- The basic formula for solving is: 1/r + 1/s = 1/h
- Let us take a case, say a person Hrithik
- Let us say that in 1 day Hrithik will do 1/20
^{th}of the work and 1 day Dhoni will do 1/30^{th}of the work. Now if they are working together they will be doing 1/20 + 1/30 = 5/60 = 1/12^{th}of the work in 1 day. Now try to analyze, if two persons are doing 1/12^{th}of the work on first day, they will do 1/12^{th}of the work on second day, 1/12^{th}of the work on third day and so on. Now adding all that when they would have worked for 12 days 12/12 = 1 i.e. the whole work would have been over. Thus the concept works in direct as well as in reverse condition. - The conclusion of the concept is if a person does a work in ‘r’ days, then in 1 day- 1/r
^{th}of the work is done and if 1/s^{th}of the work is done in 1 day, then the work will be finished in ‘s’ days. Thus working together both can finish 1/h (1/r + 1/s = 1/h) work in 1 day & this complete the task in ’h’ hours. - The same can also be interpreted in another manner i.e. If one person does a piece of work in x days and another person does it in y days. Then together they can finish that work in xy/(x+y) days
- In case of three persons taking x, y and z days respectively, They can finish the work together in xyz/(xy + yz + xz) days

Sol: Let Rita does the work in R days. Using basic work formula the equation would be 1/12 + 1/R = 1/8

⇒ 8R + 96 = 12R

⇒ 96 = 4R

⇒ 24 = R Working alone, Rita can do the job in 24 hours.

⇒ 8R + 96 = 12R

⇒ 96 = 4R

⇒ 24 = R Working alone, Rita can do the job in 24 hours.

Besides that one more approach can be applied in the work and time questions i.e. the unit approach. Time and work short tricks can be applied in this case, as the numbers used are 8 hours & 12 hours, let the work be equal to 24 units (which is the LCM of 8 & 12). Now as they finish the work in 8 hours working together, that implies together they do 24/8 = 3 units an hour. Working alone Aarti does this work in 12 hours, so Aarti alone does 24/12 = 2 units an hour. That means Rita will be doing 3 – 2 = 1 unit per hour. The total work is 24 units, which Rita can finish the work of her own in 24/1 = 24 hours.

Sol: A left the job 10 days before the completion. So, B worked alone for the last 10 days. First, we will calculate B’s 10 days work, which he did alone.

In 10 days B will do 10 × 1/40 = 1/4th of the work.

Remaining work 1 - ¼ = ¾ (Which A and B have done together). A and B can do 1/60 + 1/40 work in 1 day. Their one-day’s work is 1/60 + 1/40 = (2 + 3)/120 = 5/120 = 1/24. They can finish the work in 24 days.

They would have done three-fourth of the work in 24 × 3/4 = 18 days.

⇒ Total days = 18 + 10 = 28.

In 10 days B will do 10 × 1/40 = 1/4th of the work.

Remaining work 1 - ¼ = ¾ (Which A and B have done together). A and B can do 1/60 + 1/40 work in 1 day. Their one-day’s work is 1/60 + 1/40 = (2 + 3)/120 = 5/120 = 1/24. They can finish the work in 24 days.

They would have done three-fourth of the work in 24 × 3/4 = 18 days.

⇒ Total days = 18 + 10 = 28.

As discussed earlier in time work questions, time and work tricks like the unit approach can also be applied. In this case, as the numbers used are 60 & 40, let the work be equal to 120 units. That implies A does 120/60 = 2 units a day, whereas B alone does 120/40 = 3 units a day. That means working alone B would have done 3 × 10 = 30 units. The remaining 120 – 30 = 90 units of work has been done by them together. They do 2 + 3 = 5 units a day working together, thus they would have finished 90 units in 90/5 = 18 days. Hence the total work was finished in 18 + 10 = 28 days.

Must Read Time and Work Problems Articles

- Time and Work Concepts
- Time and Work Formula and Solved Problems

Sol: Using work formula here (1/A) + (1/B) + (1/C) = (1/8)

(1/C) = (1/8) - (1/A) - (1/B) ⇒ (1/C) = (1/8) - (1/24) - (1/20) ⇒ (1/C) = (1/30)

C can do this work in 30 days.

(1/C) = (1/8) - (1/A) - (1/B) ⇒ (1/C) = (1/8) - (1/24) - (1/20) ⇒ (1/C) = (1/30)

C can do this work in 30 days.

You can take the total work to be equal to 120 units (the LCM of 24, 20 & 8). That implies A does 120/24 = 5 units a day, B does 120/20 = 6 units a day. Together they finished the work in 8 days means they are doing 120/8 = 15 units a day. Let the units done by C per day be = c. Now as per the statement 5 + 6 + c = 15 ⇒ c = 4 units. Now if C does 4 units a day, he can finish the work in 120/4 = 30 days.

If we add all this it will give us the work of 2A, 2B and 2C in 1 day i.e. (1/36) + (1/48) + (1/72) + (1/16)

That also implies that A, B and C’s one day’s work will be half of this i.e. (1/2) x (1/16) = (1/32)

From here it can found that they will complete the work in 32 days.

Therefore, B = (5/2) x A ⇒ (5/2) x 48 = 120 days.

Suggested Action:

In this article we learned, how to solve the time and work questions by applying the basic time and work formula and by using the unit’s approach. Here using the unit’s approach, you make your calculations simple and you can solve the question without writing much. You can make it a point to use this approach in time work problems.