Time & Operate Contractor Questions Amongst Solutions For Ibps Po & Ssc Cgl 2018
December 02, 2018
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Time & Work Contractor Questions
Q1. Influenza A virus subtype H5N1 contractor undertook to create a slice of function inwards 150 days. He employs l men to consummate the work, simply later ninety days, he finds that 25% of the function had been completed. How many to a greater extent than employee should locomote added to terminate the function on scheduled time?
Solution: Let the disclose of additional employee = x
Since, M1D1/W1 = M2D2/W2
(50*90)/(1/4)=(50+x)(60)/(3/4)
X=175
Q2. Influenza A virus subtype H5N1 contractor undertook to consummate a slice of function inwards 300 days. He employs 100 men to consummate the work, simply later 180 days, he finds that one-fourth of the function had been completed. How many to a greater extent than employee should locomote added to terminate the function xxx days earlier?
Solution:
Let the disclose of additional employee = x
Since, M1D1/W1 = M2D2/W2
(100*180)/(1/4)=(100+x)(90)/(3/4)
X=500
Q3. Influenza A virus subtype H5N1 contractor undertakes to brand a route inwards xl days. He employs xvi men to consummate the work, simply later xxx days, he finds that one-third of the function had been completed. How many to a greater extent than employee should locomote added to terminate the function half-dozen days late?
Solution:
Let the disclose of additional employee = x
Since, M1D1/W1 = M2D2/W2
(16*30)/(1/3)=(16+x)(16)/(2/3)
X=44
Q4. Influenza A virus subtype H5N1 contractor undertook to create a slice of function inwards 75 days. He employs 25 men to consummate the work, simply later 45 days, he finds that 1/3 of the function had been completed. How many to a greater extent than employee should locomote added to terminate the function on scheduled time?
Solution:
Let the disclose of additional employee = x
Since, M1D1/W1 = M2D2/W2
(25*45)/(1/3)=(25+x)(30)/(2/3)
X=50
Q5. Influenza A virus subtype H5N1 contractor undertook to consummate a slice of function inwards 100 days. He employs twenty men to consummate the work, simply later 75 days, he finds that one-fifth of the function had been completed. How many to a greater extent than employee should locomote added to terminate the function v days earlier?
Solution:
Let the disclose of additional employee = x
Since, M1D1/W1 = M2D2/W2
(20*75)/(1/5)=(20+x)(20)/(4/5)
X=280