# Time and Work - Atreya Roy

• Author: Atreya Roy is pursuing his BTech From Kalyani Government Engineering College, Bengal.

Today we will discuss the concepts of Time and Work, which is a very important topic from the exam point of view and there are fair number of questions appearing in the exams from this part. We will try to ace this topic such that it creates no trouble at the time of exam.

Concept :

If A does a work in 10 days, then what is the work done by A in 1 day ? The answer is one-tenth part of the total work.

We can say that :

 Unit Time work = 1/ Total Time Taken by the worker to complete the whole work.

Note:

For all time and work sums, we will approach the sum by either of the two ways :
1. Taking Total work =k, Where K is generally the LCM of the days in which the workers can finish the job individually.
2. By simply forming equation considering the days the workers take to finish off a task. In this Approach Total work is always 1.

Example :  What can be the Total work (in units) if there are 3 workers who can do the same piece of work in 9,10,12 days respectively ?

Solution:

Simply take out the LCM as discussed earlier : LCM (9,10,12) = 180 units

Problem : If A and B can do a piece of work in 10 and 15 days respectively. In how many days can the whole work get completed if they work together ?

Solution:

Two way approach :
Total work = LCM(10,15)= 30
So A does 30 units of work in 10 days means each day he does 3 units of work
B does 30 units of work in 15 days means each day he does 2 units of work
When they work together : each day they complete 2+3= 5 units.
Total work 30. Per day work done = 5
Days taken = 30/5= 6 days

Or the other approach :
A does 1/10 part work per day
B does 1/15 part work per day
A+B = 1/10 +1/15 = 5/30
But since I said, Total work is 1. Time taken = 1/(5/30) = 30/5 = 6 days

Please try to do the sums using the total work to be something. Generally the LCM. It will ease your calculations and speed up answering.\

Not every worker can work with the same effort or get a work done in the same time as other do. Different workers have different speeds of working, some are slow and some are fast. Some are fast but are erroneous.  This factor is known as efficiency.

So what is efficiency in mathematical terms ?

Efficiency of workers:
If the efficiency of A is x% more than the efficiency of B, and B takes “B” days to complete the work then the time A will take to complete the work will be B/(100+x) *100 days
Similarly, If the efficiency of A is x% less than the efficiency of B, and B takes “B” days to complete the work then the time A will take to complete the work will be B/(100-x) *100 days

Example : Arjun can do a job 12 days, Ram is 60% more efficient than A. How many days will Ram take to complete the job alone ?

Solution :

Work done by Arjun in a day  = 1/12 = 8 ⅓ % per day

Efficiency of Ram = 100+60/100 = 160/100 = 1.6

Work done by Ram in a day =  8 ⅓ ×  1.6 = 40/3% per day
Number of days Ram needs to complete the work = 100 × 3/40 = 7.5 days

Example : CAT

A can do a work in 4 days. Efficiency of B is half the efficiency of A, efficiency of C is half the efficiency of B and efficiency of D is half the efficiency of C. After they have been grouped in two pairs it is found that the total number of days taken by one group is 2/3rd the time taken by the other group. Which of the following is a possible group?

Solution :

Number of days taken by A = 4 days
Number of days taken by B = 8 days
Number of days taken by C = 16 days
Number of days taken by D = 32 days
Assume that the total work = 32 units
So, the work done by A = 8 units
Work done by B = 4 units
Work done by C = 2 units
Work done by D = 1 units
It can be observed that the work done by B and C together in one day is 2/3rd the work done by A and D. So the groups are—AD and BC.

Another important concept of Time And work. We get different scenarios where there are a number of workers working in different situations. For comparing two different scenarios we use this concept.

According to the Concept of M-D-H We have

 M1*D1*H1/W1 = M2*D2*H2/W2

Where M= Number of men working
D = Days of work
H = hours worked per day
W= Work done
1 and 2 represents 2 different situations

Example : 5 men can complete a work in 10 days working 8 hours day. In how many days will 20 workers working 5 hours day complete twice the work?

Solution :

M1=5 , M2=20

H1 = 8, H2= 5

D1 = 10, D2 = ?

W1 = w, W2 = 2w

Hence, we get - (5x10x8 ) = (20x5xD)/2

Therefore D2 = 8 days

Example :

6 men and 3 women can complete a work working 8 hours for 12 days. In how much time will 12 women complete the same work working 6 hours every day, if the efficiency of each man is twice that of a woman?

Solution :

Let the work done by a man in a day be denoted by M and work done by a woman in a day be denoted by W.

1M= 2 W
6M = 12 W
As given
(6M+3W) x 8 x 12 = 12W x 6 X D
15W x 8 x 12 = 12W x 6 x D
OR D = 20

The concept of alternate work :

If there are 2 workers (say A and B ) And if A starts the work, On day 1 : A will work, On day 2 : B will work. Again on day 3, A will work; and on day 4 :B and this will go on until and unless the work gets completed.
Tip : If there are 2 workers. Find the work done in 2 days, If there are n workers, find the work done in “n” days first. Go as much close you can go to the total work that needs to be done taking work done by “n” workers. Then go on adding the individual work done per each day to get the actual day on which the work gets completed.

Example :

A can build a wall in 10 days while B can build the same wall in 40 days. If they work on alternate days in how many days, will the wall be completed if A start the job?

Solution :

Let total work = 40 units
1 day work of A = 4 units and 1 day work of B = 1 units; 2 days work = 1+4 = 5 units
To do 40 units, days required = 40 * 2 /5
=16 days

Suppose we have a situation where worker A is making a wall with bricks and another worker B, who does not like A’s work is destroying whatever he is making. This scenario where B is destroying the part of work is known as negative work.
Negative work means there is someone in the group of workers who destroys a part of work that is COMPLETED. There is a reason I wrote “Completed” in Capitals because in many sums people do this mistake, erasing the work that is not done with negative work.

Example:

Let A and B be 2 workers. A does the job in 10 days , B destroys the job in 20 days. If they work together in how many days will the work get completed ?

Solution :

LCM = total work = 20
Work done by A in a day = 2 units
Work destroyed by B in a day =1 units.
Since he is destroying, we will take his work to be -1 instead of +1
Hence B = -1
Work done by both in a day = +2-1 = +1 units. Hence total days required = 20/1 = 20 days

Note : For better understanding with the LCM approach, always take negative work to Negative.

Negative work is always < = Positive work (since we cannot destroy something which is not done)

Pipes and Cisterns

We have already learnt the concepts of Time and work . Pipes and Cisterns also follow the same logic as time and work. Here the work done is not building a wall or a road or a building , here work refers to filling or emptying a tank/container.

So,  work can be considered as ( Time Taken / Alone time of pipe)

Unit work done by pipe = 1/ Time taken by it to fill or empty the tank

Example:

If a pipe takes 40 hours to fill a tank. What is the time taken to fill 1/20th of the tank ?

Solution:

Work done by the pipe in one hour = 1/40 = 2.5% of the whole

1/20 parts = 100/20 = 5%

Hence, time required to fill 5% = 5/2.5 = 2 hours.

We may also use the LCM approach as it is mentioned in the time-work chapter.

Another term we will be using, is the CAPACITY of the Tank.

Capacity = Flow Rate * Time taken to fill

Example :

Water flows at a rate of 20l/min in a pipe. What is the capacity of the tank if the pipe fills the tank in 5 minutes?

Solution :

Capacity = Flow *Time = 20*5 = 100 Litres

Concept : We will take filling to be a positive quantity and Emptying to be a Negative quantity. This is similar to Positive and Negative work done.

Example :

There are 2 pipes A and B. A can fill the tank in 10 hours. B can empty the tank in 20 hours. When will the Tank be filled ?

Solution:

Let the Capacity of Tank  = LCM of 10 and 20 = 20 L

A can fill at a rate = +2 L / hour

B can empty at a rate = -1 L / hour

Total amount filled in 1 hour = +2-1 = 1 L

Hence to fill 20L, we will require 20/1 = 20 hours

Example :

A tap can fill a cistern in 12 hours. There is a leak in the bottom which can empty it in 20 hours. The cistern was empty when the tap was turned on. In how many hours will the cistern be filled?

Solution :

Let the capacity of cistern be 60
Therefore work done by tap in 1 hour = 5
Work done by leak in 1 hour = 3
So net work done = 2
Time taken = 60/2 = 30

Example:

There is a cistern of capacity 100 litres. A tap can fill the cistern in 10 hours. There is a leak at the half way mark of the cistern because of which the it took 15 hours to fill the cistern. Find the time in which the filled cistern can be emptied if there were 2 leaks at the half way mark?

Solution :

Can never be emptied, as both the leaks are at half way mark.

Pipe and Cistern sums are typical Time and Work sums and hence we use the same formulas and techniques .

Looks like your connection to MBAtious was lost, please wait while we try to reconnect.