Quantitative Aptitude
Time, Work, and Projects
198 Questions
Time and work questions test the ability to calculate the time required to complete tasks involving individual or combined worker efficiency. These problems frequently appear in competitive exams to assess logical calculation skills. Topics include pipes, cisterns, group efficiency, and hourly work rates.
Combined work efficiencyMen and days calculationsWorker efficiency ratiosWorking hours per dayLeaving and joining workers
Time, Work, and Projects Questions
A
Correct answer
Explanation
Test effort is measured in person-months, which represents the work done by testers in a month. With 5 testers working for 1 month, the test effort for the first month is 5 person-months. This is simply the number of testers allocated since effort is calculated per month of work.
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24Hrs.
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1 week.
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No limit.
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248 Days.
D
Correct answer
Explanation
Mainframe jobs can run continuously for maximum 248 days. This is a system limitation due to how job start time is tracked internally. Options of 24 hours, 1 week, or no limit are all incorrect.
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3:45 & 3:30
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3:30 & 3:45
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3:38 & 3:35
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3:38 & 3:40
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00:55
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1.00
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1.05
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3.39
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All of the above
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1 Day
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1 week
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8 Days
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3 Days
C
Correct answer
Explanation
CMR scheduling requires advance notice to ensure resource allocation and job preparation. 8 days allows sufficient time for processing, validation, and queue placement. Shorter timeframes (1-3 days) are typically insufficient for proper scheduling procedures.
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357912mins
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357812mins
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357712mins
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357612mins
A
Correct answer
Explanation
TIME=MAXIMUM allows a job to run for 357912 minutes, which is approximately 248 days. This value represents the maximum allowable time in JCL and is used for jobs that should run until completion without time restrictions.
D
Correct answer
Explanation
For skilled:
men -> days -> amount of wall
5 -> 20 -> 1
1 -> 100 -> 1
1 -> 1 -> 1/100
For semi-skilled:
men -> days -> amount of wall
8 -> 25 -> 1
1 -> 200 -> 1
1 -> 1 -> 1/200
For unskilled:
men -> days -> amount of wall
10 -> 30 -> 1
1 -> 300 -> 1
1 -> 1 -> 1/300
Amount of work 2 skilled in 1 day = 2/100
Amount of work 6 semi-skilled in 1 day = 6/200
Amount of work 5 unskilled in 1 day = 5/300
Total work in 1 day
= 2/100+6/200+5/300
= 12/600+18/600+10/600
= 1/15
So complete work will take 15 days
C
Correct answer
Explanation
Let each truck carry 100 units.
2800 = 4n + e
n = normal
3000 = 10n + e
e = excess/pending
$\therefore$$ n =\dfrac{100}{3}, e =\dfrac{8000}{3} $
5days $\Rightarrow$$500x = \dfrac{5.100}{3} + \dfrac{8000}{3}$
$\Rightarrow$$500x = \dfrac{8500}{3}17 \Rightarrow x > 5$
Minimum possible = 6
A
Correct answer
Explanation
Solving from the some algorithm solved in previous question the sum is 147 for the profit.
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All tasks are completed
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T1 and T6 are left out
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T1 and T8 are left out
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T4 and T6 are left out
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20 days
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18 days
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16 days
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15 days
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18 days
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72/7 days
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6 days
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144 days
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72/3 days
B
Correct answer
Explanation
When X is doing 40 % of work, it takes 12 days.
When X is doing 60% of work, it takes 18 days.
When X + Y + Z or X + 1/2X + 1/4X is doing 60% of work it takes
18/(1 + 1/2 + 1/4) = 72/7 days.
A
Correct answer
Explanation
Work done by Rahul in 5 days is 5/15.
Work done by Rishabh in 5 days is 5/10.
Total fraction of work done by Rahul and Rishabh together in 5 days = 5/10 + 5/15
Fraction of work left = 1 - (5/15 + 5/10) = 1/6