In Time and Work problems, a Worker does a Work in specified number of Days, we assume here time unit as a Day.

### The primary complexity in Time and work problems

The main complexity in time and work problems arises when two or more workers with different work capacity work together and we are to find how long they would take to complete the work, working together.

### Problem example 1.

A and B, working alone, can complete a job in 20 and 30 days respectively. In how many days will they complete the job if they work together?

We can’t just add 20 to 30 and say, answer is 50 days.

#### Working together concept

Two important points in working together scenario:

- First, when workers (with varying work capacity or rate of work) work together,
**they work totally independent of one another**. - Second, at the end of a day, we take stock of
**how much portion of total work each has done**, and add the portions to get how much work they did in a day together. Conceptually it seems perfectly alright to do so, isn’t it?

Finding the answer is then just one more step – applying **Unitary method**.

In our problem, if $W$ is total work, A and B does,

$\displaystyle\frac{1}{20}\text{th of W and }\displaystyle\frac{1}{30}\text{th of W}$, in a day each working alone,

Or, $\left (\displaystyle\frac{1}{20} + \displaystyle\frac{1}{30}\right)\text{ of W}$,

Or, $\displaystyle\frac{1}{12}\text{th of W}$ in 1 day working together,

Or, Whole work $W$ in 12 days working together - applying Unitary method we inverted $\displaystyle\frac{1}{12}$.

This is the **single most important basic concept in Time and Work problems**.

#### Rich concept of Work rate in Time and Work problems

We will now go one step ahead and define $A$ and $B$ as the work rates in a day (portion of work done in a day) for A and B.

In many problems this approach simplifies solution significantly.

### Problem example 2

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 how many days?

### Solution 2

From first part of the given statements we get a linear equation in $A$, $B$ and $C$ while from the second part a second linear equation in $A$ and $B$. Here, $A$, $B$ and $C$ are the portion of work done (or work rate) of A, B and C respectively.

If in each equation we make the coefficients of $A$, $B$ and $C$ same as in the other equation, we can simply eliminate $A$ and $B$ to get a third relation between $C$ and work $W$.

Applying rich concept of $A$, $B$ and $C$ as portions of work $W$ done by A, B and C respectively in 1 day,

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

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

$50(A+B) = W$,

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

Subtracting second from first,

$C = \left(\displaystyle\frac{1}{30} – \displaystyle\frac{1}{50}\right)W = \left(\displaystyle\frac{2}{150}\right)W = \left(\displaystyle\frac{1}{75}\right)W$

C then will complete the work alone in 75 days.

The fraction calculation is concentrated wholly on the single variable $W$ which is a constant.

If you want, you can easily arrive at the answer completely in mind without any writing. It is so simple, isn't it?

### Problem example 3

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 three of them work together for 10 days. In how many more days would A complete the work?

### Solution 3

Thinking from the end, we decide, we must first find out how much work is done by A, B, and C working together in 10 days. That would give us work left. We also need to find how long A takes to complete the work alone. The work left and work capacity of A would then give us required number of days.

Analyzing the three statements we notice that each variable appears exactly twice in the three resulting linear equations. As such we can’t combine the three equations as the coefficients are different.

So what do we have to do? We need to simply **make the coefficients of the variables in each relation same**.

Let us see how. Given,

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

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

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

We equalized the coefficients directly after using the * Work rate technique*.

So,

$A+B+C = \displaystyle\frac{W}{2} \left(\displaystyle\frac{1}{20}+\displaystyle\frac{1}{24}+\displaystyle\frac{1}{30}\right)$

$= \displaystyle\frac{W}{240} (6+5+4) = \displaystyle\frac{W}{16}$, we have used 120 as LCM of 20, 24, 30 on the go.

We will find A now,

$A = (A+B+C)-(B+C)$

$=\displaystyle\frac{W}{16} – \displaystyle\frac{W}{24} $

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

In 10 days, A, B, C had done work,

$10(A+B+C) = \displaystyle\frac{5W}{8}$, work left was $\displaystyle\frac{3W}{8}$.

In 1 day work done by A,

$A = \displaystyle\frac{W}{48}$

So, $18A = \displaystyle\frac{3W}{8}$

In 18 days A will complete the leftover work of $\displaystyle\frac{3W}{8}$.

### Steps done in mind

#### Step 1

Being conversant with work rate technique, without writing the individual three equations we directly formed the fraction sum. For convenience we write it,

$\displaystyle\frac{W}{2}\left(\displaystyle\frac{1}{20}+\displaystyle\frac{1}{24}+\displaystyle\frac{1}{30}\right)$.

And quickly evaluate the result $\displaystyle\frac{W}{16}$ mentally. We know this is work done together by all three in a day.

#### Step 2

In step 2, we subtract mentally $\displaystyle\frac{W}{24}$ from $\displaystyle\frac{W}{16}$, to get $\displaystyle\frac{W}{48}$ as the work done by A in 1 day.

#### Step 3

Mentally we multiply 1 day work done by all three together by 10, to get $\displaystyle\frac{5W}{8}$ as the work done by all three in 10 days, and simultaneously arrive at the factor 18 of $\displaystyle\frac{W}{48}$ that will convert $\displaystyle\frac{W}{48}$ to leftover work $\displaystyle\frac{3W}{8}$.

All in all it shouldn’t take more than 50 secs if you are used to solve problems in this way.