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30 Dec 2021, 03:45
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
83% (01:56) correct
17%
(02:15)
wrong
based on 41
sessions
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Not Attempted Yet
A contractor was required to complete a project in 40 days. The contractor employed 100 workers at the start and then employed another 100 workers after 35 days. The project was completed as required. If the contractor had not employed additional workers, then how much extra time would the project have taken?
(A) 2 days
(B) 3 days
(C) 4 days
(D) 5 days
(E) 6 days
(A) 2 days
(B) 3 days
(C) 4 days
(D) 5 days
(E) 6 days
30 Dec 2021, 11:48
Kudos
Bookmarks
Bunuel
Let the efficiency of the worker be x.
So, amount of work completed by 100 workers in 35days = 35*x*100 = 3500x
So, amount of work left = 1-3500x |eq1
Rest of the work was completed in 5 days by 100 additional workers, so the new equation will be: 5*x*200 =1000x |eq2
Equating eq1 and eq2
1-3500x = 1000x
x = 1/4500
So if no additional worker was hired let's say the work would have been completed in d days.
So, 1/4500 *100*d =1
d = 45 days.
Which is 5 more days than the assigned time.
D is the answer.
30 Dec 2021, 13:12
Kudos
Bookmarks
Assume each worker can complete x units of work each day.
Work done per day by 100 worker = 100x units
Work completed by 100 workers in 35 days = 3500x units.
Work completed by 200 workers in 05 days = 1000x units.
Total work = 4500x units.
Suppose only 100 worker would have worked for the entire duration , time taken would be =\(\frac{4500x }{ 100x}\)
= 45 days.
Additional time taken = 05 days.
IMO - D
Work done per day by 100 worker = 100x units
Work completed by 100 workers in 35 days = 3500x units.
Work completed by 200 workers in 05 days = 1000x units.
Total work = 4500x units.
Suppose only 100 worker would have worked for the entire duration , time taken would be =\(\frac{4500x }{ 100x}\)
= 45 days.
Additional time taken = 05 days.
IMO - D
