Sign up now
to enroll in courses, follow best educators, interact with the community and track your progress.
Download
ESE 2018 GS Paper Solution Part-10
284 plays

More
In this lecture Pawan Saini have discussed Engineering Services GS Paper questions solution from question no 69 to 75.

Pawan Saini
Manager PSU, Worked @ SAIL, Cleared GATE 2012, 2013, SSC JE, CDS, AFCAT, BARC. Best of luck for ESE Prelims.

U
Unacademy user
water images ki problems ni h
  1. 19.10 Engineering Services 2018 GS Paper Solution Part-10 Presented by PAWAN SAINI


  2. PAWAN SAIN ABOUT ME PAWAN SAINI Oriental Institute Of Science and Technology Bhopal ASSISTANT MANAGER IN RITES (A PSU under Ministry of Railways) Joined in 2013 posted at NTPC Sipat Project Previously Worked at SAlIL Bokaro Love to do Social Service Travelling Teaching Spiritual Activities IRIMEE DLW ICF IRITM TKD GOLDEN ROCK SAIL NTPC CIL BHEL


  3. ESE 2018 GS Paper Solution 69. An association of two organisms of different species for mutual benefit, and where the individuals may not be able to survive separately, is called (a) Commensalism (b) Parasitic (c) Non-symbiotic (d) Symbiotic


  4. Ans. d) Symbiotic relationships are a special type of interaction between species. Sometimes beneficial, sometimes harmful, these relationships are essential to many organisms and ecosystems, and they provide a balance that can only be achieved by working together. Mutualism or interspecies reciprocal altruism is a relationship between individuals of different species where both individuals benefit.


  5. Commensalism describes a relationship between two living organisms where one benefits and the other is not significantly harmed or helped. A parasitic relationship is one in which one member of the association benefits while the other is harmed. Non-symbiotic mutualism: the species do not live together, nor are dependent on each other; the relationship is faculatitive or opportunistic but does profit the organisms when together.


  6. ESE 2018 GS Paper Solution A simple project comprises of two start-to-en parallel paths, each with series, with no interpath dependencies. The a, m, b data in days) for each activity are shown in the diagra Assuming that three activities in series are enough for further computations, what will be the total project duration and its standard deviation ? 70. 4, 6, 8 5, 8, 11 2, 3, 4 6, 7, 8 12, 12, 18 9, 15, 18 14 aays 35 days and ClayYS 2 1 2 1 3 5 aays 2 days and days an days and Cb) 34 *S 13 35 a days 6 1 11 aays 34 6


  7. Ans. d) Project duration on the upper path 3+6+8-17 days. Project duration for lower path 34.5 days. Standard deviation would be - (1(8-6)/612 + [(18-12)/6]2 + [(18-9)/61211/2 = 11 /6 days.


  8. Basics of Project Management 41. This is the minimum possible time in which an activity can be completed under the most ideal condition? a). Optimistic Time(to) b]. Pessimistic Timetp) c). Most Likely Time(tm) d). None


  9. ESE 2018 GS Paper Solution 71. Crashing is (a) (b) Abandoning the project Completing the project with all possible haste Reduction of duration for a few of the activities (c) (d) Reducing the cost of the project with all needful modifications


  10. Ans. c)


  11. ESE 2018 GS Paper Solution 72. ABC analysis in materials management is a method of classifying the inventories based on the (a) Economic order quantity (b) Value of annual usage of the items (c) Volume of material consumption (d) Quantity of materials used ""


  12. Distribution of ABC Class Number of ltems Total amount Required ABC Class 5% 10% 85% 100% 70% 15% 15% 100% Total


  13. Ans. c)


  14. Comparison between PERT and CPM S.no PERT CPM Event oriented Activity oriented 2 Associated with probabilistic Associated with deterministic activities Based on three time estimate Time required to complete activities Based on single time estimate Used for repetitive work 3 4 activity is not certain Cost analysis is not considered Cost analysis is importance & priority is given to minimize the cost Used for research and development project Used for construction project and production line


  15. ESE 2018 GS Paper Solution 74. The objective function z = 3x1 + 5x2 is to be maximized subject to constraints x1 + 2x2 s 200 x1 x2 150 The values of x1 and x2 in this context are, respectively (a) 100 and 75 (b) 125 and 75 (c) 100 and 50 (a) 125 and 50