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Time & Work

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high importance~1 Q in Tier 119 formulas⚡ 15 shortcuts5 subtopics
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Subtopic 4 of 5·← Pipes & Cisterns·Joining / Leaving Mid-work, Alternate Days & Wages →

Men–Days–Hours Chain & Provisions

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Work is the product of every resource: Work=M×D×H×E\text{Work} = M \times D \times H \times E (men, days, hours/day, efficiency).

M1D1H1E1W1=M2D2H2E2W2\frac{M_1 D_1 H_1 E_1}{W_1} = \frac{M_2 D_2 H_2 E_2}{W_2}

Same work ⇒ equal products: M1D1=M2D2M_1 D_1 = M_2 D_2. More men ⇒ fewer days (inverse proportion).

Provisions problems: stock = men × days. After t days, remaining stock feeds the new headcount: days=M1DleftM2\text{days} = \frac{M_1 D_{\text{left}}}{M_2}.

Consumption questions (garrison, fodder, food stores) are men-days problems in disguise.

📖 Notes🧩 4 question types⚡ Formulas🎯 12 practice Q

Detailed notes

Work as a product of resources

The same job can be finished by many men in few days or few men in many days. The bridge between the two situations is that the total work stays the same, and total work is the product of every resource: Work=M×D×H×E\text{Work} = M \times D \times H \times E (men × days × hours per day × efficiency). Equating the products of two situations: M1D1H1E1W1=M2D2H2E2W2\frac{M_1 D_1 H_1 E_1}{W_1} = \frac{M_2 D_2 H_2 E_2}{W_2} For the same work, drop the W's. Quantities that grow with days (men, hours, efficiency) sit beside M; write the missing quantity last and solve. 18 men × 25 days = 450 man-days; with 30 men, 45030=15\frac{450}{30} = 15 days. More men → fewer days: men and days are inversely proportional.

Hours per day in the chain

12 men working 8 hours a day finish in 15 days: work =12×15×8=1440= 12 \times 15 \times 8 = 1440 man-hours. How long for 18 men at 10 hours a day? 144018×10=8\frac{1440}{18 \times 10} = 8 days. Hours behave exactly like days — more hours a day, fewer days needed.

Provisions (garrison, hostel, fodder)

A stock of food "for M men for D days" is M×DM \times D man-days of food. After t days the stock left is M(D−t)M(D - t) man-days, to be shared by the new headcount: days left=M1(D−t)M2\text{days left} = \frac{M_1 (D - t)}{M_2} 400 men have food for 25 days; after 5 days 100 more men arrive → stock left =400×20=8000= 400 \times 20 = 8000 man-days ÷ 500 = 16 days. If men leave instead, the headcount drops and the stock lasts longer. If nobody joins or leaves, the remaining days are simply D−tD - t.

Work left after some days / reinforcement mid-work

Convert the whole job into man-days first. 60 men need 40 days → job = 2400 man-days. After 10 days they have done 600, leaving 1800 for the new team. If 10 men leave, the remaining 50 need 180050=36\frac{1800}{50} = 36 days more. Always apply the chain to the remaining work, never to the whole job again.

Men, women and boys (efficiency equivalence)

"3 men or 5 women can do a work in 12 days" pins the unit rates: one man does 13×12=136\frac{1}{3 \times 12} = \frac{1}{36} per day, one woman 160\frac{1}{60} per day. Any team can then be priced: 6 men + 5 women =636+560=14= \frac{6}{36} + \frac{5}{60} = \frac14 per day → 4 days. Read "or" as "the two teams are equally big in work terms".

Efficiency inside the chain

If one person is twice as efficient, count him as 2 men. A team of 3 men and 2 boys, where a man = 2 boys, is 3×2+2=83 \times 2 + 2 = 8 boy-units. Converting everyone to one standard unit before multiplying avoids all confusion.

Common traps

  • Writing more men in the numerator (more men → FEWER days).
  • Forgetting to subtract the elapsed days from the provisions.
  • Applying the chain to the full job instead of the remaining job.
  • Counting a man's efficiency twice — E enters the product once.

Quick revision

  • M1D1H1E1=M2D2H2E2M_1D_1H_1E_1 = M_2D_2H_2E_2 for the same work.
  • Provisions: stock =M(D−t)= M(D-t) man-days ÷ new headcount.
  • Mid-work: remaining work = total − (rate × days gone).
  • Equivalence: rate of one man = 1men×days\frac{1}{\text{men} \times \text{days}}; price every team this way.
  • Sanity: more men / more hours / higher efficiency → fewer days.

Types of questions asked

Every way this subtopic shows up in exams — how to recognise it, the formula or logic to use, and a solved example.

Type 1: Men–days–hours chainvery common3 practice Q
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.

M1D1H1E1=M2D2H2E2M_1 D_1 H_1 E_1 = M_2 D_2 H_2 E_2
  1. Compute the work as M×D×H×EM \times D \times H \times E for the given scenario.
  2. Divide by the new men, hours and efficiency.
  3. Everything that grows with time stays on the left; the unknown lands alone.

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

Example: 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?

12×15×8=144012 \times 15 \times 8 = 1440 man-hours; 144018×10=8\frac{1440}{18 \times 10} = 8 days.

Type 2: Provisions of a garrison / hostelvery common2 practice Q
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.

days=M1(D−t)M2\text{days} = \frac{M_1 (D - t)}{M_2}
  1. Stock consumed per day = M units; total stock =M×D= M \times D man-days.
  2. After t days, stock left =M(D−t)= M(D-t).
  3. Divide by the new headcount for the number of remaining days.

Why: each man eats one unit a day, so the stock is simply a number of man-days.

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

Stock left =400×20=8000= 400 \times 20 = 8000 man-days → 8000500=16\frac{8000}{500} = 16 days.

Type 3: Reinforcement / men leaving mid-workcommon2 practice Q
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.

extra days=remaining work (man-days)new team size\text{extra days} = \frac{\text{remaining work (man-days)}}{\text{new team size}}
  1. Whole job in man-days = M × D.
  2. Subtract the man-days already done (M × days worked).
  3. Divide the remainder by the new number of men.

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

Example: 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?

Job =2400= 2400 man-days; done =600= 600; left =1800= 1800 for 50 men → 180050=36\frac{1800}{50} = 36 days.

Type 4: Men ↔ women ↔ boys equivalencecommon2 practice Q
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.

rate of one=1(group size)×days\text{rate of one} = \frac{1}{\text{(group size)} \times \text{days}}
  1. From each 'or' statement, write the rate of one person: 1(count)(days)\frac{1}{(\text{count})(\text{days})}.
  2. Price the asked team as a sum of unit rates.
  3. Invert for the days.

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

Example: 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?

1 man: 136\frac{1}{36}/day; 1 woman: 160\frac{1}{60}/day. Team rate =636+560=14= \frac{6}{36} + \frac{5}{60} = \frac{1}{4} → 4 days.

Formulas

MDH chain
M1D1H1E1=M2D2H2E2(W1=W2)M_1 D_1 H_1 E_1 = M_2 D_2 H_2 E_2 \quad (W_1 = W_2)
Men × days constant
M1D1=M2D2M_1 D_1 = M_2 D_2
Provisions remaining
days=M1(Dtotal−t)M2\text{days} = \frac{M_1 (D_{\text{total}} - t)}{M_2}
Piece of work
days for kn of work=knT\text{days for } \tfrac{k}{n} \text{ of work} = \frac{k}{n} T

Shortcut tricks

⚡ Multiply resources, equate products

Everything inverse-proportional lands in the numerator; direct-proportional in the denominator.

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

15×20=25×D15 \times 20 = 25 \times D ⇒ D=12D = 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:

Stock left =500×24=12000= 500 \times 24 = 12000 man-days ⇒ 12000800=15\frac{12000}{800} = 15 days.

⚡ Work left after a share is done

First find what fraction remains, then apply men-days to that remainder.

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

Left =54081=623= \frac{540}{81} = 6\frac{2}{3} days.

Where students lose marks

  • Putting more men in the numerator of the chain equation (more men ⇒ fewer days ⇒ denominator).

  • Forgetting to subtract elapsed days in provisions problems.

  • Counting efficiency twice — E belongs inside the product once.

  • Applying men-days to the whole work instead of the remaining work.

Practice sets — 12 questions

Sets of 10, mixed across the question types above. Each answer comes with a step-by-step explanation.

Topic test · 12 questions

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