UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 51-53)

UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 51-55)

Q 51. At a particular time of computation, the value of a counting semaphore is 10. Then 12 P operations and “x” V operations were performed on this semaphore. If the final value of semaphore is 7, x will be:

(A) 8
(B) 9
(C) 10
(D) 11

Answer :  (B) 9
Solution : P operation : Decrements the value of semaphore by 1
V operation : Increments the value of semaphore by 1
Initially, value of semaphore = 10
After 12 P operations, value of semaphore = 10 – 12 = – 2
Now, after x V operations, value of semaphore = 7

– 2 + xV = 7
xV = 2 + 7 = 9 

Q 52. In a paged memory, the page hit ratio is 0.40. The time required to access a page in secondary memory is equal to 120 ns. The time required to access a page in primary memory is 15 ns. The average time required to access a page is ________.

(A) 105
(B) 68
(C) 75
(D) 78

Answer :  (D) 78

Solution : Average access time = hit ratio * primary memory access time + (1 – hit ratio) * secondary memory access time

Average access time = 0.4 * 15 + 0.6 * 120←
Average access time = 6 + 72
Average access time = 78.
So, option (D) is correct.


Q 53. In a multi-user operating system, 30 requests are made to use a particular resource per hour, on an average. The probability that no requests are made in 40 minutes, when arrival pattern is a poisson distribution, is _________.
 
(A) e^-15
(B) 1 – e^-15
(C) 1 – e^-20
(D) e^-20


Answer : (D) e^-20

Solution : In One Hour = 30 Requests Sent
 in 40 minutes =   30 x 40   =  20  Requests Sent

                                        ^{60 
 Here 20 is Number of request Sent in 40 minutes. (Let 𝛌 is number of request in 40 minutes)
Therefore}  𝛌 = 20

^{But according to given statement in question Actual requests are made in 40 minutes is ZERO.
Let x is actual requests in 40 minutes. (According to Question)
i.e. x = 0} 

  

e−t/θe−t/θ
^e−t/θ