Time and Work – Quantitative Aptitude Question
Time and Work – Quantitative Aptitude Question
Question 1:
A can do a work in 10 days and B can do the same work in 15 days. How long will they take to complete the work if both work together?
Options: A) 6 days
B) 5 days
C) 8 days
D) 12 days
Answer: A) 6 days
Explanation:
- A’s work rate = 1/10 (work done per day)
- B’s work rate = 1/15 (work done per day)
Together, they can complete (1/10 + 1/15) of the work per day.
LCM of 10 and 15 = 30, so:
- A’s rate = 3/30
- B’s rate = 2/30
Together, they complete (3 + 2) = 5/30 of the work per day. Thus, they will take 30/5 = 6 days to complete the work.
Question 2:
If 8 men can do a piece of work in 12 days, how long will it take for 6 men to do the same work?
Options: A) 14 days
B) 16 days
C) 18 days
D) 20 days
Answer: B) 16 days
Explanation: The work done is inversely proportional to the number of men. So, the relationship can be represented as:Men1×Days1=Men2×Days2
Substituting the given values:8×12=6×Days2 96=6×Days2 Days2=96/6=16 day
Question 3:
A alone can complete a piece of work in 12 days, and B alone can complete the same work in 18 days. If A and B work together, in how many days will they finish the work?
A) 6 days
B) 7.2 days
C) 8 days
D) 10 days
Answer: A) 6 days
Solution:
- Work done by A in 1 day = 1/12
- Work done by B in 1 day = 1/18
- Work done by A and B together in 1 day = (1/12) + (1/18) = (3 + 2) / 36 = 5/36
- Total days to complete the work = 36/5 = 7.2 days
Question 4:
If 8 men can complete a job in 20 days, how many men will be required to complete the same job in 10 days?
A) 12
B) 16
C) 20
D) 24
Answer: B) 16
Solution:
- Work = 8 men × 20 days = 160 man-days
- Required number of men = 160 man-days / 10 days = 16 men
Question 5:
A and B can complete a task in 10 days together. A alone can complete it in 15 days. How long will B take to finish the work alone?
A) 20 days
B) 25 days
C) 30 days
D) 35 days
Answer: C) 30 days
Solution:
- Work done by A and B together in 1 day = 1/10
- Work done by A alone in 1 day = 1/15
- Work done by B alone in 1 day = (1/10) – (1/15) = (3 – 2) / 30 = 1/30
- So, B alone takes 30 days to complete the work.
Question 6:
A contractor hires 10 workers to complete a work in 40 days. After 10 days, he hires 5 more workers. In how many more days will the work be completed?
A) 20 days
B) 22 days
C) 24 days
D) 25 days
Answer: C) 24 days
Solution:
- Total work = 10 workers × 40 days = 400 worker-days
- Work done in 10 days = 10 × 10 = 100 worker-days
- Work remaining = 400 – 100 = 300 worker-days
- Now, 15 workers are working. Time required = 300/15 = 20 days
- Total additional days = 20 days
Question 6:
A, B, and C together can complete a work in 6 days. A alone can do it in 12 days, and B alone can do it in 18 days. How long will C alone take to complete the work?
A) 9 days
B) 24 days
C) 36 days
D) 48 days
Answer: C) 36 days
Solution:
- Work done by A, B, and C together in 1 day = 1/6
- Work done by A alone in 1 day = 1/12
- Work done by B alone in 1 day = 1/18
- Work done by C alone in 1 day = (1/6) – (1/12) – (1/18)
- LCM of 6, 12, and 18 = 36
- (6 – 3 – 2) / 36 = 1/36
- So, C alone will take 36 days.
A can do a piece of work in 16 days, while B can do the same work in 24 days. They started working together, but A left after 4 days. In how many more days will B complete the remaining work?
A) 10 days
B) 12 days
C) 14 days
D) 16 days
Answer: B) 12 days
Solution:
- Work done by A in 1 day = 1/16
- Work done by B in 1 day = 1/24
- Work done by A and B together in 1 day = (1/16) + (1/24) = (3 + 2) / 48 = 5/48
- Work done by A and B in 4 days = 4 × (5/48) = 20/48 = 5/12
- Work remaining = 1 – 5/12 = 7/12
- Time taken by B alone to finish 7/12 of the work = (7/12) ÷ (1/24) = (7/12) × 24 = 12 days
Question 7:
Two pipes can fill a tank in 10 hours and 15 hours, respectively. A third pipe can empty the tank in 30 hours. If all three pipes are opened together, how long will it take to fill the tank?
A) 6 hours
B) 8 hours
C) 10 hours
D) 12 hours
Answer: B) 8 hours
Solution:
- Work done by first pipe in 1 hour = 1/10
- Work done by second pipe in 1 hour = 1/15
- Work done by third pipe in 1 hour (emptying) = -1/30
- Net work done in 1 hour = (1/10) + (1/15) – (1/30)
- LCM of 10, 15, and 30 = 30
- (3 + 2 – 1) / 30 = 4/30 = 2/15
- Time required to fill the tank = 15/2 = 7.5 hours
Question 8:
A work is to be completed in 24 days, and 30 men were employed for it. But after 6 days, 10 more men were added. In how many days will the work be completed?
A) 16 days
B) 18 days
C) 20 days
D) 22 days
Answer: A) 16 days
Solution:
- Total work = 30 men × 24 days = 720 man-days
- Work done in 6 days = 30 × 6 = 180 man-days
- Work remaining = 720 – 180 = 540 man-days
- Now, total men working = 40 men
- Time required to complete remaining work = 540 / 40 = 13.5 days
- Total days taken = 6 (initial) + 13.5 = 19.5 ≈ 16 days
Sample Quantitative Aptitude Question
Conclusion
Time and Work problems are essential in quantitative aptitude, testing a candidate’s ability to handle efficiency-based calculations. Understanding concepts like individual and group work efficiency, LCM method, and inverse proportionality can significantly improve problem-solving speed. Practicing a variety of questions helps in mastering different scenarios, such as pipes and cisterns or workforce changes.
