Description : Consider a source with symbols A, B, C, D with probabilities 1/2, 1/4, 1/8, 1/8 respectively. What is the average number of bits per symbol for the Huffman code generated from above information? (A) 2 bits per symbol (B) 1.75 bits per symbol (C) 1.50 bits per symbol (D) 1.25 bits per symbol
Last Answer : (B) 1.75 bits per symbol
Description : A data file of 1,00,000 characters contains only the characters g-l, with the frequencies as indicated in table: using the variable-length code by Huffman codes, the file can be encoded with (A) 2,52,000 bits (B) 2,64,000 bits (C) 2,46,000 bits (D) 2,24,000 bits
Last Answer : (D) 2,24,000 bits
Description : The content of the accumulator after the execution of the following 8085 assembly language program, is: MVI A, 42H MVI B, 05H UGC: ADD B DCR B JNZ UGC ADI 25H HLT (A) 82 H (B) 78 H (C) 76 H (D) 47 H
Last Answer : (C) 76 H
Description : A Huffman code: A = 1, B = 000, C = 001, D = 01 ,P(A) = 0.4, P(B) = 0.1, P(C) = 0.2, P(D) = 0.3 The average number of bits per letter is A. 8.0 bit B. 2.0 bit C. 1.9 bit D. 2.1 bit
Last Answer : C. 1.9 bit
Description : A Huffman code: A = 1, B = 000, C = 001, D = 01 ,P(A) = 0.4, P(B) = 0.1, P(C) = 0.2, P(D) = 0.3 The average number of bits per letter is a. 8.0 bit b. 2.0 bit c. 1.9 bit d. 2.1 bit
Last Answer : c. 1.9 bit
Description : The initial basic feasible solution of the following transportion problem : then the minimum cost is (A) 76 (B) 78 (C) 80 (D) 82
Last Answer : (A) 76
Description : The energy consumption and production patterns in a chemical plant over a 9 month period is provided in the table below; Month 1 2 3 4 5 6 7 8 9 Production in Tonnes / month 493 ... and give your inference on the result ? ( consider 9 month data for evaluation for predicted energy consumption)
Last Answer : It is required to use the equations Y= mX + C and nC + mΣX = ΣY cΣX + mΣX2 = ΣXY Month X = Production in Tonnes / month Y =Energy Consumption MWh /month X 2 XY 1 493 78.2 ... 73.7 54756 17240.63 8 239 64.4 57121 15402.96 9 239 72.1 57121 17228.98 3387 665.3 1410683 253221
Description : A network with bandwidth of 10 Mbps can pass only an average of 12,000 frames per minute with each frame carrying an average of 10,000 bits. What is the throughput of this network ? (A) 1 Mbps (B) 2 Mbps (C) 10 Mbps (D) 12 Mbps
Last Answer : (B) 2 Mbps
Description : Juice extraction capacity of vertical three roller type bullock down sugercane crusher E. 50-70% F. 70-90% G. 90-100% H. 30-50%
Last Answer : E. 50-70%
Description : A memory management system has 64 pages with 512 bytes page size. Physical memory consists of 32 page frames. Number of bits required in logical and physical address are respectively: (1) 14 and 15 (2) 14 and 29 (3) 15 and 14 (4) 16 and 32
Last Answer : (3) 15 and 14
Description : Shift-Reduce parsers perform the following : (A) Shift step that advances in the input stream by K(K>1) symbols and Reduce step that applies a completed grammar rule to some recent parse trees, ... in the input stream and Reduce step that applies a completed grammar rule to form a single tree.
Last Answer : (B) Shift step that advances in the input stream by one symbol and Reduce step that applies a completed grammar rule to some recent parse trees, joining them together as one tree with a new root symbol.
Description : In the following graph, discovery time stamps and finishing time stamps of Depth First Search (DFS) are shown as x/y where x is discovery time stamp and y is finishing time stamp. It shows which of the following depth first forest? (A ... {a,b,e} {f,g} {c,d} {h} (D) {a,b,c,d} {e,f,g} {h}
Last Answer : Answer: A
Description : There are three events A, B and C , one of which must and only can happen. If the odds are 8:3 against A, 5:2 against B, then the odds against C must be: e) 13:7 f) 3:2 g) 43:77 h) 43:34
Last Answer : Answer : d
Description : The virtual address generated by a CPU is 32 bits. The Translation Lookaside Buffer (TLB) can hold total 64 page table entries and a 4-way set associative (i.e. with 4- cache lines in the set). The page size is 4 KB. The minimum size of TLB tag is (A) 12 bits (B) 15 bits (C) 16 bits (D) 20 bits
Last Answer : (C) 16 bits Explanation: VirtualAddress = 32 bits PageSize = 4KB = 12 bits therefore : VPNTag = 20 bits, OffsetTag = 12 bits TLBEntryLength = VPNTag = 20 bits TotalTLBEntries = 64, 4-way implies ... therefore : TLBIndex = 4 bits TLBTag = TLBEntryLength - TLBIndex = 20 - 4 = 16 bits
Description : Consider a discrete memoryless channel and assume that H(x) is the amount of information per symbol at the input of the channel; H(y) is the amount of information per symbol at the output of the channel; H(x|y) is the amount of uncertainty ... (H(x) - H(x|y))] p(x) (D) max H(x|y) p(x)
Last Answer : (D) max H(x|y) p(x)
Description : Consider a disk queue with request for input/output to block on cylinders 98, 183, 37, 122, 14, 124, 65, 67 in that order. Assume that disk head is initially positioned at cylinder 53 and moving ... and 252 cylinders (B) 640 and 236 cylinders (C) 235 and 640 cylinders (D) 235 and 252 cylinders
Last Answer : Answer: 236 and 208 cylinders Explanation: SSTF Initial head position =53 The closest queue to initial head position=65 head moves from 53 to 65=12 head moves from 65 to 67=2 head moves from 67 ... 122=24 head moves from 122 to 124=2 head moves from 124 to 183=59 Total head movement=208
Description : 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
Last Answer : (A) 2 Mbps Explanation: 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), ... second (Mbps) or gigabits per second (Gbps). Here, Throughput = 15000 x 8000/60 = 2 Mbps
Description : Assume that we need to download text documents at the rate of 100 pages per minute. A page is an average of 24 lines with 80 characters in each line and each character requires 8 bits. Then the required bit rate of the channel is ... ...... (A) 1.636 Kbps (B) 1.636 Mbps (C) 2.272 Mbps (D) 3.272 Kbps
Last Answer : Answer: Marks given to all
Description : Consider f(N) = g(N) + h(N) Where function g is a measure of the cost of getting from the start node to the current node N and h is an estimate of additional cost of getting from the current ... ? (A) A* algorithm (B) AO* algorithm (C) Greedy best first search algorithm (D) Iterative A* algorithm
Last Answer : (C) Greedy best first search algorithm
Description : Consider the fractional knapsack instance n = 4, (p1, p2, p3, p4) = (10, 10, 12, 18), (w1, w2, w3, w4) = (2, 4, 6, 9) and M = 15. The maximum profit is given by (Assume p and w denotes profit and weight of objects respectively) (A) 40 (B) 38 (C) 32 (D) 30
Last Answer : (B) 38
Description : How many symbols exist in Baudot code? A) 32 B) 116 C) 58 D) 76
Last Answer : Answer : A
Description : A grouped frequency distribution table with classes of equal sizes using 63-72 (72 included) as one of the class is constructed for the following data 30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88, -Maths 9th
Last Answer : NEED ANSWER
Last Answer : (b) We arrange the given data into groups like 13-22,23-32 103-112. (since, our data is from 14 to 112). The class width in this case is 9. Now, the given data can be arranged in tabular form as follows. Hence, the number of classes in distribution will be 10.
Description : The mass and energy equivalent to 1 a.m.u. respectively are - (1) 1.67 x 10−27 g, 9.30 MeV (2) 1.67 x 10−27kg, 930 MeV (3) 1.67 x 10−27kg, 1 MeV (4) 1.67 x 10−34 kg, 1 MeV
Last Answer : (2) 1.67 x 10−27kg, 930 MeV Explanation: The mass and energy equivalent to 1 a.m.u. respectively are 1.67 x 10−27kg, 930 MeV.
Description : Reshma purchased 120 chairs at price of Rs. 110 per chair. He sold 30 chairsat a profit of Rs. 12 per chair and 75 chairs at profit of Rs. 14 per chairs. The remaining chairs were sold at a loss of Rs. 7 per chairs ... the average profit per table? A. Rs 10.56 B. Rs 10.87 C. Rs 12.123 D. Rs 12.67
Last Answer : Answer – B (Rs 10.875) Explanation – Total C.P = Rs. (120 x 110) =Rs. 13200. Total S.P = Rs.[(30 x 110 + 30 x 12) + (75 x 110 + 75 x 14) + (15 x 110 – 15 x 7) =Rs..14505 Average profit = Rs. (14505 – 13200) /120 = Rs. 1305/120= 10.875
Description : Consider the following database table having A, B, C and D as its four attributes and four possible candidate keys (I, II, III and IV) for this table: I: {B} II: {B, C} III: {A, D} IV: {C, D} If ... the database table? (A) I and III only (B) III and IV only (C) II only (D) I only
Last Answer : (C) II only
Description : Consider the following Entity-Relationship (E-R) diagram and three possible relationship sets (I, II and III) for this E-R diagram: If different symbols stand for different values (e.g., t1 is definitely not equal to t2 ... diagram ? (A) I only (B) I and II only (C) II only (D) I, II and III
Last Answer : (A) I only
Description : Number of bits per symbol used in Baudot code is a. 7 b. 5 c. 8 d. 9
Last Answer : 5
Description : The average weather(rainfall) for the first 6 days out of 8 days recorded to be 9 cm. The rainfall on last 2 days was in the ratio 2:3. the average of 8 days was 14.2 cm. What was the rainfall on the last day? A) 7.8 B) 34.4 C) 35.76 D) 8.9
Last Answer : C) according to question (6×9+3x+2x)/8=14.2 54+5x=113.6 X=11.92 Last day rainfall=3×11.92=35.76cm
Description : Suppose the function y and a fuzzy integer number around -4 for x are given as y=(x-3)2+2 Around -4={(2,0.3), (3,0.6), (4,1), (5,0.6), (6,0.3)} respectively. Then f(Around - 4) is given by: (A) {(2,0.6), (3,0.3), ... 6), (3,1), (6,0.6), (11,0.3)} (D) {(2,0.6), (3,0.3), (6,0.6), (11,0.3)}
Last Answer : (C) {(2,0.6), (3,1), (6,0.6), (11,0.3)}
Description : Which formula shows how to work out the percentage a file has been compressed by? A. bits in (Huffman *7) /100 B. bits in ASCii - bits in Huffman C. difference in bits / bits in ASCII * 100
Last Answer : C. difference in bits / bits in ASCII * 100
Description : How do you calculate the number of bits of a body of text in ASCII? A. Number of characters * 7 B. Number of characters (including spaces) *7 C. bits in Huffman * 7 D. bits in Huffman / 7
Last Answer : B. Number of characters (including spaces) *7
Description : The basic idea behind Huffman coding is to A. compress data by using fewer bits to encode fewer frequently occuring characters B. compress data by using fewer bits to encode more ... frequently occuring characters D. expand data by using fewer bits to encode more frequently occuring characters
Last Answer : B. compress data by using fewer bits to encode more frequently occuring characters
Description : The basic idea behind Huffman coding is to a. compress data by using fewer bits to encode fewer frequently occuring characters b. compress data by using fewer bits to encode more ... frequently occuring characters d. expand data by using fewer bits to encode more frequently occuring characters
Last Answer : b. compress data by using fewer bits to encode more frequently occuring characters
Description : A source code whose average word length approaches the fundamental limit set by the entropy of a discrete memoryless source. A. Prefix code B. Huffman code C. Entropy code D. Source code
Last Answer : B. Huffman code
Description : Average length of Extended Huffman code is upper bounded by : a. R b. R+1 c. R-1 d. R+1/n
Last Answer : d. R+1/n
Description : (37÷44) + (78÷34) x (480÷85) = ? a) 11.769 b) 12.756 c) 13.796 d) 12.796 e) none
Last Answer : (37÷44) + (78÷34) x (480÷85) = ? Sol: 0.841 + 2.29412 x 5.64706 =? 0.841 + 12.955 =? ? = 13.796 Answer: c)
Description : Consider a system having ‘m’ resources of the same type. These resources are shared by three processes P1, P2 and P3 which have peak demands of 2, 5 and 7 resources respectively. For what value of ‘m’ deadlock will not occur? (A) 70 (B) 14 (C) 13 (D) 7
Last Answer : (B) 14
Description : ABCD is a parallelogram. The diagonals AC and BD intersect at the point O. If E, F, G and H are the mid-points of AO, DO, CO and BO respectively -Maths 9th
Last Answer : answer:
Description : Let G(x) be generator polynomial used for CRC checking. The condition that should be satisfied by G(x) to correct odd numbered error bits, will be: (1) (1+x) is factor of G(x) (2) (1-x) is factor of G(x) (3) (1+x2) is factor of G(x) (4) x is factor of G(x)
Last Answer : (1+x) is factor of G(x)
Description : A relation R={A,B,C,D,E,F,G} is given with following set of functional dependencies: F={AD→E, BE→F, B→C, AF→G} Which of the following is a candidate key? (A) A (B) AB (C) ABC (D) ABD
Last Answer : Answer: D
Description : The average speed of a bus in the upward journey is 50% more than that in the return journey. The bus halts for 2 hrs on reaching the destination. The total time taken for the complete to and from trip is 34 hrs ... speed of the bus in the upward journey is. a) 63.24 b) 65.67 c) 63.49 d) 62.49
Last Answer : D Let the speed in return journey be x km/hr Then speed in upward journey = 150 /100 x = (3/2 x)km/hr Average speed = (2*3x/2*x) / (3x/2 +x) =3x2 / (5x/2) =6x2 / 5x =6x/5 km/hr Therefore (1600 ... 192x = 8000 X = 41.66 Speed in upward journey = (3/2 x ) km/hr =(3/2*41.66) =62.49 km/hr
Description : The average of weight of three men A,B and C is 84 kg. Another man D joins the group and the average now becomes 80 kg. If another man E, whose weight is 3 kg.more than that of D, replaces A, then the average weight of B,C ... 79 kg. The weight of A is : a) 70 kg b) 72 kg c) 75 kg d) 80 kg e) 78 kg
Last Answer : A+B+C = 3×84=252 A+B+C+D= (4×80)=320 D = (320-252)=68 and E = (68+3)=71 Now, B+C+D+E = (4×79)=316 B+C+D=(316-71)=245 kg So, A = (320-245)=75 kg Answer: c)
Description : Let C be a binary linear code with minimum distance 2t + 1 then it can correct upto ............bits of error. (1) t + 1 (2) t (3) t - 2 (4) t/2
Last Answer : (2) t
Description : Let f and g be the functions from the set of integers to the set integers defined by f(x) = 2x + 3 and g(x) = 3x + 2 Then the composition of f and g and g and f is given as (A) 6x + 7, 6x + 11 (B) 6x + 11, 6x + 7 (C) 5x + 5, 5x + 5 (D) None of the above
Last Answer : (A) 6x + 7, 6x + 11 Explanation: fog(x)=f(g(x))=f(3x+2)=2(3x+2)+3=6x+7 gof(x)=g(f(x))=g(2x+3)=3(2x+3)+2=6x+11
Description : What is the critical rotation speed in revolutions per second, for a ball mill of 1.2 m diameter charged with 70 mm dia balls? (A) 0.5 (B) 1.0 (C) 2.76 (D) 0.66
Last Answer : (D) 0.66
Description : If an integer occupies 4 bytes and a character occupies 1 byte of memory, each element of the following structure would occupy how many bytes ? struct name { int age; char name[30]; }; A) 30 B) 32 C) 34 D) 36
Last Answer : C) 34
Description : A byte addressable computer has a memory capacity of 2 m Kbytes and can perform 2 n operations. An instruction involving 3 operands and one operator needs a maximum of (A) 3m bits (B) m + n bits (C) 3m + n bits (D) 3m + n + 30 bits
Last Answer : (D) 3m + n + 30 bits
Description : Nesha scored 120 out of 150 in English, 120 out of 180 in mathematics and 160 out of 200 in Science. Find Nesha’s score as percentage: (i) in Mathematics (ii) in all the three subjects (on the whole). a) 66 2/3, 78 22/51 b) 71 23/5, 45 7/31 c) 54 13/33, 82 3/4 d) 60 3/4, 76 32/41
Last Answer : Answer: A (i) Percentage scored in Mathematics = 120/180 100 = 12000/180 =1200/18 = 66 2 /3 % (ii) Total maximum of all the three subjects = 150 + 160 + 200 = 510 and Total score in the ... percentage on the whole = (400/510 100) = (40000/510) = 4000/51 = 78 22/51 %