Time and Work Concepts by Atreya Roy

Author: Atreya Roy is pursuing his BTech From Kalyani Government Engineering College, Bengal.
In this article 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. So we will Practice some easy, moderate and Difficult Questions and concepts . 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 onetenth part of the total work.
We can say that : Unit Time work = 1/ Total Time Taken by the worker to complete the whole work
#Concept
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.#Problem
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 ?
Simply take out the LCM as discussed earlier : LCM (9,10,12) = 180
#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 ?
Two ways to 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 unit sof work
When they work together : each day they complete 2+3= 5units.
Total work 30. Per day work done = 5
Days taken = 30/5= 6 daysOr 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 daysPlease try to do the sums using the total work to be something. Generally the LCM. It will ease your calculations . And it will be damn fast.
#Problem
A, B and C can finish a work independently in 10, 12 and 15 days. C starts the work and after 1 day, B joins him. After 1 day of B, A also joins them but leave 3 days before completion of the work, while B left two days before completion of work. What will be the total number of days taken to complete the work?
Let the work be 60 units
A= > 6 B= > 5 c=4
(X5)*6 +(x3)*5 +4x=60
15x=105
X=7#Concept
Alternate work :
The concept of alternate work is simple.
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. Then go on adding the individual work done per each day.#Problem
A can build a wall in 20 days while B can build the same wall in 30 days. If they work on alternate days in how many days, will the wall be completed if A start the job?
Let total work = 60 units
1 day work of A = 3 units and 1 day work of B = 2 units; 2 days work = 3 + 2 = 5 units
To do 60 units, days required =
60*2 /5
=24 days#Concept
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/(100x) *100 days#CAT2001 Problem
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?
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.#Concept
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#Concept
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.#Problem
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 ?
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 = +21 = +1 units. Hence total days required = 20/1 = 20 daysDid you know ?
Negative work is the opposite of alternate work.
In alternate work, people work on different days and their work gets added.
In negative work people work together or alternately, but their work gets subtracted.Thanks Guys, for reading it. Hope it will clear all your doubts. Time and Work sums ? Not more trouble anymore to you guys. Happy Learning.