Time and work: the basics

Ques­tions about time and work ask how long a job takes when peo­ple work alone or together: two painters shar­ing a wall, a team of labour­ers lay­ing a road, three friends clean­ing a hall. They look var­ied, but almost every one of them is solved with a sin­gle idea: think about how much of the job gets done in one day. This les­son sets up that idea and uses it for peo­ple work­ing alone and together.

One day's work#

Treat the whole job as $1$ (one com­plete piece of work).

Rule 1. If A can fin­ish a piece of work in $n$ days, then A's 1 day's work $\displaystyle = \frac{1}{n}$.

For exam­ple, if A can fin­ish a piece of work in $18$ days, A's 1 day's work is $\displaystyle \frac{1}{18}$ of the job. In $5$ days A does $\displaystyle \frac{5}{18}$ of it.

The rule also runs back­wards: if some­one's 1 day's work is $\displaystyle \frac{1}{n}$, they fin­ish the job in $n$ days. So if a per­son does $\displaystyle \frac{3}{20}$ of a job each day, they need $\displaystyle \frac{20}{3}$ days. To turn a day's work into a num­ber of days, take the rec­i­p­ro­cal.

Rule 2. If A is twice as good a work­man as B, then the time taken by A to fin­ish the work $\displaystyle = \frac{1}{2} \times$ (time taken by B).

A bet­ter worker takes less time. Twice as good means half the time, three times as good means a third of the time, and so on.

More work­ers, less time#

If more peo­ple share a job, it is done sooner. Sup­pose $10$ men take $20$ hours to fin­ish a piece of work. How long will $20$ men take?

The total amount of work, mea­sured in "man-hours", stays the same:

$$10 \times 20 = 20 \times x \implies x = 10$$

So $20$ men take $10$ hours. The num­ber of men is inversely pro­por­tional to the time taken: dou­ble the men, half the time.

Exam­ple 1#

If the num­ber of men doing a job is changed in the ratio $2 : 3$, in what ratio does the time taken change?

Let $M_1, M_2$ be the num­bers of men and $T_1, T_2$ the times. Since $\displaystyle M \propto \frac{1}{T}$, we have $\displaystyle M = \frac{k}{T}$ for some con­stant $k$, so

$$\displaystyle M_1 = \frac{k}{T_1}, \quad M_2 = \frac{k}{T_2} \implies \frac{M_1}{M_2} = \frac{T_2}{T_1} \implies \frac{T_1}{T_2} = \frac{M_2}{M_1} = \frac{3}{2}$$

The time changes in the ratio $3 : 2$. Inverse pro­por­tion sim­ply turns the ratio upside down.

Work­ing together#

When peo­ple work together, their day's works add.

Exam­ple 2#

A can do a piece of work in $12$ days and B alone can do it in $15$ days. How long will they take work­ing together?

$$\displaystyle \text{A's 1 day's work} = \frac{1}{12}, \qquad \text{B's 1 day's work} = \frac{1}{15}$$

$$\displaystyle (A + B)\text{'s 1 day's work} = \frac{1}{12} + \frac{1}{15} = \frac{5 + 4}{60} = \frac{9}{60} = \frac{3}{20}$$

Tak­ing the rec­i­p­ro­cal, they fin­ish together in $\displaystyle \frac{20}{3} = 6\frac{2}{3}$ days.

A use­ful check: work­ing together must be faster than the faster worker alone. Here $\displaystyle 6\frac{2}{3}$ days is less than $12$ days, as it should be.

Short­cut. If A alone takes $a$ days and B alone takes $b$ days, together they take $\displaystyle \frac{ab}{a + b}$ days. For Exam­ple 2: $\displaystyle \frac{12 \times 15}{27} = \frac{180}{27} = \frac{20}{3}$.

Exam­ple 3#

A and B together can do a piece of work in $12$ days. B alone can fin­ish it in $30$ days. In how many days can A alone fin­ish it?

Now we sub­tract instead of adding.

$$\displaystyle (A + B)\text{'s 1 day's work} = \frac{1}{12}, \qquad \text{B's 1 day's work} = \frac{1}{30}$$

$$\displaystyle \text{A's 1 day's work} = \frac{1}{12} - \frac{1}{30} = \frac{5 - 2}{60} = \frac{3}{60} = \frac{1}{20}$$

So A alone fin­ishes the work in $20$ days.

Three work­ers in pairs#

Exam­ple 4#

A and B can do a piece of work in $12$ days, B and C in $15$ days, and C and A in $20$ days. In how many days will they fin­ish it together, and in how many days can each fin­ish it alone?

$$\displaystyle (A + B)\text{'s 1 day's work} = \frac{1}{12}, \quad (B + C)\text{'s} = \frac{1}{15}, \quad (C + A)\text{'s} = \frac{1}{20}$$

Add all three. Each per­son appears twice, so the total is twice the work of A, B and C together:

$$\displaystyle 2 \times (A + B + C)\text{'s 1 day's work} = \frac{1}{12} + \frac{1}{15} + \frac{1}{20} = \frac{5 + 4 + 3}{60} = \frac{12}{60} = \frac{1}{5}$$

$$\displaystyle (A + B + C)\text{'s 1 day's work} = \frac{1}{10}$$

So A, B and C together fin­ish the work in $10$ days.

To find one per­son alone, take away the pair that leaves them out:

$$\displaystyle \text{A's 1 day's work} = \frac{1}{10} - \frac{1}{15} = \frac{3 - 2}{30} = \frac{1}{30}, \text{ so A alone takes } 30 \text{ days.}$$

$$\displaystyle \text{B's 1 day's work} = \frac{1}{10} - \frac{1}{20} = \frac{2 - 1}{20} = \frac{1}{20}, \text{ so B alone takes } 20 \text{ days.}$$

$$\displaystyle \text{C's 1 day's work} = \frac{1}{10} - \frac{1}{12} = \frac{6 - 5}{60} = \frac{1}{60}, \text{ so C alone takes } 60 \text{ days.}$$

Check: $\displaystyle \frac{1}{30} + \frac{1}{20} + \frac{1}{60} = \frac{2 + 3 + 1}{60} = \frac{6}{60} = \frac{1}{10}$. Cor­rect.

Prac­tice#

1. Ravi can paint a wall in $9$ days. What frac­tion of the wall does he paint in $1$ day, and in $4$ days?
2. $8$ men can build a wall in $15$ days. How many days will $12$ men take?
3. The num­ber of work­ers on a job is changed in the ratio $4 : 5$. In what ratio does the time change?
4. Sita can stitch a set of cur­tains in $10$ days and Gita in $15$ days. How long will they take together?
5. P is twice as good a work­man as Q, and Q takes $18$ days to fin­ish a job. How long does P take alone, and how long do they take together?
6. A and B together fin­ish a job in $8$ days. A alone takes $12$ days. How long does B take alone?
7. A, B and C can do a job alone in $20$, $30$ and $60$ days. How long will all three take together?
8. A can do a job in $16$ days and B in $12$ days. How long will they take together?
9. A and B can do a job in $24$ days, B and C in $40$ days, and C and A in $30$ days. How long will all three take together, and how long will each take alone?

Answers#