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Individual Work

If Amit can do a certain work in 10 days, then he will finish 1/10th of the work in one day.
 
Example-1
Amit can do some work in 12 days and Vinit can do the same work in 15 days. In how many days will both of them do the work when working together?
Solution
Assume total work = 1 unit
 
Work done by Amit in one day = 1/12 unit
 
Work done by Vinit in one day = 1/15 unit
 
Work done by both of them in one day when working together = (1/12) + (1/15) = 9/60 unit
 
Hence, they will be doing the whole work in 60/9 days = Description: 2206.pngdays
 
 
LCM method of solving time and work questions
 
This can be understood in terms of the above written example in the following way:
 
Let us assume total work to be equal to the LCM of the days taken by Amit and Vinit (i.e., of 10 and 15)
 
Assume work = 60 units
 
Work done by Amit in one day = 5 units
 
Work done by Vinit in one day = 4 units
 
Work done by both of them in one day when working together = 9 units
 
So, the number of days taken by both of them when working together = 60/9 = Description: 2215.pngdays
 
Alternatively, we can do this problem with the help of a graph also.
 
Description: 7714.png
 
We have created the pillars of the number of days of Amit and Vinit. Now the point of intersection of the top of the first pillar to the bottom of the second pillar and the top of second pillar and the bottom of the first pillar is the number of days taken by both of them when working together. In the above drawn graph, it can be clearly seen that the point of intersection of both the straight lines is a bit more than 6.
 
Despite graphical method appearing easier than the earlier two methods, the usage of this method should be avoided due to the complexity of denoting the points on the graph paper.
 
Example-2
A, B and C can do a piece of work individually in 8, 12 and 15 days respectively. A and B start working, but A quits after working for 2 days. After this, C joins B till the completion of work. In how many days will the work be complete?
Solution
Let us assume that the Work = LCM (8, 12, 15) = 120 units
 
So, the work done by A in one day = 15 units
 
Work done by B in one day = 10 units
 
Work done by C in one day = 8 units
 
Work done by A and B in two days = 2 × 25 = 50 units
 
Remaining work = 70 units
Work done by C and B in one day = 18 units
 
Time taken to complete the remaining work by C and Description: 2219.png
 
So, the total number of days = Description: 2228.png
 
 
Example-3
A and B together can do a work in 12 days, B and C together can do the same work in 10 days and A and C together can do the same work in 8 days. In how many days will the work be complete if A, B and C are working together?
Solution
Let us assume work = LCM of (12, 10, 8) = 120 units
 
So, A and B are doing 10 units in one day, B and C are doing 12 units in a day and A and C are doing 15 units in a day.
 
Adding all these, 2 (A + B + C) are doing 37 units in a day.
 
⇒ (A + B + C) are doing Description: 2237.png = 18.5 units in a day
 
So, time taken to complete the work = Description: 2246.png days
 
= 6.48 days
 

Individual Efficiency

Efficiency is also known as work-rate.
 
If A is taking less number of days with respect to B to complete the same work, we can say that the efficiency of A is more than the efficiency of B.
 
So, more the efficiency, less will be the number of days and less the efficiency, more will be the number of days to do a certain work. We have observed individual efficiency in case of percentage also (Product Stability ratio).
 
Now, assume A takes 20 days to complete a work and B takes 25 days to complete the same work. It means A is doing 5% (100%/20) work in one day and B is doing 4% (100%/25) work in a day. So, efficiency of A is 25% more than efficiency of B.
 
General expression correlating time taken and efficiency
 
If efficiency of A is x per cent more than the efficiency of B and B takes ‘B’ days to complete the work, then A will take Description: 2255.png days to complete the same work.
 
If efficiency of A is x% less than the efficiency of B and B takes ‘B’ days to complete the work, then A will take Description: 2264.png days to complete the same work.
 
So, if A is 20% more efficient than B and B takes ‘B’ days to complete the work, then A will take Description: 2273.png days to do the same work.
 
With this, it can also be observed that if work is constant then time taken is inversely proportional to efficiency.
 
Example-4
John is thrice as efficient as Abraham and hence, completes a work in 60 days less than the number of days taken by Abraham. What will be the number of days taken by both of them when working together?
Solution
Since John is thrice as efficient as Abraham, so the number of days taken by him will be 1/3rd the number of days taken by Abraham. If John is taking x days, then Abraham will take 3x days to complete the same work.
Now, 3x − x = 2x = 60 days
So, x = 30 days and 3x = 90 days
Let us assume that the total work = 90 units (LCM of 30 and 90)
So, the total work done by both of them in one day
= 3 + 1 = 4 units of work.
So, the total number of days = Description: 2282.pngdays = 22.5 days
 
 
Example-5
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? (CAT 2001)
  1. AB
  2. BC
  3. CD
  4. AC
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.
 
 
Example-6
In a nuts and bolts factory, one machine produces only nuts at the rate of 100 nuts per minute and needs to be cleaned for 5 minutes after production of every 1,000 nuts. Another machine produces only bolts at the rate of 75 bolts per minute and needs to be cleaned for 10 minutes after production of every 1,500 bolts. If both the machines start their production at the same time, what is the minimum duration required for producing 9,000 pairs of nuts and bolts?
  1. 130 minutes
  2. 135 minutes
  3. 170 minutes
  4. 180 minutes
Solution
Machine I
 
Number of nuts produced in one minute = 100
 
Time required to produce 1,000 nuts = 10 minutes
 
Cleaning time for nuts = 5 minutes
 
Overall time to produce 1,000 nuts = 15 minutes.
 
Over all time to produce 9,000 = 138 minutes − 5 minutes
 
= 133 minutes … (1)

Machine II
 
Time required to produce 75 bolts = 1 minute
 
Time required to produce 1,500 bolts = 20 minutes
 
Cleaning time for bolts = 10 minutes
 
Effective time to produce 1,500 bolts = 30 minutes
 
Effective time to produce 9,000 bolts = 30 × 6 − 10
 
= 170 minutes … (2)
 
From (1) and (2)
 
Minimum time = 170 minutes
 

Example-7
A, B and C are assigned a piece of work which they can complete by working together in 15 days. Their efficiencies (measured in terms of rate of doing work) are in the ratio of 1:2:3. After Description: 2291.pngrd of the work is completed, one of them has to be withdrawn due to budget constraint. Their wages per day are in the ratio of 3:5:6. The number of days in which the remaining two persons can complete the work (at optimal cost) is
  1. 18
  2. 20
  3. 15
  4. 12
Solution
A, B and C together in 15 days= A alone in 90 days, B alone in 45 days, C alone in 30 days.
 
Wages per day per unit work for A, B and C are Description: 2300.png
 
Hence, A is the least efficient and hence, must be done away with.
 
For B and C, the whole work can be finished in 18 days and hence, remaining 2/3rd of the work can be finished in 12 days only.
 




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