## 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?

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

$\therefore 8 \times x = 72$

$\implies x = 9$

Number of workers and time taken are inversely proportional (Refer

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

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

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

### Solution

**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

**lesson)***Percentages***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**: $9$ daysWhere 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?

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.

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

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

### Solution

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.