# Quant Study Notes: Time and Work

**Dear Readers,**

Today we’ll discuss about Time and Work. This topic is can fetch you marks easily but you need to know the right concepts and types of questions to practice.

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**All about Time and work**

This chapter is based on the concept of direct and inverse variations. We need to understand relation among time, work done and number of employees working.

Assuming that all employees work with the same efficiency, we can conclude that work done is directly proportional to number of employees working and number of days to complete the work is inversely proportional to number of employees working.

**☞**For example if a person does a work in 10 days then in 1 day he does only one tenth of the work.

**☞**For example it two men can complete a work working alone in 10 and 20 days respectively, then one day’s work of both the men will be

Thus the total can be completed by both of them in

**Important points:**

**➀**If A can do a piece of work in X days, then A’s one day’s work =1/Xth part of whole work.

**➁**If A’s one day’s work =1/Xth part of whole work, then A can finish the work in X days.

**➂**If A can do a piece of work in X days and B can do it in Y days then A and B working together will do the same work in

**If A, B and C can do a work in X, Y and Z days respectively then all of them working together can finish work in**

**➃****Example:**

**If X can do a work in 10 day, Y can do the same work in 20 days, Z can do double of the work in 30 days. If all 3 started working together, find the total time required to complete the work.**

**Solution:**

**(i) Unitary Method:**

**(ii) LCM Method:**

**CONCEPT of MAN DAYS**

Assuming that all men work with same efficiency, we can conclude that if the work done is constant then the number of days is inversely proportional to the number of men working. For example, if we say that 20 men can do a work in 30 days, this means that total work is 20x30 man days = 600 man days, which means that the same work can be done by 10 men in 60 days or 60 men can do in 10 days only etc. So for a constant work number

of man days is constant.

where M is the number of men

⇒ MD = constant

Thus we obtain a relationship

If work is not constant then it is directly proportional to both number of men and number of days

Where MD is equivalent to number of man days.

**Example:**

**10 men or 20 women or 30 children, can do a work in 15 days. If 10 men, 12 women and 18 children work on the same work, find the time in which work can be competed.**

**Solution:**

We see that 10 men are equivalent to 20 women that is equivalent to 30 children

Hence 1 man = 2 women = 3 children

Total work is 10 × 15 = 150 man days

10 men, 12 women and 18 children are equivalent to 10 + (12/2) + (18/3) = 22 men.

Hence time taken to complete the work

= 150/22 = 75/11 = 6 (9/11) days.

**ALTERNATE WORKING**

In the following example, we will discuss how to find the total time taken to complete the work, if the two or more workers are not working together but they are working on alternate days.

Suppose A and B are the two workers working on a project such that A and B can complete a work working alone in 20 and 12 days respectively. Now they are working on alternate day, now to find the total time required to complete the work, there can be two cases:

*(a) Starting with ‘A’*

*(b) Starting with ‘B’*

In these types of questions work done on the first day is not same as the work done on the second day but not work done in first two days is same as the work done in the next two days and that is same as the work done in the next two days and so on.

**AB AB AB ………**

Work done in 2 days

If we assume 2 days to be one cycle and total work is always considered as 1 unit, then approximate number of cycles required to complete the work =15/2=7 (integral number)

Total work done after 7 complete cycles

Hence the remaining work is 1/15

(1)

**A started the work:**In one day A completes 1/20th of the work, hence 1/15th of the work will be completed in more than one day. A will complete 1/20th part and the remaining work will be done by B.
Remaining work =

**PIPES AND CISTERNS**

This is the topic which gives the relation between the time required to fill or empty the tank with the taps opened or closed.

Till the time we have only defined the concept of positive work but in the problems related to pipes and cisterns we have to define the concept of negative work also. Concept of negative work is defined as when the work is done against the requirement.

**Example:**A tap can fill a tank in 16 minutes and another can empty it in 8 minutes. If the tank is already ½ full and both the taps are opened together, will the tank be filled or emptied? How long will it take before the tank is either filled or emptied completely?

**Solution:**

If both the pumps are opened together, then the tank will be emptied because the working efficiency of pump empting is more than that of the pump filling it. Thus in 1 min net proportion of the volume of tank filled

(Which means that tank will be emptied 1/6th in one minute.)

Hence in 8 minutes half of the tank will be emptied.

__Some Important Tricks__**➀**If A and B working together, can finish a piece of work in x days, B and C in y days, C and A in z days, then

**➁**

**☞**If A can complete a work in x days and B is k times efficient than A, then the time taken by both A and B, working together to complete the work is

**☞**If A and B, working together, can complete a work in x days and B is k times efficient than A,

then the time taken by –

A working alone to complete the work in

→ (k + 1) x

B working alone to complete the work is

**➂**If A working alone takes ‘a’ days more than A and B, and B working alone takes ‘b’ days more than A and B together, then the number of days taken by A and B, working together, to finish a job is given by

**➄**If ‘a’ men and ‘b’ women can do a piece of work in ‘n’

days, then ‘c’ men and ‘d’ women can do the work in

**Types of Questions for Time and Work**

**Practice Time and Work & Pipes and Cistern Questions**

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