Time & Work

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high importance~1 Q in Tier 122 formulas⚡ 15 shortcuts5 subtopics

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Joining / Leaving Mid-work, Alternate Days & Wages

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⏱ 5 min read🧩 5 question types🎯 12 practice Q

The idea in one minute

Every twist here — leaving, joining, alternating, wages — uses one method: count the work each person actually did.

A leaving after t days finishes $\dfrac{t}{T_A}$ of the job; whoever continues inherits the rest. Alternate days pair into a 2-day cycle of $\dfrac{1}{T_A} + \dfrac{1}{T_B}$ .

Wages split in the ratio of work done — which equals the rate ratio when everyone works the same days.

Count the work actually done

Someone leaves, someone joins late, people alternate days, wages get split — one method answers all of it. Put the job in LCM units, write each worker's daily rate, then add up the work each person actually did. Set the total equal to the whole job.

Leaving after t days

A works t of his $T_A$ days and stops. He finished $\dfrac{t}{T_A}$ of the job; the rest, $1 - \dfrac{t}{T_A}$ , belongs to whoever continues.

A (12 days) works 3 days → $\dfrac{1}{4}$ done. B (18 days) alone needs $\dfrac{3}{4} \times 18 = 13.5$ days.

Rule: Work done = rate × days worked. Whoever continues inherits only the remainder.

Leaving before the end

'A leaves 3 days before the work finishes' hides the total time T. Write both stints against T: A works $T - 3$ days, B works all T days.

\frac{T - t}{T_A} + \frac{T}{T_B} = 1

With $T_A = 12$ , $T_B = 18$ , $t = 3$ : $\dfrac{T-3}{12} + \dfrac{T}{18} = 1$ → $3(T-3) + 2T = 36$ → T = 9 days.

Tip: One unknown T, one linear equation. Multiply through by the LCM of the two times.

Joining after t days

A starts alone; B joins later. Add the solo stint to the team stint:

\frac{t}{T_A} + (\text{days together}) \times \left(\frac{1}{T_A} + \frac{1}{T_B}\right) = 1

A (20 days) works 5 days → $\dfrac{1}{4}$ . The pair then does $\dfrac{1}{20} + \dfrac{1}{15} = \dfrac{7}{60}$ a day → $\dfrac{45}{7} = 6\dfrac{3}{7}$ days more, $11\dfrac{3}{7}$ in total.

Watch: 'In how many days is the work finished?' can mean the total or the extra days after the change. Underline which one is asked.

Alternate days

Pair the days into one cycle of two days: A's day plus B's day $= \dfrac{1}{T_A} + \dfrac{1}{T_B}$ of the job. Count whole cycles, then walk the leftover day by day — the next day is the other worker's turn.

A = 8 days (3 units of 24), B = 12 days (2 units). A cycle = 5 units. Four cycles = 8 days, 20 units done. Day 9 is A's: 3 more, 1 left. Day 10 is B's at 2 a day: half a day. Total $9\dfrac{1}{2}$ days.

Careful: The tail does not always fit the pattern. Check whose turn the next day really is.

Wages follow work

Money splits in the ratio of work actually delivered: rate × days for each person. When everyone works the full time, the days cancel and the wage ratio is just the rate ratio.

A alone 10 days, B alone 15, ₹600 earned together → rates 3 : 2 → ₹360 and ₹240. An unknown member: his rate = team rate − the known rates. A (12 days), B (20 days) and C finish in 5 days → C $= \dfrac{1}{5} - \dfrac{1}{12} - \dfrac{1}{20} = \dfrac{1}{15}$ ; the shares run 5 : 3 : 4.

Read the question twice

The same numbers support four different asks: total days, days after the change, each person's share of work, each person's share of money. Mark the words total, more, alone and remaining before computing. Most lost marks here are reading losses, not arithmetic ones.

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

A leaves after t days

How to spot it:

A works a few days and leaves; B (or B and C, or the rest of the team) completes the remaining work; the remaining or total days are asked.

\text{remaining work} = 1 - \frac{t}{T_A}, \qquad \text{days} = \frac{\text{remaining}}{\text{continuing rate}}

Method

Put the job in units; find A's daily rate.
Work done by A in t days = rate × t; subtract from the job.
Divide the remainder by the daily rate of whoever continues.

Why it works:

When A stops, his units stop flowing — only the finished units leave the pool.

Try this

A can do a work in 12 days and B in 18 days. A works for 3 days and then leaves. In how many days will B alone finish the remaining work?

Show solution

A does $\dfrac{3}{12} = \dfrac{1}{4}$ .
Remaining $= \dfrac{3}{4}$ .
B needs $\dfrac{3}{4} \times 18 = 13.5$ days.

Answer

13.5 days

Type 2common2 practice Q

B joins after t days

How to spot it:

A starts alone; after some days B (or a team) joins and the rest is done together; the total time or the after-joining time is asked.

\frac{t}{T_A} + (T - t)\left(\frac{1}{T_A} + \frac{1}{T_B}\right) = 1

Method

Work done by A alone = rate × t.
Remaining work ÷ combined daily rate = days after joining.
Add the solo days for the total time.

Why it works:

The job has two phases with different daily rates; each phase is a rate × time sum.

Try this

A can finish a work in 20 days and B in 15 days. A works alone for 5 days, after which B joins him. In how many days is the whole work finished?

Show solution

A did $\dfrac{5}{20} = \dfrac{1}{4}$ .
Pair rate $= \dfrac{1}{20} + \dfrac{1}{15} = \dfrac{7}{60}$ a day.
$\dfrac{3}{4} \div \dfrac{7}{60} = 6\dfrac{3}{7}$ more → $11\dfrac{3}{7}$ total.

Answer

11 3/7 days in total (6 3/7 more)

Type 3common2 practice Q

Alternate days (A one day, B next)

How to spot it:

'A and B work on alternate days, A starting' — find the total time, or which day the work ends.

\text{per 2-day cycle} = \frac{1}{T_A} + \frac{1}{T_B}

Method

One 2-day cycle does $\dfrac{1}{T_A} + \dfrac{1}{T_B}$ of the job (use units).
Complete as many full cycles as fit; note the work left.
Walk the leftover one day at a time — the next day belongs to the other worker. A partial last day counts as a fraction.

Why it works:

Pairing the days turns a jumpy schedule into a smooth repeated cycle.

Try this

A can finish a work in 8 days and B in 12 days. Working on alternate days beginning with A, in how many days is the work finished?

Show solution

Job $= 24$ units; A $3$ /day, B $2$ /day → cycle $= 5$ units.
4 cycles $= 8$ days → $20$ units done, $4$ left.
Day 9 (A): $3$ → $1$ left. Day 10 (B): half a day. Total $9\dfrac{1}{2}$ days.

Answer

9.5 days

Type 4very common3 practice Q

Wages split by work done

How to spot it:

A total wage for a joint job is given; find one person's share — sometimes with unequal days worked or an unknown member.

w_A : w_B = (\text{rate}_A \times d_A) : (\text{rate}_B \times d_B)

Method

Write each worker's rate in units/day and multiply by his own days.
The wage ratio is the ratio of those products.
Give each worker his fraction of the total money.

Why it works:

Wages pay for work delivered, and work = rate × days.

Try this

A and B together earn ₹600 for a job. A alone can do the job in 10 days and B alone in 15 days. A's share is:

Show solution

Job $= 30$ units: rates $3 : 2$ .
A gets $\dfrac{3}{5} \times 600 = ₹360$ (B: $₹240$ ).

Answer

₹360

Type 5common

A leaves t days before the work finishes

How to spot it:

'A and B start together; A leaves t days before the work is completed' — the total time is asked.

\frac{T - t}{T_A} + \frac{T}{T_B} = 1

Method

Let the total time be T; A works $T - t$ days, B works all $T$ days.
Write A's share plus B's share equal to 1; multiply by the LCM of the times.
Solve the linear equation for T.

Why it works:

Both stints are measured against the same unknown finish line, so one equation closes the question.

Try this

A can do a work in 12 days and B in 18 days. They begin together, but A leaves 3 days before the work is finished. Find the total time taken.

Show solution

$\dfrac{T-3}{12} + \dfrac{T}{18} = 1$ ; multiply by $36$ .
$3(T-3) + 2T = 36$ , so $5T = 45$ .
$T = 9$ days (A worked $6$ , B worked $9$ ).

Answer

9 days

Formula sheet

Work done by A in t days

\frac{t}{T_A}

Leaves t days before the end

\frac{T - t}{T_A} + \frac{T}{T_B} = 1

Two-day cycle

\frac{1}{T_A} + \frac{1}{T_B} \text{ per 2 days}

Wage split

w_A = \text{total} \times \frac{1/T_A}{1/T_A + 1/T_B}

Shortcuts that save time

⚡ Subtract the finished part

Whoever continues inherits only the remainder.

Example

A can do a work in 15 days and B in 20 days. A works for 4 days and leaves. B alone finishes the rest in:

Show solution

A did $\dfrac{4}{15}$ ; left $= \dfrac{11}{15}$ .
B needs $20 \times \dfrac{11}{15} = 14\dfrac{2}{3}$ days.

Answer

14 2/3 days

⚡ Count full cycles, then the tail

Alternate days: measure progress per 2-day cycle.

Example

A takes 6 days and B takes 12 days for a job. They work on alternate days, A starting. The job is finished in:

Show solution

Per 2 days: $\dfrac{1}{6} + \dfrac{1}{12} = \dfrac{1}{4}$ .
4 cycles = 8 days, exactly the whole job.

Answer

8 days

⚡ Wages follow work, not days alone

Rate ratio × equal days = work ratio.

Example

A alone does a work in 12 days, B in 18 days. They work together and earn ₹840. A's share is:

Show solution

$A : B = \dfrac{1}{12} : \dfrac{1}{18} = 3 : 2$ .
A gets $840 \times \dfrac{3}{5} = ₹504$ .

Answer

₹504

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Splitting wages by days alone when efficiencies differ.

Wages follow work done: rate × days for each person.

Mistake 02

In alternate-day problems, forgetting who works the final partial day.

Walk the tail day by day; the next day is the other worker's turn.

Mistake 03

Applying the remaining work to both workers.

Only whoever continues inherits the remainder; A has stopped.

Mistake 04

Counting a 2-day cycle as one day.

One cycle = one day of A plus one day of B = two calendar days.

Mistake 05

Reading 'leaves 3 days before completion' as 'works 3 days'.

A works T − 3 of the T total days; set up the equation in T.

Quick revision

Read this the night before the exam.

Work done = rate × days worked; the rest inherits the remainder.
Leaves t days before the end: (T−t)/T_A + T/T_B = 1.
Alternate days: count 2-day cycles, then hand the tail to the right worker.
Wages split by work done; equal days → rate ratio.
Unknown member: rate = team rate − sum of known rates.
Mark total / more / alone / remaining before computing.

Practice: 12 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 12 questions

Suggested time 8 min · wrong answers go to your mistake notebook automatically.

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