Time and Work: Definition, Formula, Questions, Examples

Concept of 1-day Work?

In time and work problems, the concept of "1 day work" is essential. It represents the fraction of work completed by an individual or a group in one day. To find the efficiency of a person, you can use the formula: If we consider total work as 1 unit.

Efficiency = $\frac{1}{\text{Time taken}}$

For example, if a man completes a work in 10 days, then their efficiency is $\frac{1}{10}$.

This simply means that the man can do $\frac{1}{10}$th of the work in 1 day.

Basic Concepts and Definitions in Time and Work

Basic Concepts and Definitions in Time and Work form the foundation for solving related aptitude problems. In quantitative aptitude, Work refers to the task or job to be completed, Time is the duration taken to finish it, and Efficiency is the rate at which work is done. Understanding these three terms is essential before moving to formulas and problem-solving techniques.

Work, Time, and Efficiency

In quantitative aptitude, Work refers to the task or job that needs to be completed. Time is the number of hours, days, or months taken to finish the work. Efficiency is the rate at which a person, team, or machine can complete a given task. The higher the efficiency, the less time it takes to finish the work. In Time and Work aptitude problems, efficiency often plays a key role in determining how quickly a job can be completed, especially in competitive exams like SSC CGL, IBPS PO, SBI Clerk, and CAT.

Understanding Work Rate and Combined Work

The work rate is a measure of how much work is completed in a given amount of time. For example, if a person can finish a job in 5 days, their daily work rate is 1/5 of the work per day. In combined work problems, the rates of two or more workers are added together to find how long they will take to finish the work if they work simultaneously. This concept is frequently tested in SSC, Banking, and Railway exams through numerical problems that require quick mental calculation and formula application.

Concept of Man-Days and Work Units

Man-days represent the total amount of work done by one person in one day. For example, if 3 workers complete a job in 4 days, the work requires 12 man-days in total. Similarly, work units are used to represent the total quantity of work, often breaking it down into smaller, measurable parts. This method is helpful for solving Time and Work aptitude questions, especially when comparing the productivity of different workers or machines. These ideas are crucial for handling advanced problems in government exams, MBA entrance tests, and campus placements.

Important Time and Work Formula

In quantitative aptitude, mastering the fundamental formulas of Time and Work is essential for solving problems quickly and accurately, especially in exams like SSC CGL, IBPS PO, SBI Clerk, RRB NTPC, and CAT. These formulas help you calculate work, time, efficiency, and combined work rates with ease:

1. Work Formula:
   $Work = Time \times Efficiency$
   This is the basic relationship that connects the total work with the time taken and the rate at which the work is done.

2. Time Formula:
   $Time = \frac{Work}{Efficiency}$
   Used to calculate how long it will take to finish a task when efficiency and total work are known.

3. Efficiency Formula:
   $Efficiency = \frac{Work}{Time}$
   Helps determine how productive a person or machine is over a specific period.

4. Combined Work Formula:
   If A can do a job in $x$ days and B can do it in $y$ days, working together they can complete it in:
   $Time = \frac{xy}{x + y}$

5. Rate of Work Formula:

   • Work done = Time taken × Rate of work

   • Rate of work = $\frac{1}{\text{Time taken}}$

   • Time taken = $\frac{1}{\text{Rate of work}}$

6. Work Done in One Day:
   If a work is done in $n$ days, then the work done in one day = $\frac{1}{n}$

7. Total Work Formula:
   Total work done = Number of days × Efficiency

8. Efficiency-Time Relationship:
   Efficiency and Time taken are inversely proportional to each other. If one increases, the other decreases in the same ratio.

9. Men-to-Time Ratio:
   If $m:n$ is the ratio of the number of men to complete a piece of work, then the ratio of the time taken by them is $n:m$.

10. General Work Relation Formula:
    If $M_1$ number of people can do $W_1$ work in $D_1$ days, working $T_1$ hours each day, and $M_2$ number of people can do $W_2$ work in $D_2$ days, working $T_2$ hours each day, then:
    $\frac{M_1 \times D_1 \times T_1}{W_1} = \frac{M_2 \times D_2 \times T_2}{W_2}$

Types of Problems

There are various types of problems in mathematics related to ‘Time and Work’ concepts, let us discuss them one by one with proper examples.

How do you determine a person's efficiency?

To find the efficiency of a person use the formula:

Efficiency = $\frac{1}{\text{Time taken}}$

For example, if a boy completes a piece of work in 5 days,

Then his efficiency = $\frac{1}{5}$.

How to find the time taken by an individual to do a piece of work

To find the time taken by an individual to do a piece of work, use the formula:

Time taken = $\frac{1}{\text{Efficiency}}$

For example, if a person has efficiency of $\frac{1}{3}$, then time taken by him to complete a piece of work = $\frac{1}{\frac{1}{3}}$ = 3 days.

Two or more persons working together

When two or more people work together, their combined efficiency is the sum of their individual efficiencies.

Example: If Ripan alone can do a piece of work in 10 days and Kushal alone can do it in 15 days, then find the required time for both of them to complete the work together.

Solution: Here, efficiency of Ripan = $\frac{1}{10}$

And efficiency of Kushal = $\frac{1}{15}$

So, their combined efficiency = $\frac{1}{10}+\frac{1}{15}$ = $\frac{5}{30}$

Therefore, the required time taken by them

= $\frac{1}{\frac{5}{30}}$ = $\frac{30}{5}$ = 6 days.

Work on alternate days

When individuals work on alternate days, calculate the total work done by each person separately and add them up. In this, first, find the 2 days’ work and then find the total time.

Example: If A can do a piece of work in 4 days and B can do the same work in 6 days, then find the time taken by them if they worked on alternate days starting with A.

Solution: Work done by A in 1 day = $\frac{1}{4}$

Work done by B in 1 day = $\frac{1}{6}$

Work done by them together in 2 days = $\frac{1}{4}+\frac{1}{6}$ = $\frac{5}{12}$

So, work done by them together in 4 days = $\frac{5}{12}×2$ = $\frac{10}{12}$

The remaining work = $(1-\frac{10}{12})$ = $\frac{2}{12}$ = $\frac{1}{6}$

This remaining $\frac{1}{6}$ part will be done by A, for that the required time

= $\frac{\frac{1}{6}}{\frac{1}{4}}$ = $\frac{4}{6}$ = $\frac{2}{3}$ days.

Hence, the total required time = $4 + \frac{2}{3}$ = $4\frac{2}{3}$ days.

A person joins or leaves in the middle of the work

If a person joins or leaves in the middle of the work, then we have to calculate the work done by them separately for different periods and add them up.

Example: A can do a piece of work in 5 days and B can do the same work in 10 days.

They start to work together and after 2 days A leaves the work and B completes it alone. Find the total time it took to complete the work.

Solution: Efficiency of A = $\frac{1}{5}$ and efficiency of B = $\frac{1}{10}$

Now, the combined efficiency = $\frac{1}{5}+\frac{1}{10}$ = $\frac{3}{10}$

So, work done by A and B together in 1 day = $\frac{3}{10}$

After 2 days the work done = $\frac{3}{10}×2$ = $\frac{6}{10}$

The remaining work = $(1 - \frac{6}{10})$ = $\frac{4}{10}$ which is done by B alone.

The required time for B = $\frac{\frac{4}{10}}{\frac{1}{10}}$ = 4 days

Hence, the total taken to complete the work = (2 + 4) = 6 days.

Work done by a group of individuals in a certain time duration

The problems with work done by a group of individuals in a certain period can be solved by using the following formula: $\frac{M_1×D_1×T_1}{W_1}=\frac{M_2×D_2×T_2}{W_2}$

Where $M_1$ number of people can do $W_1$ work, in $D_1$ days, working $T_1$ hours each day and $M_2$ number of people can do $W_2$ work, in $D_2$ days, working $T_2$ hours each day.

Example: Working 6 hours a day, Raman can complete a piece of work in 24 days. How many hours a day should he work to finish the work in 18 days?

Solution: We know,

$\frac{M_{1}D_{1}H_{1}}{W_{1}} = \frac{M_{2}D_{2}H_{2}}{W_{2}}$, where $M$ = number of men , $D$ = number of days, $W$ = amount of work done, and $H$ = hours per day.

Here,

$M_{1} = M_{2} = 1$

$W_{1} = W_{2} = W$

$D_{1} = 24$

$D_{2} = 18$

$H_{1} = 6$

Now, applying the values in the above equation:

$1 \times 24 \times 6 = 1 \times 18 \times H_{2}$

⇒ $H_{2} = 8$

Hence, the correct answer is 8 hours.

Concept of Time and Work in Pipe and Cistern Problems

Pipe and cistern problems are a special type of time and work problem, but instead of workers completing a task, we deal with pipes filling or emptying a tank.

• Inflow pipes fill the cistern.

• Outflow pipes empty the cistern.

• The capacity of the cistern is considered as 1 unit of work (just like total work in standard time and work problems).

• The rate of work is the fraction of the tank filled or emptied in one hour (or minute).

• When both filling and emptying pipes are open together, their rates are combined by adding inflow rates and subtracting outflow rates.

Formulas

1. Rate of filling (per hour) = $\frac{1}{\text{Time taken to fill alone}}$

2. Rate of emptying (per hour) = $\frac{1}{\text{Time taken to empty alone}}$

3. Net rate when both are open = (Sum of inflow rates) − (Sum of outflow rates)

4. Total time to fill/empty = $\frac{\text{Capacity of tank}}{\text{Net rate}}$

Example

A pipe A can fill a tank in 6 hours. Another pipe B can empty it in 9 hours. If both are opened together, how long will it take to fill the tank?

Solution:

• Rate of filling by A = $\frac{1}{6}$ tank/hour

• Rate of emptying by B = $\frac{1}{9}$ tank/hour

• Net rate = $\frac{1}{6} - \frac{1}{9} = \frac{3 - 2}{18} = \frac{1}{18}$ tank/hour

Time required $= \frac{1}{1/18} = 18$ hours

Final Answer: The tank will be filled in 18 hours.

Concept of Time and Work in Work and Wages Problems

Work and wage problems are another important application of time and work concepts. These problems link the amount of work done to the wages earned, ensuring payment is proportional to work done.

• Wages are directly proportional to the amount of work done.

• If two or more people work together, their total wage is split based on their contribution (efficiency $\times$ time worked).

• If all workers complete the same task together, the ratio of their wages = the ratio of their efficiencies.

Formulas

1. Wage Formula:
   $\text{Wages earned} = \text{Total wage} \times \frac{\text{Work done by person}}{\text{Total work done}}$

2. Efficiency from Time Taken:
   Efficiency = $\frac{1}{\text{Time taken to complete work alone}}$

3. Wage Ratio:
   If time is constant,
   $\text{Wage ratio} = \text{Efficiency ratio}$

Example

A can complete a job in 10 days, and B can do it in 15 days. They work together and earn Rs.3,000. How much should each get?

Solution:

• Efficiency of A = $\frac{1}{10}$ work/day

• Efficiency of B = $\frac{1}{15}$ work/day

• Efficiency ratio = $\frac{1/10}{1/15} = \frac{15}{10} = 3:2$

Total wage = Rs.3,000
A’s share = $\frac{3}{3+2} \times 3000 = 1800$
B’s share = Rs.3,000 − Rs.1,800 = Rs.1,200

Final Answer: A gets Rs.1,800, B gets Rs.1,200.

Tips and Tricks for Time and Work Problems

1. Always use the same unit for time (hours or days).

2. Combine efficiencies when workers or pipes work together.

3. Use ratios to split wages or compare times.

4. Remember: Efficiency and Time are inversely proportional.

Practice Questions/Solved Examples on Time and Work

Q 1. A and B can do a piece of work in 72 days. B and C can do it in 120 days. A and C can do it in 90 days. In how many days can all three together do the work?

1. 80

2. 100

3. 60

4. 150

Hint: If a man can do the work in $n$ days, then one day's work of the man $=\frac{1}{n}$.

Solution: A and B can do a piece of work in 72 days.

B and C can do it in 120 days.

A and C can do it in 90 days.

(A + B)'s 1 day's work = $\frac{1}{72}$

(B + C)'s 1 day's work = $\frac{1}{120}$

(A + C)'s 1 day's work = $\frac{1}{90}$

Adding the above three equations we get

2(A + B + C)'s 1 day's work = $\frac{1}{72}+\frac{1}{120}+\frac{1}{90} = \frac{5+3+4}{360} = \frac{12}{360} = \frac{1}{30}$

⇒ (A + B + C)'s 1 day's work = $\frac{1}{60}$

$\therefore$ (A + B + C) together complete the work in 60 days.

Hence, the correct answer is option (3).

Q 2. A and B can do a piece of work in 8 days, B and C can do it in 24 days while C and A can do it in $8\frac{4}{7}$ days. In how many days can C do it alone?

1. 60

2. 40

3. 30

4. 10

Hint: If a man can do the work in $n$ days, then one day's work of the man $=\frac{1}{n}$.

Solution: A and B can do a piece of work in 8 days. B and C can do it in 24 days, and A and C can do it in $\frac{60}{7}$ days.

(A+B)'s 1 day's work = $\frac{1}{8}$

(B+C)'s 1 day's work = $\frac{1}{24}$

(A+C)'s 1 day's work = $\frac{7}{60}$

Adding the above three equations we get,

2(A+B+C)'s 1 day's work = $\frac{1}{8}+\frac{1}{24}+\frac{7}{60}=\frac{15+5+14}{120} = \frac{34}{120} = \frac{17}{60}$

⇒ (A+B+C)'s 1 day's work is $\frac{17}{120}$.

C's 1 days's work $=\frac{17}{120} - \frac{1}{8} = \frac{17–15}{120} = \frac{2}{120} = \frac{1}{60}$

So, C can alone complete the work in 60 days.

Hence, the correct answer is option (1).

Q 3. A can do a piece of work in 30 days while B can do it in 40 days. How many days can A and B when working together do it?

1. $42\frac{3}{4}$ days

2. $27\frac{1}{7}$ days

3. $17\frac{1}{7}$ days

4. $70$ days

Hint: Find the work done by A and B in a day and then solve the question.

Solution: A can do a piece of work in 30 days while B can do it in 40 days.

(A + B)'s work in 1 day $=\frac{1}{30} + \frac{1}{40}=\frac{4+3}{120} = \frac{7}{120}$

$\therefore$ (A + B) can complete the work in $=\frac{120}{7} = 17\frac{1}{7}$ days

Hence, the correct answer is option (3).

Q 4. A contractor undertook to finish a work in 92 days and employed 110 men. After 48 days, he found that he had already done $\frac{3}{5}$ part of the work, the number of men he can withdraw so that the work may still be finished in time is:

1. 45

2. 40

3. 35

4. 30

Hint: Use this formula:

$\frac{M_{1}\times D_{1}}{W_{1}}=\frac{M_{2}\times D_{2}}{W_{2}}$

Where $M_1$ and $M_2$ are men, $D_1$ and $D_2$ are days, and $W_1$ and $W_2$ are work done.

Solution: A contractor undertook to finish a work in 92 days and employed 110 men. After 48 days, he found that he had already done $\frac{3}{5}$ part of the work.

We know,

$\frac{M_{1}\times D_{1}}{W_{1}}=\frac{M_{2}\times D_{2}}{W_{2}}$

Where $M_1$ and $M_2$ are men, $D_1$ and $D_2$ are days and $W_1$ and $W_2$ are work done.

Let $n$ number of men be withdrawn.

So, $\frac{110 \times 48}{\frac{3}{5}} = \frac{(110-n) \times 44}{\frac{2}{5}}$

$⇒ 110 × 16 = (110-n) × 22$

$⇒160 = (110-n) × 2$

$⇒80 = 110-n$

$⇒n = 110 - 80$

$\therefore n = 30$

Hence, the correct answer is option (4).

Q 5. A contractor has the target of completing a work in 40 days. He employed 20 persons who completed $\frac{1}{4}$th of the work in 10 days and left. The number of persons he has to employ to finish the remaining part as per the target is:

1. 10

2. 20

3. 40

4. 30

Hint: Use the following formula:

$\frac{M_{1}\times D_{1}}{W_{1}}=\frac{M_{2}\times D_{2}}{W_{2}}$

Where $M_1$ and $M_2$ are men, $D_1$ and $D_2$ are days and $W_1$ and $W_2$ are work done.

Solution: 20 persons completed $\frac{1}{4}$th of the work in 10 days and left.

Now the contractor has 30 more days to complete the job.

We know,

$\frac{M_{1}\times D_{1}}{W_{1}}=\frac{M_{2}\times D_{2}}{W_{2}}$

Where $M_1$ and $M_2$ are men, $D_1$ and $D_2$ are days and $W_1$ and $W_2$ are work done

Let $n$ number of men should be added.

So, $\frac{20 \times 10}{\frac{1}{4}} = \frac{n \times 30}{\frac{3}{4}}$

$⇒20 \times 10 = n \times 10$

$\therefore n = 20$

Hence, the correct answer is option (2).

Q 6. A can do a piece of work in 5 days and B in 4 days. How long will they take to do the same work when working together?

1. $9$ days

2. $2\frac{2}{9}$ days

3. $4\frac{1}{3}$ days

4. $3\frac{2}{9}$ days

Hint: If a man can do the work in $n$ days, then one day's work of the man $=\frac{1}{n}$.

Solution: A can do a piece of work in 5 days and B in 4 days.

If a man can do the work in $n$ days, then one day's work of the man $=\frac{1}{n}$.

1 day work of A $=\frac{1}{5}$

1 day work of B $=\frac{1}{4}$

1 day work of (A + B) together $=\frac{1}{5} + \frac{1}{4}=\frac{4+5}{20} = \frac{9}{20}$

So, (A + B) together can complete the work in $=\frac{20}{9} = 2\frac{2}{9}$ days

Hence, the correct answer is option (2).

Q 7. A certain number of men completed a piece of work in 60 days. If there were 8 men more, the work could be finished in 10 days less. The number of men originally was:

1. 36

2. 40

3. 30

4. 32

Hint: Find the work done in a day in the case of $x$ men and $(x+8)$ men. Then use this formula:

$M_{1}D_{1}=M_{2}D_{2}$, where $M_1$ and $M_2$ are men and $D_1$ and $D_2$ are days.

Solution: Let the original no of men be $x$.

$x$ men can finish it in 60 days,

and $(x+8)$ men can finish in 50 days.

$M_{1}D_{1}=M_{2}D_{2}$, where $M_1$ and $M_2$ are men and $D_1$ and $D_2$ are days.

$⇒60x = 50(x+8)$

$⇒10x= 400$

$\therefore x =40$

So, the number of men originally was 40.

Hence, the correct answer is option (2).

Q 8. A, B, and C can complete a work in 10, 12, and 15 days respectively. They started the work together, but A left the work before five days of its completion. B also left the job two days after A left. In how many days was the work completed?

1. 4

2. 5

3. 7

4. 8

Hint: Find the one-day work of A, B, and C, then assume the number of days required to complete the whole work is $x$.

Solution: A left the work 5 days before completion.

B left the work 3 days before completion.

We know that Total work = Number of days × Efficiency of work.

Let the number of days required to complete the whole work = $x$ and the whole work is 1.

1 day work of A = $\frac{1}{10}$

1 day work of B = $\frac{1}{12}$

1 day work of C = $\frac{1}{15}$

According to the question,

$\frac{x-5}{10} + \frac{x-3}{12} + \frac{x}{15} = 1$

$⇒6(x-5)+5(x-3)+4x = 60$

$⇒6x-30+5x-15+4x =60$

$⇒15x-45 = 60$

$⇒15x=105$

$\therefore x=7$

Hence, the correct answer is option (3).

Q 9. If A and B together can do a piece of work in 20 days, and A alone can do the same work in 30 days, in how many days can B alone complete the same work?

1. 45

2. 40

3. 60

4. 50

Hint: If a man can do the work in $n$ days, then work done in one day $=\frac{1}{n}$.

Solution: A and B together can do a piece of work in 20 days.

Part of the work A and B together can do in 1 day $=\frac{1}{20}$

A alone can do the work in 30 days.

Part of the work A can do in 1 day $=\frac{1}{30}$

$\therefore$ Part of the work B alone can do in a day $=\frac{1}{20} – \frac{1}{30} = \frac{1}{60}$

So, B will complete the work in 60 days.

Hence, the correct answer is option (3).

Q 10. A worker completes $\frac{3}{5}$th of a task in 12 days. In how many days will he complete $\frac{3}{4}$th of the task?

1. 18

2. 20

3. 16

4. 15

Hint: The entire task can be completed in $\frac{12}{\frac{3}{5}}$ = 20 days.

Solution: A worker completes $\frac{3}{5}$th of a task in 12 days.

So, he completes the whole task in = $\frac{12}{\frac{3}{5}}$ = 20 days

To complete $\frac{3}{4}$th of the task, he will take $20\times \frac{3}{4}$ = 15 days

Hence, the correct answer is option (4).

Tips to Improve Speed and Accuracy in Time and Work Aptitude

If you want to score high in Time and Work aptitude questions during competitive exams, you need two things: solid concepts and the ability to solve questions quickly without mistakes. This balance comes from targeted practice, smart use of formulas, and regular revision. Below are proven strategies and tips for solving time and work aptitude problems that will help you boost both speed and accuracy.

Shortcut Practice Strategy

To solve Time and Work questions faster, you must be comfortable with short-cut solving techniques. Long, step-by-step methods are useful for learning, but in an exam, they can cost you precious time. Instead:

• Memorize key formulas like $\text{Work} = \text{Time} \times \text{Efficiency}, \quad \text{Time} = \frac{\text{Work}}{\text{Efficiency}}$ Work and the combined work formula $\frac{xy}{x+y}$ for two workers.

• Use the LCM method to convert time taken into total work units and assign efficiencies directly.

• Apply the fractional work approach where total work is assumed as 1 and each person’s work rate is calculated in fractions per day/hour.

• Remember that time and efficiency are inversely proportional doubling the efficiency halves the time needed.

Example: If A can finish work in 10 days and B in 15 days, total work = LCM(10, 15) = 30 units.
A’s rate = 3 units/day, B’s rate = 2 units/day. Together = 5 units/day

Work finished in 6 days.

Mock Test and Timed Practice

Even if you know all the formulas, exam pressure can slow you down. That’s why timed practice is crucial:

• Solve full-length mock tests at least twice a week to simulate real exam conditions.

• Set strict time limits for each question and avoid overthinking easy problems.

• Practice mixed problem sets so you can quickly switch between standard Time and Work, Pipes & Cisterns, and Work & Wages questions.

• Track your mistakes after each mock and revise the specific type of question that caused errors.

Topic-Wise Revision Plan

A structured revision plan ensures that your learning stays fresh until the exam:

1. Day 1-2: Revise basic concepts, definitions, and efficiency formulas.

2. Day 3-4: Focus on combined work, alternate work, and group efficiency problems.

3. Day 5: Practice special types: Pipes & Cisterns and Work & Wages.

4. Day 6: Attempt mixed question sets under a timer.

5. Day 7: Review mistakes, re-attempt tricky problems, and note shortcuts in a quick-reference sheet.