In this article, we are going to demystify time and work based questions. A lot of aspirants find these questions difficult. In fact, these questions are extremely straightforward once you understand the basic relationship between time and work and decide which method is required to solve the question.

We’ll start with a simple question.

If Amol can do a job in 10 days, what fraction of the work will be done by Amol in one day?

There are two ways of solving this question. You can assume the work to be x units. For the sake of convenient, always assume the work to be some multiple of the days taken to do that work. This way, you will get an integral quantity of work done per day which will make calculations easier. 

So we can assume work to be 10 units. Amol can do 10 units in 10 days which means that he is doing one unit work per day. In one day, he will do one unit and the fraction of the work done will be 1/10.

The other way is to simply say that as Amol can do a job in 10 days, in one day, he will do 1/10.

So if the question says, A can do a job in x days:

In one day: 1/x
or Total units / number of days

In n days: n/x
or n × unit / day

Remaining work to be done after n days = 1 – (n/x)
or Total units – n × unit / day

If you are clear till here, we’ll now look at two or more people working simultaneously.


If A can do a job in 10 days, B can do a job in 15 days, and C can do a job in 20 days, in how many days will they complete a job if they start working together?

Fraction method: 1/10 + 1/15 + 1/20 = 13/60. Time taken will be reciprocal of this = 60/13 days

Unit method: Let the units be LCM (10, 15, 20) = 60 units. Hence, A will do 60/10 = 6 units per day; B will do 60/15 = 4 units per day; C will do 60/20 = 3 units per day. In one day, together they will do 6 + 4 + 3 = 13 units. Hence, to do 60 units, they will take 60/13 days.

The possible variations in these questions are: Individuals working on alternate days or a certain set of individuals working for different number of days and so on. In any case, understand what the question is asking and decide the starting point.

[Note: While assuming units, always go for LCM of multiple of LCM]
We will now move to questions with negative work (destructive work). Read the following question:

A and B are building a wall whereas C loves destroying the work done by A and B. A can build the complete wall in 25 days whereas B can build the complete wall in 10 days. C can destroy the complete wall in 20 days. If all of them start working together, how much time will they take to build the wall?

Fraction method: Negative work will have a negative sign. Rest of the procedure remains the same.

1/25 + 1/10 – 1/20 = 9/100. Time taken = 100/9 days

Unit method: Let the work be 200 units. (Multiple of 100 which is the LCM. As it will cancel out with per day units, the answer will remain the same)

A will do 8 units per day, B will do 20 units per day and C will do negative work of 10 units per day. Work done in one day = 8 + 20 – 10 = 18 units. Time taken to complete the task = 200/18 = 100/9 days
Solved questions:

  1. A takes 5 days more than B to do a task. A takes 9 days more than C to do the same task. A and B together can do the same work in the same time as C. How many days will A take to do it?

To solve this question, we just need to write one equation. Let A take x days to do the task. B will take (x – 5) days and C will take (x – 9) days. It is given that work done by A and B together in one day is same as work done by C in one day.

1/x + 1/(x-5) = 1/(x-9)

(2x – 5)(x – 9) = x(x – 5)

Solving this will give you x = 15.

This question will be very difficult to solve using units as we know the time taken by individuals relative to each other.

  1. A wall can be built by A and B in 10 days, by B and C in 15 days, and by A and C in 20 days. They worked for 6 days and then A left. B and C worked for 4 more days and B left. In how many more days will C complete the remaining work of constructing the wall?

Let the work be 60 units.

A + B = 6 units/day; B + C = 4 units/day; A + C = 3 units/day.

2(A + B + C) = 13. Hence, A + B + C = 6.5 (This will give us A = 2.5, B = 3.5, C = 0.5)

In six days, they will do 6(6.5) = 39 units. Remaining work 60 – 39 = 21 units.

In 4 days, B and C will do 4(4) = 16 units. Remaining work = 21 – 16 = 5 units.

Hence, C will take 5/0.5 = 10 more days to complete the work.

  1. A can do a job in 90 days, B can do the same job in 40 days and C can do it in 12 days. They work for a day each in turn. On the first day, only A works. On the second day, only B works. On the third day, only C works and they continue in this manner. This cycle continues till the work gets done. If they get 480 INR to do this job, find their individual share.

Fraction method:

Work done in three days: 1/90 + 1/40 + 1/12 = 43/360. From here, we need to understand that the integral part of the fraction will give us cycles = 8*3 = after 24 days 43*8 = 344/360 work is done. On the next day (25th day), A will do 4/360 work. Work done = 348/360. On the next day, B will do 9/360 work and the total work done will be 357/360. On 27th day, C will do the remaining 3/360 work. Hence, the work done by A is for 9 complete days. B is 9 complete days and C is 8 complete days + 3/360

A’s share = 9*1/90 = 9/90 = 36/360*480 = 48 INR
B’s share = 9*1/40 = 9/40 = 81/360*480 = 108 INR
C’s share = 8*1/12 + 3/360 = 243/360*480 = 324 INR

Unit method:

Let the work be 360 units (LCM).In 3 days, work done = 4 + 9 + 30 = 43 units. 360/43 integral part is 8. So in the first 24 days, work done will be

A:B:C = 32:72:240 units. Remaining work = 360 – 344 = 16 units.

Day 25 = A will do 4 units
Day 26 = B will do 9 units
Day 27 = C will do 3 units.

Hence, in total work = A:B:C = 36:81:243. Share ratio 4:9:27 and individual share will be 48, 108, and 324 respectively.


Hope you are reading our 75 days to CAT series and learning important concepts every day. Have a tough time and work question? Want us to solve it? Drop a line in the comments section.

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