## Worker equivalence—for quick solution, get number of workers equivalent to number of another type of worker

The **problems involving Time and work form an important part of most of the competitive job or entrance tests** such as SSC CGL, Bank POs, Bank clerk exams, CAT, GMAT etc. The topic of these problems **are taught during middle schools.**

To solve these problems, **it is **** usually assumed that a worker, **(say a man)

**takes certain number of time units**(usually days or hours)

*T***to complete the work**. So the work rate of the worker in one time unit (a day or an hour) is expressed as

**$\displaystyle\frac{1}{T}$th portion of the total amount of work.**

The reason why this work rate in terms of work portion per unit time is the most important concept in Time and Work problems is—it makes possible * summing up of efforts of more than one type of workers working together at different work rates over unit time.* This is the

**core concept**behind the deductions of Time and Work problems of any type.

In **conventional approaches**, as in any conventional approach, this core concept is used **in a straightforward manner and directly.**

### Conventional method based on work rate per unit time—Work rate concept

**Example:** For example, in case of 1 man working alone completing a job in 96 days, and 1 woman working alone completing the same job in 48 days respectively, if we are asked to find the number of days taken to complete the job by 4 men and 4 women working together, by the conventional approach, we derive the per day work rate of 1 man and 1 woman as,

$\displaystyle\frac{1}{96}$ portion of work, and $\displaystyle\frac{1}{48}$ portion of work.

When two teams of 4 men and 4 women work together for 1 day, we would now be able to sum up result of their efforts in one day as,

$\displaystyle\frac{4}{96} + \displaystyle\frac{4}{48} = \frac{3}{24} = \frac{1}{8}$ portion of work done.

We have arrived at this crucial result using two methods (or concepts),

**work rate concept**by which we**evaluate**the**work rate**of a specific type of worker in terms of portion of work done by the worker in a day, and**working together method**by which we can add up work rates of two types of workers to derive**combined work rate**in terms of**work portion done by the two teams of workers in a day.**

The result of combined work portion done in a day is a fraction.

As work done is directly proportional to number of days (*more days the same workers worked, the more portion of work gets done*), here we get the number of days the two teams take to complete the job working together by just dividing 1 by the fraction of combined work, or by just inverting it. This is application of the nearly invisible but highly important **unitary method**.

The final answer is then, 8 number of days.

This approach seems to be a bit complex and time-consuming **as it deals with inverses**, but this is the usual method followed.

### Improvements to the basic conventional method

We have improved this situation **in our last two posts, *** How to solve time and work problems in simpler steps type 1, *and

*How to solve more time and work problems in simpler steps type 2,*

- by introducing new
work rate techniquein which variable $A$ is defined directly as the per day work rate of worker A, instead of conventional approach of defining, worker A takes $A$ number of days to complete the work and hence, per day work rate of A becomes $\displaystyle\frac{1}{A}$, an inverse of $A$. By our new definition, we increased the comfort level of deductions by eliminating all inverse work rate variables, and- by defining the powerful
mandays techniquein whichtotal work amountis expressed as $md$mandays, the product of number of men $m$ who take $d$ number of days to complete the work of amount $W$. The power of this concept and technique arises from its ability to solve different types of time and work problems quickly and cleanly.

Today we would introduce **Worker equivalence**, a third powerful technique, that helps to solve new types of difficult time and work problems elegantly and quickly.

We would explain the use of the technique by solving **three problems of two types**,

- groups of different types of workers working together to complete the same job in different time duration, and
- different types of workers with different efficiencies working on a job; these are problems involving
**Worker efficiency.**

We will also show how the two types of problems can be solved **conventionally** by using the basic concept of work rate of a worker per unit time. This will help you to appreciate the effectiveness of this improved worker equivalence method.

### Third improvement—the Worker equivalence method

First we will take up a simple problem to explain the worker equivalence method.

#### Example problem

If 4 men working independently complete a job in 8 days and 4 women working independently complete the same job in 12 days, in how days of working together will 6 men and 15 women complete the job?

#### Example solution

By problem statement, work amount, say, $W$ in terms of mandays is,

$4\times{8}=32$ mandays.

The **real meaning** of this result is, 1 man working for a day (that's why the **name manday for the work unit**) will complete, $\displaystyle\frac{1}{32}$ portion of work $W$.

In the same way, work amount in terms of womanday (manday is a general term; we will use womanday or boyday as work unit for women or boys working) is,

$W=48$ womandays.

As both the men and women work and complete the same amount of job, these work amounts are equal.

So,

$32\text{ mandays}=48\text{ womandays}$,

Or, $2\text{ mandays}=3\text{ womandays}$.

In other words,

Work done by 2 men is equivalent to work done by 3 women in a day, or in any same time period. This is the core concept in worker equivalence method.

When groups of men and women work together, by this worker equivalence **we can replace** every 2 men by 3 women, or every 3 women by 2 men **depending on how much simplification** is achieved by the replacement.

Coming back to our example problem where 6 men and 15 women work together, we will replace 6 men by equivalent number of 9 women. This makes the team strength as 24 women.

As the total work amount is 48 womandays, it is a simple matter to conclude that 24 women will take 2 days to complete the work.

No fractions involved and all deductions can be done in mind if you apply the concepts and methods properly.

Let us solve a larger problem, the first of the three problems that we'll solve today.

#### Problem example 1: Two groups of men and women work on a job

18 men can complete a piece of work in 24 days and 12 women can complete the same piece of work in 32 days. 18 men start working and after a few days, 4 men leave the job and 8 women join. If the remaining work was completed in $15\displaystyle\frac{15}{23}$ days, after how many days did the four men leave?

- 2 days
- 5 days
- 4 days
- 6 days
- 8 days

**Solution problem example 1: Key pattern identification and Worker equivalence method**

On a brief analysis of the total work amount in mandays and womandays we identified a simple relationship between 1 manday and 1 womanday. This is the **key pattern** as, by this * worker equivalence* we would be able to substitute a suitable group of women by an equivalent number of men thereby simplifying the evaluation greatly. Usually in this type of problem where groups of men and women complete a job in specific number of days, one must try to use worker equivalence.

Let us explain the key pattern of simple relation between capacity of work of a man and a woman. This plays a key role in quick solution of the problem.

First we equate the total work amount as mandays and womandays,

$\text{Total work}=18\times{24}\text{ mandays}=12\times{32}\text{ womandays}$,

Or, $9\text{ mandays}=8\text{ womandays}$.

In other words, **work done in a specific period by 9 men equals work done in the same period by 8 women**. This is **worker equivalence**.

In the final phase of the work, substituting 8 women by 9 men, effectively 23 men worked. This awkward number cancels out the denominator in the number of days. This is the **second key pattern**, and the concepts and techniques applied form the **key strategy**.

#### Solution problem example 1: Applying Worker equivalence and Work portion to work time direct proportionality

In the last fragment of statement, 14 men and 8 women work for $15\frac{15}{23}=\displaystyle\frac{360}{23}$ days, we can now substitute 8 women by 9 men. So effectively, a total of $14+9=23$ men work during this period and complete,

$\displaystyle\frac{360}{18\times{24}}=\frac{5}{6}$th of the work.

Rest, $\displaystyle\frac{1}{6}$th of the work was then completed by 18 men earlier.

As same number of 18 men take 24 days to complete the whole work, to complete $\displaystyle\frac{1}{6}$th of the whole work,

$\displaystyle\frac{24}{6}=4$ days will be required (by work portion done to work days required proportionality, less the work less the days required).

**Answer:** Option c: 4 days.

** Key concepts used:** *Problem analysis -- Key pattern identification -- Work amount as mandays -- Worker equivalence -- Work time proportionality -- Mathematical reasoning -- Solving in mind.*

We try not to use any formula as it loads the memory. Instead, we use concepts, patterns and methods along with problem analysis that determines the strategies and methods that need to be applied most effectively.

This problem could be solved in mind in a few tens of seconds.

Let's now solve a second problem of similar nature, but with a twist. This will further consolidate the understanding.

#### Problem example 2: Groups of men and women work in phases

18 women complete a project in 24 days and 24 men complete the same project in 15 days. 16 women worked for 3 days and then they left. 20 men work for next 2 days and then they are joined by 16 women. In how many days will they complete the remaining work?

- $7\displaystyle\frac{3}{5}$
- $5\displaystyle\frac{2}{5}$
- $9\displaystyle\frac{1}{5}$
- $6\displaystyle\frac{1}{5}$
- None of the above

#### Problem example 2: Solution by Work amount as mandays and Worker equivalence concepts

By given statements, total work amount for women in womandays and for men in mandays will be related as,

$18\times{24}\text{ womandays}=24\times{15}\text{ mandays}$

Or, $6\text{ womandays}=5\text{ mandays}$.

In other words, 5 men can be substituted by 6 women (and vice versa) as their work done in same work period are same.

So in the first phase of 48 womandays of work can be replaced by 40 mandays work. In the second phase also 40 mandays work is done, totalling 80 mandays work done in first two phases. This is equivalent to,

$\displaystyle\frac{80}{24\times{15}}=\frac{2}{9}$th of the total work.

The leftover work is then, $\displaystyle\frac{7}{9}$th of total work.

In the last phase of work, replacing 20 men by 24 women now (instead of replacing women by men, as this reverse replacement simplifies situation much more) effectively 40 women work to finish the leftover $\displaystyle\frac{7}{9}$th of total work. This period will then be,

$\displaystyle\frac{7}{9}\times{\displaystyle\frac{18\times{24}}{40}}$

$=\displaystyle\frac{42}{5}$

$=8\frac{2}{5}$ days.

**Answer:** Option e: None of the above.

Now we will show the solution by conventional method based on work rate per day for each worker type.

**Problem example 2: Conventional solution based on work rate as work portion completed in a day**

Work rate of a woman is, $\displaystyle\frac{1}{18\times{24}}$ portion of work per day, as whole work amount is $18\times{24}$ womandays.

Work rate of a man is, $\displaystyle\frac{1}{24\times{15}}$ portion of work per day, as whole work amount is $24\times{15}$ mandays.

So, 16 women working for 3 days completed $48\times{\displaystyle\frac{1}{18\times{24}}}=\displaystyle\frac{1}{9}$ portion of work, 48 womandays multiplied by one woman work rate per day.

Similarly, 20 men doing 2 days work complete further, $40\times{\displaystyle\frac{1}{24\times{15}}}=\displaystyle\frac{1}{9}$ portion of work, 40 mandays multiplied by one man work rate per day.

Rest, $\displaystyle\frac{7}{9}$ portion of work will be done by 20 men and 16 women in, say, $x$ number of days, where,

$20x\times{\displaystyle\frac{1}{24\times{15}}}+16x\times{\displaystyle\frac{1}{18\times{24}}}=\displaystyle\frac{7}{9}$

Or, $\displaystyle\frac{x}{18}+\displaystyle\frac{x}{27}=\displaystyle\frac{7}{9}$,

Or $5x=42$,

Or, $x=8\frac{2}{5}$ days.

**Answer: **Option e: None of the above.

**Key concepts used:** **Work rate concept -- Work amount as mandays -- Leftover work concept -- Working together concept -- Delayed evaluation technique.**

To ensure accuracy with fair amount of calculations, we needed to write down the intermediate products using **delayed evaluation technique** cancelling out the common factors at the end. This improved speed of solution.

Nevertheless, it will be interesting to compare the efficiency and clarity of two the solutions. Try it yourself.

We will now solve a different type of problem involving **Worker efficiency** using the same worker equivalence method, but in a different situation.

**Truth:** A method is more powerful if it helps to solve more than one type of problem quickly and elegantly.

### Worker efficiency problems by Worker equivalence method

In a worker efficiency problem, work capacity of one worker is expressed in terms of work capacity of a second worker.

**Example 1:** B is twice as efficient as A who completes a job in 30 days. In how many days will A and B working together complete the same job?

**Solution 1:** By the problem statement, work done by B is twice that of A in same time duration. So when two of them work together we can replace per day work of B by 2 times per day work of A. This transforms the team strength to 3A working so that the job will be finished by the team in, $\displaystyle\frac{30}{3}=10$ days.

Try solving the problem using conventional approach. Hint: B completes the job in 15 days.

Let's solve a larger problem of worker efficiency type.

#### Problem example 3: One type of worker 20% more efficient than a second type of worker

A alone can finish a work in 42 days. B is 20% more efficient than A and C is 40% more efficient than B. In how many days B and C working together can finish the same piece of work?

- $12\displaystyle\frac{11}{12}$ days
- $11\displaystyle\frac{5}{12}$ days
- $15\displaystyle\frac{1}{12}$ days
- $14\displaystyle\frac{7}{12}$ days
- $13\displaystyle\frac{5}{12}$ days

**Problem example 3: Problem analysis and solution by worker equivalence concept**

By the given condition, B is equivalent to 1.2 of A (converting percentage to decimal).

Similarly C will be equivalent to $1.4\text{ of B}=1.4\times{1.2}\text{ of A}$.

So when B and C work together, effectively they will be equivalent to, $1.2+1.4\times{1.2}=1.2\times{2.4}\text{ of A}$ working.

Using unitary method on inverse relation between worker and time to complete,

1 of A completes the work in 42 days,

So $1.2\times{2.4}$ of A will complete the work in,

$\displaystyle\frac{42}{1.2\times{2.4}}=\frac{7}{0.48}$

$=\displaystyle\frac{700}{48}$

$=\displaystyle\frac{175}{12}$

$=14\displaystyle\frac{7}{12}$ days.

**Answer:** Option d: $14\displaystyle\frac{7}{12}$ days.

**Key concepts used:** * Worker equivalence concept -- Worker efficiency concept -- Percentage to decimal conversion* --

*--*

**Worker to time inverse proportionality**

**Unitary method on inverse proportionality -- Solving in mind.**The problem could be solved in mind primarily because of converting worker efficiency to worker equivalence and use of percentage to decimal conversion.

We'll end this session now with the conventional solution of this third problem.

#### Conventional solution problem example 3: by work rate per day concept

A completes the job in 42 days. As B is 20% more efficient than A, he will complete $120\text{%}=\displaystyle\frac{120}{100}=1.2$ times amount of work A completes in same time period. Then B completes the job done in 42 days by A in,

$\displaystyle\frac{42}{1.2}=35$ days,

Or work rate of B as portion of work done in a day is,

$\displaystyle\frac{1}{35}$.

As C is 40% more efficient than B, he will complete $140\text{%}=1.4$ times of work completed by B in same time duration. When B completes the work in the problem in 35 days, C will complete it then in,

$\displaystyle\frac{35}{1.4}=25$ days.

Work rate of C is then,

$\displaystyle\frac{1}{25}$ portion of work per day.

Thus, when B and C work together, portion of work completed by them in a day is,

$\displaystyle\frac{1}{35}+\displaystyle\frac{1}{25}=\displaystyle\frac{12}{175}$,

Or, B and C together will complete the job in,

$\displaystyle\frac{175}{12}=14\frac{7}{12}$ days.

Compare the two solutions to appreciate the difference.

### Useful resources to refer to

#### Guidelines, Tutorials and Quick methods to solve Work Time problems

**7 steps for sure success in SSC CGL Tier 1 and Tier 2 competitive tests**

**How to solve Arithmetic problems on Work-time, Work-wages and Pipes-cisterns**

**Basic concepts on Arithmetic problems on Speed-time-distance Train-running Boat-rivers**

**How to solve a hard CAT level Time and Work problem in a few confident steps 3**

**How to solve a hard CAT level Time and Work problem in a few confident steps 2**

**How to solve a hard CAT level Time and Work problem in few confident steps 1**

**How to solve Work-time problems in simpler steps type 1**

**How to solve Work-time problem in simpler steps type 2 **

**How to solve a GATE level long Work Time problem analytically in a few steps 1**

**How to solve difficult Work time problems in simpler steps, type 3**

#### SSC CGL Tier II level Work Time, Work wages and Pipes cisterns Question and solution sets

**SSC CGL Tier II level Solution set 26 on Time-work Work-wages 2**

**SSC CGL Tier II level Question set 26 on Time-work Work-wages 2**

**SSC CGL Tier II level Solution Set 10 on Time-work Work-wages Pipes-cisterns 1**

**SSC CGL Tier II level Question Set 10 on Time-work Work-wages Pipes-cisterns 1**

#### SSC CGL level Work time, Work wages and Pipes cisterns Question and solution sets

**SSC CGL level Solution Set 72 on Work time problems 7**

**SSC CGL level Question Set 72 on Work time problems 7**

**SSC CGL level Solution Set 67 on Time-work Work-wages Pipes-cisterns 6 **

**SSC CGL level Question Set 67 on Time-work Work-wages Pipes-cisterns 6**

**SSC CGL level Solution Set 66 on Time-Work Work-Wages Pipes-Cisterns 5**

**SSC CGL level Question Set 66 on Time-Work Work-Wages Pipes-Cisterns 5**

**SSC CGL level Solution Set 49 on Time and work in simpler steps 4**

**SSC CGL level Question Set 49 on Time and work in simpler steps 4**

**SSC CGL level Solution Set 48 on Time and work in simpler steps 3**

**SSC CGL level Question Set 48 on Time and work in simpler steps 3**

**SSC CGL level Solution Set 44 on Work-time Pipes-cisterns Speed-time-distance**

**SSC CGL level Question Set 44 on Work-time Pipes-cisterns Speed-time-distance**

**SSC CGL level Solution Set 32 on work-time, work-wage, pipes-cisterns**

*SSC CGL level Question Set 32 on work-time, work-wages, pipes-cisterns*

#### SSC CHSL level Solved question sets on Work time

**SSC CHSL Solved question set 2 Work time 2**

**SSC CHSL Solved question set 1 Work time 1**

#### Bank clerk level Solved question sets on Work time

**Bank clerk level solved question set 2 work time 2**

**Bank clerk level solved question set 1 work time 1**