Time and Work

Importance of time and work problems

Arithmetical questions based on time and work are included in a student’s curriculum well ahead of his preparation for many competitive examinations. In course of time, there are certain variations in the question pattern. Although short answer-type questions are common from this section of quantitative aptitude, students can expect an occasional variation of descriptive problems involving time and work.

Usually, we notice cutthroat competition in the case of Government and other competitive examinations. In the preliminary exams, aspirants are eliminated in bulk. To secure the edge over other competitors an examinee must devote his precious time to thoroughly learn the various concepts of quantitative aptitude. Time and work problems are easy to solve after repetitive practice over time.

Time and Work – brief description

Time and work problems inform us about the total time needed by a person to finish a job. They may ask us to find the requisite time to complete the work with greater efficiency. At times time and work questions deal with a group of people instead of a single person. 

Time and Work – Probable Question Patterns 

A set pattern of questions is asked under this section. Let us understand the different types. 

• The examiner is asked to evaluate the person’s or group’s efficiency in performing a specific work. 
• Time is required by a man to complete a task. 
• Time is required by a crew of workers to complete a work. 
• Work performed within a specified time limit by an individual. 
• Work performed within a specified time limit by a group. 

The time and work sums typically interrogate any one or more than one of the above questions. 

Time and work formula – discussion 

One can relate to a specific formula after reading the problem once if they have memorized them well. Thus, we need to remember the following formulas to solve the numerical problems in a simplified fashion. 

List of time and work formulas 

• Work performed = rate of work x time required for completion of the work

• Work rate = 1 / time required for completion of the work

• Time required to do a task = 1 / work rate

The efficiency of a person or group gets reduced if he/ they need(s) more time to finish a task. Therefore, time and efficiency are in inverse proportion. 

Important note: A man will complete 1/ W part of a work in a single day if he takes W days to finish the whole task. 

• Total work produced = efficiency x Number of days

Let the ratio of the people required to complete a job be M: N. Then they will require N:M time to do the whole work. 

• If M number of men can do a task T1 in D1 days while working H1 hours each day, then two separate groups can complete another task T2 in D2 days if they work for H2 hours on a single day. 

We can draw an equation as M1 x H1 x D1/ T1 = M2 x H2 x D2/ T2

Time and work problems 

Problems involving time and work must be solved before the examination. Otherwise without any practical execution memorizing the formulas will be of no use. Repeated practice will reduce the time required to find the correct solution. 

Example 1

An industrialist hires three workers in his new factory at Muzaffarpur. The workers spent 20, 30, and 60 days to finish their share of the assignment. How many days will be sufficient for the first worker to do the entire task if he is assisted by the second and third coworkers on each third day? 

In a single shift, the first worker completes 1/20th part of the job. 

Similarly, the second and third workers will do 1/30th and 1/60th part of the job in one whole day. 

Part of the work done by the first worker in the first two days is 2 x (1/20) = 1/10

The portion of the task completed in three days = work completed by the first worker in 2 days + total work contribution from each worker on the 3rd day. 

 ∴ portion of task done in 3 days = (1/30 + 1/60 + 1/20) + 1/10

= (2 + 1 + 3) /60 + 1/ 10

= 6/60 + 1/10

= 1/10 + 1/10

= 2/10

= 1/5

∴ 1/5 amount of work is completed in 3 days. 

So, the whole work will be done in 3 ÷ 1/5 days or, 3 x 5 = 15 days. 

Example 2

I completed my homework in 20 days. How many fewer days will my elder sister require if she has 25% better efficiency? 

Let the number of days taken by my elder sister be D. 

The time ratio incurred by me and my elder sister is 100: 125 or 4 : 5

∴ 4: 5: D: 20

or, D = (20 x 4) ÷ 5

⇒ D = 16

Therefore, my elder sister will need (20 – 16) = 4 less days to finish the homework. 

Conclusion 

Time and work problems are easy to solve and carry a significant weightage in examination papers. The rules are simple and clear-cut. One major rule is the inverse proportionality of efficiency and time.