Twenty workers can finish a piece of work in 30 days

rate of 1 man per day=1/(20*30)=1/600
let d=number of days with 20 workers
20d/600+[15(35-d)]/600=1
d=15 days
D

We are given that 20 workers can complete a job in 30 days. Since rate = work/time, the rate of the 20 workers is 1/30. We can let x be the number of days it would take 5 fewer workers (i.e., 15 workers) to complete the job (from beginning to end); thus the rate of the 15 workers would be 1/x. Let’s determine x using the following proportion:

20/(1/30) = (15)/(1/x)

600 = 15x

40 = x

Thus, the number of days it takes for the 15 workers to complete the job (from beginning to end) is 15 and their rate is 1/40.

However, we need to determine at what point in time 5 of the 20 workers should leave the job so that the work is completed in 35 days. We can let this be t, and note that the work done by the 20 workers in these t days plus the work done by the 15 workers after these t days (i.e., in 35 - t days) will be the complete job. Recall that work = rate x time, so we have:

(1/30)t + (1/40)(35 - t) = 1

t/30 + (35 - t)/40 = 1

Multiplying the equation by 120, we have:

4t + 3(35 - t) = 120

4t + 105 - 3t = 120

t + 105 = 120

t = 15

Answer: D

20 workers finish the job in 30 days. i.e total work = 20*30 = 600 worker days

Not suppose after x days 5 workers left the job. i.e 20 workers worked for x days and 15 workers worked for (35 -x) days to finish the job.

600 = 20*x + 15*(35-x)

x = 15

Ans 15 days i.e option D

1 worker takes 600 days to finish the job. So rate for each worker is 1/600 per day.
rate of 15 workers is 15/600. Job done by 15 workers in 5 days is 75/600.
How many days would it have taken 5 workers to do 75/600 units of the job?
time taken by 5 workers to complete 75/600 of the job = 75/600 / rate of 5 workers= (75/600)/(5/600) = 75/5 = 15
workers left after 30-15=15 days

ans = 15 days

let the rate of work done by each worker be x and the total work be T. Also the no of days after 5 workers leave be d
20*30*x = T
20*d*x + 15*(35-d)*x = T = 20*30*x

solving for d we get d = 15 days

Let, assume 5 workers will leave after X days in 30 days,
basically, the x days are the time spent by all the workers from estimated 30 days dead line.

so, 30+5-x= \(\frac{20*(30-X)}{(20-5)}\)

>35- x= \(\frac{20*(30-X)}{15}\)

>20x-15x= 600 - 525

>X= 15 days
so, the ans is (D) 15 days.

20 workers worked for 30 days, so there are 600 units worked in total

After x days, there will be 5 worker left, therefore total amount of units worked up to that point = 20*x

Time allocated to the remaining 15 workers is (35- x) (since the work need to be completed in 35 days)

Together we have the following equation: 20*x + (35-x)*15 = 600 (still work the same units)

Solve for x we got x = 15 => Answers D

1. 20 workers in 30 days do 600 units of work (20*30=600). And each worker does 600/20= 30 units of work every day.
2. If 20 workers do work for 35 days they'll finish 35*20=1050 units of work.
3. Extra work is 1050-600=450 units of of work.
4. Hence 450/(30units of work per day)= 15 days

Bunuel
@veristasprepkarishma
Is this approach correct?

Total work = 20*30 = 600 units
If 5 workers leave, the number of remaining workers = 15, who work for 35 days.
So, these 15 do 15*35 = 525 units in 35 days.
Remaining units = 600-525 = 75 units
The number of days should 5 workers leave after = 75/5 = 15 days