Time & Work

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

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Men–Days–Hours Chain & Provisions

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

The idea in one minute

Total work is the product of the resources: $W = M \times D \times H \times E$ — men, days, hours per day, efficiency.

For the same job, $M_1 D_1 H_1 E_1 = M_2 D_2 H_2 E_2$ : more men means fewer days.

Provisions are the same arithmetic: stock = men × days, and after t days the remaining stock feeds the new headcount for $\dfrac{M_1(D - t)}{M_2}$ days.

Work is a product

Many men in few days or few men in many days — the job keeps the same size. Total work is the product of every resource:

W = M \times D \times H \times E

men × days × hours a day × efficiency. For the same job, equate two situations: $M_1 D_1 H_1 E_1 = M_2 D_2 H_2 E_2$ . 18 men × 25 days = 450 man-days; with 30 men the same job needs $\dfrac{450}{30} = 15$ days.

Rule: For a fixed job, more men means fewer days. Men and days are inversely proportional.

Hours join the chain

Hours a day behave exactly like men. 12 men working 8 hours a day for 15 days put in $12 \times 15 \times 8 = 1440$ man-hours. With 18 men at 10 hours a day the job takes $\dfrac{1440}{180} = 8$ days.

Write the missing quantity last, then divide. No proportion table is needed.

Tip: Convert the whole job to one labour unit first — man-days or man-hours. The answer is then a single division.

Provisions

Food 'for M men for D days' is $M \times D$ man-days of stock. After t days the stock left is $M(D - t)$ , and the new headcount consumes it:

\text{days left} = \frac{M_1(D - t)}{M_2}

400 men hold 25 days of food. After 5 days, 100 more men arrive: $400 \times 20 = 8000$ man-days ÷ 500 men = 16 days. Men leaving works the same way, and then the stock lasts longer.

Watch: Subtract the elapsed days before anything else. Feeding the new headcount from the full D is the standard error.

Mid-work changes

A team works for a while, then men join or leave. Convert the whole job to man-days, subtract what is already done, and hand the remainder to the new team.

60 men need 40 days → 2400 man-days. After 10 days, 600 are done. Ten men leave, and the remaining 1800 man-days need $\dfrac{1800}{50} = 36$ days more.

Careful: Apply the chain to the remaining work only. Restarting from the full job double-counts the finished part.

Men, women and boys

'3 men or 5 women can do it in 12 days' fixes both unit rates, because both teams finish equal work. So one man does $\dfrac{1}{36}$ per day and one woman $\dfrac{1}{60}$ per day.

Price any mixed team by adding rates: 6 men + 5 women $= \dfrac{6}{36} + \dfrac{5}{60} = \dfrac{1}{4}$ → 4 days.

Tip: Read 'or' as 'the two teams are equal in work terms'. The two unit rates then fall out together.

Scaling the work itself

Sometimes the work grows instead of the crew. Work then scales directly with men and days:

W_2 = W_1 \times \frac{M_2}{M_1} \times \frac{D_2}{D_1}

8 men reap 20 hectares in 12 days. In 9 days, 20 men reap $20 \times \dfrac{20}{8} \times \dfrac{9}{12} = 37.5$ hectares. Multiply for resources that grew, divide for those that shrank.

Efficiency inside the product

A worker twice as efficient counts as two men. A team of 3 men and 2 boys, where one man equals 2 boys in output, is $3 \times 2 + 2 = 8$ boy-units. Convert everyone to one standard unit before multiplying — never mix raw headcounts with a separate efficiency factor.

Question types you will see

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

Type 1very common3 practice Q

Men–days–hours chain

How to spot it:

One scenario of men, days and (maybe) hours per day is given; a second scenario changes some of them and one value is asked.

M_1 D_1 H_1 E_1 = M_2 D_2 H_2 E_2

Method

Compute the work as $M \times D \times H \times E$ for the given scenario.
Divide by the new men, hours and efficiency.
The unknown lands alone on one side.

Why it works:

The total job is a fixed number of man-hours, so the two products must be equal.

Try this

If 12 men working 8 hours a day can complete a work in 15 days, in how many days will 18 men working 10 hours a day complete it?

Show solution

Work $= 12 \times 15 \times 8 = 1440$ man-hours.
New crew supplies $18 \times 10 = 180$ man-hours a day.
$1440 \div 180 = 8$ days.

Answer

8 days

Type 2very common2 practice Q

Provisions of a garrison / hostel

How to spot it:

Food or fodder for M men for D days; after t days some men join or leave; how much longer the stock lasts is asked.

\text{days} = \frac{M_1 (D - t)}{M_2}

Method

Total stock = M × D man-days; each man eats one unit a day.
After t days, stock left $= M(D - t)$ .
Divide by the new headcount for the remaining days.

Why it works:

The stock is simply a number of man-days, whoever eats them.

Try this

A garrison of 400 men has provisions for 25 days. After 5 days, 100 more men join. The provisions will now last:

Show solution

Stock left $= 400 \times (25 - 5) = 8000$ man-days.
New headcount $= 500$ .
$8000 \div 500 = 16$ days.

Answer

16 days

Type 3common2 practice Q

Reinforcement / men leaving mid-work

How to spot it:

A team works for some days, then men join or leave; the extra days needed for the remaining work are asked.

\text{extra days} = \frac{\text{remaining man-days}}{\text{new team size}}

Method

Whole job in man-days = M × D.
Subtract the man-days already done (M × days worked).
Divide the remainder by the new number of men.

Why it works:

Only the unfinished man-days remain to be shared among whoever is still on the job.

Try this

60 men can complete a piece of work in 40 days. They work for 10 days, after which 10 men leave. In how many days will the remaining work be completed?

Show solution

Job $= 60 \times 40 = 2400$ man-days.
Done $= 60 \times 10 = 600$ ; left $= 1800$ for $50$ men.
$1800 \div 50 = 36$ days.

Answer

36 days

Type 4common2 practice Q

Men, women and boys equivalence

How to spot it:

'3 men or 5 women can do a work in 12 days' — a mixed team of men and women (or boys) is asked about.

\text{rate of one} = \frac{1}{\text{group size} \times \text{days}}

Method

From each 'or' statement, write one person's rate: $\dfrac{1}{\text{count} \times \text{days}}$ .
Price the asked team as a sum of unit rates.
Invert for the days.

Why it works:

'3 men or 5 women in 12 days' means the two teams do equal work per day, fixing both unit rates.

Try this

If 3 men or 5 women can do a piece of work in 12 days, in how many days will 6 men and 5 women together do it?

Show solution

1 man: $\dfrac{1}{36}$ /day; 1 woman: $\dfrac{1}{60}$ /day.
Team rate $= \dfrac{6}{36} + \dfrac{5}{60} = \dfrac{1}{4}$ per day.
$4$ days.

Answer

4 days

Type 5common

Scaling the work itself

How to spot it:

A crew produces a measured output — hectares, chairs, kilometres of road; a new crew and time give a new output.

W_2 = W_1 \times \frac{M_2}{M_1} \times \frac{D_2}{D_1}

Method

Put each change as a fraction: new ÷ old.
Multiply the old output by both fractions.
Reduce the fractions first to keep the arithmetic small.

Why it works:

Output grows in direct proportion to men and to days — the chain equation read forward.

Try this

If 8 men can reap 20 hectares in 12 days, how many hectares can 20 men reap in 9 days?

Show solution

Men factor $= \dfrac{20}{8}$ ; days factor $= \dfrac{9}{12}$ .
$20 \times \dfrac{20}{8} \times \dfrac{9}{12} = 37.5$ hectares.

Answer

37.5 hectares

Formula sheet

MDH chain

M_1 D_1 H_1 E_1 = M_2 D_2 H_2 E_2 \quad (W_1 = W_2)

Men and days constant

M_1 D_1 = M_2 D_2

Provisions remaining

\text{days} = \frac{M_1 (D_{total} - t)}{M_2}

Work scaling

W_2 = W_1 \times \frac{M_2}{M_1} \times \frac{D_2}{D_1}

Shortcuts that save time

⚡ Multiply resources, equate products

Everything that grows the work sits beside M; the missing quantity lands alone.

Example

15 men complete a work in 20 days. In how many days will 25 men complete the same work?

Show solution

$15 \times 20 = 25 \times D$ .
$D = 300 \div 25 = 12$ days.

Answer

12 days

⚡ Provisions after reinforcement

Stock left = original men × days left; then divide by the new headcount.

Example

A garrison of 500 men has provisions for 27 days. After 3 days, 300 more men join. The provisions will now last:

Show solution

Stock left $= 500 \times 24 = 12000$ man-days.
$12000 \div 800 = 15$ days.

Answer

15 days

⚡ Work left after a share is done

First find the remaining man-days, then apply them to the new team.

Example

45 men start a job they would finish in 16 days. After 4 days, 36 more men join. The remaining work now takes:

Show solution

Job $= 720$ man-days; done $= 180$ ; left $= 540$ .
$540 \div 81 = 6\dfrac{2}{3}$ days.

Answer

6 2/3 days

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Writing more men on the wrong side (more men, more days).

For a fixed job men and days are inverse: M₁D₁ = M₂D₂.

Mistake 02

Forgetting to subtract the elapsed days in provisions.

Stock left = M(D − t); only that feeds the new headcount.

Mistake 03

Applying the chain to the whole work after a mid-work change.

Subtract the man-days already done; chain only the remainder.

Mistake 04

Counting efficiency twice.

E enters the product once. Convert everyone to one standard unit first.

Mistake 05

Using two different labour units in one chain.

Convert boys and women to man-equivalents (or one common unit) before multiplying.

Quick revision

Read this the night before the exam.

M₁D₁H₁E₁ = M₂D₂H₂E₂ for the same work.
Provisions: stock left = M(D − t) man-days ÷ new headcount.
Mid-work: remaining = total man-days − done; chain only the remainder.
Equivalence: one person's rate = 1/(group size × days).
Work scaling: W₂ = W₁ × (M₂/M₁) × (D₂/D₁).
Sanity: more men, more hours, higher efficiency → fewer days.

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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