Time and Work Unitary Method and Time and Work with 3 Variables

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How to solve Time and Work problems in simpler steps, type 1

Time and work unitary method and Time and work with 3 variables

Time and work unitary method, Time and work formula, Work rate technique and Time and work with 3 variables

Time and work unitary method, Time and work formula, Improved work rate technique for working together and Time and work with 3 variables solved are covered.

You will learn,

  1. What are Time and work problems
  2. Most basic time and work concepts of proportionality
  3. Working together concept and Time and work formula
  4. Improved Work rate technique
  5. Time and work problems with 3 variables.

What are time and work problems

Time and work problems are an important category of problems in competitive exams.

Three elements form a time and work problem,

  1. Workers who do a job.
  2. Time in days (or hours) to do a job by a worker.
  3. The work or job that usually is mentioned simply as THE WORK or THE JOB.

An example of a time and work problem

A and B can complete a job independently in 3 and 5 days respectively. How long will they take to complete half the job working together?

Solution to example problem on time and work

A and B are workers, 3 and 5 days are the time duration in this case. Work amount is FIXED. It is completed either by A or by B working alone in 3 and 5 days respectively.

Most basic time and work concepts of proportionality

  1. Work done is DIRECTLY PROPORTIONAL to number of workers. For example, if 1 laborer produces 5 kgs of stone chips a day, 2 laborers will produce $2\times{5}=10$ kgs of stone chips in a day. Conversely, number of workers needed is directly proportional to amount of work to be completed. Assumption: Each worker works at the same constant rate, And time duration is constant.
  2. Time to complete a work is DIRECTLY PROPORTIONAL to amount of work. For example, if a worker completes a job in 8 days, in 1 day the worker will complete $\frac{1}{8}$th of the whole work. Assumption: Number of workers is constant.
  3. Time to complete a work is INVERSELY PROPORTIONAL to the number of workers. It makes sense. More the number of workers, less will be the time needed to do the work. Assumption: Work amount is constant.

Frequently, workers of different work abilities work together on one job. For example, 3 men and 4 women may work together to complete a job.

Working together concept

In our example problem, A and B are two types of workers of different work abilities. How to determine the amount of work done by the two working together is the issue.

The Trick is,

Add up the work done by A in 1 day with work done by B in 1 day to get the portion of work done by the two working together in 1 day.

This is where unitary method is applied on the basis of the proportional relations.

In 3 days, A completes the whole work, say W.

$\Rightarrow$ In 1 day, A will complete $\frac{1}{3}$rd of the whole work W.

In 5 days, B complete the whole work W,

$\Rightarrow$ In 1 day, B will complete $\frac{1}{5}$th of the whole work W.

Add up to get portion of whole work done by A and B working together in 1 day as,

$\left(\displaystyle\frac{1}{3}+\displaystyle\frac{1}{5}\right)W=\displaystyle\frac{8}{15} W$..........................(1)

Time and work formula

This is time and work formula for two different types of workers working together.

Apply time and work unitary method a second time.

A and B together do $\displaystyle\frac{8}{15}$th of whole work in 1 day.

$\Rightarrow$ A and B together will do $\displaystyle\frac{1}{2}$ of whole work in,

$\displaystyle\frac{1}{\displaystyle\frac{8}{15}}\times{\displaystyle\frac{1}{2}}=\displaystyle\frac{15}{16}$ days.

Answer to example problem: $\displaystyle\frac{15}{16}$ days.

Equation 1 is one of the pillars in solving time and work problems.

Growing up in confidence, we'll solve now a few test level time and work problems in 3 variables. It means three different types of workers will be working together in these problems.

But before that we'll learn how and why to use the improved work rate technique instead of conventional method.

Improved Work rate technique with worker variables as work rates for each type of worker

To solve time and work problems, usually it is assumed that a worker takes certain number of days (time in units) to complete the work.

There is nothing wrong with this conventional approach except that it results in dealing with inverses of all the variables involved (for each type of worker) which is more time consuming and error prone.

We would instead adopt a more direct approach that we call Work rate technique.

Advantages of this new approach are,

  • It simplifies solution of Time and work problems by reducing fraction arithmetic significantly as well as,
  • it is conceptually simpler to understand.

In Work rate technique,

We will assume a worker variable as portion of work done by the worker in one time unit.

By this approach, if $A$ and $B$ be the work rates per day of two workers, we can directly add the two variables to get work done by the two together in 1 day as, $A+B$.

If workers A and B complete a work in 5 days, we can express it simply as, $5(A+B)=W$, $W$ being the total work amount; fractions eliminated in the relationship altogether.

A few time and work problems in 3 variables are next solved using the direct approach of improved work rate technique.

First Time and work problem with 3 variables

If A, B and C together complete a piece of work in 30 days and A and B together complete the same work in 50 days, then C working alone completes the work in,

  1. 75 days
  2. 60 days
  3. 150 days
  4. 80 days

Solution to first time and work problem with 3  variables: by improved Work rate technique

Let $A$, $B$ and $C$ be the portion of work completed by A, B and C respectively in a day.

This is the key assumption of direct per day work rate of each worker eliminating the need to use inverse of any variable.

With this assumption, by the first statement we have the total work,

$W = 30A + 30B + 30C$,

and by the second statement,

$W = 50A + 50B$,

Or, $\displaystyle\frac{3}{5}W = 30A + 30B$.

Subtracting this equation from the first, eliminating both $A$ and $B$ we get the per day work rate of $C$ as,

$30C = \displaystyle\frac{2}{5}W$,

Or, $C = \displaystyle\frac{1}{75}W$.

To complete the work $W$ then C will take $75$ days.

Answer: Option a: 75 days.

To Skip the following nearly philosophical discourse, click here. 


A form of Base equalization technique

It is a bit surprising that even this basic technique of dealing with time and work problems is nothing but a form of Base equalization technique that we had elaborately covered earlier starting from its application in solving a type of indices problems, all fraction arithmetic and lastly in more versatile ways.

Let us restate the abstract definition of the Base equalization technique in a series of general steps,

  • In a set of components each having multiple entities, identify the base to be equalized.
  • Identify the target value to be equalized to.
  • Equalize the bases of each component in the set.
  • Establish direct relationship between the other entities of all components in the set.
  • Carry out the desired operations on these entities that was the objective of the whole series of operations.

In our problems of interest here, that is, the time and work problems, the core technique used is to equalize the unit of rate of work of each work-agent. Here we have equalized the unit as the portion of whole work done by each work-agent per unit time, which is 1 day here.

Unless we bring work capacity of each work-agent to this same base of portion of work done in a day, we won't be able to consider the effect of all the work-agents working together. This is the key technique in this problem type area.

We went one step further to build this technique right into the work capacity variables while defining them, eliminating thus the need to deal with inverse of the variables altogether.

The core technique used is same but its manner of use is different making the environment simpler and more comfortable compared to the conventional approach.

The reason why it is possible to use Base equalization technique in so many different topic areas of math (and other real life problems) lies in the concept of abstraction.

Abstraction means expressing in more general terms.

The special characteristic of the base equalization technique is its amenability to abstraction. It is possible to express the common core abstract base equalization technique as a series of general steps as above. And these steps are independent of any specific topic area. This is Abstraction technique working in its full flow.


Conventional approach to solve first Time and work problem with 3 variables

Usually, in conventional solutions it is assumed that,

Let A, B and C can complete the work in $A$, $B$ and $C$ days. In this case we would need an extra step of evaluating the per day work rates of A, B and C as,

$\displaystyle\frac{1}{A}$, $\displaystyle\frac{1}{B}$ and $\displaystyle\frac{1}{C}$, each as a portion of work $W$.

In this case then, to assess the effect of them working together we have to deal with inverses as,

$\displaystyle\frac{30}{A} + \displaystyle\frac{30}{B} + \displaystyle\frac{30}{C} = W$, and

$\displaystyle\frac{50}{A} + \displaystyle\frac{50}{B} = W$,

Or, $\displaystyle\frac{30}{A} + \displaystyle\frac{30}{B} = \frac{3}{5}W$.

Again eliminating $A$ and $B$ by simple subtraction of this equation from the first we get,

$\displaystyle\frac{30}{C} = \frac{2}{5}W$,

Or, $\displaystyle\frac{75}{C} = W$, which means C takes 75 days to complete the work according to initial definition.

In this conventional approach we deal with inverse of variables $A$, $B$ and $C$, whereas in our direct approach we deal with no inverse of any variable at all. There lies the simplicity of the direct approach.

It should always be your endeavor to reduce dealing with inverses as far as possible.

It makes deductions psychologically more comfortable, conceptually simpler, and consequently faster more accurate.

We will use the same concept in another time and work problem that is a bit more complex.

Second Time and work problem with 3  variables

C and A can do a piece of work in 20 days, B and C in 24 days and A and B in 30 days. B and C leave after all three of them work together for 10 days. In how many more days A would complete the work?

  1. 36 days
  2. 24 days
  3. 30 days
  4. 18 days

Solution to the second time and work problem with 3 variables: By improved Work rate technique

Again $A$, $B$, and $C$ are assumed as the per day whole work portion of W done by A, B and C respectively.

Thus in the three given situations we get,

$20C + 20A = W$,

$24B + 24C = W$, and

$30A + 30B = W$.

Converting the coefficients of $A$, $B$ and $C$ in all three expressions to 10, and summing up the three equations we get,

$20A + 20B + 20C = W\left(\displaystyle\frac{1}{2} + \displaystyle\frac{5}{12} + \displaystyle\frac{1}{3}\right) = \displaystyle\frac{5}{4}W$,

Or, $10A + 10B + 10C = \displaystyle\frac{5}{8}W$.

This has been the work completed by all three working together for 10 days. Work left to be done is $\displaystyle\frac{3}{8}W$. We have to find out the rate of work of A alone to know the desired number of days taken by A to complete this leftover work.

We have,

$24B + 24C = W$ and so, $10B + 10C = \displaystyle\frac{5}{12}W$.

Using this result with the result of $10A + 10B + 10C = \displaystyle\frac{5}{8}W$ we have,

$10A + \displaystyle\frac{5}{12}W = \displaystyle\frac{5}{8}W$,

Or, $A = \displaystyle\frac{1}{10}\left(\displaystyle\frac{5}{8} - \displaystyle\frac{5}{12}\right)W = \displaystyle\frac{1}{48}W$

So the rest of the work $\displaystyle\frac{3}{8}W = 18A$, that is 18 days' of A's work.

Answer: Option d: 18 days.

Key concepts used:

With the conventional approach, it is usually assumed A, B and C each working alone takes $A$, $B$ and $C$ number of days to complete the work $W$. In this approach all along we have to deal with inverses of A, B and C.

Instead, with our recommended approach, we don't have to deal with inverse of any variable at all thus making the process much easier and more comfortable to execute or understand.

The old adage applies here,

It is always easier to deal with direct entities than with indirect entities.

We will end with the conventional solution to the problem.

Conventional solution of second time and work problem with 3  variables

Let A, B and C take $A$, $B$ and $C$ days respectively to complete the work working alone. Thus we have,

$\displaystyle\frac{20}{C} + \displaystyle\frac{20}{A} = W$,

$\displaystyle\frac{24}{B} + \displaystyle\frac{24}{C} = W$, and

$\displaystyle\frac{30}{A} + \displaystyle\frac{30}{B} = W$.

In the same way as before, transforming each equation to make the coefficient of each variable equal to 10 and summing up the three equations we get,

$\left(\displaystyle\frac{10}{A} + \displaystyle\frac{10}{B} + \displaystyle\frac{10}{C}\right) = \displaystyle\frac{1}{2}W\left(\displaystyle\frac{1}{2} + \displaystyle\frac{5}{12} + \displaystyle\frac{1}{3}\right) = \displaystyle\frac{5}{8}W$.

This is the work done in 10 days by A, B and C working together.

Work left again is,  $\displaystyle\frac{3}{8}W$.

From, $\displaystyle\frac{24}{B} + \displaystyle\frac{24}{C} = W$, we get,

$\displaystyle\frac{10}{B} + \displaystyle\frac{10}{C} = \frac{5}{12}W$.

Putting this in $\left(\displaystyle\frac{10}{A} + \displaystyle\frac{10}{B} + \displaystyle\frac{10}{C}\right) = \displaystyle\frac{5}{8}W$, we have,

$\left(\displaystyle\frac{10}{A} + \displaystyle\frac{5}{12}W\right) = \displaystyle\frac{5}{8}W$,

Or, $\displaystyle\frac{1}{A} = \displaystyle\frac{1}{10}\left(\displaystyle\frac{5}{8} - \displaystyle\frac{5}{12}\right)W = \displaystyle\frac{1}{48}W$

So the rest of the work $\displaystyle\frac{3}{8}W = \displaystyle\frac{18}{A}$, that is 18 days' of A's work.

Interesting, isn't it? In this approach, we had to use all along inverse of $A$, $B$ and $C$ that was not really necessary and could have been avoided with subtle variation in the definition of the variables itself.

Finally, we can then name this approach in terms of tuning of the definition itself - more formally - Definition tuning approach.


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

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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

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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

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

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

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

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

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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

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Bank clerk level Solved question sets on Work time

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