Work Equivalent Concept is a based on the universal concept of measuring work in two parameters; Agent who is doing work and the duration the work takes to get completed. Let us understand with the help of an example.
Suppose the time taken to complete a task by 4 men all of the equal efficiency is 8 days. But say we need to get the work done in 4 days only. The logical deduction is that we need to increase the manpower. But how much more men required? Yes, everyone can answer that it will require 8 men to complete the work in 4 days. BY applying simple logic we can solve this question, but we will see the mathematical calculation behind this reasoning.
From the first scenario, i.e. 4 men can complete the task in 8 days; we measured the work in terms of effort and time required. That is,
Work = Number of Men \( \times \) Number of days (we will keep the unit of measurement of work as man-days)
So, when the number of days decreases to half, to the keep the measurement of the work same, we have to double the manpower.
Some Standard Results:
- When work (W) is constant then, the number of men (M) is inversely proportional to the number of days (D) required to complete the work. i.e., \(M \propto \frac{1}{D}\).
Example: As we saw in the above example, in both cases work remained the same, Number of men was inversely proportional to the number of days
- If the number of days (D) is constant, then work is directly proportional to the number of men working. i.e., more men more work and vice versa.
- If the number of men (M) is constant, then work is directly proportional to the number of days. i.e., more days more work and vice versa.
Let us see some simple solved examples based on work equivalence:
Example 1: If 5 men take 90 days to complete a piece of work then how many days will 15 men take to complete the same work?
Solution:
Here the total work can be simply expressed as 5\( \times \)90 man-days
Now to finish the same job, the measurement of the work should be the same, i.e. 5\( \times \)90 man-days.
Let 15 men take x days to finish then by using the concept of work equivalence
15x = 5\( \times \). 90 or x = 30 days.
Alternate approach: As work is constant, \(M \propto \frac{1}{D}\). Here the number of men going up three times (from 5 to 15), hence the number of days required will be 1/3rd of 90 = 30 days.
Example 2: It takes 6 technicians a total of 10 hours to build a new server from Direct Computer, with each working at the same rate. If six technicians start to build the server at 11 am and one technician per hour is added beginning at 5 pm, at what time will the server be complete? (CAT 2002)
Solution:
Application of work equivalence concept:
Total work = 6\( \times \)10 = 60 man-hours
From 11 am to 5 pm, i.e. for 6 hours 6 technicians worked. The portion of work completed by them = 6\( \times \)6 = 36 man-hours. So, 60-36 = 24 man-hours work is pending.
From 5 pm onwards, at every interval of one hour, one more technician joins them.
Hence amount of work done between 5 pm and 6 pm = 7 men\( \times \) 1 hour = 7 man-hours.
Similarly,
Amount of work done between 6 pm and 7 pm = 8 men\( \times \) 1 hour = 8 man-hours and between 7 pm and 8 pm = 9 men\( \times \) 1 hour = 9 man-hours.
Observe that 7 + 8 +9 = 24 man-hours. Therefore, the task gets completed by 8 pm.
Different persons with different efficiencies working together
There is a type of problem in which different persons working together have different efficiencies. In such questions, the approach to solve them varies slightly. Let us take an example to understand.
Example 3: 2 men and 3 women can complete a work in 16 days, while 3 men and 7 women can complete it in 10 days. In how many days will 5 women complete it?
Solution:
Let each man and each woman work x and y units of work per day respectively.
From the first condition, units of work done in one day are 2x + 3y.
So the total units of work = (2x + 3y)\( \times \)16
Similarly, from the second condition, total units of work = (3x + 7y)\( \times \)10.
Since both are same work we get
(2x + 3y)\( \times \)16 = (3x + 7y)\( \times \)10
Or 32x + 48y = 30x + 70y
=> 2x = 22y
=> \(\frac{x}{y} = \frac{{11}}{1}\) so let us assume that x = 11 and y= 1.
Therefore, the total units of work = (2x + 3y)\( \times \)16 = (2\( \times \)11 + 3\( \times \)1)\( \times \)16=400 units.
Hence 5 women can complete the work in 400/5 = 80 days.
Constantly increasing work with the increase in time
Example 4: 29 horses can graze a uniformly growing grass field in 7 days, 25 horses can do the same in 9 days. In how many days can 32 horses graze the same field?
Solution:
Let the amount of grass on day one was x units and y units of grass grow per day.
From the first condition, the field was completely grazed in 7 days. Therefore, amount of work done = x + 7y. And to complete this work, 29 horses took 7 days, hence the relation we get is:
x + 7y = 29\( \times \)7 …. eq (1)
Similarly, from the second condition we get:
x + 9y = 25\( \times \)9 …. eq (2)
Solving both the equations for x and y we get, x = 126 and y = 11.
Now, let us take the number of days in which 32 horses graze the same field is N.
Therefore, the required relation is:
x + Ny = 32N
Or 126 + 11N = 32N
Or N = 6 days
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