# CAT Quant – Time and Work – Important Formulas and Concepts

## Basic Assumptions

- Work to be done is usually considered as a whole 1 unit.
- If a person does some work in a certain number of days, we assume (unless explicitly stated) he does the work uniformly i.e same amount of work every day.
**Example –**A does some work in —> 15 days; He does —> (1/15)th of the work in 1 day.

- If there is more than 1 person (workforce) carrying out the work, w assume that each person does the same amount of work everyday (unless otherwise specified). They share the work equally.
**Example –**2 People can do a work in —> 8 days; 1 man do it in —> 16 days. Thus each person can do —> (1/16)th of work per day.

## Unitary Method

- The method used in “Time and Work” is Unitary Method i.e “time taken per unit method”.
- Number of persons required to complete Unit Work.
- Work completed by unit person in Unit Time

- Note – A useful approach to look at time and work is in terms of Percentage(%) as it gives a direct comparison and a clear picture.

## LCM Method

- Another approach to solving Time and Work problem is through LCM. Different cases using LCM to solve time and work problems are explained below with the help of examples.
**Case 1 –**If A completes a work in 10 days and B completes a work in 15 days. How much time will they together take to complete the work. $$ A = 10 \ B =15 \longrightarrow LCM = 30 \\ A’s Work {\LARGE[} \frac{30}{15} {\LARGE]} = 3 \ ; \ B’s Work {\LARGE[} \frac{30}{15} {\LARGE]} = 2 \\ (A + B) \ Together = (3+5) = 5 \\ Time \ taken \ to \ complete \ the \ work \ together = {\LARGE[} \frac{30}{5} {\LARGE]} = 6 \ days$$**Case 2 –**If A and B completes a work in 15 days and B alone completes the work in 20 days. How much time will A take to complete the work alone. $$(A+B) = 15 \ B =20 \longrightarrow LCM = 60 \\ (A+B)’s Work {\LARGE[} \frac{60}{15} {\LARGE]} = 4 \ ; \ B’s Work {\LARGE[} \frac{60}{20} {\LARGE]} = 3 \\ [(A+B) – B] = (4-3) = 1 \\ Time \ taken \ by \ A \ to \ complete \ the \ work \ alone = {\LARGE[} \frac{30}{5} {\LARGE]} = 6 \ days$$

## Formulas

- The basic formula that applies to Time and Work problems are as – $$Work \ done = Men \ (Work \ Rate) * Time \\ W = M * T$$
- The following Conclusions can be drawn out –
- When “Time” is constant —> W \( \infty \) M
- When “Men” is constant —> W \( \infty \) T
- When “Work” is constant —> M \( \infty \) 1/T

**Man-days –**The number of men multiplied by the number of days they take to complete the work is the Man-Days required to do the work.**Efficiency =**1/ Time Taken**Case 1 –**If M_{1}men can do W_{1}work in D_{1}days working H_{1}hours per day and M_{2}men can do W_{2}work in D_{2}days working H_{2}hours per day (where all men work at the same rate) then, $$\frac{M_1D_1H_1}{W_1} = \frac{M_2D_2H_2}{W_2}$$**Case 2 –**If A can do a piece of work in p days and B can do it in q days then A and B together can complete the same work in – $$\frac{1}{p} + \frac{1}{q} = \frac{pq}{p+q} \ days $$

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