56. Match the following
    

    List-I (Problem areas)	List-II (Techniques)
    A. JIT	1. CRAFT
    B. Computer assisted	2. PERT layout
    C. Scheduling	3. Johnson's rule
    D. Simulation	4. Kanbans
    	5. EOQ rule
    	6. Monte Carl

1. 1. A –4, B–5, C–3, D–1.

   
   
   

   2. A –1, B–5, C–3, D–4.

   
   
   

   3. A –4, B–1, C–3, D–6

   
   
   

   4. A –5, B–2, C–3, D–6

   
   
   

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1. View Hint View Answer Discuss in Forum

   (a) JIT – (4) Kanbans (b) Computer –(1) CRAFT Assisted Layout (c) Scheduling –(3) John son’s Rule (d) Simulations – (6) Monte carl
   ⇒ A –4, B–1, C–3, D–6.
   

   Correct Option: C

   (a) JIT – (4) Kanbans (b) Computer –(1) CRAFT Assisted Layout (c) Scheduling –(3) John son’s Rule (d) Simulations – (6) Monte carl
   ⇒ A –4, B–1, C–3, D–6.
   

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57. The standard deviation of the critical path of the project is
    

1. 1. √151 days
      

   
   
   

   2. √155 days
      

   
   
   

   3. √200 days
      

   
   
   

   4. √238 days

   
   
   

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1. View Hint View Answer Discuss in Forum

   √25 + 81 + 36 + 9 = √151 days.
   

   Correct Option: A

   √25 + 81 + 36 + 9 = √151 days.
   

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58. The expected time (t_e) of a PERT activity in terms of optimistic time (t_o), pessimistic time (t_p) and most likely time (t_l) is given by
    

1. 1. te =	to + 4tl + tp
      	6

      
      

   
   
   

   2. te =	to + 4tp + tl
      	6

      
      

   
   
   

   3. te =	to + 4tl + tp
      	3

      
      

   
   
   

   4. te =	to + 4tp + tl
      	3

      
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: A

   NA

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59. A residential school stipulates the study hours as 8 : 00 pm to 10 : 30 pm. Warden makes random checks on a certain student 11 occasions a day during the study hours over a period of 10 days and observes that he is studying on 71 occasions. Using 95% confidence interval, the estimated minimum hours of his study during that 10 day period is
    

1. 1. 8.5 hours
      

   
   
   

   2. 13.9 hours
      

   
   
   

   3. 16.1 hours
      

   
   
   

   4. 18.4 hours

   
   
   

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1. View Hint View Answer Discuss in Forum

   Number of total observation in 10 days = 11 × 10 = 110
   No. of observation when studying = 71
   

   ∴ P =	71	= 0.6455
   	110

   
   (Probability of studying)
   Total studying hour in 10 days = (2.5 hr) × 10
   Hence, Minimum no. of hours of studying in 10 days = (25 hr) × p = 25 × 0.6455 = 16.13 hrs.

   Correct Option: C

   Number of total observation in 10 days = 11 × 10 = 110
   No. of observation when studying = 71
   

   ∴ P =	71	= 0.6455
   	110

   
   (Probability of studying)
   Total studying hour in 10 days = (2.5 hr) × 10
   Hence, Minimum no. of hours of studying in 10 days = (25 hr) × p = 25 × 0.6455 = 16.13 hrs.

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60. The expected completion time of the project is
    

1. 1. 238 days
      

   
   
   

   2. 224 days
      

   
   
   

   3. 171 days
      

   
   
   

   4. 155 days

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   ⇒ a – c – e – f
   ⇒ 30 + 60 + 45 + 20 = 155 days

   Correct Option: D

   
   ⇒ a – c – e – f
   ⇒ 30 + 60 + 45 + 20 = 155 days

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