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A group of workers was put on a job. From the second day onwards, one worker was withdrawn each day. The job was finished when the last worker was withdrawn. Had no worker been withdrawn at any stage, the group would have finished the job in two-thirds the time. How many workers were there in the group?
- a)2
- b)3
- c)5
- d)11
Correct answer is option 'B'. Can you explain this answer?
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A group of workers was put on a job. From the second day onwards, one...
Given:
- A group of workers was put on a job.
- From the second day onwards, one worker was withdrawn each day.
- The job was finished when the last worker was withdrawn.
- Had no worker been withdrawn at any stage, the group would have finished the job in two-thirds the time.
To find: How many workers were there in the group?
Solution:
Let's assume that the total number of workers in the group is 'x'.
Case 1: When one worker was withdrawn each day
- On the first day, all 'x' workers worked.
- On the second day, 'x-1' workers worked.
- On the third day, 'x-2' workers worked.
- Similarly, on the nth day, 'x-(n-1)' workers worked.
Using the formula for the sum of an arithmetic progression:
- Total work done in 'n' days = n/2 * [2x - (n-1)]
Since the job was finished when the last worker was withdrawn:
- Total work done in 'x' days = x/2 * [2x - (x-1)] = x^2
Case 2: When no worker was withdrawn
- Let's assume that the time taken to finish the job when no worker was withdrawn is 't'.
- In this case, all 'x' workers worked for 't' days.
Therefore, the total work done is:
- Total work done in 't' days = x*t
As per the given condition:
- Total work done in 'x' days = 2/3 * Total work done in 't' days
- x^2 = 2/3 * x*t
- t = 3/2 * x
Substituting this value of 't' in the equation for total work done in 't' days:
- Total work done in 't' days = x * t = x * 3/2 * x = 3/2 * x^2
Equating this with the total work done in 'x' days (from case 1):
- 3/2 * x^2 = x^2
- x = 2
Therefore, there were originally '2' workers in the group.
Answer: Option B) 3
- A group of workers was put on a job.
- From the second day onwards, one worker was withdrawn each day.
- The job was finished when the last worker was withdrawn.
- Had no worker been withdrawn at any stage, the group would have finished the job in two-thirds the time.
To find: How many workers were there in the group?
Solution:
Let's assume that the total number of workers in the group is 'x'.
Case 1: When one worker was withdrawn each day
- On the first day, all 'x' workers worked.
- On the second day, 'x-1' workers worked.
- On the third day, 'x-2' workers worked.
- Similarly, on the nth day, 'x-(n-1)' workers worked.
Using the formula for the sum of an arithmetic progression:
- Total work done in 'n' days = n/2 * [2x - (n-1)]
Since the job was finished when the last worker was withdrawn:
- Total work done in 'x' days = x/2 * [2x - (x-1)] = x^2
Case 2: When no worker was withdrawn
- Let's assume that the time taken to finish the job when no worker was withdrawn is 't'.
- In this case, all 'x' workers worked for 't' days.
Therefore, the total work done is:
- Total work done in 't' days = x*t
As per the given condition:
- Total work done in 'x' days = 2/3 * Total work done in 't' days
- x^2 = 2/3 * x*t
- t = 3/2 * x
Substituting this value of 't' in the equation for total work done in 't' days:
- Total work done in 't' days = x * t = x * 3/2 * x = 3/2 * x^2
Equating this with the total work done in 'x' days (from case 1):
- 3/2 * x^2 = x^2
- x = 2
Therefore, there were originally '2' workers in the group.
Answer: Option B) 3
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Community Answer
A group of workers was put on a job. From the second day onwards, one...
Let the total number of workers be n
First day: n workers will be working
Second day: n-1 workers will be working
Third day: n-2 workers will be working
......
Last day: 1 worker will be working
Hence, it takes n days to finish the work
Total in all n days, the number of workers worked = n + (n-1) + (n-2) + … + 3 + 2 + 1
= n(n+1)/2
If all the n workers are working together, then they will finish it in (2/3)n days
Hence, work done = n X (2/3)n
Therefore, n(n+1)/2 = 2n2/3
3n + 3 = 4n or n = 3
Hence, the number of workers = 3
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A group of workers was put on a job. From the second day onwards, one worker was withdrawn each day. The job was finished when the last worker was withdrawn. Had no worker been withdrawn at any stage, the group would have finished the job in two-thirds the time. How many workers were there in the group?a)2b)3c)5d)11Correct answer is option 'B'. Can you explain this answer?
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A group of workers was put on a job. From the second day onwards, one worker was withdrawn each day. The job was finished when the last worker was withdrawn. Had no worker been withdrawn at any stage, the group would have finished the job in two-thirds the time. How many workers were there in the group?a)2b)3c)5d)11Correct answer is option 'B'. Can you explain this answer? for CAT 2024 is part of CAT preparation. The Question and answers have been prepared according to the CAT exam syllabus. Information about A group of workers was put on a job. From the second day onwards, one worker was withdrawn each day. The job was finished when the last worker was withdrawn. Had no worker been withdrawn at any stage, the group would have finished the job in two-thirds the time. How many workers were there in the group?a)2b)3c)5d)11Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for CAT 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A group of workers was put on a job. From the second day onwards, one worker was withdrawn each day. The job was finished when the last worker was withdrawn. Had no worker been withdrawn at any stage, the group would have finished the job in two-thirds the time. How many workers were there in the group?a)2b)3c)5d)11Correct answer is option 'B'. Can you explain this answer?.
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