Work and Wages - Solved Questions and Answers

Last Updated : 25 Jun, 2025

Question 1: To complete the work, Person A takes 10 days and Person B takes 15 days. If they work together, how much time will they take to complete the work?

Solution:

Method 1 :
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.
Method 2 (Short Method):
Let the total work be LCM (10, 15) = 30 units
=> A's efficiency = 30/10 = 3 units / day
=> B's efficiency = 30/15 = 2 units / day
Combined efficiency of A and B = 3+2 = 5 units / day
=> In one day, A and B working together can finish of 5 units of work, out of the given 30 units.
Therefore, time taken to complete total work = 30 / 5 = 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: 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 (efficiency of A, B, 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 4: 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 5: 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 6: 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 7: 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 8: 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
Alternate Method
Here, we can use the formula for comparison of work and efficiency
M1 D1 H1 E1 / W1 = M2 D2 H2 E2 / W2
Here, M1 = 45 (initial number of men)
D1 = 6 (number of days 45 men work)
W1 = 270 (work done by 45 men in 6 days)
E1 = E2 = 1 (efficiency of each man)
We assume H1 = H2 = Number of working hours in a day
M2 = 75 (number of men after 6 days)
D2 = Number of days 75 men work or Number of more days required
W2 = 450 (work to be done by 75 men)

So, we have (45 x 6 x 1) / 270 = (75 x D2 x 1) / 450
Therefore, D2 = 6 days
Hence, 6 more days are required to complete the work.

Question 9: 2 Men and 3 Women working together can finish a job in 10 days. It takes 8 days to finish the same job if 3 Men and 2 Women are employed. If only 2 Men and 1 Woman are employed, find the time they would take to complete the job.

Solution:

Here, we need to use the summation formula for the comparison of work and efficiency
∑(Mi Ei) D1 H1 / W1 = ∑(Mj Ej) D2 H2 / W2
Here, ∑(Mi Ei) = 2M + 3W, where M is the efficiency of each Man and W is the efficiency of each Woman
∑(Mj Ej) = 3M + 2W
D1 = 10
D2 = 8
Also, H1 = H2 and W1 = W2
So, we have (2M + 3W) x 10 = (3M + 2W) x 8
=> M:W = 7:2
Assume the constant of proportionality to be 'k' here.
=> M = 7k and W = 2k

Now, we again apply the summation formula with LHS being any of the given set of values and RHS being the set of values corresponding to 2 Men and 1 Woman.
Therefore, (2M + 3W) x 10 = (2M + 1W) x D, where D is the number of days required to complete the work if 2 Men and 1 woman are employed.
=> 20k x 10 = 16k x D
=> D = 12.5 days

Question 10: To complete a job, A alone takes 2 more days than A and B together. B alone takes 18 more days than A and B together. Find the time taken if they work together.

Solution:

Let the time required if A and B work together be 'n' days.
=> A alone takes n+2 days
=> B alone takes n+18 days
So, work done by A in one day alone = 1 / (n+2)
Work done by B in one day alone = 1 / (n+18)
Total work done by both A and B in one day alone = 1/(n+2) + 1/(n+18)
But, total work done in one day if both A and B work together = 1/n
Therefore, 1/(n+2) + 1/(n+18) = 1/n
=> (2n + 20) / [(n+2) x (n+18)] = 1/n
=> 2n² + 20n = n² + 20n + 36
=> n² = 36
=> n = 6 (Since 'n' is the number of days and cannot be negative)
Therefore, time taken to complete the job if both A and B work together = 6 days

Short Method
In these type of questions, we can simply do as :
n² = d1 x d2, where d1 is the additional days required by A and d2 is the additional days required by B.
(NOTE : This short cut is applicable if only two people are working on a job)
So, n² = 2 x 18 = 36
=> n = 6.
Therefore, time taken to complete the job if both A and B work together = 6 days

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