CAT Time and Work Questions & Solutions

A sample of real CAT Time and Work past-year questions, each with a worked solution and the correct answer highlighted. AthenaPrep has 45 Time and Work questions in all — sign up free to practise them timed.

Q1.CAT 2025
Arun, Varun and Tarun, if working alone, can complete a task in 24, 21, and 15 days, respectively. They charge Rs 2160, Rs 2400, and Rs 2160 per day, respectively, even if they are employed for a partial day. On any given day, any of the workers may or may not be employed to work. If the task needs to be completed in 10 days or less, then the minimum possible amount, in rupees, required to be paid for the entire task is
- 38400
- 38880
- 34400
- 47040
Show solution
Let's assume that Total work = 1

If the work is entirely carried by each one of them it would cost

Arun: $2160 \times 24 = 51840$

Varun: $2400 \times 21 = 50400$

Tarun: $2160 \times 15 = 32400$

The cheapest worker is Tarun. He can finish the work in 15 days. But we need the work completed in 10 days.

Tarun does $10 \times \dfrac{1}{15} = \dfrac{2}{3}$

$1 - \dfrac{2}{3} = \dfrac{1}{3}$

Complete the remaining (1/3) work using the next cheapest worker, which is Varun.

Days needed: $\dfrac{x}{21} = \dfrac{1}{3} \quad\Rightarrow\quad x = 7$

Total cost = Amount paid for Tarun + Amount paid for Varun = $10 \times 2160 + 7 \times 2400 = 38400$
Q2.CAT 2025
Ankita is twice as efficient as Bipin, while Bipin is twice as efficient as Chandan. All three of them start together on a job, and Bipin leaves the job after 20 days. If the job got completed in 60 days, the number of days needed by Chandan to complete the job alone, is
Answer: 340
Show solution
Let Chandan’s one-day work = x.
Then Bipin = 2x and Ankita = 4x.

All three work together for the first 20 days. Their combined daily work = $x + 2x + 4x = 7x$ .
Work done in 20 days = $20\times7x = 140x$ .

Bipin leaves after 20 days; Ankita and Chandan continue for the remaining (60-20=40) days.
Their combined daily work = $4x + x = 5x$ . Work done in 40 days = $40\times5x = 200x$ .

Total work = $140x+200x=340x$ .

We know that Chandan does x work in a day. So, the number of days it takes him to finish the work is 340.

Answer: 340 days.
Q3.CAT 2023
The amount of job that Amal, Sunil and Kamal can individually do in a day, are in harmonic progression. Kamal takes twice as much time as Amal to do the same amount of job. If Amal and Sunil work for 4 days and 9 days, respectively, Kamal needs to work for 16 days to finish the remaining job. Then the number of days Sunil will take to finish the job working alone, is
Answer: 27
Show solution
Let us assume the efficiencies of Amal, Sunil, and Kamal are a, s, and k, respectively.

Given that they are in H.P.

=> $\dfrac{2}{s}=\dfrac{1}{a}+\dfrac{1}{k}$ ---(1)

Also, given that Kamal takes twice as much time as Amal to do the same amount of job

=> a = 2k

Given that when Amal and Sunil work for 4 days and 9 days, respectively, Kamal needs to work for 16 days to finish the remaining job.

=> If W is the total work => 4a + 9s + 16k = W.

from (1) $\dfrac{2}{s}=\dfrac{1}{a}+\dfrac{2}{a}$ => $a=\dfrac{3}{2}s$ and $k=\dfrac{3}{4}s$

=> $4\left(\dfrac{3s}{2}\right)+9s+16\left(\dfrac{3s}{4}\right)=W$

=> $6s+9s+12s=W$

=> $27s=W\ =>\ s\ =\ \dfrac{W}{27}$

=> Sunil will take 27 days to finish the work when working alone.
Q4.CAT 2022
Working alone, the times taken by Anu, Tanu and Manu to complete any job are in the ratio 5 : 8 : 10. They accept a job which they can finish in 4 days if they all work together for 8 hours per day. However, Anu and Tanu work together for the first 6 days, working 6 hours 40 minutes per day. Then, the number of hours that Manu will take to complete the remaining job working alone is
Answer: 6
Show solution
Let the time taken by Anu, Tanu and Manu be 5x, 8x and 10x hours.

Total work = LCM(5x, 8x, 10x) = 40x

Anu can complete 8 units in one hour

Tanu can complete 5 units in one hour

Manu can complete 4 units in one hour

It is given, three of them together can complete in 32 hours.

32(8 + 5 + 4) = 40x

x = $\frac{68}{5}$

It is given,

Anu and Tanu work together for the first 6 days, working 6 hours 40 minutes per day, i.e. 36 + 4 = 40 hours

40(8 + 5) + y(4) = 40x

4y = 24

y = 6

Manu alone will complete the remaining work in 6 hours.
Q5.CAT 2021
Amar, Akbar and Anthony are working on a project. Working together Amar and Akbar can complete the project in 1 year, Akbar and Anthony can complete in 16 months, Anthony and Amar can complete in 2 years. If the person who is neither the fastest nor the slowest works alone, the time in months he will take to complete the project is
Answer: 32
Show solution
Let the total work be 48 units. Let Amar do 'm' work, Akbar do 'k' work, and Anthony do 'n' units of work in a month.

Amar and Akbar complete the project in 12 months. Hence, in a month they do $\frac{48}{12}$ =4 units of work.

m+k = 4.

Similarly, k+n = 3, and m+n = 2.

Solving the three equations, we get $m=\frac{3}{2},\ k=\frac{5}{2},\ n=\frac{1}{2}$ .

Here, Amar works neither the fastest not the slowest, and he does 1.5 units of work in a month. Hence, to complete the work, he would take $\frac{48}{1.5}=32$ months.
Q6.CAT 2021
One day, Rahul started a work at 9 AM and Gautam joined him two hours later. They then worked together and completed the work at 5 PM the same day. If both had started at 9 AM and worked together, the work would have been completed 30 minutes earlier. Working alone, the time Rahul would have taken, in hours, to complete the work is
- 11.5
- 10
- 12.5
- 12
Show solution
Let Rahul work at a units/hr and Gautam at b units/hour
Now as per the condition :
8a+6b =7.5a+7.5b
so we get 0.5a=1.5b
or a=3b
Therefore total work = 8a +6b = 8a +2a =10a
Now Rahul alone takes 10a/10 = 10 hours.
Q7.CAT 2019
Anil alone can do a job in 20 days while Sunil alone can do it in 40 days. Anil starts the job, and after 3 days, Sunil joins him. Again, after a few more days, Bimal joins them and they together finish the job. If Bimal has done 10% of the job, then in how many days was the job done?
- 12
- 13
- 15
- 14
Show solution
Let the total work be LCM of 20, 40 = 40 units

Efficiency of Anil and Sunil is 2 units and 1 unit per day respectively.

Anil works alone for 3 days, so Anil must have completed 6 units.

Bimal completes 10% of the work while working along with Anil and Sunil.

Bimal must have completed 4 units.

The remaining 30 units of work is done by Anil and Sunil

Number of days taken by them 30/3=10

The total work is completed in 3+10=13 days
Q8.CAT 2018
When they work alone, B needs 25% more time to finish a job than A does. They two finish the job in 13 days in the following manner: A works alone till half the job is done, then A and B work together for four days, and finally B works alone to complete the remaining 5% of the job. In how many days can B alone finish the entire job?
- 20
- 22
- 16
- 18
Show solution
Let us assume that A can complete 'a' units of work in a day and B can complete 'b' units of work in a day.
A works alone till half the work is completed.
A and B work together for 4 days and B works alone to complete the last 5% of the work.
=> A and B in 4 days can complete 45% of the work.

Let us assume the total amount of work to be done to be 100 units.
4a + 4b = 45 ---------(1)
B needs 25% more time than A to finish a job.
=> 1.25*b = a ----------(2)

Substituting (2) in (1), we get,
5b+4b = 45
9b = 45
b = 5 units/day

B alone can finish the job in 100/5 = 20 days.
Therefore, option A is the right answer.
Q9.CAT 2018
Ramesh and Ganesh can together complete a work in 16 days. After seven days of working together, Ramesh got sick and his eﬃciency fell by 30%. As a result, they completed the work in 17 days instead of 16 days. If Ganesh had worked alone after Ramesh got sick, in how many days would he have completed the remaining work?
- 14.5
- 11
- 13.5
- 12
Show solution
Let 'R' and 'G' be the amount of work that Ramesh and Ganesh can complete in a day.

It is given that they can together complete a work in 16 days. Hence, total amount of work = 16(R+G) ... (1)

For first 7 days both of them worked together. From 8th day, Ramesh worked at 70% of his original efficiency whereas Ganesh worked at his original efficiency. It took them 17 days to finish the same work. i.e. Ramesh worked at 70% of his original efficiency for 10 days.

$\Rightarrow$ 16(R+G) = 7(R+G)+10(0.7R+G)

$\Rightarrow$ 16(R+G) = 14R+17G

$\Rightarrow$ R = 0.5G ... (2)

Total amount of work left when Ramesh got sick = 16(R+G) - 7(R+G) = 9(R+G) = 9(0.5+G) = 13.5G

Therefore, time taken by Ganesh to complete the remaining work = $\dfrac{13.5G}{G}$ = 13.5 days.
Q10.CAT 2017
Amal can complete a job in 10 days and Bimal can complete it in 8 days. Amal, Bimal and Kamal together complete the job in 4 days and are paid a total amount of Rs 1000 as remuneration. If this amount is shared by them in proportion to their work, then Kamal's share, in rupees, is
- 100
- 200
- 300
- 400
Show solution
Let the time take by kamal to complete the task be x days.
Hence we have $\frac{1}{10} + \frac{1}{8} + \frac{1}{x} = \frac{1}{4}$
=> x = 40 days.
Ratio of the work done by them = $\frac{1}{10} : \frac{1}{8} : \frac{1}{40}$ = 4 : 5 : 1
Hence the wage earned by Kamal = 1/10 * 1000 = 100
Q11.CAT 2001
A can complete a piece of work in 4 days. B takes double the time taken by A, C takes double that of B, and D takes double that of C to complete the same task. They are paired in groups of two each. One pair takes two-thirds the time needed by the second pair to complete the work. Which is the first pair?
- A and B
- A and C
- B and C
- A and D
Show solution
A takes 4 days to complete the work.
So, B takes 8 days to complete the same work.
C takes 16 days to complete the work.
D takes 32 days to complete the same work.

In order to measure, let the total work be of 64 units. Hence, the speed of working of each of the four persons is given below.

A - 16 units/hr
B - 8 units/hr
C - 4 units/hr
D - 2 units/hr

From the given options, we need to find two pairs in such a way that their speeds are in the ratio 3:2. Note that A+D=18 while B+C=12 and the ratio is 3:2

Hence, the first pair is A and D and the second pair is B and C
Passage
The Weirdo Holiday Resort follows a particular system of holidays for its employees. People are given holidays on the days where the first letter of the day of the week is the same as the first letter of their names. All employees work at the same rate.
Q12.CAT 1997
Starting on February 25, 1996 (Sunday), if Raja had finished his job on April 2, 1996, when would T and S together likely to have completed the job, had they started on the same day as Raja?
- March 15, 1996
- March 14, 1996
- March 22, 1996
- Data insufficient
Show solution
The number of days taken by Raja to complete the work is 5+31+2 = 38
So, cumulative number of days needed by T and S to complete the work is 38.
Both of them take two days off in a week as S takes of on Saturday and Sunday and T takes off on Tuesday and Thursday.
So, total number of man-working days per week by the duo is 10.
Hence, after three weeks, they finish 30 man working days.
i.e by end of 17th March 1996 (Sunday), 30 man working days are finished.
Both of them work on Monday,
S works on Tuesday
Both of them work on Wednesday
S works on Thursday
Both of them work on Friday and the remaining 8 man working days are also over.
Hence, the required date is 17+5 = 22 March 1996 (Friday)

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CAT Time and Work Questions & Solutions

Q1.CAT 2025

Arun, Varun and Tarun, if working alone, can complete a task in 24, 21, and 15 days, respectively. They charge Rs 2160, Rs 2400, and Rs 2160 per day, respectively, even if they are employed for a partial day. On any given day, any of the workers may or may not be employed to work. If the task needs to be completed in 10 days or less, then the minimum possible amount, in rupees, required to be paid for the entire task is

38400
38880
34400
47040

Show solution

Let's assume that Total work = 1

If the work is entirely carried by each one of them it would cost

Arun: $2160 \times 24 = 51840$

Varun: $2400 \times 21 = 50400$

Tarun: $2160 \times 15 = 32400$

The cheapest worker is Tarun. He can finish the work in 15 days. But we need the work completed in 10 days.

Tarun does $10 \times \dfrac{1}{15} = \dfrac{2}{3}$

$1 - \dfrac{2}{3} = \dfrac{1}{3}$

Complete the remaining (1/3) work using the next cheapest worker, which is Varun.

Days needed: $\dfrac{x}{21} = \dfrac{1}{3} \quad\Rightarrow\quad x = 7$

Total cost = Amount paid for Tarun + Amount paid for Varun = $10 \times 2160 + 7 \times 2400 = 38400$

Q2.CAT 2025

Ankita is twice as efficient as Bipin, while Bipin is twice as efficient as Chandan. All three of them start together on a job, and Bipin leaves the job after 20 days. If the job got completed in 60 days, the number of days needed by Chandan to complete the job alone, is

Answer: 340

Show solution

Let Chandan’s one-day work = x.
Then Bipin = 2x and Ankita = 4x.

All three work together for the first 20 days. Their combined daily work = $x + 2x + 4x = 7x$ .
Work done in 20 days = $20\times7x = 140x$ .

Bipin leaves after 20 days; Ankita and Chandan continue for the remaining (60-20=40) days.
Their combined daily work = $4x + x = 5x$ . Work done in 40 days = $40\times5x = 200x$ .

Total work = $140x+200x=340x$ .

We know that Chandan does x work in a day. So, the number of days it takes him to finish the work is 340.

Answer: 340 days.

Q3.CAT 2023

The amount of job that Amal, Sunil and Kamal can individually do in a day, are in harmonic progression. Kamal takes twice as much time as Amal to do the same amount of job. If Amal and Sunil work for 4 days and 9 days, respectively, Kamal needs to work for 16 days to finish the remaining job. Then the number of days Sunil will take to finish the job working alone, is

Answer: 27

Show solution

Let us assume the efficiencies of Amal, Sunil, and Kamal are a, s, and k, respectively.

Given that they are in H.P.

=> $\dfrac{2}{s}=\dfrac{1}{a}+\dfrac{1}{k}$ ---(1)

Also, given that Kamal takes twice as much time as Amal to do the same amount of job

=> a = 2k

Given that when Amal and Sunil work for 4 days and 9 days, respectively, Kamal needs to work for 16 days to finish the remaining job.

=> If W is the total work => 4a + 9s + 16k = W.

from (1) $\dfrac{2}{s}=\dfrac{1}{a}+\dfrac{2}{a}$ => $a=\dfrac{3}{2}s$ and $k=\dfrac{3}{4}s$

=> $4\left(\dfrac{3s}{2}\right)+9s+16\left(\dfrac{3s}{4}\right)=W$

=> $6s+9s+12s=W$

=> $27s=W\ =>\ s\ =\ \dfrac{W}{27}$

=> Sunil will take 27 days to finish the work when working alone.

Q4.CAT 2022

Working alone, the times taken by Anu, Tanu and Manu to complete any job are in the ratio 5 : 8 : 10. They accept a job which they can finish in 4 days if they all work together for 8 hours per day. However, Anu and Tanu work together for the first 6 days, working 6 hours 40 minutes per day. Then, the number of hours that Manu will take to complete the remaining job working alone is

Answer: 6

Show solution

Let the time taken by Anu, Tanu and Manu be 5x, 8x and 10x hours.

Total work = LCM(5x, 8x, 10x) = 40x

Anu can complete 8 units in one hour

Tanu can complete 5 units in one hour

Manu can complete 4 units in one hour

It is given, three of them together can complete in 32 hours.

32(8 + 5 + 4) = 40x

x = $\frac{68}{5}$

It is given,

Anu and Tanu work together for the first 6 days, working 6 hours 40 minutes per day, i.e. 36 + 4 = 40 hours

40(8 + 5) + y(4) = 40x

4y = 24

y = 6

Manu alone will complete the remaining work in 6 hours.

Q5.CAT 2021

Amar, Akbar and Anthony are working on a project. Working together Amar and Akbar can complete the project in 1 year, Akbar and Anthony can complete in 16 months, Anthony and Amar can complete in 2 years. If the person who is neither the fastest nor the slowest works alone, the time in months he will take to complete the project is

Answer: 32

Show solution

Let the total work be 48 units. Let Amar do 'm' work, Akbar do 'k' work, and Anthony do 'n' units of work in a month.

Amar and Akbar complete the project in 12 months. Hence, in a month they do $\frac{48}{12}$ =4 units of work.

m+k = 4.

Similarly, k+n = 3, and m+n = 2.

Solving the three equations, we get $m=\frac{3}{2},\ k=\frac{5}{2},\ n=\frac{1}{2}$ .

Here, Amar works neither the fastest not the slowest, and he does 1.5 units of work in a month. Hence, to complete the work, he would take $\frac{48}{1.5}=32$ months.

Q6.CAT 2021

One day, Rahul started a work at 9 AM and Gautam joined him two hours later. They then worked together and completed the work at 5 PM the same day. If both had started at 9 AM and worked together, the work would have been completed 30 minutes earlier. Working alone, the time Rahul would have taken, in hours, to complete the work is

11.5
10
12.5
12

Show solution

Let Rahul work at a units/hr and Gautam at b units/hour
Now as per the condition :
8a+6b =7.5a+7.5b
so we get 0.5a=1.5b
or a=3b
Therefore total work = 8a +6b = 8a +2a =10a
Now Rahul alone takes 10a/10 = 10 hours.

Q7.CAT 2019

Anil alone can do a job in 20 days while Sunil alone can do it in 40 days. Anil starts the job, and after 3 days, Sunil joins him. Again, after a few more days, Bimal joins them and they together finish the job. If Bimal has done 10% of the job, then in how many days was the job done?

Show solution

Let the total work be LCM of 20, 40 = 40 units

Efficiency of Anil and Sunil is 2 units and 1 unit per day respectively.

Anil works alone for 3 days, so Anil must have completed 6 units.

Bimal completes 10% of the work while working along with Anil and Sunil.

Bimal must have completed 4 units.

The remaining 30 units of work is done by Anil and Sunil

Number of days taken by them 30/3=10

The total work is completed in 3+10=13 days

Q8.CAT 2018

When they work alone, B needs 25% more time to finish a job than A does. They two finish the job in 13 days in the following manner: A works alone till half the job is done, then A and B work together for four days, and finally B works alone to complete the remaining 5% of the job. In how many days can B alone finish the entire job?

Show solution

Let us assume that A can complete 'a' units of work in a day and B can complete 'b' units of work in a day.
A works alone till half the work is completed.
A and B work together for 4 days and B works alone to complete the last 5% of the work.
=> A and B in 4 days can complete 45% of the work.

Let us assume the total amount of work to be done to be 100 units.
4a + 4b = 45 ---------(1)
B needs 25% more time than A to finish a job.
=> 1.25*b = a ----------(2)

Substituting (2) in (1), we get,
5b+4b = 45
9b = 45
b = 5 units/day

B alone can finish the job in 100/5 = 20 days.
Therefore, option A is the right answer.

Q9.CAT 2018

Ramesh and Ganesh can together complete a work in 16 days. After seven days of working together, Ramesh got sick and his eﬃciency fell by 30%. As a result, they completed the work in 17 days instead of 16 days. If Ganesh had worked alone after Ramesh got sick, in how many days would he have completed the remaining work?

14.5
11
13.5
12

Show solution

Let 'R' and 'G' be the amount of work that Ramesh and Ganesh can complete in a day.

It is given that they can together complete a work in 16 days. Hence, total amount of work = 16(R+G) ... (1)

For first 7 days both of them worked together. From 8th day, Ramesh worked at 70% of his original efficiency whereas Ganesh worked at his original efficiency. It took them 17 days to finish the same work. i.e. Ramesh worked at 70% of his original efficiency for 10 days.

$\Rightarrow$ 16(R+G) = 7(R+G)+10(0.7R+G)

$\Rightarrow$ 16(R+G) = 14R+17G

$\Rightarrow$ R = 0.5G ... (2)

Total amount of work left when Ramesh got sick = 16(R+G) - 7(R+G) = 9(R+G) = 9(0.5+G) = 13.5G

Therefore, time taken by Ganesh to complete the remaining work = $\dfrac{13.5G}{G}$ = 13.5 days.

Q10.CAT 2017

Amal can complete a job in 10 days and Bimal can complete it in 8 days. Amal, Bimal and Kamal together complete the job in 4 days and are paid a total amount of Rs 1000 as remuneration. If this amount is shared by them in proportion to their work, then Kamal's share, in rupees, is

Show solution

Let the time take by kamal to complete the task be x days.
Hence we have $\frac{1}{10} + \frac{1}{8} + \frac{1}{x} = \frac{1}{4}$
=> x = 40 days.
Ratio of the work done by them = $\frac{1}{10} : \frac{1}{8} : \frac{1}{40}$ = 4 : 5 : 1
Hence the wage earned by Kamal = 1/10 * 1000 = 100

Q11.CAT 2001

A can complete a piece of work in 4 days. B takes double the time taken by A, C takes double that of B, and D takes double that of C to complete the same task. They are paired in groups of two each. One pair takes two-thirds the time needed by the second pair to complete the work. Which is the first pair?

A and B
A and C
B and C
A and D

Show solution

A takes 4 days to complete the work.
So, B takes 8 days to complete the same work.
C takes 16 days to complete the work.
D takes 32 days to complete the same work.

In order to measure, let the total work be of 64 units. Hence, the speed of working of each of the four persons is given below.

A - 16 units/hr
B - 8 units/hr
C - 4 units/hr
D - 2 units/hr

From the given options, we need to find two pairs in such a way that their speeds are in the ratio 3:2. Note that A+D=18 while B+C=12 and the ratio is 3:2

Hence, the first pair is A and D and the second pair is B and C

Passage

The Weirdo Holiday Resort follows a particular system of holidays for its employees. People are given holidays on the days where the first letter of the day of the week is the same as the first letter of their names. All employees work at the same rate.

Q12.CAT 1997

Starting on February 25, 1996 (Sunday), if Raja had finished his job on April 2, 1996, when would T and S together likely to have completed the job, had they started on the same day as Raja?

March 15, 1996
March 14, 1996
March 22, 1996
Data insufficient

Show solution

The number of days taken by Raja to complete the work is 5+31+2 = 38
So, cumulative number of days needed by T and S to complete the work is 38.

Both of them take two days off in a week as S takes of on Saturday and Sunday and T takes off on Tuesday and Thursday.

So, total number of man-working days per week by the duo is 10.
Hence, after three weeks, they finish 30 man working days.
i.e by end of 17th March 1996 (Sunday), 30 man working days are finished.

Both of them work on Monday,
S works on Tuesday
Both of them work on Wednesday
S works on Thursday
Both of them work on Friday and the remaining 8 man working days are also over.

Hence, the required date is 17+5 = 22 March 1996 (Friday)

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