Time and Work MCQ Quiz - Objective Question with Answer for Time and Work - Download Free PDF

Last updated on Mar 19, 2025

Time and Work Questions is an essential part of Competitive Exams such as Bank exams, Insurance exams, SSC and Railway Exams, etc. To help you master this section, Testbook has brought a set off an intermediate level of Time and Work Questions. Time and Work MCQ Quiz come with their solutions and explanations followed by some tips and tricks. To begin your preparation for entrance exams, read this article and solve these questions.

Latest Time and Work MCQ Objective Questions

Time and Work Question 1:

"Rahul" and "Sohan" have respective completion times of 15 and 25 days for a task. They began working together on the project, but after a few days, "Rahul" left and "Priya," who can finish the entire task on her own in 12 days, joined. Three days before the job was finished, "Priya" left. Determine the amount of work completed by "Sohan" if "Rahul" and "Priya" worked for a similar number of days.

  1. 7/5
  2. 9/25
  3. 2/5
  4. 3/25
  5. None of these

Answer (Detailed Solution Below)

Option 2 : 9/25

Time and Work Question 1 Detailed Solution

Rahul's work rate: 1/15 task per day

Sohan's work rate: 1/25 task per day

Priya's work rate: 1/12 task per day

Let:

x = Number of days Rahul and Priya worked.

Total time taken to complete the task = T days.

Rahul: Worked for x days.

Work done = x/15.

Sohan: Worked for the entire T days.

Work done = T/25

Priya: Worked for x days but 3 days before the task was completed.

Work done = (x - 3)/12

The total work done is equal to 1 the entire task. So:

x/15 + T/25 + (x - 3)/12 = 1

Since Rahul and Priya worked for the same number of days x, and Priya 3 days before the task was completed:

T = x + (x - 3) = 2x - 3

Substitute T = (2x - 3) into the Work Equation

x/15 + (2x - 3)/25 + (x - 3)/12 = 1

Multiply through by 300 LCM of 15, 25, and 12 to eliminate denominators:

20x + 12(2x - 3) + 25(x - 3) = 300

20x + 24x - 36 + 25x - 75 = 300

69x - 111 = 300

69x = 411

x = 411 / 69 = 6

T = 2x - 3 = 2(6) - 3 = 9

Sohan worked for the entire T = 9 days.

Work done by Sohan = 9/25

The amount of work completed by Sohan is 9/25.

Time and Work Question 2:

A person can walk a certain distance in 15 days when they rest 6 hours a day. How long will they take to walk three times the distance, 1.5 times as fast, and resting twice as long each day?

  1. 41 days
  2. 42 days
  3. 45 days
  4. 22(1/2) days
  5. 21 days

Answer (Detailed Solution Below)

Option 3 : 45 days

Time and Work Question 2 Detailed Solution

Original Distance: Let the distance be D.

Original Time: 15 days.

Resting Time: 6 hours per day.

Walking Time per Day: 24 - 6 = 18 hours.

Total walking time in 15 days = 15 × 18 = 270 hours.

Speed = D/270 distance per hour.

New Distance: 3D.

New Speed: 1.5 × D/270 = 1.5D/270 = D/180

New Resting Time: 2 × 6 = 12 hours per day.

New Walking Time per Day: 24 - 12 = 12 hours.

Time required to walk 3D at speed D/180:

Time = 3D / (D/180) = 540 hours.

Number of days required:

Days = 540 / 12 = 45 days.

The person will take 45 days to walk three times the distance, 1.5 times as fast, and resting twice as long each day.

Time and Work Question 3:

A and B together complete a work in 12 day. With the help of C, they complete the work in 6 days. Ratio of efficiency of A and B is 3:1. If C complete the work alone in [R + 8] days find in how many days A alone complete the work?

  1. R + 8
  2. R + 16
  3. R + 12 
  4. R + 10
  5. R + 14

Answer (Detailed Solution Below)

Option 3 : R + 12 

Time and Work Question 3 Detailed Solution

[A + B]'s one days work is 1/12

[A +B+C]' one days work is 1/6

so, C's one days work is 1/6 - 1/12 = 1/12

so, C alone complete the work alone in 12 days.

12 = R + 8

so, R = 4 

Let A and B alone complete the work in x and 3x days respectively.

so, 1/x + 1/3x = 1/12

or, 4/3x = 1/12

or, x = 16

so, A alone complete the work in 16 days.

16 = R + 12

Time and Work Question 4:

A specific task always begins at 8:00 a.m. on Mondays and can only be carried out between 8:00 a.m. to 12:00 noon and from 1:00 p.m. to 5:00 p.m. on any given day. When completed by A alone, the work finishes at 5:00 p.m. on the following day, i.e., Tuesday. If B alone works on it, the task is completed during the morning session on Tuesday. Given that today is a Monday and both A and B start working together, at what time can they likely finish the work?

  1. Monday, 1:30 pm
  2. Monday, 4:30 pm
  3. Tuesday, 9:15 am
  4. Monday, 3:00 pm

Answer (Detailed Solution Below)

Option 4 : Monday, 3:00 pm

Time and Work Question 4 Detailed Solution

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Time taken by A to complete the job: \( 16 \) hours

Time taken by B to complete the job: \( 8 \text{ to } 12 \) hours (Since it is mentioned that B completes the work in the morning session, it can be at \( 8:00 \) a.m., \( 12:00 \) noon, or any time in between.)

Maximum time taken by A and B together:

\( \frac{1}{16} + \frac{1}{12} = \frac{7}{48} \Rightarrow 6 \text{ hours } 51 \text{ minutes} \)

Minimum time taken by A and B together:

\( \frac{1}{16} + \frac{1}{8} = \frac{3}{16} \Rightarrow 5 \text{ hours } 20 \text{ minutes} \)

Thus, together, A and B can complete the work at any time between \( 2:20 \) p.m. and \( 3:51 \) p.m. on the same day.

Only option (4), satisfies the given time frame.

Time and Work Question 5:

10 pipes of the same diameter can fill a tank in 24 minutes. If 2 pipes go out of order, how long will the remaining pipes take to fill the tank?

  1. 30 min.
  2. 40 min.
  3. 45 min.
  4. 19 min.

Answer (Detailed Solution Below)

Option 1 : 30 min.

Time and Work Question 5 Detailed Solution

Given:

10 pipes can fill a tank in 24 minutes.

2 pipes go out of order.

Formula used:

Time taken is inversely proportional to the number of pipes.

Calculation:

Initially, 10 pipes take 24 minutes to fill the tank.

⇒ Total time with 10 pipes = 24 minutes

Let the time taken by the remaining 8 pipes be x minutes.

⇒ 10 pipes × 24 minutes = 8 pipes × x minutes

⇒ 10 × 24 = 8 × x

⇒ x = (10 × 24) / 8

⇒ x = 30 minutes

∴ The correct answer is option 1.

Top Time and Work MCQ Objective Questions

A cistern has two pipes one can fill it with water in 16 hours and other can empty it in 10 hours. In how many hours will the cistern be emptied if both the pipes are opened together when 1/5th of the cistern is already filled with water?

  1. 11.4 hours
  2. 3.66 hours
  3. 5.33 hours
  4. 8.33 hours
  5. None of these

Answer (Detailed Solution Below)

Option 3 : 5.33 hours

Time and Work Question 6 Detailed Solution

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Shortcut Trick

F2 Shraddha Vaibhav (Black Diag) 06.04.2021 D9

If both pipes are open, total efficiency = (A + B) = 5 + (-8) = -3 units

According to question,

Amount of water in the tank = (1/5) × 80 = 16 units

Time taken to empty the tank = work/efficiency = 16/((-3)) = 5.33 hours

Alternate Method

GIVEN :

Time by which pipe A can fill the tank = 16 hours

Time by which pipe B can empty the tank = 10 hours 

The cistern is (1/5)th full.

CONCEPT :

Total work = time × efficiency

CALCULATION :

Work Time Efficiency
A 16 80/16 =  5
B 10 80/10 = (-8)

total work (LCM)

80  


Negative efficiency indicates pipe B is emptying the tank.

If both pipes are open, total efficiency = (A + B) = 5 + (-8) = -3 units

From the total efficiency it is clear that when both are opened, the tank is being emptied. 

Amount of water in the tank = (1/5) × 80 = 16 units

The water level will not rise as the total action is emptying when both are opened together.

Time taken to empty the tank = work/efficiency = 16/((-3)) = 5.33 hours

∴ Time taken to empty the tank is 5.33 hours.

A and B together can do a piece of work in 50 days. If A is 40% less efficient than B, in how many days can A working alone complete 60% of the work?

  1. 70
  2. 110
  3. 80
  4. 105

Answer (Detailed Solution Below)

Option 3 : 80

Time and Work Question 7 Detailed Solution

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Given:

A and B together can do a piece of work in 50 days.

A is 40% less efficient than B

Concept used:

Total work = Efficiency of the workers × time taken by them

Calculation:

Let the efficiency of B be 5a

So, efficiency of A = 5a × 60%

⇒ 3a

So, total efficiency of them = 8a

Total work = 8a × 50

⇒ 400a

Now,

60% of the work = 400a × 60%

⇒ 240a

Now,

Required time = 240a/3a

⇒ 80 days

A can complete 60% of the work working alone in 80 days.

Shortcut Trick

​We know 40% = 2/5, Efficiency of B = 5 and A = 3

So, Total work = (5 + 3) × 50 = 400 units

So, 60% of the total work = 60% of 400 = 240 units

So A alone can do the work in 240/3 = 80 days

A can finish a work in 15 days, B can finish the same work in 25 days. They work together for 5 days. The rest of the work is finished by A and C in 4 days. Then C alone can finish the work in:

  1. 18 days
  2. 24 days
  3. 20 days
  4. 21 days

Answer (Detailed Solution Below)

Option 3 : 20 days

Time and Work Question 8 Detailed Solution

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Given:

A can finish in 15 days, B can finish it in 25 days.

They work together for 5 days.

Concept used:

Efficiency = (Total work / Total time taken)

Efficiency = work done in a single day 

Calculation:

Let total work be 75 units ( LCM of 15 and 25 is 75)

The efficiency of A

 75 /15 = 5 units

The efficiency of B  

 75 / 25 = 3 units

The efficiency of A+B,

⇒ (5 + 3) units = 8 units

In 5 days total work done is 8 × 5 = 40 units

Remaining work 75 - 40 = 35 units

In the last 4 days, A does 4 × 5 = 20 units

Remaining work 35 - 20 = 15 units done by C in 4 days

So C does 75 units in (75 / 15) × 4 = 20 days

∴ The correct option is 3

23 people could do a piece of work in 18 days. After 6 days 8 of the workers left. How many days from then will it take to complete the work? 

  1. 17.6
  2. 18.4
  3. 20.4
  4. 16.8

Answer (Detailed Solution Below)

Option 2 : 18.4

Time and Work Question 9 Detailed Solution

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Given:

23 people could do a piece of work in 18 days.

After 6 days 8 of the workers left. 

Concept used:

Total work = Men needed × Days needed to finish it entirely

Calculation:

Total work = 23 × 18 = 414 units

In 6 days, total work done = 23 × 6 = 138 units

Remaining work = (414 - 138) = 276 units

Time taken to complete the remaining work = 276 ÷ (23 - 8) = 18.4 days

∴ 18.4 days it will take to finish the work.

Two pipes, when working one at a time, can fill a cistern in 3 hours and 4 hours, respectively while a third pipe can drain the cistern empty in 8 hours. All the three pipes were opened together when the cistern was 1/12 full. How long did it take for the cistern to be completely full?

  1. 2 hours
  2. 1 hour 45 minutes
  3. 2 hour 11 minutes
  4. 2 hour 10 minutes

Answer (Detailed Solution Below)

Option 1 : 2 hours

Time and Work Question 10 Detailed Solution

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Given:

First pipe can fill the cistern = 3 hours

Second pipe can fill the cistern = 4 hours

Third pipe can drain the cistern = 8 hours

Calculation:

Let the total amount of work in filling a cistern be 24 units. (LCM of 3, 4 and 8)

Work done by pipe 1 in 1 hour = 24/3 = 8 units.

Work done by pipe 2 in 1 hour = 24/4 = 6 units.

Work done by pipe 3 in 1 hour = 24/ (-8) = -3 units

Total work done in 1 hour = 8 + 6 – 3 = 11 units

The time required to complete 11/12th of the work = 11/12 × 24/ 11 = 2 hours

∴ The correct answer is 2 hours.

Harish and Bimal can complete a task in 20 days. They worked at it for 15 days and then Bimal left. The remaining work was done by Harish alone, in 10 days. Harish alone can complete the entire task in:

  1. 40 days
  2. 30 days
  3. 35 days
  4. 45 days

Answer (Detailed Solution Below)

Option 1 : 40 days

Time and Work Question 11 Detailed Solution

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Given:

No of days taken by Harish and Bimal = 20 

Formula used:

No of days taken = Work/Efficiency 

Calculation:

Let the total work be = 1 

One day work done by Harish and Bimal = 1/20 

Work done by Harish and Bimal in 15 days = 1/20 × 15 = 3/4

⇒ Remaining work = 1 - 3/4 = 1/4 

Harish did remaining work in 10 days alone.

⇒ One day work done by Harish  = 1/4 ÷ 10 = 1/40 

∴ Time taken by Harish to do the entire task alone = 1 ÷ 1/40 = 40 days

 Shortcut TrickFraction of work done by Harish & Bimal in 15 days = 15/20 = 3/4 

The remaining 1/4th (25%) of work was done by Harish in 10 days. 

∴ The 100% work would be done by Harish in (10 × 4) 40 days.

The efficiency of A, B, and C is 2 : 3 : 5. A alone can complete a work in 50 days. They all work together for 5 days and then C left the work, in how many days A and B together can complete the remaining work?

  1. 50 days
  2. 30 days
  3. 20 days
  4. 10 days

Answer (Detailed Solution Below)

Option 4 : 10 days

Time and Work Question 12 Detailed Solution

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Given:

Efficiency of A, B and C = 2 : 3 : 5

A alone can complete the work in = 50 days

Formula:

Total work = Efficiency × Time

Calculation:

Let efficiency of A be 2 units/day

Efficiency of A, B and C = 2 : 3 : 5

Total work = 2 × 50 = 100 units

Work done by A, B and C in 5 days = (2 + 3 + 5) × 5 = 10 × 5 = 50 units

Remaining work = 100 – 50 = 50 units

∴ Time taken by A and B to complete the remaining work = 50/(2 + 3) = 50/5 = 10 days

A and B together are supposed to do 13/15 of the work and B and C together 11/20 of the work. If the difference between wages of A and C is Rs. 7600, then the total wages of A and C is:

  1. Rs. 14000
  2. Rs. 36000
  3. Rs. 18000
  4. Rs. 56000

Answer (Detailed Solution Below)

Option 1 : Rs. 14000

Time and Work Question 13 Detailed Solution

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Given:

Difference between wages of A and C = Rs. 7600

Formula Used:

Share in wages = Work done/Total work × Total wages

Calculation:

Let total work be 60 unit,

Work done by A and B = 13/15 × 60 = 52 unit

⇒ Work done by C = 60 – 52 = 8 unit

Work done by B and C = 11/20 × 60 = 33 unit

⇒ Work done by A = 60 – 33 = 27 unit

Work done by B = 60 – 27 – 8 = 25 unit

According to the question,

27 – 8 = 19 unit = 7600

⇒ 1 unit = 400

Total wages of A and C = (27 + 8) = 35 units = 35 × 400 = Rs. 14000

A,B and C can do a piece of work in 30 days, 40 days and 50 days, respectively. Beginning with A, if A, B and C do the work alternatively then in how many days will the work be finished?

  1. \(38\frac{1}{12}\)
  2. \(36\frac{1}{12}\)
  3. 36
  4. \(39\frac{1}{12}\)

Answer (Detailed Solution Below)

Option 1 : \(38\frac{1}{12}\)

Time and Work Question 14 Detailed Solution

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Given:

A can do a piece of work = 30 days

B can do a piece of work = 40 days

C can do a piece of work = 50 days

Formula used:

Total work = efficiency × time

Calculation:

Efficiency Person Time Total work
20 A 30 600
15 B 40
12 C 50

According to the question:

⇒ (20 + 15 + 12) = 47 units = 3 days

⇒ 47 × 12 = 564 units = 3 × 12 = 36 days

⇒ (564 + 20 + 15) = 599 units = 38 days

Total work = 600 units = 38 + (1/12) = 38\(1\over12\) days.

∴ The correct answer is 38\(1\over12\) days.

If 'A' is 6 times more efficient than 'B', 'B' takes 32 days to complete the task, then find the number of days required to finish the whole work by 'A' and 'B'  working together.

  1. 2 days
  2. 4 days
  3. 6 days
  4. 8 days

Answer (Detailed Solution Below)

Option 2 : 4 days

Time and Work Question 15 Detailed Solution

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Given:

A is 6 times more efficient than B, & B takes 32 days to complete the task.

Formula used:

Total work = Efficiency × Time taken

Calculation:

A is 6 times more efficient than B

Efficiency of A ∶ Efficiency of B = 7 ∶ 1

Total work = Efficiency of B × Time taken 

⇒ 1 × 32 = 32 units

Number of days required to finish the whole work by (A + B)  = Total work/Efficiency of (A+ B)

⇒ 32/8

⇒ 4 

∴ The total number of days required to finish the whole work by (A + B) is 4 days.

There is a difference in "Efficient" and " More efficient"

A is 6 times efficient than B means if B is 1 then, A will be 6

A is 6 times more efficient than B means if B is 1 then, A will be (1 + 6) = 7

In the question, it is given that A is 6 times more efficient which means if B is 1, then A will (1 + 6) times = 7 times efficient

So, Total efficiency of A and B = (1 + 7) = 8 units/day

Time taken to complete the work together = 32/8 days

⇒ 4 days and this is the answer. 

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