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ISC Class XII Analysis Of Pupil Performance 2018 : Computer Science

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Analysis of Pupil Performance COMPUTER SCIENCE Research Development and Consultancy Division Council for the Indian School Certificate Examinations New Delhi Year 2018 __________________________________________________________________________________ Published by: Research Development and Consultancy Division (RDCD) Council for the Indian School Certificate Examinations Pragati House, 3rd Floor 47-48, Nehru Place New Delhi-110019 Tel: (011) 26413820/26411706 E-mail: council@cisce.org Copyright, Council for the Indian School Certificate Examinations All rights reserved. The copyright to this publication and any part thereof solely vests in the Council for the Indian School Certificate Examinations. This publication and no part thereof may be reproduced, transmitted, distributed or stored in any manner whatsoever, without the prior written approval of the Council for the Indian School Certificate Examinations. FOREWORD This document of the Analysis of Pupils Performance at the ISC Year 12 and ICSE Year 10 Examination is one of its kind. It has grown and evolved over the years to provide feedback to schools in terms of the strengths and weaknesses of the candidates in handling the examinations. We commend the work of Mrs. Shilpi Gupta (Deputy Head) of the Research Development and Consultancy Division (RDCD) of the Council and her team, who have painstakingly prepared this analysis. We are grateful to the examiners who have contributed through their comments on the performance of the candidates under examination as well as for their suggestions to teachers and students for the effective transaction of the syllabus. We hope the schools will find this document useful. We invite comments from schools on its utility and quality. Gerry Arathoon Chief Executive & Secretary October 2018 i PREFACE The Council has been involved in the preparation of the ICSE and ISC Analysis of Pupil Performance documents since the year 1994. Over these years, these documents have facilitated the teaching-learning process by providing subject/ paper wise feedback to teachers regarding performance of students at the ICSE and ISC Examinations. With the aim of ensuring wider accessibility to all stakeholders, from the year 2014, the ICSE and the ISC documents have been made available on the Council s website www.cisce.org. The documents include a detailed qualitative analysis of the performance of students in different subjects which comprises of examiners comments on common errors made by candidates, topics found difficult or confusing, marking scheme for each answer and suggestions for teachers/ candidates. In addition to a detailed qualitative analysis, the Analysis of Pupil Performance documents for the Examination Year 2018 have a component of a detailed quantitative analysis. For each subject dealt with in the document, both at the ICSE and the ISC levels, a detailed statistical analysis has been done, which has been presented in a simple user-friendly manner. It is hoped that this document will not only enable teachers to understand how their students have performed with respect to other students who appeared for the ICSE/ISC Year 2018 Examinations, but also provide information on how they have performed within the Region or State, their performance as compared to other Regions or States, etc. It will also help develop a better understanding of the assessment/ evaluation process. This will help teachers in guiding their students more effectively and comprehensively so that students prepare for the ICSE/ ISC Examinations, with a better understanding of what is required from them. The Analysis of Pupil Performance document for ICSE for the Examination Year 2018 covers the following subjects: English (English Language, Literature in English), Hindi, History, Civics and Geography (History and Civics, Geography), Mathematics, Science (Physics, Chemistry, Biology), Commercial Studies, Economics, Computer Applications, Economic Applications, Commercial Applications. Subjects covered in the ISC Analysis of Pupil Performance document for the Year 2018 include English (English Language and Literature in English), Hindi, Elective English, Physics (Theory), Chemistry (Theory), Biology (Theory), Mathematics, Computer Science, History, Political Science, Geography, Sociology, Psychology, Economics, Commerce, Accounts and Business Studies. I would like to acknowledge the contribution of all the ICSE and the ISC examiners who have been an integral part of this exercise, whose valuable inputs have helped put this document together. I would also like to thank the RDCD team of, Dr. M.K. Gandhi, Dr. Manika Sharma, Mrs. Roshni George and Mrs. Mansi Guleria who have done a commendable job in preparing this document. Shilpi Gupta Deputy Head - RDCD October 2018 ii CONTENTS Page No. FOREWORD i PREFACE ii INTRODUCTION 1 QUANTITATIVE ANALYSIS 3 QUALITATIVE ANALYSIS 10 INTRODUCTION This document aims to provide a comprehensive picture of the performance of candidates in the subject. It comprises of two sections, which provide Quantitative and Qualitative analysis results in terms of performance of candidates in the subject for the ISC Year 2018 Examination. The details of the Quantitative and the Qualitative analysis are given below. Quantitative Analysis This section provides a detailed statistical analysis of the following: Overall Performance of candidates in the subject (Statistics at a Glance) State wise Performance of Candidates Gender wise comparison of Overall Performance Region wise comparison of Performance Comparison of Region wise performance on the basis of Gender Comparison of performance in different Mark Ranges and comparison on the basis of Gender for the top and bottom ranges Comparison of performance in different Grade categories and comparison on the basis of Gender for the top and bottom grades The data has been presented in the form of means, frequencies and bar graphs. Understanding the tables Each of the comparison tables shows N (Number of candidates), Mean Marks obtained, Standard Errors and t-values with the level of significance. For t-test, mean values compared with their standard errors indicate whether an observed difference is likely to be a true difference or whether it has occurred by chance. The t-test has been applied using a confidence level of 95%, which means that if a difference is marked as statistically significant (with * mark, refer to t-value column of the table), the probability of the difference occurring by chance is less than 5%. In other words, we are 95% confident that the difference between the two values is true. t-test has been used to observe significant differences in the performance of boys and girls, gender wise differences within regions (North, East, South and West), gender wise differences within marks ranges (Top and bottom ranges) and gender wise differences within grades awarded (Grade 1 and Grade 9) at the ISC Year 2018 Examination. The analysed data has been depicted in a simple and user-friendly manner. 1 Given below is an example showing the comparison tables used in this section and the manner in which they should be interpreted. The table shows comparison between the performances of boys and girls in a particular subject. The t-value of 11.91 is significant at Comparison on the basis of Gender Gender Girls Boys *Significant at 0.05 level N 2,538 1,051 Mean 66.1 60.1 SE 0.29 0.42 t-value 11.91* 0.05 level (mentioned below the table) with a mean of girls as 66.1 and that of boys as 60.1. It means that there is significant difference between the performance of boys and girls in the subject. The probability of this difference occurring by chance is less than 5%. The mean value of girls is higher than that of boys. It can be interpreted that girls are performing significantly better than boys. The results have also been depicted pictographically. In this case, the girls performed significantly better than the boys. This is depicted by the girl with a medal. Qualitative Analysis The purpose of the qualitative analysis is to provide insights into how candidates have performed in individual questions set in the question paper. This section is based on inputs provided by examiners from examination centres across the country. It comprises of question wise feedback on the performance of candidates in the form of Comments of Examiners on the common errors made by candidates along with Suggestions for Teachers to rectify/ reduce these errors. The Marking Scheme for each question has also been provided to help teachers understand the criteria used for marking. Topics in the question paper that were generally found to be difficult or confusing by candidates, have also been listed down, along with general suggestions for candidates on how to prepare for the examination/ perform better in the examination. 2 STATISTICS AT A GLANCE Total Number of Candidates: 21,225 Mean Marks: Highest Marks: 100 82.3 Lowest Marks: 03 3 PERFORMANCE (STATE-WISE & FOREIGN) West Bengal 82.1 Uttarakhand 82.9 Uttar Pradesh 82.2 Tripura 76.4 Telangana 83.9 Tamil Nadu 86.1 Sikkim 79.6 Rajasthan 80.7 Punjab 87.8 Odisha 75.8 New Delhi 91.4 Meghalaya 69.0 Maharashtra 88.3 Madhya Pradesh 79.7 Kerala 85.6 Karnataka 88.5 Jharkhand 79.2 Himachal Pradesh 78.9 Haryana 90.4 Gujarat 78.1 Chhattisgarh 70.3 Chandigarh 90.9 Bihar 83.9 Assam 97.6 Andhra Pradesh 85.6 Foreign 91.3 The States/ UTs of Assam, New Delhi and Chandigarh secured highest mean marks. Mean marks secured by candidates studying in schools abroad were 91.3. 4 GENDER-WISE COMPARISON BOYS GIRLS Mean Marks: 82.8 Mean Marks: 82.0 Candidates: 7,210 Candidates: 14,015 Number of Number of Comparison on the basis of Gender Gender Girls Boys *Significant at 0.05 level N Mean SE t-value 7,210 14,015 82.8 82.0 0.18 0.14 3.53* Girls performed significantly better than boys. 5 REGION-WISE COMPARISON East North Mean Marks: 80.8 Mean Marks: 82.3 Number of Candidates: 6,557 Number of Candidates: 12,051 Highest Marks: 100 Lowest Marks: 06 Highest Marks: 100 Lowest Marks: 03 REGION Mean Marks: 86.1 Mean Marks: 84.2 Number of Candidates: 1,744 Number of Candidates: 796 Highest Marks: 100 Lowest Marks: 27 South Mean Marks: 91.3 Number of Candidates: 77 Highest Marks: 100 Lowest Marks: 52 Foreign 6 Highest Marks: 100 Lowest Marks: 14 West Mean Marks obtained by Boys and Girls-Region wise 96.7 89.9 86.8 82.8 82.0 North 81.4 85.8 83.4 80.6 East South 84.6 West Foreign Comparison on the basis of Gender within Region Region North (N) East (E) South (S) West (W) Foreign (F) Gender Girls Boys Girls Boys Girls Boys Girls Boys Girls Boys N Mean SE 4,345 7,706 2,025 4,532 581 1,163 243 553 16 61 82.8 82.0 81.4 80.6 86.8 85.8 83.4 84.6 96.7 89.9 0.22 0.18 0.36 0.27 0.51 0.39 1.12 0.71 2.34 1.51 *Significant at 0.05 level The performance of girls was significantly better than that of boys in the northern and foreign region. In other regions no significant difference was observed. 7 t-value 2.81* 1.88 1.54 -0.88 2.43* MARK RANGES : COMPARISON GENDER-WISE Comparison on the basis of gender in top and bottom mark ranges Marks Range Top Range (81-100) Bottom Range (0-20) *Significant at 0.05 level Gender Girls Boys Girls Boys N Mean SE 4,653 8,842 1 25 92.1 92.6 17.0 15.2 0.08 0.06 0 1.02 Boys Marks Range (81-100) Girls t-value -4.71* 1.77 All Candidates 92.6 81 - 100 92.1 92.5 71.7 61 - 80 71.8 71.7 Marks Range (81-100) 52.3 41 - 60 53.2 52.5 35.5 21 - 40 35.6 35.5 15.2 0 - 20 17.0 15.3 8 GRADES AWARDED : COMPARISON GENDER-WISE Comparison on the basis of gender in Grade 1 and Grade 9 Grades Gender Girls Boys Girls Boys Grade 1 Grade 9 *Significant at 0.05 level N Mean SE 3,104 6,154 17 80 95.6 95.8 24.3 22.8 0.06 0.04 0.78 0.70 Boys Grade 1 t-value -3.67* 1.41 Girls All Candidates 95.8 95.6 95.7 1 84.9 84.8 84.9 2 74.9 74.8 74.8 3 Grade 1 64.6 65.0 64.7 4 57.1 57.1 57.1 5 52.1 52.2 52.1 6 Grade 9 47.1 47.2 47.1 7 39.6 39.9 39.7 8 9 9 22.8 24.3 23.1 Part I (20 Marks) Answer all questions. While answering questions in this Part, indicate briefly your working and reasoning, wherever required. Question 1 (a) State the Commutative law and prove it with the help of a truth table. [1] (b) Convert the following expression into its canonical POS form: [1] F (X, Y, Z) = (X+Y ). (Y +Z) (c) Find the dual of: [1] (A + B). (1 +B ) = A + B (d) Verify the following proposition with the help of a truth table (P Q) (P Q) = P [1] (e) If [1] F (A, B, C) = A (BC + B C), then find F Comments of Examiners (a) Some candidates only drew the truth table without mentioning the Law. A few candidates were confused with the Associative Law. Many candidates used the 3-variable input truth table instead of the 2-variable input. (b) Some candidates wrote the direct answer without showing the working. A few candidates got confused with the term Canonical and reduced the expression instead of converting it. (c) Most of the candidates answered this question correctly. The common mistakes observed were: Compliments were also changed and 0 s and 1 s were not interchanged. In a few answer scripts, only the definition was written. (d) Some candidates interchanged the symbols and . Some candidates verified the proposition with Boolean laws instead of Truth Table while a few candidates used the 3-variable truth table instead of the 2-variable truth table. (e) Some of the candidates wrote the answer directly without showing the working. Change of operators was not done properly by some of the candidates. A few candidates simplified the equation as they were probably not clear with the application of De Morgan s law. 10 Suggestions for teachers Advise students to practice all the laws of Boolean Algebra and Propositional logic. Emphasise proving of laws, especially by Truth Table. Give adequate practice in the conversion of SOP and POS and viceversa. Explain both Canonical and Cardinal forms laying stress on working/steps. Give adequate practice on the principle of duality to show that there exists another equation in every equation. Explain, with examples, the difference between compliment and duality. Explain the use of the symbols , , ~, => and <=> in a proposition. Give sufficient practice on complementation using De Morgan s law. MARKING SCHEME Question 1 (a) Law: Commutative Law (i) A + B = B + A (ii) A . B = B. A Truth Table: A B A+B B+A 0 0 0 0 0 1 1 1 1 0 1 1 1 1 1 1 (b) F (X, Y, Z) (c) Equation: (A + B). (1 +B ) = A + B Dual = A . B + 0.B = A . B (d) Verify: (P Q) (P Q) = P = (X+Y ) . (Y +Z) = (X+Y +0). (Y +Z+0) = (X+Y + (Z.Z )). (Y +Z+ (X.X )) = (X+Y +Z). (X+Y +Z ). (X+Y +Z). (X +Y +Z) = (X+Y +Z). (X+Y +Z ). (X +Y +Z) P 0 0 1 1 (e) F (A, B,C) Q 0 1 0 1 P Q 0 0 0 1 P Q 0 0 1 0 (P Q) (P Q) 0 0 1 1 = A (BC +B C) F = A + (BC + B C) = A+ (BC ) . (B C) = A + (B + C). (B + C ) Question 2 (a) What are Wrapper Classes? Give any two examples. [2] (b) A matrix A[m][m] is stored in the memory with each element requiring 4 bytes of storage. If the base address at A [1][1] is 1500 and address of A [4][5] is 1608, determine the order of matrix when it is stored in Column Major Wise. [2] (c) Convert the following infix notation to postfix form: [2] A + (B C * (D / E) * F) (d) Define Big O notation. State the two factors which determine the complexity of an algorithm. [2] (e) What is exceptional handling? Also, state the purpose of finally block in a try catch statement. [2] 11 Comments of Examiners (a) The definition written by small number of candidates Suggestions for teachers was vague. Considerable number of candidates wrote Revise the initial topics, with only the examples without the definition while a few examples, of Class XI syllabus in class others wrote the definition without examples. In a few XII. answer scripts the names of wrapper classes were Elaborate on the basic concepts of Java mentioned in the lower case. along with programs. (b) Only a few candidates wrote the answer directly Give adequate amount of practice to without showing the working/formula. Numerous calculate addresses using Row major candidates made calculation errors while a few used the and Column major wise. Row major instead of the Column major to determine Explain the different terms used in the order of the matrix. address calculations. (c) Several candidates wrote the correct answer without Train students to solve equations. showing the working. Some candidates applied the Give examples to practice the postfix correctly but were unable to derive the final conversion of Infix to Postfix notation, answer due to wrong operator precedence. A few the order of precedence. Also teach Polish Stack method. candidates followed order of operation BODMAS Train students to practice error while in some answer scripts the concept of left-tohandling codes. right was not followed. Illustrate to the students, all the three (d) Some candidates wrote examples to illustrate the cases of complexities with their factors answers. Some mentioned the types of complements, that influence them. while a few other candidates mentioned the cases of Lay emphasis on the difference complexity (best case, average case and worst case) as between factors effecting complexities examples of factors. and types of cases of complexities. (e) Various answers were given by the candidates for this question. Some candidates wrote syntax to explain exceptional handling. A few candidates were unable to understand and state the purpose of finally block in a try catch statement and hence, did not attempt writing the answer. MARKING SCHEME Question 2 (a) Wrapper Class: A class which defines a primitive data type. A Wrapper class is a class whose object wraps or contains a primitive data types. Wrapper class in java provides the mechanism to convert primitive into object and object into primitive. Example: Integer, Character, Double etc. (b) Column Major Wise: P[i][j] = BA + W [ (j l c )* rows + (i l r ] Putting the values: 1608 = 1500 + 4[ (4*m + 3)] 1608 = 1500 + 16m + 12 16m = 1608-1512 m=6 (c) Infix to Postfix: (d) Big O Notation: The unit of measurement of efficiency of an algorithm. A + (B C * (D / E) * F) = A + (B C * DE/ * F) = A + (B- CDE/* * F) =A + (BCDE/*F*-) Ans: ABCDE/*F*-+ 12 (e) A theoretical measure of the execution of an algorithm, usually the time or memory needed, given the problem size n, which is usually the number of items. Two factors are Time and Space(memory) Exceptional Handling: The way to handle anomalous situations during run time. Finally, is a block of code which is executed with or without exceptions. Question 3 The following is a function of some class which checks if a positive integer is a Palindrome number by returning true or false. (A number is said to be Palindrome if the reverse of the number is equal to the original number.) The function does not use modulus (%) operator to extract digit. There are some places in the code marked by?1?, ?2?, ?3?, ?4?, ?5? which may be replaced by a statement / expression so that the function works properly. boolean PalindromeNum( int N ) { int rev= ?1?; int num=N; while(num>0) { int f= num/ 10; int s = ?2?; int digit = num - ?3?; rev= ?4? + digit; num /= ?5?; } if ( rev= =N ) return true; else return false; } (i) (ii) (iii) (iv) (v) What What What What What is the statement is the statement is the statement is the statement is the statement or expression at ?1? or expression at ?2? or expression at ?3? or expression at ?4? or expression at ?5? 13 Comments of Examiners In sub-parts (i), (ii) and (iii), most of the candidates were able to write the correct answer. Some candidates used the % operator to extract the digit. In sub-part (iv), several candidates wrote only rev instead of rev * 10. In part (v) a few candidates wrote num/10 instead of only 10. Some candidates were not clear in separating digits without the % operator and attempted the code using their own logic and not as per the question. Suggestions for teachers Give adequate practice to the students on program using conditions / looping/recursion and other output related programs. Show the dry run/ working of program and instruct the students that working is necessary to get full credit. Lay stress on the practice of standard programs. MARKING SCHEME Question 3 (i) (ii) (iii) (iv) (v) 0 f*10 s OR f*10 rev*10 10 PART II (50 Marks) Answer six questions in this part, choosing two questions from Section A, two from Section B and two from Section C. SECTION - A Answer any two questions. Question 4 (a) Given the Boolean function F (A, B, C, D) = (0,2,4,8,9,10,12,13). (i) (ii) (b) Reduce the above expression by using 4-variable Karnaugh map, showing the various groups (i.e. octal, quads and pairs). Draw the logic gate diagram for the reduced expression. Assume that the variables and their complements are available as inputs. Given the Boolean function: F (A, B, C, D) = (3,4,5,6,7,10,11,14,15). [4] [1] (i) Reduce the above expression by using 4-variable Karnaugh map, showing the various groups (i.e. octal, quads and pairs). (ii) Draw the logic gate diagram for the reduced expression. Assume that the [1] variables and their complements are available as inputs. 14 [4] Comments of Examiners (a) (i) Many candidates were unable to draw the Suggestions for teachers correct K-Map for the SOP expression while some candidates drew the POS K-Map instead Teach students to reduce SOP and of SOP. Several candidates used different POS expressions using K-Map variables to draw the K-Map instead of the one simultaneously. given in the question. A few candidates Give adequate practice to students included redundant groups in the final to draw the K-Map, to fill the KMap with 0 s and 1 s, to mark the expression which was not required. groups and to reduce the groups. (ii) Some candidates drew the logic circuit using Train students to draw logic circuits NAND gates while some others drew vague using basic gates and with universal diagrams with different shapes instead of gates. standard logic gates. Labeling and flow lines Emphasise on arranging the were not shown by a few candidates. variables in proper order and the (b) (i) Most candidates answered this question importance of cell values correctly. However, the common anomalies corresponding with the variables. found in some answer scripts were: Explain clearly how groups are Errors in place value and putting variables framed and reduced. Tell students in K-Map. that redundant groups are not to be included in the final reduced Groups were reduced by laws. expression. Incorrect drawing of K-Map. Including the redundant group in the final expression. Drawing the SOP K-Map instead of POS. (ii) Some candidates drew the logic circuit using NOR gates while some others drew vague diagrams. Labeling and flow lines were not shown by a few candidates. MARKING SCHEME Question 4 (a) F(A, B,C,D) = ( 0, 2 , 4, 8, 9, 10, 12, 13 ) C D A B A B AB AB There are three quads: 0 4 12 8 1 1 1 1 C D 1 5 13 9 0 0 1 1 15 3 7 15 11 CD 0 0 0 0 CD 2 6 14 10 1 0 0 1 Quad 1 (m 0+ m 4+ m 12+ m 8 ) = C D Quad 3 (m 8+ m 9 + m 12 + m 13 ) = AC Quad2 ( m 0+ m 2+ m 8+ m 10 ) = B D Hence F (A, B, C, D) = C D + B D + AC C D B D A C 4(b) F(A,B,C,D) = ( 3 , 4 , 5 , 6 , 7 , 10 , 11 , 14 , 15 ) A+B A+B A +B A +B 0 4 12 8 C+D 1 0 1 1 There are three quads: Quad 1: (M 4 M 5 M 6 M 7 ) Quad 3: (M 10 M 11 M 14 M 15 ) 1 5 13 9 C+D 3 1 7 0 15 1 11 1 = A + B = A + C C +D 0 0 0 0 2 C +D 6 14 10 1 0 0 0 Quad 2: ( M 3 M 7 M 11 M 15 ) = C + D Hence F (A, B,C,D) = ( A + B ) . (C + D ). (A + C ) 16 A B C D A C Question 5 (a) A training institute intends to give scholarships to its students as per the criteria given below: [5] The student has excellent academic record but is financially weak. OR The student does not have an excellent academic record and belongs to a backward class. OR The student does not have an excellent academic record and is physically impaired. The inputs are: INPUTS A F C I Has excellent academic record Financially sound Belongs to a backward class Is physically impaired (In all the above cases 1 indicates yes and 0 indicates no). (b) (c) Output: X [1 indicates yes, 0 indicates no for all cases] Draw the truth table for the inputs and outputs given above and write the SOP expression for X (A, F, C, I). Using the truth table, state whether the following proposition is a tautology, contingency or a contradiction: ~(A B) V ( ~A => B) Simplify the following expression, using Boolean laws: A (A + B) C (A + B) 17 [3] [2] Comments of Examiners (a) Some candidates did not write the final expression. Some candidates were confused with the SOP/POS expression and interchanged 0 s and 1 s and viceversa. A few candidates reduced the expression using K-Map which was not required. Decoding criteria was not clear to some of the candidates. (b) Some candidates were confused by the symbols , , and =>. Some candidates used the truth table for 3-variables instead of 2-variables. In a few answer scripts, one column (=>) was incorrect and so the result was also incorrect. Some candidates proved using law instead of truth table while a few did not mention the proposition being a tautology. (c) Some candidates either gave only the final answer without showing the working/steps or mentioning the laws or lacked clarity in the order of precedence of operators. Suggestions for teachers - Give adequate practice to the students in truth table with 4 input variables (i.e. 16 combinations). - Explain propositional logic to find the criteria for the output. - Ask students to write the final expression in either Canonical or Cardinal form for both SOP and POS expressions. - Advise students to practice all the laws of Boolean algebra and Propositional logic. - Emphasize proving of all the laws. - Explain the use of the symbols , V, ~, => and <=> in a proposition. Make students practice the concept of tautology, contingency or a contradiction and truth tables for conditional (=>) and bi-conditional (<=>). - Ensure sufficient practice to students in reducing / simplifying expressions involving maximum laws. - Encourage students to mention laws along with their working. MARKING SCHEME Question 5 (a) A 0 0 0 0 0 0 0 0 1 1 1 F 0 0 0 0 1 1 1 1 0 0 0 1 1 1 1 1 C 0 0 1 1 0 0 1 1 0 0 1 I 0 1 0 1 0 1 0 1 0 1 0 0 1 1 1 1 1 0 0 1 1 18 X (OUTPUT) 0 1 1 1 0 1 1 1 1 1 1 1 0 1 0 1 1 0 0 0 0 OR SOP Expression: X (A ,F, C, I) = (1, 2, 3, 5, 6, 7, 8, 9, 10, 11) X = A F C I + A F CI + A F CI + A FC I + A FCI + A FCI + AF C I + AF C I + AF CI + AF CI (b) (A B) ( A=>B) is a Tautology, Contradiction or Contingency A 0 0 1 1 B 0 1 0 1 A 1 1 0 0 B 1 0 1 0 (A B) 1 1 1 0 A=>B 0 1 1 1 (A B) ( A=>B) 1 1 1 1 Hence, it is a Tautology (c) Simplify : A.(A +B).C.(A+B) = (AA + AB).(AC + BC) = (0 + AB).(AC + BC) = ABC + ABC = ABC Question 6 (a) What is an Encoder? Draw the Encoder circuit to convert A-F hexadecimal numbers to binary. State an application of a Multiplexer. [5] (b) Differentiate between Half Adder and Full Adder. Draw the logic circuit diagram for a Full Adder. [3] (c) Using only NAND gates, draw the logic circuit diagram for A + B. [2] Comments of Examiners (a) (b) (c) Some candidates defined decoder instead of encoder. A few candidates drew the entire hexadecimal encoder instead of only A F. Some candidates, neither named nor labeled the input lines. Several candidates drew the multiplexer diagram to illustrate examples. In a few answer scripts, multiplexer was defined as conversion from one form to another . In some answer scripts, either the differences were vague or the term adds was missing in the difference. Some candidates drew the truth table and expression for full adder which was not required. A few candidates drew two different diagrams for sum and carry. Some candidates, instead of drawing the logic circuit, drew the block diagram. Some candidates converted the expression and then drew the circuit using NAND gates. Some 19 Suggestions for teachers - Give sufficient practice to the students so that they are conversant with the circuit diagram, truth table, expression, definition, and use of all applications of Boolean algebra i.e. Half adder, Full adder. Encoders, Decoders etc. - Explain the purpose and working of the gates for the said applications. - Train students to get basic gates using only Universal gates (NAND and NOR). - Ensure in-depth knowledge of Universal gates. candidates took the inputs directly as A and B. A few candidates simply replaced the basic gates with NAND gates. MARKING SCHEME Question 6 (a) An Encoder is a combinational circuit which inputs 2n or fewer lines and outputs n lines. Circuit diagram of a A F Encoder: A B D E C F Application of a Multiplexer: 1. Used for Routing of signals 2. Data Transmission 3. Telephone Exchanges / TV etc (b) Half Adders : It is a combinational circuit which adds two input binary bits and outputs two binary bits. Full Adders : It is a combinational circuit which adds three input binary bits and outputs two binary bits. Circuit of a Full Adder : SUM=A B C CARRY = AB + BC + AC (c) Using NAND - circuit of A + B 20 A A + B B SECTION B Answer any two questions. Question 7 Design a class Perfect to check if a given number is a perfect number or not. [ A number [10] is said to be perfect if sum of the factors of the number excluding itself is equal to the original number] Example : 6 = 1 + 2 + 3 (where 1, 2 and 3 are factors of 6, excluding itself) Some of the members of the class are given below: Class name : Data members/instance variables: num : Methods/Member functions: Perfect (int nn) : Perfect to store the number parameterized constructor to initialize the data member num=nn int sum_of_factors(int i) : returns the sum of the factors of the number(num), excluding itself, using recursive technique void check() : checks whether the given number is perfect by invoking the function sum_of_factors() and displays the result with an appropriate message Specify the class Perfect giving details of the constructor (), int sum_of_factors(int) and void check (). Define a main () function to create an object and call the functions accordingly to enable the task. Comments of Examiners Some candidates were not clear on the concept of recursion and the parameters in the method sum of factors (). A few candidates used the additional instance variables which was not in accordance with the question. In some answer scripts, the recursive case was improper as several candidates attempted the recursive method without using the recursive technique and instead the used iteration (loops). Some candidates missed the base case. Some had problem in check() method as they did not invoke the sum of factors() method properly, although the other methods including the constructor were well answered. Some candidates either did not write the 21 Suggestions for teachers Give lots of practice to the students to solve programs using recursive techniques. Explain thoroughly recursion and its techniques with examples. Ensure that the students are well conversant with the base case and recursive case for every program using recursive technique. main () method or were unable to properly do the object creation and method calling in the main() function. Documentation/comments were missing in a few answer scripts. - - Train students Invoking function within another function. Explain the difference between iteration and recursion. Ensure that the students have understood instance variables and their accessibility Advise students to read the question carefully and answer according to its requirement. MARKING SCHEME Question 7 import java.util.*; class Perfect { int num; Perfect(int nn) { num=nn; } int sum_of_factors(int i) { if (i == num) return 0; else if (num%i==0) return i + sum_of_factors(i+1); else return sum_of_factors(i+1); } void check() { int s=sum_of_factors(1); if (s==num) System.out.println(num+"is a perfect number"); else System.out.println(num+"is not a perfect number"); } public static void main() { Scanner sc=new Scanner(System.in); System.out.println("Enter a number"); int no=sc.nextInt(); Perfect ob=new Perfect(no); // OR Perfect ob=new Perfect(6); ob.check(); // ob.check(); } } 22 Question 8 Two matrices are said to be equal if they have the same dimension and their corresponding [10] elements are equal. For example, the two matrices A and B given below are equal: Matrix A 1 2 3 2 4 5 Matrix B 3 5 6 1 2 3 2 4 5 3 5 6 Design a class EqMat to check if two matrices are equal or not. Assume that the two matrices have the same dimension. Some of the members of the class are given below: Class name Data members/instance variables: a[ ][ ] m n Member functions/methods: EqMat(int mm, int nn) : EqMat : : : to store integer elements to store the number of rows to store the number of columns : parameterised constructor to initialise the data members m = mm and n = nn to enter elements in the array checks if the parameterized objects P and Q are equal and returns 1 if true, otherwise returns 0 void readarray( ) int check(EqMat P, EqMat Q) : : void print( ) : displays the array elements Define the class EqMat giving details of the constructor( ), void readarray( ), int check(EqMat, EqMat) and void print( ). Define the main( ) function to create objects and call the functions accordingly to enable the task. Comments of Examiners Most candidates defined Constructor properly. However, some candidates did not declare memory to the array using the new operator and a few others passed two different arrays instead of objects. In a few answer scripts, the passing of objects was not done correctly. Several candidates checked the size of the matrix which was not required. Some candidates either did not properly do the object creation in the main () method or created three objects instead of two in the main (). Even calling the method check () was incomplete in some answer scripts. Some candidates found the returning type of the method check () a bit confusing and returned true/false instead of 0 or 1. 23 Suggestions for teachers - Give a lot of practice in passing of objects to a function through parameters - Explain the working on one dimensional and two-dimensional arrays with various examples. - Illustrate pass by value and pass by reference and make the students practice it. - Teach dynamic binding concepts and dot (.) operator - Explain the use of flag variables. MARKING SCHEME Question 8 import java.util.*; class EqMat { int a[][]; int m; int n; static Scanner sc=new Scanner(System.in); EqMat(int mm,int nn) { m=mm; n=nn; a=new int[m][n]; } void readarray() { System.out.println("enter" + (m*n) + "elements" ); for (int i=0;i<m;i++) for (int j=0;j<n;j++) a[i][j]=sc.nextInt(); } int check(EqMat P,EqMat Q) { for (int i=0;i<P.m;i++) for (int j=0;j<P.n;j++) { if (P.a[i][j]!=Q.a[i][j]) return 0; } return 1; } void print() { for (int i=0;i<m;i++) { System.out.println(); for (int j=0;j<n;j++) System.out.print(a[i][j]+"\t"); } } public static void main() { EqMat ob1=new EqMat(3,3); EqMat ob2=new EqMat(3,3); System.out.println("enter nos for the Ist Matrix"); ob1.readarray(); System.out.println("enter nos for the 2nd Matrix"); ob2.readarray(); if (ob1.check(ob1,ob2)==1) { System.out.println("Equal Matrix"); ob1.print(); ob2.print(); } else System.out.println("not equal"); } } 24 Question 9 A class Capital has been defined to check whether a sentence has words beginning with a [10] capital letter or not. Some of the members of the class are given below: Class name : Capital sent : to store a sentence freq : stores the frequency of words beginning with a capital letter Capital( ) : default constructor void input() : to accept the sentence boolean isCap(String w) : checks and returns true if word begins with a capital letter, otherwise returns false void display () : displays the sentence along with the frequency of the words beginning with a capital letter Data variable: member/instance Member functions/methods: Specify the class Capital, giving the details of the constructor( ), void input( ), boolean isCap(String) and void display( ). Define the main( ) function to create an object and call the functions accordingly to enable the task. Comments of Examiners Some candidates used different methods/logic to split the sentence into words. Several candidates used blank space, while some others used the split () method to separate the words in the sentence. Different logics were used to check the upper case. In a few answer scripts, the ASCII values were incorrect. Some candidates did the splitting in the main () method while some others did it in isCap () method itself. 25 Suggestions for teachers - Give practice to the students to extract characters from words, words from sentences and sentences from paragraphs. - Adopt different methods /logic so that the students get a wider exposure to string manipulation related programs. - Ensure that the students have the knowledge of constructors to initialize a string and other data members and have been given adequate practice in this topic. - Give adequate practice in conversion of string into characters and concatenating of strings. MARKING SCHEME Question 9 import java.util.*; class Capital { String sent; int freq; Capital() { sent=""; freq=0; } void input() { Scanner sc=new Scanner(System.in); System.out.println("Enter the sentence"); sent=sc.nextLine(); } boolean isCap(String w) { char c=w.charAt(0); if (Character.isUpperCase(c) ) return true; else return false; } void display() { System.out.println("sentence="+sent); StringTokenizer ob=new StringTokenizer(sent); int count=ob.countTokens(); for (int i=0;i<count;i++) { String wd=ob.nextToken(); if (isCap(wd)) freq=freq+1; } System.out.println("Freqency of title word="+freq); } OR void display() { System.out.println("sentence="+sent); String b[]=sent.split(" "); for(int i=0;i<b.length;i++) { if(isCap(b[i])) freq++; } System.out.println("Freqency of title word="+freq); } OR void display() 26 { System.out.println("sentence="+sent); String b=" "+sent; char c; for(int i=0;i<b.length();i++) { c=b.charAt(i); if (c==' ') { if(Character.isUpperCase(b.charAt(i+1)) ) freq++; } } System.out.println("Freqency of title word="+freq); } public static void main() { Capital ob=new Capital(); ob.input(); ob.display(); } } SECTION C Answer any two questions. Question 10 A super class Number is defined to calculate the factorial of a number. Define a sub class [5] Series to find the sum of the series S = 1! + 2! + 3! + 4!+ ......+n! The details of the members of both the classes are given below: Class name Data member/instance variable: n Member functions/methods: Number(int nn) : Number : to store an integer number : parameterized constructor to initialize the data member n=nn int factorial (int a) : returns the factorial of a number (factorial of n = 1 2 3 n) void display() : displays the data members Class name: Data member/instance variable: sum Member functions/methods: Series( ) void calsum( ) Series : to store the sum of the series : parameterized constructor to initialize the data members of both the classes calculates the sum of the given series : 27 void display( ) : displays the data members of both the classes Assume that the super class Number has been defined. Using the concept of inheritance, specify the class Series giving the details of the constructor( ),void calsum( ) and void display( ). The super class, main function and algorithm need NOT be written. Comments of Examiners Most of the candidates wrote the program correctly. However, the common anomalies found in some answer scripts were: Vague concept of Inheritance. Missing keywords extends and super. Constructor with inheritance was not answered correctly. the data members were not declared properly. The function calsum( ) in the derived class was not answered properly. No loop being used to calculate the sum of the series. Algorithm being written instead of a program. Some candidates were not clear about accessing the members of the super class by the derived class. A few candidates declared the base class also, which was not required. Suggestions for teachers - - - - - Explain the importance of the keywords extends and super in inheritance and Concept of overriding in inheritance with examples. Give adequate practice to the students on the concept of Inheritance. Illustrate the use of constructor of the base class. Explain to the students the different visibility modes and their accessing capability. Ensure that the students have the knowledge of calling the member function from the super class to the derived class. Instruct students to read the question attentively and answer accordingly. must be explained with examples. MARKING SCHEME Question 10 class Series extends Number { long sum; Series(int nn) { super(nn); sum=0; } void calsum() { for (int i=1;i<=n;i++) sum=sum+factorial(i); OR void calsum() { int f=1; for (int i=1;i<=n;i++) 28 } { f=f*i ; sum=sum + f ; } } void display() { super.display(); System.out.println("sum of the series="+sum); } } Question 11 Register is an entity which can hold a maximum of 100 names. The register enables the user to add and remove names from the top most end only. Define a class Register with the following details: Class name Data members / instance variables: stud[ ] Register array to store the names of the students cap stores the maximum capacity of the array top to point the index of the top end Member functions: Register (int max) (a) : constructor to initialize the data member cap = max, top = 1 and create the string array void push(String n) to add names in the register at the top location if possible, otherwise display the message OVERFLOW String pop() removes and returns the names from the top most location of the register if any, else returns $$ void display() displays all the names in the register Specify the class Register giving details of the functions void push(String) and String pop( ). Assume that the other functions have been defined. [4] The main function and algorithm need NOT be written. (b) Name the entity used in the above data structure arrangement. 29 [1] Comments of Examiners (a) The concept of stack was not clear to most of the candidates. The common errors made by some of the candidates were: the condition / logic for underflow and overflow was not answered correctly. increment / decrement of top index was not done properly. used the stack to store integers instead of string. the methods push() and pop() were found to be difficult. A few candidates used all the methods which were not required. (b) Some candidates were unable to name the entity and wrote vague answers. A few candidates over looked this part while some mentioned the principles LIFO and FIFO as entities. Suggestions for teachers - Give more practice to the students in data structure programs like the stacks, queues, dequeue etc. - Do the working as to how the stack or a queue performs and support the workings with examples. - Explain the concept of LIFO and FIFO with real life examples. - Explain the concept taking the base as an array. - Emphasise the implementation of stacks, queues and de queues using arrays. - Illustrate that an array related program cannot be manipulated by shifting / inserting or initializing by any value since these data structures require pointers. - Clarify students that pointers are not supported in java. Hence, the array is used to show the working of a stack, queue or a dequeue. MARKING SCHEME Question 11 (a) class Register { void push(String n) { if (top<cap-1) stud[++top]=n; else System.out.println("OVERFLOW"); } String pop() { if (top>=0) return stud[top--]; else return " $$"; } } (b) The entity used in the above data structure is Stack. 30 Question 12 (a) A linked list is formed from the objects of the class Node. The class structure of the [2] Node is given below: class { Node int n; Node link; } Write an Algorithm OR a Method to search for a number from an existing linked list. The method declaration is as follows: void FindNode( Node str, int b ) (b) Answer the following questions from the diagram of a Binary Tree given below: A E G B C D H F (i) Write the inorder traversal of the above tree structure. [1] (ii) State the height of the tree, if the root is at level 0 (zero). [1] (iii) List the leaf nodes of the tree. [1] Comments of Examiners (a) Many candidates answered this part correctly. However, some candidates had problem in moving the pointer to the next node and checking for null. Several candidates did not create the temporary pointer. Loop was also missing in a few answer scripts. Some candidates wrote the algorithm in simple English language covering all the main steps. (b) (i) Some candidates wrote the other orders instead of inorder traversal of the given tree. In a few answer scripts, one or two nodes were not placed correctly. (ii) Some candidates wrote the height of the tree as 4 instead of 3. (iii) Most of the candidates answered this part correctly barring a few exceptions who could not list the leaf nodes accurately. 31 Suggestions for teachers Allow students to practice more methods / algorithms with link list and binary tree data structure. Train students to use the diagrams to illustrate the link list and the Binary Trees. Ask students to revise frequently the concepts related to temporary pointer, checking for null condition and moving pointer to the next node etc. Illustrate the concept of root, height, depth, size, degree, siblings, nodes (internal and external), levels, tree traversals etc. thoroughly by using a binary tree diagram. MARKING SCHEME Question 12 (a) (b) ALGORITHM: Step 1. Start Step 2. Set temporary pointer to the first node Step 3. Repeat steps 4 and 5 until the pointer reaches null. Display number not found Step 4. check for number , if found display, exit Step 5. Move pointer to the next node Step 6. End algorithm METHOD: void FindNode(Node str, int b) { Node temp=str; while(temp!=null) { if (temp.n == b) {System.out.prinln(b+ is found ); break; } temp=temp.link; } } (i) G,E,H,C,A,B,F,D (ii) 3 (iii) G,H,F 32 Topics found difficult by candidates Concepts in which candidates got confused Suggestions for candidates Wrapper classes Extracting digits from a number without using the % operator. K-MAPS (grouping, redundant groups, map-rolling, place value). Recursive technique: Passing objects to functions. Stack operations for push() and pop ( ) methods. The symbols ( => , and v )in a proposition. Complexity and factors. Exceptional handling and finally block in a try catch statement. Output without using the % operator. Passing objects to functions. Splitting words from a sentence. Passing of objects. Use of single instance variable for multiple operations in various functions. Link list and Stacks. Height of tree. Refer to the reliable resources [text books, surf the net (Wikipedia etc.)] for latest concepts and improved techniques of programming. Download directly proper definitions, output programs, algorithms etc. from the net. Prepare summary for each chapter. Underline/highlight important words, terms, definitions etc. Practise on the system on a regular basis to understand the syntax and to correct errors. Mention the Laws while reducing a Boolean Expression. Practise one form of K-Map with proper place value for both SOP and POS. Documentation is compulsory in programming and should be mentioned with each program. Declare the class with data members and member functions. Expand or define each function according to the instructions given by the side of each function. Understand the logic of writing a program. Practise constructors with every program. Treat each function of a class as a separate program. Scope of the syllabus should be followed. Prepare section wise following the scope of the syllabus. Solve previous years ISC question papers. Workings, wherever required, should be shown on the side of each question. Always answer 33 to any question as per the requirement of the question.

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Additional Info : ISC Class XII Analysis Of Pupil Performance 2018 : Computer Science
Tags : ISC Board, Class 11th, Class 12th, NDA/NA Entrance Examination

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