540 man-days of work can be completed by a certain number of men in some days. If the numbers of people (men) are increased by 27, then the number of day required to complete the same work is decreased by 15. The number of days are required to complete the three times work (than the previous/actual work) b7 27 men?
Answers
Answer: 60 days
Step-by-step explanation:
This can be solved by using the MDW formula, therefore,
[M1*D1]/W1 = [M2*D2]/W2 …. (i)
Step 1:
Let the no. of men completing the work “W1” be “M1” & the certain no. of days be “D1”. Therefore based on the formula given above we can write
∴ [M1 * D1]/W1 = 540 …. (ii)
Step 2:
Now, here the work is 3 times the previous work is done by 27 men.
Therefore,
W2 = 3 (considering W1 = 1 work done) and M2 = 27 men … (iii)
Let “D2” be the required no. of days to complete 3 times work by 27 men.
So, by substituting the values from (ii) & (iii) in (i), we get
The required no. of days as,
[M1*D1]/W1 = [27*D2]/3*W1
⇒ 540 = [27 * D2] / 3
⇒ D2 = [540 * 3] / 27 = 60 days
Thus, 60 days are required to complete three times the previous work by 27 men.
Answer:
60 Days
Step-by-step explanation:
Work given = 540 man days
three time of the work will be = 3 * 540 man days
=> Total Work would be = 1620 Man days
number of men working = 27
Number of Days required = Total Work to be done / total people
=> Number of Days required = 1620 Man Days / 27 Men
= 60 Days