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RahatRafi
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A group of workers was put on a job. From second day onwards, one work
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 Updated on: 08 Jan 2026, 09:23
Originally posted by RahatRafi on 19 Apr 2020, 05:59.
Last edited by Bunuel on 08 Jan 2026, 09:23, edited 1 time in total.
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A group of workers was put on a job. From second day onwards, one worker was withdrawn each day. The job was finished when the last worker was withdrawn. Had no worker been withdrawn at any stage, the group would have finished the job in 55% of the time. How many workers were there in the group?

A. 15
B. 14
C. 12
D. 10
E. 8
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A group of workers was put on a job. From second day onwards, one work
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19 Apr 2020, 06:40
Assuming n is the number of workers:
On day 1 they will produce n units of work.
On day 2 they will produce n-1 units of work
[..]
On day n they will produce 1 unit of work (and finish the job in n days)

The total amount of work produced is a sum of k with k = 1 to n. This is equal to n(n+1)/2

If no workers are removed. The production will be n units per day (constant).
It will take n(n+1)/2n days to finish and we know this will be equal to 55% of the previous time (which is n)

Hence we have: n(n+1)/2n = 55n/100
Which simplifies to 100n + 100 = 110n
Hence n = 10

Answer D!

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A group of workers was put on a job. From second day onwards, one work
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22 Apr 2020, 01:22
It took me 3 minutes and 30 seconds to solve this question. After I list the formula, it took me quite some time to solve it. I should be able to improve on the speed.

Assuming there are n workers in the group. For the next n days, (n) + (n-1) + (n-2) + ... 3 + 2 + 1 = n/2 * (n+1) worker-days are needed to complete the task.

The total worker-days can also be solved with 0.55 * n * n.

Now, 0.55 * n * n = n/2 * (n+1)
And n = 10
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A group of workers was put on a job. From second day onwards, one work
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23 Apr 2020, 09:50
RahatRafi
A group of workers was put on a job. From second day onwards, one worker was withdrawn each day.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 55% of the time. How many workers were there in the group?

A. 15

B. 14

C. 12

D. 10

E. 8
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EMPOWERgmatRichC
Hello,
May I know how to solve this question with easiest way?
Thank you..

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A group of workers was put on a job. From second day onwards, one work
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23 Apr 2020, 13:30
Hi All,

To start, this question is a bit vague in how it's worded, but the 'intent' is ultimately that at the end of each work day, one of the workers leaves (meaning that there is one fewer worker at the start of the next work day). In addition, we are meant to assume that all workers work at an equal (and consistent) rate for the entirety of the job. This question can be solved rather easily by TESTing THE ANSWERS.

Since the prompt refers to 55% - which is a rather 'exact' percentage - and not anything stranger (such as 57.33333% or an ugly fraction such as 7/13), it's likely that we'll be starting off with a "round number" of workers. Let's TEST Answer D first...

Answer D: 10 workers

Since 1 worker is removed on each day after the 1st day, the total number of worker-days of effort would be:

Day 1: 10
Day 2: 9
Day 3: 8
etc.

10+9+8....+3+2+1 = 55 total worker-days of effort to complete the job. The prompt tells us that the job is finished at the end of the last day (when the last worker would leave), so we now know what we are measuring against: the entire job takes 55 worker-days of effort.... and with one worker leaving each day, we would need 10 workers to get the job done under these circumstances.

If ALL 10 of the workers STAYED for each day though, how many days would it take to complete 55 worker-days of effort? That's actually a pretty easy calculation:
55/10 = 5.5 days

What percent is 5.5 relative to 10..... 5.5/10 = 55%. This is an exact match for what we were told in the prompt, so this MUST be the answer!

Final Answer:
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D

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A group of workers was put on a job. From second day onwards, one work
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27 Apr 2020, 02:54
RahatRafi
A group of workers was put on a job. From 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 55% of the time. How many workers were there in the group?

A. 15

B. 14

C. 12

D. 10

E. 8
Show more

Let n be the number of workers. Since one worker was withdrawn each day from the second day onwards and there was one worker left on the last day, the job took n days. However, since the job would have been completed in 55% of the time if no workers had been withdrawn at any stage, it took 0.55n days to complete the job. Therefore, the rate of the n workers was 1/(0.55n) of the job per day and the rate of each worker was [1/(0.55n)]/n = 1/(0.55n^2) of the job per day. Now, if we let r = the rate of each worker, we can create the equation:

nr + (n - 1)r + (n - 2)r + … + 2r + r = 1

r[n + (n - 1) + (n - 2) + … + 2 + 1] = 1

Note that the sum indicated inside the brackets is the sum of n consecutive integers, which can be expressed as n(n + 1)/2.

r[n(n + 1) / 2] = 1

r = 2 / [n(n + 1)]

Since r, the rate of each worker, is also 1/(0.55n^2), we can create the equation:

2/[n(n + 1)] = 1/(0.55n^2)

2(0.55n^2) = n(n + 1)

1.1n^2 = n^2 + n

0.1n^2 = n

Dividing both sides by n (since n can’t be 0), we have:

0.1n = 1

n = 10

Answer: D

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A group of workers was put on a job. From second day onwards, one work
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12 Oct 2024, 00:16
Here's how I solved it in less than 2 minutes:

Since the ratio of their time is 55:100 = 11:20 (no worker withdrawn:withdrawn), so the ratio of their rates must be 20:11. So the combined rate of all of them should be 20x.

Now, combined rate = (individual rate)(no. of workers) = multiple of 20. Only 2*10 works so individual rate = 2 and no. of workers = 10.
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A group of workers was put on a job. From second day onwards, one work
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14 Oct 2024, 19:25
I am struggling to understand how the LHS has n^2

My logic was - 0.55 * n * r = n/2 * (n+1) * r -> This obviously gives me wrong answer, but I am not able to figure out my miss.

zhanbo
A group of workers was put on a job. From 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 55% of the time. How many workers were there in the group?
A. 15

B. 14

C. 12

D. 10

E. 8

It took me 3 minutes and 30 seconds to solve this question. After I list the formula, it took me quite some time to solve it. I should be able to improve on the speed.

Assuming there are n workers in the group. For the next n days, (n) + (n-1) + (n-2) + ... 3 + 2 + 1 = n/2 * (n+1) worker-days are needed to complete the task.

The total worker-days can also be solved with 0.55 * n * n.

Now, 0.55 * n * n = n/2 * (n+1)
And n = 10
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A group of workers was put on a job. From second day onwards, one work
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15 Oct 2024, 02:08
glagad
I am struggling to understand how the LHS has n^2

My logic was - 0.55 * n * r = n/2 * (n+1) * r -> This obviously gives me wrong answer, but I am not able to figure out my miss.

zhanbo
A group of workers was put on a job. From 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 55% of the time. How many workers were there in the group?
A. 15

B. 14

C. 12

D. 10

E. 8

It took me 3 minutes and 30 seconds to solve this question. After I list the formula, it took me quite some time to solve it. I should be able to improve on the speed.

Assuming there are n workers in the group. For the next n days, (n) + (n-1) + (n-2) + ... 3 + 2 + 1 = n/2 * (n+1) worker-days are needed to complete the task.

The total worker-days can also be solved with 0.55 * n * n.

Now, 0.55 * n * n = n/2 * (n+1)
And n = 10
Show more

If no worker were withdrawn, all n workers would work for n days. Thus, the LHS has n^2 because it represents the total number of worker-days with no withdrawals. With n workers working for n days, the total is n * n = n^2 worker-days.

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A group of workers was put on a job. From second day onwards, one work
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11 Feb 2025, 10:55
RahatRafi
A group of workers was put on a job. From 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 55% of the time. How many workers were there in the group?

A. 15

B. 14

C. 12

D. 10

E. 8
Show more
So what i did to solve this question was to take time as 200 days or 400 days for total numbers of days took for doing the work with withdraw so the amount of work required to do the job without withdraw is 200*0.55= 110 days and 400*0.55 = 220 days. so now the only number that divides 300, 110 and 400, 220 both is 10 so hence that is the answer.
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A group of workers was put on a job. From second day onwards, one work
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08 Jan 2026, 09:18
ZIX
Here's how I solved it in less than 2 minutes:

Since the ratio of their time is 55:100 = 11:20 (no worker withdrawn:withdrawn), so the ratio of their rates must be 20:11. So the combined rate of all of them should be 20x.

Now, combined rate = (individual rate)(no. of workers) = multiple of 20. Only 2*10 works so individual rate = 2 and no. of workers = 10.
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Very interesting!

But wouldn't 4 x 15 also work (60 is a multiple of 20), and similarly 12 x 5 (60) and 8 x 5 (40)?
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