PGT Computer Science (Level-3), (HTET Practice MCQs) (Part-VI)Total Questions: 10051. Quadrature Amplitude Modulation means changing both:A. Frequency and phase of the carrier.B. Frequency and Amplitude of the carrier.C. Amplitude and phase of the carrier.D. Amplitude and Wavelength of the carrier.Correct Answer: C. Amplitude and phase of the carrier.52. If a file consisting of 50,000 characters takes 40 seconds to send, then the data rate is ________ .A. 1 kbpsB. 1.25 kbpsC. 2 kbpsD. 10 kbpsCorrect Answer: D. 10 kbps53. Match the following: List-I List-II (a) Data link layer 1. Encryption (b) Network layer 2. Connection control (c) Transport layer 3. Routing (d) Presentation layer 4. FramingCodes: Option (a) (b) (c) (d) A. 4 3 1 2 B. 3 4 2 1 C. 4 2 3 1 D. 4 3 2 1A.B.C.D.Correct Answer: D.54. The address of a class B host is to be split into subnets with a 6-bit subnet number. What is the maximum number of subnets and maximum number of hosts in each subnet?A. 62 subnets and 1022 hostsB. 64 subnets and 1024 hostsC. 62 subnets and 254 hostsD. 64 subnets and 256 hostsCorrect Answer: A. 62 subnets and 1022 hosts55. Which of the following statements are true?(a) The fragmentation fields in the base header section of IPv4 have moved to the fragmentation extension header in IPv6. (b) The authentication extension header is new in IPv6. (c) The record route option is not implemented in IPv6. Codes:A. (a) and (b) onlyB. (b) and (c) onlyC. (a) and (c) onlyD. (a), (b) and (c)Correct Answer: D. (a), (b) and (c)56. Consider a full binary tree with n internal nodes, internal path length i, and external path length e. The internal path length of a full binary tree is the sum, taken over all nodes of the tree, of the depth of each node. Similarly, the external path length is the sum, taken over all leaves of the tree, of the depth of each leaf.Which of the following is correct for the full binary tree?A. e = i+nB. e = i + 2nC. e = 2i + nD. e = 2ⁿ + iCorrect Answer: B. e = i + 2n57. You are given a sequence of n elements to sort. The input sequence consists of n/k bsequences, each containing k elements. The elements in a given subsequence are all smaller than the elements in the succeeding subsequence and larger than the elements in the preceding subsequence. Thus, all that is needed to sort the whole sequence of length n is to sort the k elements in each of the n/k subsequences. The lower bound on the number of comparisons needed to solve this variant of the sorting problem is:A.B.C.D.Correct Answer: C.58. Consider the recurrence relation:T(n) = 8T (n/2) + Cn, if n >1 = b if n = 1 Where b and c are constants. The order of the algorithm corresponding to above recurrence relation is:A. nB. n²C. n lg nD. n³Correct Answer: D. n³59. Consider the following two sequences:X = < B, C, D, C, A, B, C > and Y = < C, A, D, B, C, B > The length of longest common subsequence of X and Y is:A. 5B. 3C. 4D. 2Correct Answer: C. 460. A text is made up of the characters a, b, c, d, e each occurring with the probability 0.11, 0.40, 0.16, 0.09 and 0.24 respectively. The optimal Huffman coding technique will have the average length of:A. 2.40B. 2.16C. 2.26D. 2.15Correct Answer: B. 2.16Submit Quiz« Previous12345678910Next »