CBSE UGC NET Computer Science Paper III Solved December 2015 - Part 2

11.       A network with bandwidth of 10 Mbps can pass only an average of 15,000 frames per minute with each frame carrying an average of 8,000 bits. What is the throughput of this network ?
(A) 2 Mbps        (B) 60 Mbps
(C) 120 Mbps    (D) 10 Mbps
Answer: A
In data transmission, throughput is the amount of data moved successfully from one place to another in a given period of time, and typically measured in bits per second (bps), or in megabits per second (Mbps) or gigabits per second (Gbps).
Here, Throughput = 15000 x 8000/60 = 2 Mbps
12.       Consider a subnet with 720 routers. If a three-level hierarchy is choosen with eight clusters, each containing 9 regions of 10 routers, then total number of entries in the routing table is ...................
(A) 25                 (B) 27
(C) 53                 (D) 72
Answer: A
Each router needs 10 entries for local routers, 8 entries for routing to other regions within its own cluster, and 7 entries for distant clusters, for a total of 25 entries.
13.       In a classful addressing, the IP addresses with 0 (zero) as network number:
(A) refers to the current network
(B) refers to broadcast on the local network
(C) refers to broadcast on a distant network
(D) refers to loopback testing
Answer: A
14.       In electronic mail, which of the following protocols allows the transfer of multimedia
(A) IMAP            (B) SMTP
(C) POP 3         (D) MIME
Answer: D
15.       A device is sending out data at the rate of 2000 bps. How long does it take to send a file of 1,00,000 characters ?
(A) 50                 (B) 200
(C) 400              (D) 800
Answer: C
1,00,000 characters = 1,00,000 x 8 bits = 8,00,000 bits
8,00,000 bits/2000 bps = 400 seconds

16.       In Activity-Selection problem, each activity i has a start time si and a finish time fi where si≤fi. Activities i and j are compatible if:
(A) si≥fj                           (B) sj≥fi
(C) si≥fj or sj≥fi               (D) si≥fj and sj≥fi
Answer: C
17.       Given two sequences X and Y:
X = <a, b, c, b, d, a, b>
Y = <b, d, c, a, b, a>
The longest common subsequence of X and Y is:
(A) <b, c, a>                  (B) <c, a, b>
(C) <b, c, a, a> (D) <b, c, b, a>
Answer: D
A subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.
A sequence G is said to be a common subsequence of X and Y, if Z is a subsequence of both X and Y.
Here X = <a, b, c, b, d, a, b>, the sequences <b,c,a>, <c,a,b>, <b,c,b,a>  are subsequences of X.
Given a second sequence of symbols Y = <b, d, c, a, b, a>, then <b,c,a>, <c,a,b>, <b,c,b,a>  are common subsequences to both X and Y.
However, the longest common subsequence of X and Y is <b,c,b,a>.
18.       If there are n integers to sort, each integer has d digits and each digit is in the set {1,2, ..., k}, radix sort can sort the numbers in:
(A) O(d n k)                   (B) O(d nk)
(C) O((d+n)k)                (D) O(d(n+k))
Answer: D
19.       The solution of the recurrence relation
is :
(A) O(lg n)         (B) O(n)
(C) O(n lg n)     (D) None of the above
Answer: D
20.    Floyd-Warshall algorithm utilizes ............... to solve the all-pairs shortest paths problem on a directed graph in ................ time.
(A) Greedy algorithm, θ(V3)                (B) Greedy algorithm, θ(V2 lgn)
(C) Dynamic programming, θ(V3)     (D) Dynamic programming, θ(V2 lgn)
Answer: C

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