## Define the variables directly in terms of rate of work and eliminate troublesome inverses

The **problems involving Time and work form an important part of most of the competitive job tests** such as SSC CGL, Bank POs etc. These problems are taught during middle schools. To work out these problems, **usually it is assumed that a work-agent takes certain number of time 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 the agents working) which is awkward and more time consuming and error prone.

* We would instead adopt a more direct approach *that we call

**Work rate technique.****Advantages of this new approavh are significant,**

- It simplifies solution of Time and work problems by largely eliminating fractions as well as,
- it is conceptually simpler to understand.

**This is the efficient time and work problem solving concept.**

In **Work rate technique,**

We will assume

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 worker variable as portion of work done by the worker in one time unit.

Let us illustrate the direct approach with two problem examples.

We would show in each case the alternative conventional approach of solving these problems also.

**Q1.** 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,

- 75 days
- 60 days
- 150 days
- 80 days

**Solution:**

* 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 work-agent eliminating the need to use inverse of any variable altogether.**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.

#### Key concepts used:

Working with per day work rate of each of A, B and C, we can easily express the effect of working together in two statements as two simple expressions of per day work rate (in terms of portion of work completed by each in a day), and thereby eliminate the two variables $A$ and $B$ leaving only $C$ with $W$.

The * key concept in Time and Work* that has to be used

*(or time unit). Only then if two or three individuals*

**is to consider how much portion of work an individual completes in a day***we can*

**working at different rates work together***(or time unit). This is the*

**assess the effect of their working together in a day**

**most imprtant concept in this topic area.**### 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***and lastly in*

**all fraction arithmetic***.*

**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

*in its full flow.*

**Abstraction technique working**#### Conventional approach

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.

**Q2.** 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?

- 36 days
- 24 days
- 30 days
- 18 days

**Solution:**

Again * we assume $A$, $B$, and $C$ to be 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

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**

**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**