20 Time and Work Aptitude Questions with Answers and Solutions
20 Time and Work Aptitude Questions with Answers and Solutions
Time and Work is an important quantitative aptitude topic frequently asked in placement tests, competitive examinations, recruitment assessments, and entrance tests. These questions test your ability to understand work efficiency, individual and combined work rates, time required to complete a task, and the relationship between workers and days.
This practice set contains 20 original Time and Work Aptitude Questions with detailed answers and step-by-step solutions. The questions begin with basic work-rate calculations and gradually move toward combined work, efficiency, alternate working, and workers-and-days problems.
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Try to solve all 20 questions before checking the answers.
Detailed answers and solutions are provided after Question 20.
Time and Work Aptitude Test
Question 1
If Juhi can complete a piece of work in 12 days, what fraction of the work does she complete in one day?
A. (1/6)
B. (1/10)
C. (1/12)
D. (1/24)
Question 2
Cherry can complete a job in 15 days. How much of the work will she complete in 5 days?
A. (1/5)
B. (1/3)
C. (2/5)
D. (1/2)
Question 3
Juhi can complete a job in 10 days and Cherry can complete the same job in 15 days. How many days will they take if they work together?
A. 5 days
B. 6 days
C. 7 days
D. 8 days
Question 4
Charlie can complete a task in 18 days and Kanchan can complete it in 36 days. How long will they take working together?
A. 9 days
B. 10 days
C. 12 days
D. 15 days
Question 5
Juhi and Nyra together can complete a job in 8 days. If Juhi alone can complete it in 12 days, how many days would Nyra take working alone?
A. 18 days
B. 20 days
C. 24 days
D. 30 days
Question 6
A worker completes (3/5) of a job in 12 days. At the same rate, how many days will the worker require to complete the entire job?
A. 18 days
B. 20 days
C. 22 days
D. 24 days
Question 7
12 workers can complete a job in 20 days. How many days will 15 workers take to complete the same job, assuming every worker has equal efficiency?
A. 14 days
B. 16 days
C. 18 days
D. 25 days
Question 8
18 workers can complete a task in 24 days. How many workers are required to complete the same task in 16 days?
A. 24
B. 27
C. 30
D. 32
Question 9
Juhi is twice as efficient as Cherry. If Cherry can complete a job in 24 days, how many days will Juhi require to complete the same job?
A. 8 days
B. 10 days
C. 12 days
D. 16 days
Question 10
Charlie is 50% more efficient than Nyra. If Nyra completes a job in 18 days, how many days will Charlie take?
A. 10 days
B. 12 days
C. 14 days
D. 15 days
Question 11
Juhi can complete a job in 20 days and Cherry in 30 days. They work together for 6 days. What fraction of the work remains unfinished?
A. (1/3)
B. (1/2)
C. (3/5)
D. (2/3)
Question 12
Kanchan can complete a job in 16 days. She works alone for 4 days, after which Nyra joins her. Together they complete the remaining work in 6 days. How many days would Nyra alone take to complete the entire job?
A. 12 days
B. 16 days
C. 24 days
D. 32 days
Question 13
Juhi, Cherry, and Charlie can individually complete a job in 12, 18, and 36 days respectively. How many days will they take if all three work together?
A. 5 days
B. 6 days
C. 7 days
D. 8 days
Question 14
Juhi can complete a job in 8 days and Cherry can complete it in 12 days. They work on alternate days, starting with Juhi. In how many days will the work be completed?
A. 9 days
B. (9\frac{1}{3}) days
C. (9\frac{1}{2}) days
D. 10 days
Question 15
A group of 20 workers can complete a project in 18 days. After working for 6 days, 5 workers leave. How many additional days will the remaining workers need to complete the unfinished work?
A. 12 days
B. 14 days
C. 16 days
D. 18 days
Question 16
Juhi can do a piece of work in 24 days and Cherry can do it in 16 days. They work together for 4 days, after which Juhi leaves. How many more days will Cherry need to finish the remaining work?
A. 7 days
B. 8 days
C. 9 days
D. 10 days
Question 17
A contractor estimates that 25 workers can complete a job in 32 days. After 8 days, only 20 workers remain. How many more days will the remaining workers require to finish the job?
A. 25 days
B. 28 days
C. 30 days
D. 32 days
Question 18
Juhi and Cherry together can complete a job in 10 days, Cherry and Charlie together in 12 days, and Charlie and Juhi together in 15 days. How many days will all three take working together?
A. 6 days
B. 8 days
C. 10 days
D. 12 days
Question 19
Juhi completes half of a job in 8 days. Cherry completes one-third of the same job in 6 days. How long will they take to complete the entire job if they work together?
A. 8 days
B. (8\frac{8}{17}) days
C. 9 days
D. (9\frac{1}{2}) days
Question 20
24 workers working 7 hours per day can complete a job in 15 days. How many workers are required to complete the same job in 10 days if they work 9 hours per day at the same efficiency?
A. 24
B. 26
C. 28
D. 30
Answers and Detailed Solutions
Answer 1: C — (1/12)
If Juhi completes the whole work in 12 days, then her one-day work is:
[
\frac{1}{12}
]
Therefore:
Correct Answer: (1/12)
The basic rule is:
[
\text{One-day work}=\frac{1}{\text{Number of days required}}
]
Answer 2: B — (1/3)
Cherry completes the entire work in 15 days.
Her one-day work is:
[
\frac{1}{15}
]
Work completed in 5 days:
[
5\times\frac{1}{15}
]
[
=\frac{5}{15}
]
[
=\frac{1}{3}
]
Therefore:
Correct Answer: (1/3)
Answer 3: B — 6 days
Juhi’s one-day work:
[
\frac{1}{10}
]
Cherry’s one-day work:
[
\frac{1}{15}
]
Together:
[
\frac{1}{10}+\frac{1}{15}
]
LCM of 10 and 15 is 30.
[
=\frac{3}{30}+\frac{2}{30}
]
[
=\frac{5}{30}
]
[
=\frac{1}{6}
]
Therefore, together they complete (1/6) of the job each day.
Time required:
[
6\text{ days}
]
Correct Answer: 6 days
Answer 4: C — 12 days
Charlie’s one-day work:
[
\frac{1}{18}
]
Kanchan’s one-day work:
[
\frac{1}{36}
]
Together:
[
\frac{1}{18}+\frac{1}{36}
]
[
=\frac{2}{36}+\frac{1}{36}
]
[
=\frac{3}{36}
]
[
=\frac{1}{12}
]
Therefore:
Time required = 12 days
Correct Answer:
12 days
Answer 5: C — 24 days
Juhi and Nyra together complete:
[
\frac{1}{8}
]
of the work per day.
Juhi alone completes:
[
\frac{1}{12}
]
per day.
Therefore, Nyra’s one-day work is:
[
\frac{1}{8}-\frac{1}{12}
]
LCM = 24.
[
=\frac{3}{24}-\frac{2}{24}
]
[
=\frac{1}{24}
]
Therefore, Nyra alone requires:
[
24\text{ days}
]
Correct Answer: 24 days
Answer 6: B — 20 days
The worker completes:
[
\frac{3}{5}
]
of the job in 12 days.
If total time is (x):
[
\frac{3}{5}x=12
]
Alternatively:
Time for whole work:
[
12\times\frac{5}{3}
]
[
=20
]
Therefore:
Correct Answer: 20 days
Answer 7: B — 16 days
For the same amount of work:
[
\text{Workers}\times\text{Days}=\text{Constant}
]
Therefore:
[
12\times20=15\times x
]
[
240=15x
]
[
x=16
]
Therefore:
Correct Answer: 16 days
This is an example of inverse proportion.
More workers require fewer days to complete the same work.
Answer 8: B — 27 workers
Total work:
[
18\times24
]
For 16 days:
[
18\times24=x\times16
]
Therefore:
[
x=\frac{18\times24}{16}
]
[
=27
]
Therefore:
Correct Answer: 27 workers
Answer 9: C — 12 days
Juhi is twice as efficient as Cherry.
This means:
[
\text{Juhi’s efficiency}:\text{Cherry’s efficiency}=2:1
]
Time is inversely proportional to efficiency.
Therefore:
[
\text{Juhi’s time}:\text{Cherry’s time}=1:2
]
Cherry requires 24 days.
Therefore Juhi requires:
[
24\div2=12
]
days.
Correct Answer: 12 days
Answer 10: B — 12 days
Nyra’s efficiency can be represented as 100%.
Charlie is 50% more efficient.
Therefore Charlie’s efficiency is:
[
150%
]
Efficiency ratio:
[
Charlie:Nyra=150:100
]
[
=3:2
]
Time is inversely proportional to efficiency.
Therefore:
[
Charlie’s\ Time:Nyra’s\ Time=2:3
]
Nyra takes 18 days.
Charlie takes:
[
18\times\frac{2}{3}
]
[
=12
]
Therefore:
Correct Answer: 12 days
Answer 11: B — (1/2)
Juhi’s one-day work:
[
\frac{1}{20}
]
Cherry’s one-day work:
[
\frac{1}{30}
]
Together:
[
\frac{1}{20}+\frac{1}{30}
]
LCM = 60.
[
=\frac{3}{60}+\frac{2}{60}
]
[
=\frac{5}{60}
]
[
=\frac{1}{12}
]
Work completed in 6 days:
[
6\times\frac{1}{12}
]
[
=\frac{1}{2}
]
Therefore, remaining work:
[
1-\frac{1}{2}
]
[
=\frac{1}{2}
]
Correct Answer: (1/2)
Answer 12: C — 24 days
Kanchan completes the whole work in 16 days.
Her one-day work:
[
\frac{1}{16}
]
Work completed by Kanchan in 4 days:
[
4\times\frac{1}{16}
]
[
=\frac{1}{4}
]
Remaining work:
[
1-\frac{1}{4}
]
[
=\frac{3}{4}
]
Kanchan and Nyra together complete this remaining (3/4) work in 6 days.
Therefore, their combined one-day work is:
[
\frac{3/4}{6}
]
[
=\frac{3}{24}
]
[
=\frac{1}{8}
]
Kanchan’s one-day work:
[
\frac{1}{16}
]
Therefore Nyra’s one-day work:
[
\frac{1}{8}-\frac{1}{16}
]
[
=\frac{1}{16}
]
So Nyra alone would take:
[
16\text{ days}
]
Therefore:
Correct Answer: B — 16 days
Answer 13: B — 6 days
Juhi’s one-day work:
[
\frac{1}{12}
]
Cherry’s:
[
\frac{1}{18}
]
Charlie’s:
[
\frac{1}{36}
]
Combined work:
[
\frac{1}{12}+\frac{1}{18}+\frac{1}{36}
]
LCM = 36.
[
=\frac{3}{36}+\frac{2}{36}+\frac{1}{36}
]
[
=\frac{6}{36}
]
[
=\frac{1}{6}
]
Therefore:
Correct Answer: 6 days
Answer 14: B — (9\frac{1}{3}) days
Juhi’s one-day work:
[
\frac{1}{8}
]
Cherry’s one-day work:
[
\frac{1}{12}
]
In two days, they complete:
[
\frac{1}{8}+\frac{1}{12}
]
LCM = 24.
[
=\frac{3}{24}+\frac{2}{24}
]
[
=\frac{5}{24}
]
In 8 days, there are four complete two-day cycles.
Work completed:
[
4\times\frac{5}{24}
]
[
=\frac{20}{24}
]
[
=\frac{5}{6}
]
Day 9 belongs to Juhi.
She completes:
[
\frac{1}{8}
]
Therefore total after Day 9:
[
\frac{5}{6}+\frac{1}{8}
]
LCM = 24.
[
=\frac{20}{24}+\frac{3}{24}
]
[
=\frac{23}{24}
]
Remaining work:
[
\frac{1}{24}
]
Cherry’s full-day work is:
[
\frac{1}{12}
]
Therefore, the fraction of a day Cherry needs is:
[
\frac{1/24}{1/12}
]
[
=\frac{1}{2}
]
So the total time is:
[
9\frac{1}{2}\text{ days}
]
Therefore:
Correct Answer: C — (9\frac{1}{2}) days
Answer 15: C — 16 days
Total work in worker-days:
[
20\times18=360
]
Work completed during the first 6 days:
[
20\times6=120
]
Remaining work:
[
360-120=240
]
Five workers leave.
Remaining workers:
[
20-5=15
]
Additional days required:
[
\frac{240}{15}
]
[
=16
]
Therefore:
Correct Answer: 16 days
Answer 16: B — 8 days
Juhi’s one-day work:
[
\frac{1}{24}
]
Cherry’s one-day work:
[
\frac{1}{16}
]
Together:
[
\frac{1}{24}+\frac{1}{16}
]
LCM = 48.
[
=\frac{2}{48}+\frac{3}{48}
]
[
=\frac{5}{48}
]
Work completed in 4 days:
[
4\times\frac{5}{48}
]
[
=\frac{20}{48}
]
[
=\frac{5}{12}
]
Remaining work:
[
1-\frac{5}{12}
]
[
=\frac{7}{12}
]
Cherry alone completes:
[
\frac{1}{16}
]
per day.
Time required for the remaining work:
[
\frac{7/12}{1/16}
]
[
=\frac{7}{12}\times16
]
[
=\frac{28}{3}
]
[
=9\frac{1}{3}\text{ days}
]
Therefore:
Correct Answer: C — approximately 9⅓ days
For a clean MCQ, change Option C to:
C. (9\frac{1}{3}) days
Answer 17: C — 30 days
Total work:
[
25\times32=800
]
worker-days.
Work completed in first 8 days:
[
25\times8=200
]
Remaining work:
[
800-200=600
]
Remaining workers:
20
Additional days:
[
600\div20=30
]
Therefore:
Correct Answer: 30 days
Answer 18: B — 8 days
Let the one-day work rates of Juhi, Cherry, and Charlie be:
[
J,\ C,\ H
]
We know:
[
J+C=\frac{1}{10}
]
[
C+H=\frac{1}{12}
]
[
H+J=\frac{1}{15}
]
Add all three equations:
[
2(J+C+H)
\frac{1}{10}+\frac{1}{12}+\frac{1}{15}
]
LCM = 60.
[
2(J+C+H)
\frac{6+5+4}{60}
]
[
=\frac{15}{60}
]
[
=\frac{1}{4}
]
Therefore:
[
J+C+H=\frac{1}{8}
]
Hence all three together complete the work in:
[
8\text{ days}
]
Correct Answer: 8 days
Answer 19: B — (8\frac{8}{17}) days
Juhi completes half the job in 8 days.
Therefore Juhi’s one-day work:
[
\frac{1/2}{8}
]
[
=\frac{1}{16}
]
Cherry completes one-third of the job in 6 days.
Therefore Cherry’s one-day work:
[
\frac{1/3}{6}
]
[
=\frac{1}{18}
]
Together:
[
\frac{1}{16}+\frac{1}{18}
]
LCM = 144.
[
=\frac{9}{144}+\frac{8}{144}
]
[
=\frac{17}{144}
]
Time required:
[
\frac{144}{17}
]
[
=8\frac{8}{17}
]
days.
Therefore:
Correct Answer: (8\frac{8}{17}) days
Answer 20: C — 28 workers
For the same amount of work:
[
\text{Workers}\times\text{Hours per Day}\times\text{Days}
]
remains constant.
Original arrangement:
[
24\times7\times15
]
New arrangement:
[
x\times9\times10
]
Therefore:
[
24\times7\times15=x\times9\times10
]
[
2520=90x
]
[
x=28
]
Therefore:
Correct Answer: 28 workers
Quick Answer Key
| Question | Answer | Question | Answer |
|---|---|---|---|
| 1 | C | 11 | B |
| 2 | B | 12 | B |
| 3 | B | 13 | B |
| 4 | C | 14 | C |
| 5 | C | 15 | C |
| 6 | B | 16 | C* |
| 7 | B | 17 | C |
| 8 | B | 18 | B |
| 9 | C | 19 | B |
| 10 | B | 20 | C |
*For Question 16, use (9\frac{1}{3}) days as Option C before publishing.
How Did You Score?
| Correct Answers | Performance |
|---|---|
| 18–20 | Excellent — You have a strong understanding of Time and Work. |
| 15–17 | Very Good — Your fundamentals are strong; practice alternate-work and changing-workforce problems. |
| 11–14 | Good — Review combined work rates and efficiency concepts. |
| 6–10 | Needs Practice — Revise one-day work and workers-days relationships. |
| 0–5 | Beginner — Start with basic individual and combined work-rate questions. |
Important Time and Work Formulas
One-Day Work
If a person completes a job in (x) days:
[
\text{One-day work}=\frac{1}{x}
]
For example, if a person completes a job in 10 days:
[
\text{One-day work}=\frac{1}{10}
]
Work Completed in Several Days
If one-day work is:
[
\frac{1}{x}
]
then work completed in (d) days is:
[
\frac{d}{x}
]
Combined Work
If Juhi completes a job in (x) days and Cherry completes it in (y) days:
[
\text{Combined one-day work}
\frac{1}{x}+\frac{1}{y}
]
Therefore:
[
\text{Time Together}
\frac{xy}{x+y}
]
when both work continuously at constant rates.
Efficiency and Time
Efficiency and time are inversely proportional.
If:
[
\text{Efficiency of A}:\text{Efficiency of B}=2:3
]
then:
[
\text{Time taken by A}:\text{Time taken by B}=3:2
]
A more efficient worker requires less time.
Workers and Days
For the same amount of work and equal worker efficiency:
[
\text{Workers}\times\text{Days}=\text{Constant}
]
Therefore:
[
W_1D_1=W_2D_2
]
For example:
12 workers × 20 days
is equivalent to:
15 workers × 16 days.
Workers, Days, and Hours
When working hours also change:
[
W_1\times D_1\times H_1
W_2\times D_2\times H_2
]
where:
- (W) = workers
- (D) = days
- (H) = hours worked per day
This assumes equal efficiency.
A Useful LCM Method
Some Time and Work questions become easier if the total work is assumed to be the LCM of the individual completion times.
Suppose Juhi takes 12 days and Cherry takes 18 days.
LCM:
[
LCM(12,18)=36
]
Assume total work = 36 units.
Juhi’s daily work:
[
36\div12=3\text{ units}
]
Cherry’s daily work:
[
36\div18=2\text{ units}
]
Together:
[
3+2=5\text{ units/day}
]
Time required:
[
\frac{36}{5}=7.2\text{ days}
]
This method avoids repeatedly adding fractions.
Work and Efficiency Relationship
Suppose Juhi is three times as efficient as Cherry.
Then in the same time:
Juhi completes 3 units of work for every 1 unit completed by Cherry.
Efficiency ratio:
[
3:1
]
Time ratio:
[
1:3
]
Therefore, if Cherry takes 30 days:
[
Juhi’s\ time=30\div3=10\text{ days}
]
Common Mistakes in Time and Work Questions
Adding completion times directly
If one person takes 10 days and another takes 15 days, they do not take:
[
10+15=25
]
days together.
You must add their work rates.
Confusing efficiency and time
Higher efficiency means less time, not more time.
Efficiency and time are inversely proportional.
Forgetting completed work
If workers complete part of a job before someone joins or leaves, first calculate the work already completed.
Then calculate the remaining work.
Ignoring changing worker numbers
When workers leave or join midway, divide the problem into separate stages.
For example:
Stage 1: Original workers × days worked
Stage 2: Remaining work ÷ new workforce
Assuming equal efficiency without checking
Worker-day formulas such as:
[
W_1D_1=W_2D_2
]
assume all workers have equal efficiency unless stated otherwise.
Topics Covered in This Test
This Time and Work aptitude test covered:
- One-day work
- Work completed in several days
- Combined work
- Individual work rates
- Remaining work
- Workers and days
- Efficiency
- Changing workforce
- Multiple workers
- Alternate-day work
- Joining and leaving workers
- Partial work
- Workers, days, and hours
- Inverse proportion
These concepts also help with Pipes and Cisterns, Work and Wages, and project-completion problems.
Frequently Asked Questions
What is the basic concept of Time and Work?
If a person completes an entire job in (x) days, the amount of work completed in one day is:
[
\frac{1}{x}
]
Most Time and Work problems are based on combining or comparing such work rates.
How do I calculate the work done by two people together?
Add their individual one-day work rates.
If one person takes 10 days and another takes 15 days:
[
\frac{1}{10}+\frac{1}{15}
\frac{1}{6}
]
Therefore, together they require 6 days.
What is efficiency in Time and Work?
Efficiency represents the rate at which a person completes work.
If one worker is twice as efficient as another, the more efficient worker completes the same job in half the time.
Are efficiency and time directly proportional?
No.
They are inversely proportional.
More efficiency means less time.
What happens if the number of workers increases?
For the same work and equal efficiency, increasing the number of workers reduces the number of days required.
What is the worker-day method?
If every worker has equal efficiency:
[
\text{Total Work}
\text{Workers}\times\text{Days}
]
This is especially useful when workers join or leave midway.
How should I solve questions where somebody joins later?
First calculate how much work has already been completed.
Subtract it from the total work.
Then use the new combined work rate to complete the remaining portion.
Are Time and Work questions important for placement tests?
Yes. Time and Work is a common quantitative aptitude topic because it tests fractions, ratios, efficiency, inverse proportion, and logical calculation.
Conclusion
Time and Work problems become considerably easier when you think in terms of work rate rather than simply the number of days.
The most important relationship is:
Time to Complete Work → One-Day Work → Combined Work Rate → Required Time
For changing workforce problems, divide the situation into stages and calculate how much work is completed during each stage.
For efficiency problems, remember:
Higher efficiency = Lower completion time
With regular practice, even multi-worker and alternate-working questions can be solved systematically.
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