Time and Work Notes | Study Quantitative Techniques for CLAT - CLAT

Introduction 

Work is defined as something which has an effect or outcome; often the one desired or expected. The basic concept of Time and Work is similar to that across all Arithmetic topics, i.e. the concept of Proportionality.

Efficiency is inversely proportional to the Time taken when the amount of work done is constant.

To solve problems on ‘time and work’ the following simple facts should be remembered:

• If a person can do work in 5 days, then in one day he will do 1/5^th of the whole work. Conversely, if a man can do 1/5^th of the work in one day, he will complete the whole work in 5 days.
• If the number of men engaged to do a piece of work be increased in a certain ratio, the time required to do the same work will be decreased in the same ratio and vice versa. Thus if the number of men is changed in the ratio of 3:7, the time required will be changed in the ratio of 7:3.

• If A is twice as good a workman as B then A will take one-half of the time taken by B to do a certain piece of work.
• To sum up if M₁ persons working T₁ hours a day can do W₁ work in D₁ days and M₂ persons working H₂ hours a day can do W₂ work in D₂ days, then the following equation will hold good.
  

Solved Examples

Example 1. A does work in 10 days and B does the same work in 15 days. How many days they will take to do the work together.

Sol. A does the work in 10 days, so A’s one-day work = 1/10

B does the work in 15 days, so B’s one day work = 1/15

Work done by A and B in one day = 1/10 + 1/15 = 5/30 = 1/6

Thus, A and B together will be able to finish the work in 6 days.

Example 3. A tyre has two punctures. The first puncture alone can empty the tyre in 9 minutes and 2^nd puncture alone can empty the tyre in 6 minutes. How long will both the punctures take to flat the tyre. 

Sol. The first puncture takes 9 minutes to flatten the tyre, so first puncture’s one-minute work = 1/ 9  

The second puncture takes 6 minutes to flatten that tyre, so second puncture’s one-minute work = 1/6

So work done by both punctures in one minute = 1/9 + 1/6 = 5/18

Thus, both punctures together will take 18/5 minutes to flatten the tyre.

Example 5. A and B together can finish a job in 15 days. If A alone can finish the job in 25 days, in how many days can B alone finish the job. 

Sol. A and B together can finish the job in 15 days, so their one day work = 1/15
A alone can finish the job in 25 days, so A’s one day work = 1/25

B’s one day work =  

Thus, B alone can finish the job in 75/2 days.

Example.6. A and B together can build a house in 25 days. They work together for 15 days and then B goes away. A finishes the rest of the work in 20 days. How long will each take to finish the job working separately?

Sol. A+B can finish the work in 25 days, so one day work = 1/25
They work together for 15 days, so the work done by A+B in 15 days =

Remaining job =   

To complete 2/5 of work in 20 days, A can complete one work in  = 50 days 

B’s one day work = 1/25 – 1/50 = 1/50    

So B can finish the whole work in 50 days.

Ex.7. A can finish a job in 10 days, B in 12 days and C in 10 days. In how many days will they finish the job if they work together. 

Sol. A’s one day work = 1/10

B’s one day work = 1/12

C’s one day work = 1/10

A+B+C’s one day work = 1/10 + 1/12 + 1/10 = (6+5+6)/60 = 17/60

Thus, together they can complete the work in 60/17 days.

Questions on Pipes and Cisterns

Example 9. Tap A can fill a tank in 8 hours. Outlet B can empty the tank in 12 hours. If both are kept open, how long will it take to fill the tank? 

Sol. Tap A can fill the tank in 8 hours, so Tap A’s one-hour work = 1/8

Tap B can empty the tank in 12 hours, so Tap B’s one hour work = 1/12

Tap A and Tap B’s one hour work = 1/8 – 1/12 = (3 – 2) / 24  = 1/24

Thus, the tank will be full after 24 hours.

Example 10. 8 Taps can fill a reservoir in 90 minutes. In how much time 12 taps can fill up the same reservoir if all the taps have equal capacity. 

Sol. 8 taps fill the reservoir in 90 minutes

One taps fill the reservoir in 90 x 8 minutes.

12 taps fill the reservoir in = (90 x 8) / 12 = 60 minutes.

Example 11. Pipe A can fill a tank in 3 hours. Pipe B can fill it in 4 hours. An outlet pipe C can empty the filled-in tank in 6 hours. If all the three pipes are kept open simultaneously, in how many hours will the tank be half full? 

Sol. Pipe A can fill the tank in 3 hours, so Pipe A will fill in one hour =  1/3

Pipe B can fill the tank in 4 hours, so pipe B will fill in one hour = 1/4

Pipe C can empty the tank in 6 hours, so in one-hour pipe C will empty = 1/6

Pipe A + Pipe B + Pipe C = 1/3 + 1/4 - 1/6 =

So three pipes can fill the tank in 12/5 hours.

Thus, ½ tank will be filled in ½ x 12/5 = 6/5 hours.