Men, women, boys and hours

So far each worker had a name and a speed. Many ques­tions instead talk about groups: so many men, women or boys, work­ing so many hours a day for so many days. The key idea here is that a job needs a fixed amount of labour, which we can count in man-days or man-hours. Change the num­ber of work­ers, the hours or the days, and the prod­uct stays the same. This les­son also shows how to com­pare work­ers of dif­fer­ent kinds by turn­ing them all into one kind.

Man-days and man-hours#

If $10$ men take $6$ days to fin­ish a job, the job needs $10 \times 6 = 60$ man-days of labour. What­ever the team, the same job needs the same $60$ man-days: $20$ men would take $3$ days, $5$ men would take $12$ days.

For the same job: $M_1 \times D_1 = M_2 \times D_2$. If hours per day also change: $M_1 \times D_1 \times H_1 = M_2 \times D_2 \times H_2$.

This is the inverse pro­por­tion between work­ers and time from the first les­son, writ­ten for groups.

Exam­ple 1#

A cer­tain num­ber of men can com­plete a piece of work in $60$ days. If there were $8$ men more, the work could be fin­ished in $10$ days less. How many men were there orig­i­nally?

Let the orig­i­nal num­ber of men be $x$.

The num­ber of men is inversely pro­por­tional to the num­ber of days, so the prod­ucts are equal:

$$\begin{aligned} x \times 60 &= (x + 8) \times 50 \\ 6x &= 5x + 40 \\ x &= 40 \end{aligned}$$

(The sec­ond line divides both sides by $10$.) There were $40$ men orig­i­nally.

Work­ers of dif­fer­ent kinds#

When men and boys (or men and women) work together, first find how many of one kind equal one of the other. Then turn the whole team into a sin­gle kind.

Exam­ple 2#

If $4$ men or $6$ boys can fin­ish a piece of work in $20$ days, in how many days can $6$ men and $11$ boys fin­ish it?

"$4$ men or $6$ boys" means $4$ men do the same work as $6$ boys, so $4$ men $= 6$ boys. Then

$$\displaystyle 6 \text{ men} = \frac{6}{4} \times 6 = \frac{3}{2} \times 6 = 9 \text{ boys}$$

So the team is $9 + 11 = 20$ boys.

$$6 \times 20 = 20 \times x \implies x = 6$$

They fin­ish it in $6$ days.

Exam­ple 3#

$2$ men and $3$ women, work­ing $7$ hours a day, fin­ish a job in $5$ days. $4$ men and $4$ women, work­ing $3$ hours a day, do the same job in $7$ days. In how many days will $7$ men alone, work­ing $4$ hours a day, do it?

Count the total man-hours and woman-hours in each case.

Case (i): each day gives $2 \times 7$ man-hours, so in $5$ days there are $2 \times 5 \times 7 = 70$ man-hours; like­wise $3 \times 5 \times 7 = 105$ woman-hours.

Case (ii): $4 \times 7 \times 3 = 84$ man-hours and $4 \times 7 \times 3 = 84$ woman-hours.

It is the same job, so the two amounts of labour are equal. Writ­ing $m$ for one man-hour and $w$ for one woman-hour:

$$\begin{aligned} 70m + 105w &= 84m + 84w \\ 21w &= 14m \\ 3w &= 2m \end{aligned}$$

So $3$ women do the work of $2$ men. The first team, $2$ men and $3$ women, is there­fore worth $2 + 2 = 4$ men.

$$4 \times 7 \times 5 = 7 \times 4 \times x \implies 140 = 28x \implies x = 5$$

$7$ men work­ing $4$ hours a day fin­ish the job in $5$ days.

Earn­ings for groups#

Exam­ple 4#

$2$ men and $3$ boys together earn ₹252 in $7$ days, and $4$ men and $7$ boys together earn ₹624 in $8$ days. In what time will $6$ men and $8$ boys earn ₹1020?

Let one man earn ₹$x$ a day and one boy ₹$y$ a day.

$$(2x + 3y) \times 7 = 252 \implies 2x + 3y = 36 \quad (1)$$

$$(4x + 7y) \times 8 = 624 \implies 4x + 7y = 78 \quad (2)$$

Dou­ble (1): $4x + 6y = 72$. Sub­tract this from (2): $y = 6$. Then $2x + 18 = 36$, so $x = 9$.

A team of $6$ men and $8$ boys earns $6(9) + 8(6) = 54 + 48 = 102$ rupees a day. If they need $n$ days,

$$102n = 1020 \implies n = 10$$

They take $10$ days.

When the team shrinks#

Exam­ple 5#

$25$ men are employed to do a piece of work which they could fin­ish in $20$ days, but $5$ men leave at the end of every $10$ days. In what time will the work be com­pleted?

The job needs $25 \times 20 = 500$ man-days.

After two stages, $250 + 200 = 450$ man-days are done, leav­ing $50$. Fif­teen men need $\displaystyle \frac{50}{15} = \frac{10}{3} = 3\frac{1}{3}$ days for that, which is less than $10$, so no one else leaves.

Total time $\displaystyle = 10 + 10 + 3\frac{1}{3} = 23\frac{1}{3}$ days.

Prac­tice#

1. $12$ men can fin­ish a job in $30$ days. If $6$ more men join at the start, how many days will the job take?
2. A group of men can fin­ish a job in $36$ days. With $6$ more men, it would take $9$ days less. How many men are in the group?
3. $3$ men or $5$ women can do a job in $24$ days. In how many days can $6$ men and $5$ women do it?
4. $15$ men work­ing $8$ hours a day fin­ish a job in $12$ days. How many days will $18$ men work­ing $10$ hours a day take?
5. $2$ men and $1$ boy together earn ₹1100 in $5$ days, and $3$ men and $4$ boys earn ₹2400 in $6$ days. In how many days will $4$ men and $2$ boys earn ₹1760?
6. $30$ men could fin­ish a job in $10$ days, but $5$ men leave at the end of every $4$ days. In how many days is the job fin­ished?
7. $5$ men and $2$ women work­ing $6$ hours a day fin­ish a job in $8$ days. If $1$ man does the work of $2$ women, how many days will $4$ men work­ing $8$ hours a day take?

Answers#