Time & Work Contractor Questions
Q1. A contractor
undertook to do a piece of work in 150 days. He employs 50 men to complete the work,
but after 90 days, he finds that 25% of the work had been completed. How many
more employee should be added to finish the work on scheduled time?
Solution:
Let the number of additional employee = x
Since, M1D1/W1 = M2D2/W2
(50*90)/(1/4)=(50+x)(60)/(3/4)
X=175
Q2. A contractor
undertook to complete a piece of work in 300 days. He employs 100 men to complete the work, but after 180
days, he finds that one-fourth of the work had been completed. How many more employee
should be added to finish the work 30 days earlier?
Solution:
Let the number of additional employee = x
Since, M1D1/W1 = M2D2/W2
(100*180)/(1/4)=(100+x)(90)/(3/4)
X=500
Q3. A contractor
undertakes to make a road in 40 days. He employs 16 men to complete the work,
but after 30 days, he finds that one-third of the work had been completed. How
many more employee should be added to finish the work 6 days late?
Solution:
Let the number of additional employee = x
Since, M1D1/W1 = M2D2/W2
(16*30)/(1/3)=(16+x)(16)/(2/3)
X=44
Q4. A contractor
undertook to do a piece of work in 75 days. He employs 25 men to complete the work,
but after 45 days, he finds that 1/3 of the work had been completed. How many
more employee should be added to finish the work on scheduled time?
Solution:
Let the number of additional employee = x
Since, M1D1/W1 = M2D2/W2
(25*45)/(1/3)=(25+x)(30)/(2/3)
X=50
Q5. A contractor
undertook to complete a piece of work in 100 days. He employs 20 men to complete the work, but after 75
days, he finds that one-fifth of the work had been completed. How many more employee
should be added to finish the work 5 days earlier?
Solution:
Let the number of additional employee = x
Since, M1D1/W1 = M2D2/W2
(20*75)/(1/5)=(20+x)(20)/(4/5)
X=280
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