## How to solve time and work difficult problems and work efficiency problems by work equivalence method

How to solve difficult time and work problems is explained by solving selected time and work problems with groups of men and women working.

You may go through the** concept sessions** covered in details earlier,

- Time and work formula in
**How to solve Time and work problems in Simpler steps Type 1.** - Work rate technique and time and work with 3 variables in
**How to solve Time and work problems in Simpler steps Type 1.** - Man hour or Man days work concept in
**How to solve more****Time and Work problems in simpler steps Type 2.**

In this **third concept session** on how to solve time and work problems **you'll learn two more concepts in details.** The new concepts will be explained while solving selected test level difficult time and work problems.

The main sections are,

**Work equivalence**concept.**How to solve time and work problems**on**work equivalence**and**work efficiency.**

### How to solve time and work problems on work equivalence and work efficiency

Two new types of work problems are with,

- Groups of different types of workers work together to complete the same job in different time duration; these are the work problems on
**work equivalence.** - Different types of workers with different efficiencies work on a job; these are the problems on
**work efficiency.**

**We'll show how to solve these two types of work problems in simple steps.**

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 work equivalence method.

### The concept and method of Work equivalence

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

### Example problem on work equivalence

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?

### How to solve time and work problem example on work equivalence

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

$4\times{8}=32$ man days.

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

In the same way, work amount in terms of woman day (man day is a general term; we will use woman day or boy day as work unit for women or boys working) is,

$W=48$ woman days.

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

So,

$32\text{ man days}=48\text{ woman days}$,

Or, $2\text{ man days}=3\text{ woman days}$.

In other words,

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

When groups of men and women work together, by this work 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 becomes homogeneous with only 24 women and no men.

As the total work amount is 48 woman days, 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 and more difficult time and work problem, the first of the three problems that we'll solve today.

### Time and work difficult 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 to time and work difficult problem example 1: Stage 1: Key pattern identification and Work equivalence method**

On a brief analysis of the total work amount in man days and woman days a simple relationship between 1 man day and 1 woman day is identified.

This is the **key pattern based on work equivalence for 9 men and 8 women. **Based on the **work equivalence concept,** a group of women can be replaced by an equivalent number of men. This will simplify the evaluation greatly.

In this type of problem where groups of men and women complete a job in specific number of days, one must try to find

work equivalenceof the worker groups.

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 equate the total work amount in man days with total work amount in woman days,

$\text{Total work}=18\times{24}\text{ man days}=12\times{32}\text{ woman days}$,

Or, $9\text{ man days}=8\text{ woman days}$.

In other words,

work done in a specific period by 9 men equals work done in the same period by 8 women. This iswork 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 to time and work difficult problem example 1: Stage 2: Applying Work equivalence and Work portion to work time direct proportionality

In the last fragment of statement where 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, 23 cancelling out.

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 man days -- Work 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. This naturally is a more difficult time and work problem.

### Time and work difficult 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

### Solution to time and work difficult Problem example 2: By Work amount as man days and Work equivalence concepts

By given statements, total work amount for women in woman days and for men in man days will be related as,

$18\times{24}\text{ woman days}=24\times{15}\text{ man days}$

Or, $6\text{ woman days}=5\text{ man days}$.

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 woman days of work can be replaced by 40 man days work. In the second phase also 40 man days work is done, totaling 80 man days 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.

**Time and work difficult 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 $18\times{24}$ woman days.

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

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 man days -- 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.

### Time and work difficult Problem example 3: Work problem on Work equivalence based on man days concept

18 women do a job in 24 days. 24 men do it in 15 days. 16 women worked for 6 days and left. 20 men work for next 4 days. Then 16 women join them. How long will they take to complete the work?

- 5 days
- 6 days
- 7 days
- 8 days

### Solution to Time and work difficult Problem example 3 on Work equivalences

Work amount in terms of women working is,

$18\times{24}$ woman days.

Same work amount in terms of man days is,

$24\times{15}=360$ man days.

Equating the two,

$18\times{24} \text{ woman days}=24\times{15} \text{ man days}$,

Or, Work of 6 women = Work of 5 men.

This is the work equivalence of two groups of workers.

First group of 16 women worked for 6 days to do a work amount of 96 woman days.

This is equivalent to $5\times{16}=80$ man days.

Next 20 men worked for 4 days to complete 80 man days of more work.

Work left,

$360-(80+80)=200$ man days.

In 6 more days, 20 men will do 120 man days work.

In 6 more days, 16 women will do 96 woman days equivalent to 80 man days work.

The two amounts add up to remaining 200 man days work.

In 6 more days remaining work will be completed by 20 men and 16 women working together.

**Answer:** Option b: 6 days.

Can easily be solved in mind.

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

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

### Solving Work efficiency problem by Work equivalence concept

In a work 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 work efficiency type.

### Time and work difficult Problem example 4: 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

**Time and work difficult Problem example 4: Problem analysis and solution by worker efficiency based on work 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:** * Work equivalence concept -- Work 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 work efficiency to work 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 4: 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**