts the time by 20%, not 25%.

Add the work stretch by stretch: (rate of those working) × days.
Leaves after n days: together n days, then the rest alone.
Leaves k days before the end: call the total T and give that person T − k days. A in 36 days, B in 54, A leaves 4 days early: 3(T − 4) + 2T = 108, so T = 24.
Joins later: the starter works the whole time.
Trap: the total time and the time after the change are both in the options.

$\frac{M_1 D_1 H_1}{W_1} = \frac{M_2 D_2 H_2}{W_2}$ : more men or hours, fewer days; more work, more days.
21 men, 10 days, 8 hours a day for one wall. Two walls, 28 men, 6 hours: $D = \frac{21 \times 10 \times 8 \times 2}{28 \times 6}$ = 20 days.
Midway change: work left = total − done, divided by the new team.
On time: men needed = work left ÷ days left; extra = needed − present.
Trap: the question usually wants the extra men.

Turn everyone into one unit. 3 men or 5 women: 1 man = $\frac{5}{3}$ women.
4 men or 6 women in 21 days: work 126 woman-days. 2 men + 9 women = 12 women, so $10\frac{1}{2}$ days.
Two mixed groups doing the same job: set the two work totals equal to get the ratio.
Per person: 6 men in 11 days means 1 man in 66.
Trap: a man and a woman are not equal workers unless the question says so.

Find one round (2 days for A and B in turns, 3 days for a helper every third day) and its units.
Run whole rounds, then go day by day; the last day can be a part day.
A in 9 days, B in 24, on alternate days: work 72, a round gives 11. A first: 66 after 12 days, and day 13 needs 6 of A's 8, so $12\frac{3}{4}$ days. B first: $13\frac{3}{8}$ days.
Trap: (work ÷ round) × round length fails when the last round is cut short.

A breaker has a minus sign: net = builder − breaker.
Builder 7 days, breaker 12, at once: $\frac{1}{7} - \frac{1}{12} = \frac{5}{84}$ , so $16\frac{4}{5}$ days.
On alternate days the job ends on a builder's day. Builder first: wall 84 units, +12 then −7, so +5 a round. After 15 rounds (30 days) it has 75; day 31 needs 9 of his 12, so $30\frac{3}{4}$ days.
Trap: 84 ÷ 5 rounds × 2 days = $33\frac{3}{5}$ days overshoots.

Pay is shared in the ratio of work done, never of days or times.
Same days together: ratio of efficiencies, the inverse of times. 16 and 40 days give 5 : 2, so ₹3,500 splits as ₹2,500 and ₹1,000.
Different days: each person's work = rate × days worked, and the share is that fraction of the pay.
A helper's share = the part of the job the others left for him.
Trap: splitting in the ratio of times swaps the shares.

Inlet +, outlet or leak −. Time = tank ÷ net rate.
Fills in 21 hours, empties in 70: $\frac{1}{21} - \frac{1}{70} = \frac{1}{30}$ , so 30 hours.
Leak found late: a hours without it, b hours with it, so the leak alone takes $\frac{ab}{b - a}$ . 4 hours becoming 4 hours 48 minutes gives 24 hours.
Part-full tank: part to fill ÷ net rate.
Trap: product ÷ sum is for two filling pipes; product ÷ difference is for a pipe against a leak.

Go stretch by stretch: (rate of the open pipes) × time.
Closed after t hours: both for t hours, then the other pipe alone.
Opened later: the first pipe runs alone first; count from the first opening unless told otherwise.
Closed so the tank fills in T hours: the pipe left open runs all T. A 16 hours, B 20, full in 15: $\frac{t}{16} + \frac{15}{20} = 1$ , so A ran 4 hours.
Trap: the total and the time after the change both appear as options.

One round = one hour of each pipe in turn. Count the units each round adds.
Two inlets: run whole rounds, then hour by hour; the last hour can be a part hour.
Inlet and outlet in turns (the last-hour trap): the tank fills during an inlet hour. Inlet 3 hours, outlet 8: tank 24 units, +8 then −3. After 4 rounds (8 hours) the tank has 20, and hour 9 needs only 4 of the inlet's 8. Answer $8\frac{1}{2}$ hours, not $9\frac{3}{5}$ .
Trap: tank ÷ gain per round × 2 overshoots.

Put every flow in one unit (litres a minute or an hour) before mixing it with fractions of the tank.
Capacity = litres ÷ fraction. Outlet 8 litres a minute; inlet fills in 4 hours; both open, 6 hours. Outlet = $\frac{1}{4} - \frac{1}{6} = \frac{1}{12}$ of the tank an hour = 480 litres, so 5,760 litres.
Pipes d litres a minute apart, taking a and b minutes (a > b): capacity = $\frac{d \times a \times b}{a - b}$ .
Trap: litres a minute against hours.