Work Efficiency-RRB

Work Efficiency:

Where there is a comparison of work and efficiency, we use the formula
M1 D1 H1 E1 / W1 = M2 D2 H2 E2 / W2, where
M = Number of workers
D = Number of days
H = Number of working hours in a day
E = Efficiency of workers
W = Units of work

In case we have more than one type of workers, then the formula modifies to
∑(Mi Ei) D1 H1 / W1 = ∑(Mj Ej) D2 H2 / W2, where ‘i’ and ‘j’ may vary as per the number of workers.

If a person A is ‘n’ times more efficient than person B, then
Ratio of work done by A and B in one day (Ratio of efficiencies) = n : 1
Ratio of time taken by A and B = 1 : n

Total work = No. of Days x Efficiency

If a group of people are given salary for a job they do together, their individual salaries are in the ratio of their individual efficiencies if they work for same number of days. Otherwise, salaries are divided in the ratio of units of work done.

Question 1 : To complete a work, a person A takes 10 days and another person B takes 15 days. If they work together, in how much time will they complete the work ?
Solution :
A’s one day work (efficiency) = 1/10
B’s one day work (efficiency) = 1/15
Total work done in one day = 1/10 + 1/15 = 1/6
Therefore, working together, they can complete the total work in 6 days.

Question 2 : 

Two friends A and B working together can complete an assignment in 4 days. If A can do the assignment alone in 12 days, in how many days can B alone do the assignment ?

Solution :

Let the total work be LCM (4, 12) = 12
=> A’s efficiency = 12/12 = 1 unit / day
=> Combined efficiency of A and B = 12/4 = 3 units / day
Therefore, B’s efficiency = Combined efficiency of A and B – A’s efficiency = 2 units / day
So, time taken by B to complete the assignment alone = 12/2 = 6 days

Question 3:

Two friends A and B are employed to do a piece of work in 18 days. If A is twice as efficient as B, find the time taken by each friend to do the work alone.

Solution :

Let the efficiency of B be 1 unit / day.
=> Efficiency of A = 2 unit / day.
=> Combined efficiency of A and B = 2+1 = 3 units / day
=> Total work = No. of Days x Efficiency = 18 days x 3 units / day = 54 units
Therefore, time required by A to complete the work alone = 54/2 = 27 days
Time required by B to complete the work alone = 54/1 = 54 days

Question 4: 

Three people A, B and C are working in a factory. A and B working together can finish a task in 18 days whereas B and C working together can do the same task in 24 days and A and C working together can do it in 36 days. In how many days will A, B and C finish the task working together and working separately?

Solution : Let the total work be LCM (18, 24, 36) = 72
=> Combined efficiency of A and B = 72/18 = 4 units / day
=> Combined efficiency of B and C = 72/24 = 3 units / day
=> Combined efficiency of A and C = 72/36 = 2 units / day
Summing the efficiencies,
2 x (Combined efficiency of A, B and C) = 9 units / day
=> Combined efficiency of A, B and C = 4.5 units / day
Therefore, time required to complete the task if A, B and C work together = 72/4.5 = 16 days

Also, to find the individual times, we need to find individual efficiencies. For that, we subtract the combined efficiency of any two from combined efficiency of all three.
So, Efficiency of A = Combined efficiency of A, B and C – Combined efficiency of B and C = 4.5 – 3 = 1.5 units / day
Efficiency of B = Combined efficiency of A, B and C – Combined efficiency of A and C = 4.5 – 2 = 2.5 units / day
Efficiency of C = Combined efficiency of A, B and C – Combined efficiency of A and B = 4.5 – 4 = 0.5 units / day.

Therefore, time required by A to complete the task alone = 72/1.5 = 48 days
Time required by B to complete the task alone = 72/2.5 = 28.8 days
Time required by C to complete the task alone = 72/0.5 = 144 days

Question 5:

 Three workers A, B and C are given a job to paint a room. At the end of each day, they are given Rs. 800 collectively as wages. If A worked alone, the work would be completed in 6 days. If B worked alone, the work would be completed in 8 days.If C worked alone, the work would be completed in 24 days. Find their individual daily wages.

Solution : 

Let the total work be LCM (6, 8, 24) = 24 units.
=> A’s efficiency = 24/6 = 4 units / day
=> B’s efficiency = 24/8 = 3 units / day
=> C’s efficiency = 24/24 = 1 unit / day
We know that ratio of efficiencies = Ratio of wages
=> Ratio of daily wages of A, B, C = 4:3:1
Also, it is given that they get Rs. 800 collectively at the end of each day.
Therefore, A’s daily wages = Rs. 400
B’s daily wages = Rs. 300
C’s daily wages = Rs. 100

Question 6: 

Two workers A and B are employed to do a cleanup work. A can clean the whole area in 800 days. He works for 100 days and leaves the work. B working alone finishes the remaining work in 350 days. If A and B would have worked for the whole time, how much time would it have taken to complete the work?

Solution : 

Let the total work be 800 units.
=> A’s efficiency = 800/800 = 1 unit / day
=> Work done by A in 100 days = 100 units
=> Remaining work = 700 units
Now, A leaves and B alone completes the remaining 700 units of work in 350 days.
=> Efficiency of B = 700/350 = 2 units / day
Therefore, combined efficiency of A and B = 3 units / day
So, time taken to complete the work if both A and B would have worked for the whole time = 800 / 3 = 266.667 days.

Question 7: 

45 men can dig a canal in 16 days. Six days after they started working, 30 more men joined them. In how many more days will the remaining work be completed ?

Solution : 

Let the efficiency of each man be 1 unit / day.
Let the total work = 45 x 16 = 720 units
=> Work done in 6 days by 45 men = 45 x 6 = 270 units
=> Remaining work = 720-270 = 450 units
Now, we have 75 men with efficiency 1 unit / day each to complete the work.
Thus, More days required to complete the work = 450/75 = 6 days.

Question 8: 

A person A can do a piece of work in 9 days, whereas another person B can do the same piece of work in 12 days. Because of busy schedule, they decide to work one day alternately. If B is the first one to start, find the time required for the work to be completed. Consider that if a part of day is used, the whole day is to be counted.

Solution : 

Let the total work be LCM (9, 12) = 36 units
=> A’s efficiency = 36/9 = 4 units / day
=> B’s efficiency = 36/12 = 3 units / day
Now, since they work alternately, they would complete 7 units of work in two days.
=> In 5 such cycles of alternate working, i.e., 10 days, they would have completed 35 units of work.
Now, work left = 1 unit
Now, B would do that in less than one day but we have to take into account one full day even if work goes on for some part of the day.
Therefore, time required for the work to be completed = 10+1 = 11 days.

Question 9:

A is 30% more efficient than B. How much time will they, working together, take to complete a job which A alone could have done in 23 days?

Solution:

Ratio of times taken by A and B = 100 : 130 = 10 : 13.

Suppose B takes x days to do the work.

Then, 10 : 13 :: 23 : x

=> x = (23 x 13) /10

=> x = 299/10

A’s 1 day of work = 1/23

B’s 1 day of work = 10/299

(A + B)’s 1 days work = (1/23 + 10/299)

= 23/299

= 1/13

Therefore, A and B can together complete the work in 13 days.

Question 10:

A is twice as good a workman as B and is therefore able to finish a piece of work in 30 days less than B.In how many days they can complee the whole work; working together?

Solution:

Ratio of times taken by A and B = 1 : 2.

The time difference is (2 – 1) 1 day while B take 2 days and A takes 1 day.

If difference of time is 1 day, B takes 2 days.

If difference of time is 30 days, B takes 2 x 30 = 60 days.

So, A takes 30 days to do the work.

A’s 1 day’s work = 1/30

B’s 1 day’s work = 1/60

(A + B)’s 1 day’s work = 1/30 + 1/60 = 1/20

A and B together can do the work in 20 days.