Time and Work: Definition, Formula, Questions, Examples

Time and work problems are a fundamental part of quantitative aptitude tests. These problems involve calculating the amount of work done based on the time taken or determining the time required to complete a task given the rate of work. Key concepts in these problems include efficiency, rate of work, time and work formula, and the relationship between the number of workers and the time taken. Additionally, for exam preparation, resources like 'Time and work questions for bank exams', ‘Time and work questions for SSC Exam’, 'Time and work MCQ', ‘Time and work questions class 8’, and 'Time and work mock test' are essential. These resources often provide 'time and work questions with solutions pdf' to aid in understanding and practising the concepts.

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.

Important Time and Work Formula

• Work done = Time taken × Rate of work

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

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

• If a work is done in $n$ days, then the work done in one day = $\frac{1}{n}$

• Total work done = Number of days × Efficiency

• Efficiency and Time taken are inversely proportional to each other.

• 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 $n:m$.

• 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 the relation between them will be:

$\frac{M_1×D_1×T_1}{W_1}=\frac{M_2×D_2×T_2}{W_2}$

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.

1. 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 = $\frac{1}{5}$.

2. 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.

3. 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.

4. 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.

5. 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.

6. 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 work in Pipe and Cistern

Pipe and cistern problems are a specialized subset of time and work problems that involve calculating the time required to fill or empty a tank using one or more pipes. These problems typically consider the rates at which water flows into or out of the cistern and require an understanding of how different flow rates combine when multiple pipes are working together or alternately. The key concepts include the rate of inflow and outflow, the capacity of the cistern, and the combined effect of multiple pipes.

Concept of Time and Work in Work and Wages

Work and wage problems are a significant subset of time and work problems. These problems involve determining the wages earned based on the amount of work done and the time taken to complete a task. The key concepts include understanding the relationship between the total work, the time taken, and the wages paid. The core idea is to establish a fair compensation system where wages are proportional to the work done and the efficiency of the workers.

Tips and Tricks

• Always convert time into the same units before solving.

• Combine efficiencies for multiple workers.

• Use ratios to simplify problems.

• Double-check the inverse relationships.

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:

Given: 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:

Given: 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:

Given,

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:

Given,

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:

Given: 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:

Given: 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:

Given:

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:

Given: 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).