General Aptitude Quick Revision: Important Concepts of Time & Work

By Neetesh Tiwari|Updated : March 23rd, 2023

Are you looking for some short and reliable notes during your CSIR-NET preparations? Then, you have come to a perfect place!

Candidates preparing for their CSIR NET exam might need to get some short study notes and strategies to apply while preparing for the key exam of their life. At this point, We at Gradeup come up with short notes on the Important Concepts of Time & Workwhich comes under the Quant section of the General Aptitude syllabus

Our experienced subject-matter experts have meticulously designed this set of short notes on the Important Concepts of Time & Work to give you the most standard set of study materials to focus on. In this cut-throat competitive world, students need to prepare themselves with the best study materials to help them learn for their future. So, here we are offering the best study notes that are reliable and can be used by students during their preparations for the upcoming CSIR-NET 2023 exam.


Important Concepts of Time & Work

This chapter explains the concepts and provides exercises at different levels of difficulty. All concepts have been explained using examples liberally so that students can understand the concepts well.

In this chapter we study:

  • Time and work problems
  • People working together
  • Machines producing goods
  • Pipes and cisterns problems
  1. Time and Work

Basic Concepts

(i) If A can do a piece of work in n days, then work done by him in 1 day = 1/n.

(ii) If A's 1 day's work = 1/n, then A can finish the work whole work in n days.

Illustration 1. Shikha can weave a sweater in 8 days.

So, her 1 day's work = 1/8 sweater.

(iii) If A is twice as good a workman as B, then:

The ratio of work done by A and B = 2:1.

The ratio of time taken by A and B = 1:2.


(iv) If the number of men to do a certain work is changed in the ratio m:n, then the ratio of time taken to finish the work, changes in the ratio n:m.

1/ time1 + 1 / time2 = 1 / total time

The basic formula for solving work problems is 1/r + 1/s = 1/h where r and s are the number of hours it takes two people R and S respectively, to complete a job when working alone, and h is the number of hours it takes R and S to do the job when working together.

It may be noted that the relationship between time and work is inverse.

Students usually make the mistake of forgetting this.

Hence, if we are given the following sum:


Q. If 2 persons can do some work in 4 days, 4 persons will do the same work in:

  1. 1 day                         2. 2 days                      3. 4 days                     4. 8 days


The tendency of many students is to apply direct proportion and get the answer as 8 days.

This is wrong. More persons should do the work in less time, not more.

By inverse relationship, we should get half the time or 2 days.



Concept 1: Two people working together

Example 1: A does a job in 8 days and B does it in 12 days. How much time will it take to complete the job if A and B work together?

  1. 24/5           2. 5/24                        3. 16/3                        4. 3/16


Solution: One day’s work for A and B is 1/8 and 1/12 respectively.

If they work together it will take them: (1/8 +1/12) = (3 +2)/24 = 5/24

Hence the work will b completed in 24/5 days (inverse).

Example 2: A is twice as good as B. It takes 14 days if they work together. When can A alone finish the job?

  1. 17 days     2. 19 days                    3. 21 days                    4. None of these


Solution: Since A is twice as good as B, A will take half as many days as B.

Let A take x days and B take 2x days.

Then 1/x +1/2x = 1/14

On solving, we get: 3/2x = 1/14, or x = 21.

Hence A takes 21 days.

Concept 2: Using percentage

Example 3: X can do a piece of work in 48 days and with the help of Y, he finishes the job in 16 days. Y is what % faster than X?

  1. 50% 2. 75%                                     3. 100%                       4. 200%


Solution: If they both work together, then 1/48 + 1/Y = 1/16.

Hence 1/Y = 1/16 - 1/48 =2/48 .

Thus, while X takes 48 days, Y takes 24 days, which is half the time taken by X, and hence 100% faster.

Concept 3: More than 2 people working together

Example 4: A and B working together takes 12 days to do a job. B and C take 15 days while C and A take 18 days to complete the same job. In how many days can C alone complete it?

  1. 360/13     2. 360/5                      3. 360/17                    4. 360/7


Solution: Adding up the information together, we get: 2(A + B + C) = 1/12 + 1/15 + 1/18 = 37/180. Hence A + B + C = 37/360

C = (A + B + C) – (A + B) = (37/360 – 1/12) = 7/360

Hence C takes = 360/7 days.


Illustration 3: A can do a piece of work in 12 days and B in 10 days but with the help of C they finished the work in 4 days. C alone can do the work in how many days?

Solution. Here all three complete the work in 4 hours:

Hence C can do this work in 15 days.


Illustration 4: A and B can do a piece of work in 18 days. B and C in 24 days. A and C can do this work in 36 days. At what time can they do it all working together?

Solution. A and B’s one day’s work = 1/18.

B and C’s one day’s work = 1/24.

C and A’s one day’s work = 1/36.

If we add all this it will give us the work of 2A, 2B, and 2C in 1 day i.e.

A, B, and C’s one day’s work

They can complete the work in 16 days.

Concept 4: When people join/leave after some days

Example 5: A does a job in 40 days. B can do the same job in 50 days. A started the job and after 5 days, B joined. After 10 days of the start, C joined and the job was over in 20 days. How much time will C alone take to complete the job?

  1. 40 days 2. 50 days                    3. 45 days                    4. 55 days


Solution: A starts the job, B works for 5 days less and C works for 10 days less.

Hence A works for 20 days, B for 15 days, and C for 10 days.

We can thus get the equation: 20/40 + 15/50 + 10/C = 1, where 1/C is C’s one day’s work.

Solving this, we get C = 50 days.

Illustration 2: A can do a piece of work in 30 days, which B can do in 20 days. Both started the work but A left 5 days before the completion of the work. It took how many days to complete the work?


Solution. Left the job 5 days before the completion means for the last 5 days only B worked. First calculate B’s five days work, which he did alone.

In 5 days B will do 5 ´ 1/20 = ¼ th of the work. The remaining work 1- ¼ = ¾, which A and B have done together.

A and B can do 1/20 + 1/30 work in 1 day.

Their one-day work is.

They can finish the work in 12 days. Three-fourths of the work they would have done in 12 ´ ¾ = 9 days.

Total days = 5 + 9 = 14.

Concept 5: Men, women, and children working together

Example 6: If 4 men or 7 boys do work in 29 days, 12 men and 8 boys will do the same work in:

  1. 7 days     2. 8 days                      3. 11 days                    4. 12 days


Solution: We are given that 4m = 7b; hence we can say that 12m = 21b.

Now the problem becomes: if 7b can do the work in 29 days, then how many days will (12m + 8b) take?

Now (12m + 8b) = (21b + 8b) = 29b.

If 7b do it in 29 days, one boy does it in 29 ´ 7 days.

Hence 29 b will do it in (29 ´ 7)/29 = 7 days.

Concept 6: Machines producing goods

Illustration 1: If machine X can produce 1,000 bolts in 4 hours and machine Y can produce 1,000 bolts in 5 hours, in how many hours can machines X and Y, working together at these constant rates, produce 1,000 bolts?


Solution. Both machines will take:     1/4 + 1/5 = 1/h            Þ 9/20 = 1/h  

Working together, machines X and Y can produce 1,000 bolts in 20/9 or 2 2/9 hours.


II. Pipes and Cisterns

This is similar to Time and Work problems. The only difference is that here rates of filling or emptying a tank are given whereas in the previous section rates of work of persons are given. The methodology remains the same.

Illustration 5: Two pipes A and B fill a tank in 20 minutes and 40 minutes respectively. A pipe C at the bottom can empty the tank in 60 minutes. If all three pipes were open simultaneously, how long does it take to fill the empty tank?


Solution. Pipe A fills 1/20th of the tank in a minute, Pipe B fills 1/40th of the tank in a minute, Pipe C drains 1/60th of the tank in a minute.

Therefore, if all three are open the net effect = of the tank will be filled in a minute

i.e. th of the tank will be filled in a minute.

Therefore, the tank will be filled in minutes

Illustration 6: Two pipes A and B can fill a cistern in 20 and 24 minutes respectively. Both pipes being opened, find when the first pipe must be turned off, so that the cistern may be filled in 12 minutes.


Solution. Since the cistern is to be filled in 12 minutes, the Second pipe can fill only 12/24 = ½ of the cistern in total time. This means the other half must be filled by the first pipe. The first pipe can fill the whole tank in 20 minutes, so half of the tank can fill in half of the 20 minutes i.e. 10 minutes. Now the first pipe is opened from the beginning, it should be turned off after 10 minutes.

Illustration 7: Pipes A and B fill a tank in 30 minutes and 15 minutes. Pipe C drains 12 liters of water in a minute. If all of them are kept open when the tank is full, the tank empties in 30 minutes. How much water can the tank hold?

Solution. Pipe A fills 1/30th of the tank and pipe B fills 1/15th of the tank in 1 minute.

Pipe C drains 12 liters in a minute in 30 minutes it drains 12 ´ 30 = 360 liters.

So, if Q is the total capacity, then:

Full tank + input in 30 min = output in 30 min        Q + ( + ) ´ 30 = 360. Q = 90 liters.


Example 7: If two taps can fill a tank in 40 & 30 min respectively and a third pipe can empty it in 20 minutes. If the three are opened together the tank will fill in:

  1. 60 min     2. 120 min                  3. 150 min.                  4. 45 min.


Solution: As explained, the basic concepts are the same as time and work.

In this case, all three taps together result in 1/40 + 1/30 – 1/20 = 1/120

Hence it would take 120 min if all three pipes were turned on together.

Illustration 8: A cistern is filled in 9 hours and it takes 10 hours when there is a leak in its bottom. If the cistern is full, in what time shall the leak empty it?

Solution. A cistern filled in 1 hour by the filling pipe = 1/9.

A cistern filled by the leak and the filling pipe in 1 hour = 1/10.

Cistern emptied by the leak in one hour = 1/9 -1/10 = 1/90.

Hence the leak can empty the tank in 90 hours.

Illustration 9: Three pipes A, B, and C can fill a cistern in 6 hours. After working at it for 2 hours, C is closed and A and B can fill it in 7 hours more. How many hours will C alone take to fill the cistern?

Solution. In 2 hours A, B, and C would have filled  2 ´ 1/6 = 1/3 of the cistern.

The remaining part is 1 - 1/3 = 2/3, which A and B have filled in 7 hours.

Þ A and B can fill the whole cistern in 7 ´  3/2 = 21/2 hours.

Now A and B can fill this cistern in 21/2 hours.                     

A B and C can fill the tank in 6 hours.

C can fill the cistern alone in     Hence the answer is 14 hours.


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