SSC CGL Maths · Time and Work

Man-Days and Man-Hours

A job needs a fixed amount of labour, counted as workers times days (times hours a day), so more workers finish sooner in exact proportion.

All Time and Work notes

Why this matters

Nineteen questions in the bank, eleven EASY and eight MODERATE, five of them from the 2025 shifts: more men for an earlier deadline, a change in hours a day, men added partway, or a few men leaving every day. Every one rests on a single product, workers × days × hours, which stays the same for the same work.

Concept 1 of 3: Workers times days is fixed for a job

10 men working for 6 days put in 60 man-days of labour. The job needs that same labour whoever does it, so 20 men need 3 days and 5 men need 12.

Definition

Man-days of a job $=$ workers $\times$ days. It is the same for any team of equally able workers.
Fewer days need more workers: workers needed $=$ man-days $\div$ days.
A different amount of work: man-days scale with the work. Twice the work needs twice the man-days; a fraction $f$ of the job needs $f$ of them.
The full rule: $\frac{M_1 D_1}{W_1} = \frac{M_2 D_2}{W_2}$ .
"How many more" workers $=$ workers needed $-$ workers already there.

Man-days rule

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

$M_1, M_2$ numbers of workers
$D_1, D_2$ days taken
$W_1, W_2$ amounts of work (1 for the whole job)

Worked example

18 men can build a wall in 15 days. How many more men are needed to build it in 10 days?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 22 Sep 2025, Shift 1 · Q62Easy

Example 1 · Time and Work · Man-Days and Man-Hours

35 men can do a piece of work in 8 days. Find the time required by 16 men to do four times the work.

Working the proportion the wrong way

More workers means fewer days. If 6 men take 20 days, 10 men take

20 \times \frac{6}{10} = 12

days, not

20 \times \frac{10}{6}

Giving the total when "how many more" is asked

If 27 men are needed and 18 are already working, the answer to "how many more" is 9, and 27 is usually printed among the options too.

Drill 8 more on workers times days is fixed for a job

Concept 2 of 3: Hours a day and part of a job

A working day of 8 hours is more labour than a day of 6. When the hours a day differ, count labour in man-hours instead of man-days.

Definition

Man-hours $=$ workers $\times$ days $\times$ hours a day.
The rule: $\frac{M_1 D_1 H_1}{W_1} = \frac{M_2 D_2 H_2}{W_2}$ . Put the part of the job each team does (40%, the remaining 60%) in $W$ .
One worker's hours: if A needs 5 hours a day for 8 days, the job takes A 40 hours in all. For two workers, add their hourly rates, then divide the total hours by the hours worked each day.
Efficiency and hours together: units done a day $=$ efficiency per hour $\times$ hours a day.

Man-hours rule

\frac{M_1 D_1 H_1}{W_1} = \frac{M_2 D_2 H_2}{W_2}

$M_1, M_2$ numbers of workers
$D_1, D_2$ days worked
$H_1, H_2$ hours worked each day
$W_1, W_2$ parts of the job done

Worked example

12 men working 10 hours a day finish 30% of a job in 6 days. How many days will 14 men working 8 hours a day take to finish the rest?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

SSC CGL · 2023 · Tier I, 17 Jul 2023, Shift 2 · Q70Moderate

Example 2 · Time and Work · Man-Days and Man-Hours

Working 5 hours a day, A can complete a task in 8 days and working 6 hours a day, B can finish the same task in 10 days. working 8 hours a day, they can jointly complete the task in ______.

Ignoring a change in hours a day

When the hours a day change, man-days are not equal; man-hours are. 10 men for 12 days at 6 hours is 720 man-hours, the same labour as 9 men for 10 days at 8 hours.

Putting the part of the job on the wrong side

Divide each team's labour by the work that team did. The 30% goes under the labour that did the 30%, and the remaining 70% under the labour that does the rest.

Drill 3 more on hours a day and part of a job

Concept 3 of 3: Workers added or leaving partway

Labour already put in is banked. Work out what is left in man-days, then share it over the new team.

Definition

Man-days left $=$ total man-days $-$ workers $\times$ days already worked.
Days for the rest $=$ man-days left $\div$ the new number of workers. If the hours a day change too, use man-hours.
A fixed number leave every day: the workers present form a sequence $n, n - k, n - 2k, \ldots$ . Over $d$ days they put in $dn - k(0 + 1 + \cdots + (d - 1)) = dn - \frac{kd(d - 1)}{2}$ man-days. Set that equal to the planned man-days.
Fast route for the sequence: total $=$ number of days $\times$ the average of the first and last day's workers.

Changing workforce

\text{Days left} = \frac{n(D - t)}{n'} \qquad dn - \frac{kd(d - 1)}{2} = Dn

nworkers at the start
Ddays the job would take them
tdays already worked
n'workers after the change
kworkers who leave at the end of each day
ddays the job actually takes

Worked example

20 men can finish a job in 18 days. After 6 days, 4 more men join. How many more days do they need?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 20 Sep 2025, Shift 1 · Q54Moderate

Example 3 · Time and Work · Man-Days and Man-Hours

A contractor employs 15 men to complete a work in 24 days, working 8 hours per day. After six days, he adds five more men but also reduces the working hours to six per day from the seventh day onwards. If all men work with the same efficiency throughout, how many more days will be required to finish the remaining work?

Starting the leaving sequence one day early

k

men leave after each day, day 1 still has all

n

men and the last of

d

days has

n - (d - 1)k

. The amount taken away is

k(0 + 1 + \cdots + (d - 1))

, not

k(1 + 2 + \cdots + d)

Drill 5 more on workers added or leaving partway

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Workers times days is fixed for a job
Man-days rule
$M_1 D_1 = M_2 D_2 \qquad \frac{M_1 D_1}{W_1} = \frac{M_2 D_2}{W_2}$
Hours a day and part of a job
Man-hours rule
$\frac{M_1 D_1 H_1}{W_1} = \frac{M_2 D_2 H_2}{W_2}$
Workers added or leaving partway
Changing workforce
$\text{Days left} = \frac{n(D - t)}{n'} \qquad dn - \frac{kd(d - 1)}{2} = Dn$

Watch out for (5)

Working the proportion the wrong way
→ Workers times days is fixed for a job
Giving the total when "how many more" is asked
→ Workers times days is fixed for a job
Ignoring a change in hours a day
→ Hours a day and part of a job
Putting the part of the job on the wrong side
→ Hours a day and part of a job
Starting the leaving sequence one day early
→ Workers added or leaving partway

Test yourself on Time and Work

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