Time & Work

Time & Work Introduction

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What is Work?

Technically speaking, Work is the quantity of energy transferred from one system to another but for question based on this topic, Work is defined as the amount of job assigned or the amount of job actually done.


Analogy with Time-Speed-Distance

Problem on work are based on the application of concept of ratio of time and speed. Work is always considered as a whole or one. There exists an analogy between the time-speed-distance problems and work.

Work based problem are more or less related to time speed and distance.

Above mentioned definition of work throws light on three important points.

Work = 1 (It is always measured as a whole) = Distance

Rate at which work is done = speed

Number of days required to do the work = Time

Example: If Ram and Raman can do a job in 10 days and 15 days independently, how many days would they take to complete the same job working simultaneously?

Solution: 

If total work is W, Ram’s rate of working = W/10 per day and that of Raman = W/15 per day.

Thus when working simultaneously, rate of work done = W/10+W/15

and thus time taken:

=W/{W/10+W/15}=15*10/15+10=6 Days

Alternatively above problem can be worked out assuming work to be just 1 unit and thus eliminating use of W.

Assume the total work to be the LCM of the days taken individually i.e. LCM of 10 & 15 i.e. 3o units of work.

Thus Ram’s rate of working = 3 units per day and Raman’s rate of working = 2 units per day.

When working simultaneously, 3+2=5 units of work is done every day and thus it would take

30/5=6 Days.

 

Note: ALL problems or work can be solved in either way and both ways take almost the same time as there are exactly the same numbers of calculations involved. However if you are not comfortable with fraction, the approach using LCM may seem better to you.


Important Formulae:

Work from Days

If A can do a piece of work in n days, then A’s 1 day’s work =1/n

Days from Work:

If A’s 1 day’s work =1/n

, then A can finish the work in n days.

Ratio

If A is thrice as good a workman as B, then:

Ratio of work done by A and B =3:1.

Ratio of times taken by A and B to finish a work =1:3

If A is ‘x’ times as good a workman as B, then he will take(1/x)th of the time by B to do the same work.

A and B can do a piece of work in ‘a’ days and ‘b’ days respectively, then working together, they will take xy/x+y days to finish the work and in one day, they will finish (x+y/xy)th part of work.


Number of days 
required to complete the work
Work that can be done per day Efficiency in Percent
N 1/n 100/n
1 1/1 100%
2 1/2 50%
3 1/3 33.33%
4 ¼ 25%
5 1/5 20%
6 1/6 16.66%
7 1/7 14.28%
8 1/8 12.5%
9 1/9 11.11%
10 1/10 10%
11 1/11 9.09%

 

Now let’s solve questions with this trick:

Question – A take 5 days to complete a job and B takes 10 days to complete the same job. In how much time they will complete the job together ?

 

Solution – A’s efficiency = 20%, B’s efficiency = 10%. If they work together they can do 30% of the job in a day. To complete the job they need 3.33 days.

 

Question – A is twice as efficient as B and can complete a job 30 days before B. In how much they can complete the job together

Solution – Let efficiency percentage as x

A’s efficiency = 2x and B’s efficiency = x

A is twice efficient and can complete the job 30 days before B. So,

A can complete the job in 30 days and B can complete the job in 60 days

 

A’s efficiency = 1/30 = 3.33%

B’s efficiency = 1/60 = 1.66%

Both can do 5% ( 3.33%  + 1.66% ) of the job in 1 day.

 

So the can complete the whole job in 20 days (100/5)

 

Question – A tank can be filled in 20 minutes. There is a leakage which can empty it in 60 minutes. In how many minutes tank can be filled?

 

Solution –
Method 1
⇒ Efficiency of filling pipe = 20 minutes = 1/3 hour = 300%

⇒ Efficiency of leakage = 60 minutes = 100%

 

We need to deduct efficiency of leakage so final efficiency is 200%. We are taking 100% = 1 Hour as base so answer is 30 minutes.

Method 2
⇒ Efficiency of filling pipe = 100/20 = 5%
⇒ Efficiency of leakage pipe = 100/60 = 1.66%
⇒ Net filling efficiency = 3.33%
So tank can be filled in = 100/3.33% = 30 minutes

 

Pipe and Cistern problems are similar to time and work problems. A pipe is used to fill or empty the tank or cistern.

Inlet Pipe: A pipe used to fill the tank or cistern is known as Inlet Pipe.

Outlet Pipe: A pipe used to empty the tank or cistern is known as Outlet Pipe.


Some Basic Formulae

  1. If an inlet pipe can fill the tank in x hours, then the part filled in 1 hour = 1/x
  2. If an outlet pipe can empty the tank in y hours, then the part of the tank emptied in 1 hour = 1/y
  3. If both inlet and outlet valves are kept open, then the net part of the tank filled in 1 hour is (1/x)-(1/y)

Some Shortcut Methods

Rule 1: Two pipes can fill (or empty) a cistern in x and y hours while working alone. If both pipes are opened together, then the time taken to fill (or empty) the cistern is given by (xy/(x+y))

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