Time and Work
- A can do a piece of work in 18 days and B in 12 days. They began the work together, but B left the work 3 days before its completion. In how many days, in all, was the work completed?
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Let the work be finished in x days.
According to the question, A worked for x days while B worked for (x – 3) days∴ x + x - 3 = 1 18 12 ⇒ 2x + 3x - 9 = 1 36
⇒ 5x – 9 = 36
↠ 5x = 45⇒ x = 45 = 9 5
Hence, the work was completed in 9 days.
Aliter :
Using Rule 8,
Here, x = 18, y = 12, m = 3Total time taken = 
y + m 
x x + y = 12 + 3 × 18 = 9 days 18 + 12 Correct Option: D
Let the work be finished in x days.
According to the question, A worked for x days while B worked for (x – 3) days∴ x + x - 3 = 1 18 12 ⇒ 2x + 3x - 9 = 1 36
⇒ 5x – 9 = 36
↠ 5x = 45⇒ x = 45 = 9 5
Hence, the work was completed in 9 days.
Aliter :
Using Rule 8,
Here, x = 18, y = 12, m = 3Total time taken = y + m x x + y = 12 + 3 × 18 = 9 days 18 + 12
- 40 men can complete a work in 40 days. They started the work together. But at the end of each 10th day, 5 men left the job. The work would have been completed in
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For the first 10 days 40 men worked.
Now, 40 men can complete the work in 40 days
∴ 1 man will complete the same work in 1600 days∴ 1 man’s 1 day’s work 1 1600
∴ Part of work done in first 10 days= 1 4
For the next 10 days 35 men worked.
Part of the work done= 1 × 35 × 10 = 7 1600 16
For the next 10 days, 30 men worked
Part of the work done= 30 × 10 = 3 1600 16
For the next 10 days, 25 men worked. Part of the work done= 25 × 10 = 5 1600 32
Similarly, part of the work done by 20 men in next 10 days= 20 × 10 = 1 1600 8
Work done in 50 days= 1 + 7 + 3 + 5 + 1 4 32 16 32 8 = 8 + 7 + 3 + 5 + 4 = 30 = 15 32 32 16
∴ Remaining work= 1 - 15 = 1 16 16
Now 15 men remain to work 15 men’s 1 day’s work= 15 1600 ∴ Time taken to complete 1 part of work 16 = 1600 × 1 15 16 = 20 = 6 2 days 3 3 ∴ Total time = 50 + 6 2 = 56 2 days 3 3 Correct Option: A
For the first 10 days 40 men worked.
Now, 40 men can complete the work in 40 days
∴ 1 man will complete the same work in 1600 days∴ 1 man’s 1 day’s work 1 1600
∴ Part of work done in first 10 days= 1 4
For the next 10 days 35 men worked.
Part of the work done= 1 × 35 × 10 = 7 1600 16
For the next 10 days, 30 men worked
Part of the work done= 30 × 10 = 3 1600 16
For the next 10 days, 25 men worked. Part of the work done= 25 × 10 = 5 1600 32
Similarly, part of the work done by 20 men in next 10 days= 20 × 10 = 1 1600 8
Work done in 50 days= 1 + 7 + 3 + 5 + 1 4 32 16 32 8 = 8 + 7 + 3 + 5 + 4 = 30 = 15 32 32 16
∴ Remaining work= 1 - 15 = 1 16 16
Now 15 men remain to work 15 men’s 1 day’s work= 15 1600 ∴ Time taken to complete 1 part of work 16 = 1600 × 1 15 16 = 20 = 6 2 days 3 3 ∴ Total time = 50 + 6 2 = 56 2 days 3 3
- A can complete a piece of work in 10 days, B in 15 days and C in 20 days. A and C worked together for two days and then A was replaced by B. In how many days, altogether, was the work completed ?
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Work done by (A + C) in 2 days
= 2 
1 + 1 10 20 = 2 2 + 1 = 6 = 3 20 20 10 Remaining work = 1 - 3 = 7 10 10
(B + C)’s 1 day’s work= 1 + 1 = 4 + 3 = 7 15 20 60 60
∴ Time taken by (B + C) to finish7 part of the work 10 = 60 × 7 = 6 days 7 10
∴ Total time = 2 + 6 = 8 daysCorrect Option: D
Work done by (A + C) in 2 days
= 2 1 + 1 10 20 = 2 2 + 1 = 6 = 3 20 20 10 Remaining work = 1 - 3 = 7 10 10
(B + C)’s 1 day’s work= 1 + 1 = 4 + 3 = 7 15 20 60 60
∴ Time taken by (B + C) to finish7 part of the work 10 = 60 × 7 = 6 days 7 10
∴ Total time = 2 + 6 = 8 days
- 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 5 days of its completion. B also left the work 2 days after A left. In how many days was the work completed?
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Let the work be completed in x days.
According to the questionx - 5 + x - 3 + x = 1 10 12 15 ⇒ 6x - 30 + 5x - 15 + 4x = 1 60
⇒ 15x – 45 = 60⇒ 15x = 105 ⇒ x = 105 = 7 15
Hence, the work will be completed in 7 days.Correct Option: C
Let the work be completed in x days.
According to the questionx - 5 + x - 3 + x = 1 10 12 15 ⇒ 6x - 30 + 5x - 15 + 4x = 1 60
⇒ 15x – 45 = 60⇒ 15x = 105 ⇒ x = 105 = 7 15
Hence, the work will be completed in 7 days.
- A and B can complete a piece of work in 12 and 18 days respectively. A begins to do the work and they work alternatively one at a time for one day each. The whole work will be completed in
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A’s 1 day’s work = 1 12 B’s 1 day’s work = 1 18
Part of work done by A and B in first two days= 1 + 1 = 3 + 2 = 5 12 18 36 36
Part of work done by A and B in 14 days= 35 36
[14 days to be taken randomly]Remaining work = 1 - 35 = 1 36 36
Now A will work for 15th dayA will do the 1 work in 1 × 12 36 36 = 1 day 3
∴ Total Work will be done in14 1 day. 3 Correct Option: A
A’s 1 day’s work = 1 12 B’s 1 day’s work = 1 18
Part of work done by A and B in first two days= 1 + 1 = 3 + 2 = 5 12 18 36 36
Part of work done by A and B in 14 days= 35 36
[14 days to be taken randomly]Remaining work = 1 - 35 = 1 36 36
Now A will work for 15th dayA will do the 1 work in 1 × 12 36 36 = 1 day 3
∴ Total Work will be done in14 1 day. 3
