- If a person can do a piece of work in
**‘n’**days, then in one day, the person will do**‘1/n’**work. Conversely, if the person does**‘1/n’**work in one day, the person will require**‘n’**days to finish the work. - In questions where there is a comparison of work and efficiency, we use the formula

**M**, where_{1}D_{1}H_{1}E_{1}/ W_{1}= M_{2}D_{2}H_{2}E_{2}/ W_{2}

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

**∑(M**, where ‘i’ and ‘j’ may vary as per the number of workers._{i}E_{i}) D_{1}H_{1}/ W_{1}= ∑(M_{j}E_{j}) D_{2}H_{2}/ W_{2} - 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.

In this topic, the way of attempting the questions is the deciding factor for getting accurate answer in less time. We shall try to cover all the types of questions asked in this topic with detailed explanation of way of attempting them.

### Sample Problems

**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 : **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 (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 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

**M _{1} D_{1} H_{1} E_{1} / W_{1} = M_{2} D_{2} H_{2} E_{2} / W_{2}**

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 D

_{2}x 1) / 450

Therefore, D

_{2}= 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 comparison of work and efficiency

**∑(M**

_{i}E_{i}) D_{1}H_{1}/ W_{1}= ∑(M_{j}E_{j}) D_{2}H_{2}/ W_{2}Here, ∑(M

_{i}E

_{i}) = 2M + 3W, where M is the efficiency of each Man and W is the efficiency of each Woman

∑(M

_{j}E

_{j}) = 3M + 2W

D

_{1}= 10

D

_{2}= 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

=> 2n

^{2}+ 20n = n

^{2}+ 20n + 36

=> n

^{2}= 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 :

n

^{2}= d

_{1}x d

_{2}, where d

_{1}is the additional days required by A and d

_{2}is the additional days required by B.

(

**NOTE :**This short cut is applicable if only two people are working on a job)

So, n

^{2}= 2 x 18 = 36

=> n = 6.

Therefore, time taken to complete the job if both A and B work together = 6 days

### Problems on Work and Wages | Set-2

### Quiz on Work and Wages

This article has been contributed by **Nishant Arora**

Please write comments if you have any doubts related to the topic discussed above, or you are facing difficulty in any question or if you would like to discuss a question other than those mentioned above.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready.