Time & Work

One method: count the work actually done

Every "twist" question in this family — someone leaves, someone joins late, people work on alternate days, wages are divided — is solved the same way. Put the job in LCM units, write each worker's rate per day, then add up the work each worker actually did and set the total equal to the whole job.

A leaves after t days

A can do the job in $T_A$ days and works only t days: he completes $\frac{t}{T_A}$ of it. The rest, $1 - \frac{t}{T_A}$, is finished by whoever continues — alone or in a team — at their own rate. In units this is even simpler: job 30 units, A does 3/day, works 2 days → 6 units done, 24 left. A (12 days) works 3 days, B (18 days) finishes alone: A did $\frac14$; B needs $\frac34 \times 18 = 13.5$ days.

B joins after t days

A works alone first. Add his solo work to the team work: if A starts and B joins after t days, $\frac{t}{T_A} + \left(T_{\text{team}} - t\right)\left(\frac{1}{T_A} + \frac{1}{T_B}\right) = 1$ A (20 days) starts; B (15 days) joins after 5 days. A alone did $\frac14$; the pair does $\frac{1}{20}+\frac{1}{15} = \frac{7}{60}$ per day → $\frac{45}{7}$ days more, about 14.4 total. Read carefully: the question may ask for the total days or the days after joining.

Alternate days

Pair the days into a cycle. Working one day each, starting with A, one 2-day cycle does $\frac{1}{T_A} + \frac{1}{T_B}$ of the job. Divide the job by the cycle output to get the number of complete cycles, then walk the leftover days one at a time — the next day is the other worker's turn. A = 8 days (3 units of 24), B = 12 days (2 units): cycle = 5 units per 2 days. Four cycles = 8 days, 20 units done, 4 left. Day 9 is A's: 3 units, 1 left. Day 10 is B's: 2 units/day → half a day. Total $9\frac12$ days. Never assume the leftover finishes inside the cycle pattern — check whose turn day 9 actually is.

Wages follow work done

The wage is payment for work, so money splits in the ratio of the work each person actually did: $w_A : w_B = (\text{rate}_A \times \text{days}_A) : (\text{rate}_B \times \text{days}_B)$ When everyone works the full time together, the days cancel and the wage ratio is just the rate ratio. A alone 10 days, B alone 15, together they earn ₹600: rates 3 : 2 → shares ₹360 and ₹240. If one worker's time is unknown, first find his rate from the group rate (his rate = team rate − others). A (12 days), B (20 days) and C together finish in 5 days for ₹1200: team rate $\frac15$; C's rate $= \frac15 - \frac{1}{12} - \frac{1}{20} = \frac{1}{15}$; ratio $5:3:4$ → C gets ₹400.

Reading the question twice

The same numbers support four different questions: how many days in total, how many days after the change, what fraction of work each did, and whose share is what. Mark the words total, more, alone and remaining before computing — most lost marks here are reading losses, not arithmetic ones.

Quick revision

• Work done = rate × days worked; whoever continues inherits the remainder.
• Alternate days: count 2-day cycles, then hand the tail to the correct worker.
• Wages = work done ratio; equal days → rate ratio.
• Unknown member's rate = team rate − sum of known rates.
• Sanity: the stronger worker earns the bigger share.