Time & Work

1. Introduction

In the entrance tests, about $1$ to $2$ questions are asked from this topic. This chapter is similar to Time, Speed and Distance. To draw parallels, the amount of work to be completed is similar to Distance, while the efficiency or rate of completion is similar to Speed. Time is a common element across both.

Typical variables that we come across in this chapter are as follows.

1) Work: Work can be defined as anything – making a toy, building a wall, completing an office project, filling a tank with water, etc.

2) Efficiency or rate of completion: This is the amount of work completed by a person in a unit of time. This is used in section 3. Unitary Method.

3) Number of people: Questions tend to include number of people or number of pipes doing a certain work. In certain questions the efficiencies or rates of completion are different.

Questions from this lesson can be solved using one of the following approaches.

1) Worker-Days Method: This method is used when there are many people in a group and they work at the same rate (or efficiency). Depending on the group and time period provided in the question, we will have to form units such as man-days, woman-days, man-hours, woman-hours, child-hours, etc.

2) Unitary Method: This method can be used for solving all other types of questions. In this method, we assign the total work to be done as $1$. This helps us ascertain the efficiency at ease. For instance, if John takes $4$ days to complete a piece of work, then his efficiency or rate of completion is $\dfrac{1}{4}^{\text{th}}$ of the work in $1$ day. Other extended concepts such as percentage completion and parts completion are extension of this method.

2. Worker-Day Method

This method applies for group(s) of workers, where each worker works at the same constant rate or efficiency. In this approach, work is defined as the number of days $1$ worker takes to complete the work.

For instance, if 20 workers can complete a work in 5 days, then it means that it takes $20 \times 5 =$ 100 worker-days to complete the work. In other words, the work can be completed by 1 worker in 100 days or 100 workers in 1 day.

The questions of this type can be asked with different worker-types (like men, women, children, workers, typists, etc.) and different time periods (like months, weeks, hours, minutes, etc.)

Note: Unless stated otherwise, the efficiency or rate of completion of each worker is assumed to be the same.

We could also use variation and product constancy in solving these type of questions. Note that these two approaches will save time only when $2$ sets of variables are involved (for instance, workers and time taken).

Example 1

If $6$ men can complete a piece of work in $12$ days, how long does it take $8$ men to complete the work?


Standard Worker-Days Approach

Work to be completed requires $6 \times 12 =$ 72 man-days.

Let $x$ be the number of days it takes $8$ men to complete the work.

$\therefore 8 \times x = 72$

$\implies x = 9$

Alternatively (Variation)

Number of workers and time taken are inversely proportional (Refer Proportion & Variation lesson). As the work is constant, increase in workers reduces the time required to complete the work and vice-versa.

$ \dfrac{6}{8} = \dfrac{x}{12} \implies x = 9$

Alternatively (Product Constancy)

Work completed (as man-days) is a constant and is the product of number of men and the number of days they work for. (For product constancy method refer Percentages lesson)

Increase in men by $ \dfrac{1}{3}$, will result in a decrease in days of $ \dfrac{1}{3+1} = \dfrac{1}{4}$.

Days taken by $8$ men $= 12 - \dfrac{1}{4} \times 12 = 9$ days

Answer: $9$ days

Where more than two variables are involved, like in the following example, the variation and product constancy methods might be time consuming. Please use the standard worker-day method in these cases.

Example 2

$6$ workers, working $8$ hours a day, take $8$ days to assemble a car. How many days would $9$ workers working $4$ hours a day take to assemble $3$ cars?


We shall use worker-hours as the unit for work.

Assembling a car $= 6$ workers $\times 8$ days $\times 8 \space \dfrac{\text{hours}}{\text{day}} = 6 \times 8 \times 8$

Assembling $3$ cars $= 3 \times 6 \times 8 \times 8$ worker-hours

It is given that $9$ workers work $4$ hours/day to assemble $3$ cars. Let $d$ be the number of days taken by them.

$\therefore 9 \times x \times 4 = 3 \times 6 \times 8 \times 8$

$\implies x = \dfrac{3 \times 6 \times 8 \times 8}{9 \times 4} = 32$

Answer: $32$

Note: Multiplying $3 \times 6 \times 8 \times = 1152$, and then dividing by $ 9 \times 4 = 36$ would have taken more time.