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ISC Class XII Analysis Of Pupil Performance 2017 : Computer Science

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Analysis of Pupil Performance Research Development and Consultancy Division Council for the Indian School Certificate Examinations New Delhi Year 2017 __________________________________________________________________________________ Published by: Research Development and Consultancy Division (RDCD) Council for the Indian School Certificate Examinations Plot No. 35-36, Sector VI Pushp Vihar, Saket New Delhi-110017 Tel: (011) 29564831/33/37 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) and the Research Development and Consultancy Division (RDCD) of the Council 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 November 2017 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 document includes 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 2017 have a new 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 2017 Examinations, 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 them 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 2017 covers the following subjects: English (English Language, Literature in English), Hindi, History, Civics and Geography (History & Civics, Geography), Mathematics, Science (Physics, Chemistry, Biology), Commercial Studies, Economics, Computer Applications, Economics Applications, Commercial Applications. Subjects covered in the ISC Analysis of Pupil Performance document for the Year 2017 include English (English Language and Literature in English), Hindi, Elective English, Physics (Theory and Practical), Chemistry (Theory and Practical), Biology (Theory and Practical), 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. Manika Sharma, Dr. M.K. Gandhi, Ms. Mansi Guleria and Mrs. Roshni George, who have done a commendable job in preparing this document. The statistical data pertaining to the ICSE and the ISC Year 2017 Examinations has been provided by the IT section of the Council for which I would like to thank Col. R. Sreejeth (Deputy Secretary - IT), Mr. M.R. Felix, Education Officer (IT) ICSE and Mr. Samir Kumar, Education Officer (IT) - ISC. Shilpi Gupta Deputy Head - RDCD November 2017 ii CONTENTS Page No. FOREWORD i PREFACE ii INTRODUCTION 1 QUANTITATIVE ANALYSIS 3 QUALITATIVE ANALYSIS 10 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 2017 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 2017 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. 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 The t-value 11.91* table shows comparison between the performances of boys and girls in a particular subject. The t-value of 11.91 is significant at 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 The results have also been depicted pictographically. In this case, the girls performed significantly better than the than that of boys. It can be interpreted that girls are performing significantly better than boys. 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: 19,709 Mean Marks: Highest Marks: 100 81.5 Lowest Marks: 06 3 PERFORMANCE (STATE-WISE & FOREIGN) West Bengal 80.6 Uttarakhand 83.4 Uttar Pradesh 79.8 Tripura 62.1 Tamil Nadu 88.4 Telangana 74.0 Sikkim 73.6 Rajasthan 82.7 Punjab 86.7 Odisha 75.1 Madhya Pradesh 78.3 Maharashtra 88.7 Kerala 84.7 Karnataka 87.6 Jharkhand 78.6 Himachal Pradesh 78.4 Haryana 93.0 Gujarat 79.4 Delhi 84.7 Chandigarh 87.4 Chattisgarh 75.1 Bihar 78.8 Assam 95.6 Arunachal Pradesh 66.4 Andhra Pradesh 89.7 Foreign 78.7 The States of Assam, Haryana and Andhra Pradesh secured highest mean marks. Mean marks secured by candidates studying in schools abroad were 78.7. 4 GENDER-WISE COMPARISON BOYS GIRLS Mean Marks: 81.5 Mean Marks: 81.5 Candidates: 6,560 Candidates: 13,149 Number of Number of Comparison on the basis of Gender Gender Girls Boys N Mean SE t-value 6,560 13,149 81.5 81.5 0.19 0.14 0.13 5 REGION-WISE COMPARISON East North Mean Marks: 79.4 Mean Marks: 81.9 Number of Candidates: 5,799 Number of Candidates: 11,707 Highest Marks: 100 Lowest Marks: 06 Highest Marks: 100 Lowest Marks: 06 REGION Mean Marks: 85.1 Mean Marks: 84.9 Number of Candidates: 1,404 Number of Candidates: 750 Highest Marks: 100 Lowest Marks: 31 South Mean Marks: 89.6 Number of Candidates: 49 Highest Marks: 100 Lowest Marks: 48 Foreign 6 Highest Marks: 100 Lowest Marks: 41 West Mean Marks obtained by Boys and Girls-Region wise 81.8 82.0 North 79.2 85.6 79.5 East 86.2 84.8 South 96.2 84.2 West 87.9 Foreign Comparison on the basis of Gender within Region Region Gender North (N) East (E) South (S) West (W) Foreign (F) Girls Boys Girls Boys Girls Boys Girls Boys Girls Boys N Mean SE 4,110 7,597 1,741 4,058 438 966 261 489 10 39 81.8 82.0 79.2 79.5 85.6 84.8 86.2 84.2 96.2 87.9 0.23 0.18 0.40 0.27 0.61 0.45 0.74 0.66 1.17 2.16 *Significant at 0.05 level The performance of girls was significantly better than that of boys in the western and foreign region. In other regions no significant difference was observed. 7 t-value -0.65 -0.64 0.99 2.10* 3.37* 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 4,012 8,227 1 11 91.8 92.1 15.0 12.6 0.08 0.06 0.00 1.34 Boys Girls t-value -3.50* 1.76 All Candidates Marks Range (81-100) 92.1 81 - 100 91.8 92.0 71.3 61 - 80 71.6 71.4 Marks Range (81-100) 51.1 41 - 60 51.7 51.3 33.4 Marks Range (0 -20) 21 - 40 35.7 33.9 12.6 0 - 20 15.0 12.8 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 N Mean SE 2,782 5,894 11 59 94.6 94.8 29.1 26.8 1.79 1.24 8.14 3.57 Boys In Grade 1 and Grade 9 no Girls t-value -0.09 0.26 All Candidates 94.8 94.6 94.8 1 significant difference was 84.8 84.8 84.8 2 observed between the average performance of girls 74.7 74.6 74.7 3 and boys. 64.8 64.9 64.8 4 57.1 57.0 57.0 5 52.1 52.2 52.1 6 47.1 47.0 47.1 7 42.7 42.7 42.7 8 9 9 26.8 29.1 27.1 Part I (20 marks) Answer all questions. While answering questions in this Part, indicate briefly your working and reasoning, wherever required. Question 1 (a) (b) (c) (d) (e) State the law represented by the following proposition and prove it with the help of a truth table: PVP=P State the Principle of Duality. Find the complement of the following Boolean expression using De Morgan s law: F(a,b,c) = (b + c) + a Draw the logic diagram and truth table for a 2 input XNOR gate. If (~P => Q) then write its: (i) Inverse (ii) Converse [1] [1] [1] [1] [1] Comments of Examiners (a) Most of the candidates answered this part well. Some mentioned the laws involving addition while some did not. Some confused the symbol V with the symbol . A few candidates used the truth table using 3 variables instead of 2 variables. Others proved by Boolean law instead of the truth table. (b) Some candidates gave an example to illustrate the Principle of duality. Others did not mention that the complements remain unchanged. (c) Several candidates wrote the answer directly without showing the working. Change of operators was not properly done by some of the candidates. (d) A number of candidates drew the circuit instead of the gate symbol. In some cases, XOR gate was drawn instead of XNOR. (e) Some candidates were confused with the symbols = > and ~ while others interchanged the answers. A few candidates proved it with the help of Boolean laws. 10 Suggestions for teachers Candidates should be told to practice all the laws of Boolean algebra and Propositional logic. Proving of all the laws must be emphasized. The use of the symbols , V, ~, => and <=> in a proposition must be explained. Difference between complement and duality must be explained with examples. More practice on complementation using De Morgan s law should be given for such type of questions. All the gates of Boolean algebra must be practiced with their respective gate symbols, truth table, use, performance and expression. Proportional logic should be taught using all terms that are required. The symbols used in proportions must be explained. MARKING SCHEME Question 1 (a) Law : Idempotent Law. It states that P P = P and P P = P . Truth Table: P P P P 0 0 0 1 1 1 (b) Principle of Duality states that, to every Boolean Equation there exists another equation which is dual to the original equation. To achieve this, the AND s (.) are converted to OR s (+) and vice-versa, 0 s to 1 s and vice versa, however, the complements remain unchanged. (c) Complement of : F( a,b,c ) = (b + c ) +a = (b + c) . a = b . c .a = b . c . a (d) XNOR gate : (e) If (~ P => Q ) then ; (i) Inverse : ( P => ~ Q ) OR P + Q (ii) Converse : ( Q => ~ P ) OR Q + P 11 A B X 0 0 1 0 1 0 1 0 0 1 1 1 Question 2 (a) What is an interface? How is it different from a class? [2] (b) Convert the following infix expression to postfix form: [2] P * Q / R + (S + T) (c) A matrix P[15][10] is stored with each element requiring 8 bytes of storage. If the base address at P[0][0] is 1400, determine the address at P[10][7] when the matrix is stored in Row Major Wise. [2] (d) (i) [2] What is the worst case complexity of the following code segment: for (int x = 1; x <=a; x++) { statements; } for (int y = 1; y <=b; y++) { for (int z = 1; z <=c; z++) { statements; } } (ii) (e) How would the complexity change if all the three loops went to N instead of a, b and c? Differentiate between a constructor and a method of a class. 12 [2] Comments of Examiners (a) This part was answered well by most of the candidates. Some wrote vague definitions of interface. Others explained using examples. Some candidates used the keywords extends and implements to differentiate an interface from a class. (b) Most candidates were able to solve this problem correctly. Several candidates wrote the correct answer without showing the working. Some applied the postfix correctly, but could not derive the final answer due to wrong operator precedence. BODMAS was followed in some cases instead of left-to-right. (c) Some candidates wrote the answer directly without showing the working/formula. Calculation mistakes were also observed. Others did by Column major instead of Row major. (d) (i) This part was well answered by almost all candidates. Only a few did not mention O(a) in the final complexity O(a + bc). Dominant term was not clear in some cases. (ii) A few candidates were not able to answer the change in complexity. The dominant term was not clear in some cases. A few candidates wrote N3 instead of N2 in the final answer. (e) Various answers were given by candidates. Some explained with the help of examples. Others wrote the differences. 13 Suggestions for teachers Inheritance with interface and classes must be given more practice. The concept of multiple inheritance must be explained using an interface. Examples need to be practiced with conversion of Infix to Postfix notation, the order of precedence. The Polish Stack method must also be taught. More practice should be given to calculate addresses using Row major and Column major wise. The different terms used in address calculations must be explained. Complexity and Big O notation must be given more practice. Examples using loops, nested loops and conditional statements must be solved and explained. The difference between the two terms constructor and method must be clarified. This will enable students to understand the concepts and their differences clearly. definition of both, without mentioning the MARKING SCHEME Question 2 (a) Interface is a non primitive data type which has static and final data members and prototype of functions ( i.e. functions are not defined ) Difference : Interface supports multiple inheritance whereas a Class does not support multiple Inheritance (b) Infix to postfix : P*Q/R+(S+T) = P * Q / R + ST+ = P Q* / R + ST+ = PQ*R/ + ST+ Ans : P Q * R / S T + + (c) Row Major Wise: P[i][j] = BA + W [ (i l r )* column + (j l c ] = 1400 + 8[ (10-0)*10 + (7-0)] = 1400 + 856 P[10][7] = 2256 (d) (i) (ii) O(a) + O(b x c) = O(a + bc) O(N) + O(N2) = O(N2) taking the dominant term. (e) Constructor has the same name of the class where as a method has a different name. There is no returning type, not even void in constructor where as in a method it can return a value. 14 Question 3 The following function magicfun( ) is a part of some class. What will the function magicfun( ) return, when the value of n=7 and n=10, respectively? Show the dry run/working: [5] int magicfun( int n) { if ( n= = 0) return 0; else return magicfun(n/2) * 10 + (n % 2); } Comments of Examiners Suggestions for teachers A number of candidates answered this question correctly. Common errors made by candidates: the concept of recursion was not clear to some candidates; some had problems in calling the recursive function; the concept of LIFO (Last In First Out) was not clear and the last digit was missing in some cases. Some candidates did not show the working and gave the answer directly. More practice should be given in solving programs using recursive techniques. Attention should be paid by teachers towards recursion and its techniques with examples. Knowledge of base case and recursive case should be given to students for every program using recursive technique. Output program using recursive technique should be given more practice. Memory blocks must be used to show the concept of Stack (LIFO). Students must be told to show the working where ever required. MARKING SCHEME Question 3 (i) when n=7 OUTPUT : magicfun(7) magicfun(3) * 10 + 1 magicfun(1) *10 + 1 magicfun(0) *10 + 1 0 = 111 (ii) when n=10 OUTPUT : magicfun(10) magicfun(5) * 10 + 0 magicfun(2) *10 + 1 magicfun(1) *10 + 0 magicfun(0) *10 + 1 0 = 1010 15 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) = (2,3,4,5,6,7,8,10,11). (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(P, Q, R, S) = (0,1,2,4,5,6,8,10). [4] [1] (i) Reduce the above expression by using 4-variable Karnaugh map, showing the various groups (i.e. octal, quads and pairs). [4] (ii) Draw the logic gate diagram for the reduced expression. Assume that the variables and their complements are available as inputs. [1] Comments of Examiners Suggestions for teachers (a) (i) Most candidates fared well in this part. Some candidates were not able to draw the K-Map for the Make students reduce SOP and POS expressions using K-Map SOP expression correctly. Different variables were simultaneously. Students should be told used to draw the K-Map instead of those given in the question paper. For a number of candidates the not to include the redundant group in the Map rolling concept was not very clear. In some final expression. Practice should be cases, redundant groups were also included in the given in drawing the K-Map, filling the final expression which was not required. K-Map with 0 s and 1 s, marking the (ii) Most of the candidates answered correctly. groups and reducing the groups. Some drew the logic circuit using NAND gates More practice should be given in while some others drew vague diagrams with drawing logic circuits using basic gates different shapes instead of the standard logic gates. and also with universal gates. (b) (i) Some candidates made errors in place value and Emphasize on arranging the variables in putting variables in K-Map. In some cases the proper order and the importance of cell groups were reduced by laws. A few candidates values corresponding with the variables. drew the K-Map incorrectly. Several candidates included the redundant group in the final Explain clearly how the groups are expression. framed and reduced. Redundant groups (ii) A number of candidates drew the logic circuit are not to be included in the final using NOR gates while some others drew vague reduced expression. diagrams. 16 MARKING SCHEME Question 4 (a) F(A,B,C,D) = ( 2 , 3 , 4, 5 , 6 , 7 , 8 , 10 , 11 ) C D A B A B AB AB 0 4 12 8 There are two quads and one pair: Quad 1 (m 2+ m 3+ m 10 m 11 ) Pair (m 8+ m 10 ) = AB D 0 1 0 1 C D 1 5 13 9 = B C 0 1 0 0 CD 3 7 15 11 1 1 0 1 CD 2 6 14 10 1 1 0 1 Quad2 ( m 4+ m 5+ m 6+ m 7 ) = A B Hence F (A, B, C, D) = B C + A B + AB D 17 4(b) F(P,Q,R,S) = ( 0 , 1 , 2 , 4 , 5 , 6 , 8 , 10 ) 0 P+Q R+S 4 P+Q 12 P +Q 8 P +Q There are three quads : Quad 1 : ( M 0 M 1 M 4 M 5 ) Quad 3 : ( M 0 M 2 M 8 M 10 ) 0 0 1 0 1 5 13 9 R+S 0 0 1 1 = P+R = Q+S 3 7 15 11 R +S 1 1 1 1 2 6 14 10 R +S 0 0 1 0 Quad 2 : ( M 0 M 2 M 4 M 6 ) = P + S Hence F(P,Q,R,S) = ( P + R) . (P + S) . (Q + S) P R P S Q S 18 Question 5 (a) A school intends to select candidates for an Inter-School Essay Competition as per the criteria given below: The student has participated in an earlier competition and is very creative. OR The student is very creative and has excellent general awareness, but has not participated in any competition earlier. OR The student has excellent general awareness and has won prize in an inter-house competition. [5] The inputs are: INPUTS A participated in a competition earlier B is very creative C won prize in an inter-house competition D has excellent general awareness (In all the above cases 1 indicates yes and 0 indicates no). 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 POS expression for X(A,B,C,D). (b) State the application of a Half Adder. Draw the truth table and circuit diagram for a Half Adder. [3] (c) Convert the following Boolean expression into its canonical POS form: [2] F(A,B,C) = ( B + C ) (A + B) 19 Comments of Examiners (a) While a number of candidates answered this part well, some did not mention the final expression. Several candidates were confused with the POS expression and took the output with 1 s instead of 0 s. Some reduced the expression using K-Map which was not required. (b) Some candidates drew the block diagram while some others drew the Full adder instead of Half adder. The truth table and logic circuit for the Partial sum and Carry were interchanged in a few cases. (c) Candidates used various methods to convert the expression. Some were not clear with the term canonical. Working/steps were not shown in many cases. Suggestions for teachers Truth table with 4 input variables (i.e. 16 combinations) must be given for practice. Propositional logic must be explained to find the criteria for the output. Candidates should be told to write the final expression in either Canonical or Cardinal form for both SOP and POS expressions. More practice should be given so that students know the circuit diagram, truth table, expression, definition and use for all applications of Boolean algebra i.e. Half adder, Full adder, Encoders, Decoders, etc. Boolean expression with SOP and POS must be practiced in both canonical and cardinal form along with their differences. Their inter-conversion must also be practiced. MARKING SCHEME Question 5 (a) A 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 B 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 C 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 D 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 20 X (OUTPUT) 0 0 0 1 0 1 0 1 0 0 0 1 1 1 1 1 POS Expression: X (A ,B, C, D) = (0,1, 2, 4, 6, 8, 9, 10) X(A ,B, C, D) = (A+B+C+D) . (A+B+C+D ) . (A+B+C +D) . (A+B +C+D) . (A+B +C +D) .(A +B+C+D) . (A +B+C+D ) . (A +B+C +D) (b) Application of Half Adder is to perform partial addition of two bits. Truth table of Half Adder : A 0 0 1 1 B 0 1 0 1 PS 0 1 1 0 CARRY 0 0 0 1 Circuit diagram for a Half Adder : (c) Convert to Canonical form: F(A,B,C)=(B+C ) . (A +B) = (B+C +0) . (A +B+0) = (B+C +(A.A )) . (A +B+(C.C )) = (A+B+C ) . (A +B+C ) . (A +B+C) Question 6 (a) What is a Multiplexer? How is it different from a decoder? Draw the circuit diagram for a 8:1 Multiplexer. [5] (b) Prove the Boolean expression using Boolean laws. Also, mention the law used at each step. F = (x + z) + [ (y + z) (x + y) ] = 1 [3] (c) Define maxterms and minterms. Find the maxterm and minterm when: [2] P = 0, Q = 1, R = 1 and S = 0 21 Comments of Examiners (a) The definition was well answered by most of the Suggestions for teachers candidates. Some were confused with logic diagram of a multiplexer and drew the diagram of a Most practice should be given in decoder instead. Some used the OR gate instead of drawing Multiplexer, Encoder and AND gate in the logic diagram. A few candidates Decoder. Use of proper connector drew the block diagram for a multiplexer. In some and gates must be explained. Uses cases, the input signals were missing, while some for each application along with their others drew the 4:1 MUX instead of 8:1 MUX. differences should be clarified. (b) A number of candidates did not mention the laws. More practice must be given in Some used lengthy methods for reducing and reducing expression using laws. wasted their time. Others used the algebraic method Mention of laws while reducing the to prove. expression must be encouraged. (c) A number of candidates did not give a proper More practice must also be given in definition. Finding the maxterm and minterm was L.H.S. = R.H.S. type of questions in well attempted by many candidates but some Boolean algebra. interchanged the answers. A few candidates wrote Students must be told to read the the definition of SOP and POS instead of question properly and answer maxterms and minterms. accordingly. MARKING SCHEME Question 6 (a) A Multiplexer is a combinational circuit which inputs parallel data and outputs one serial data where as, a Decoder is a combinational circuit which inputs n lines and outputs 2n or fewer lines. Circuit diagram of a 8:1 Multiplexer : 22 (b) Prove: F = (x +z) + [ (y +z).(x +y) ] =1 = (x +z) + (y +z) + (x +y) = x +z + y.z + xy = (x +x)(x +y ) + (z+z )(z+y) = x +y +z+y = 1 (as y+y =1) (c) ( De Morgan Law) (Distributive Law) (Identity Law) Maxterm : It is the sum of all its literals. Minterms : It is the product of all its literals. When P= 0 , Q = 1 , R = 1 , S = 0 Maxterm = P + Q + R + S Minterms = P Q R S 23 SECTION B Answer any two questions. Question 7 A class Palin has been defined to check whether a positive number is a Palindrome number [10] or not. The number N is palindrome if the original number and its reverse are same. Some of the members of the class are given below: Class name : Palin Data members/instance variables: num : integer to store the number revnum : integer to store the reverse of the number Methods/Member functions: Palin( ) : constructor to initialize data members with legal initial values void accept( ) : to accept the number int reverse(int y) : reverses the parameterized argument y and stores it in revnum using recursive technique void check( ) : checks whether the number is a Palindrome by invoking the function reverse( ) and display the result with an appropriate message Specify the class Palin giving the details of the constructor ( ), void accept( ),int reverse( int ) and void check( ). Define the main( ) function to create an object and call the functions accordingly to enable the task. Comments of Examiners Additional instance variables were used by a number of candidates. Instance variables were declared again in the constructor by some of the candidates. The concept of recursion was not clear to many candidates. Some did not use the parameters in the function reverse( ), others wrote the function reverse( ) without using the recursive technique. Several candidates had problems in check( ) function, where some candidates did not invoked the reverse( ) function. The other function including the constructor was well answered. Object creation and method calling was not done properly in the main( ) function in some cases. A few candidates did not write the main( ) function. 24 Suggestions for teachers More practice should be given to solve programs using recursive techniques. Knowledge of base case and recursive case should be given to the students for every program using recursive technique. Invoking function within another function should be given more practice. The difference between iteration and recursion must be explained. Knowledge of instance variables and their accessibility in the class must be emphasized. Candidates must be advised to read the question and answer accordingly what is required. MARKING SCHEME Question 7 import java.util.*; public class Palin { int num,revnum; static Scanner x=new Scanner(System.in); Palin() { num=0;revnum=0; } void accept() { System.out.println( "Enter a number"); num=x.nextInt(); } int reverse(int y) { if(y>0) { revnum =revnum * 10 + y%10; return reverse(y/10); } else return revnum; } void check() { int p=num; if( num==reverse(p)) System.out.println("palindrome"); else System.out.println("not a palindrome"); } static void main() { Palin obj=new Palin(); obj.accept(); obj.check(); } } 25 Question 8 A class Adder has been defined to add any two accepted time. Example: Time A - 6 hours 35 minutes Time B - 7 hours 45 minutes Their sum is - 14 hours 20 minutes ( where 60 minutes = 1 hour) [10] The details of the members of the class are given below: : Adder : integer array to hold two elements (hours and minutes) Adder( ) : constructor to assign 0 to the array elements void readtime( ) : to enter the elements of the array [Class name Data member/instance variable: a[ ] Member functions/methods: void addtime( : Adder X, Adder Y) adds the time of the two parameterized objects X and Y and stores the sum in the current calling object void disptime( ) displays the array elements with an appropriate message (i.e. hours = and minutes = ) : Specify the class Adder giving details of the constructor( ), void readtime( ), void addtime(Adder, Adder) and void disptime( ). Define the main( ) function to create objects and call the functions accordingly to enable the task. Comments of Examiners The addtime( ) function was not done properly by some candidates. Various methods/techniques were used add the time. Several candidates did it directly without using the parameterized object. A number of candidates had problem with the passing of object to the function. In some cases the candidates failed to store the sum of the two objects in the current object. Constructor and the main() method was largely answered properly. Suggestions for teachers 26 Passing of objects to a function through parameters must be given more practice. Working on one-dimensional and twodimensional arrays must be explained with various examples. Pass by value and pass by reference must be practiced and explained in detail with examples. Candidates must be advised to adhere to the rubric of the question and answer accordingly. MARKING SCHEME Question 8 import java.util.*; public class Adder { int a[]=new int[2]; static Scanner x=new Scanner(System.in); Adder() { a[0]=0;a[1]=0; } void readtime() { System.out.println("Enter hours and minutes"); a[0]=x.nextInt(); a[1]=x.nextInt(); } void disptime() { System.out.println("Hours=" + a[0]); System.out.println("Minutes=" + a[1]); } void addtime(Adder X,Adder Y) { a[1]=X.a[1] + Y.a[1]; a[0]=a[1]/60; a[1]=a[1]%60; a[0] += X.a[0] + Y.a[0]; } static void main() { Adder a=new Adder(); Adder b=new Adder(); Adder c=new Adder(); a.readtime(); b.readtime(); c.addtime(a,b); c.disptime(); } } 27 Question 9 A class SwapSort has been defined to perform string related operations on a word input. Some of the members of the class are as follows: Class name Data members/instance variables: wrd len swapwrd sortwrd Member functions/methods: SwapSort( ) : SwapSort : : : : to store a word integer to store length of the word to store the swapped word to store the sorted word : default constructor to initialize data members with legal initial values to accept a word in UPPER CASE to interchange/swap the first and last characters of the word in wrd and stores the new word in swapwrd sorts the characters of the original word in alphabetical order and stores it in sortwrd void readword( ) void swapchar( ) : : void sortword( ) : void display( ) : displays the original word, swapped word and the sorted word Specify the class SwapSort, giving the details of the constructor( ), void readword( ), void swapchar( ), void sortword( ) and void display( ). Define the main( ) function to create an object and call the functions accordingly to enable the task. Comments of Examiners Different methods / logic were used swap characters in swapchar( ) function and to sort in alphabetical in sortword( ) function. Some candidates included local variables in the functions and shared it with other functions. Others used the Character array to sort the word instead of doing it directly. In a few cases, the replace function was used which was not required. Some candidates were confused in extracting the middle characters in swapchar( ) function. A number of candidates were not able to display the required output in the display( ) function. The main( ) function and constructor were not answered properly by some of the candidates. 28 Suggestions for teachers Practice should be given in extracting characters from words, words from sentences and sentences from paragraphs. Different methods /logic should be adopted so that wider exposure to string manipulation related programs is given to students. Knowledge of constructors to initialize a string and other data members should be given. Conversion of string into characters and concatenating of strings must be given more practice. [10] MARKING SCHEME Question 9 import java.util.*; public class SwapSort { String wrd,swapwrd,sortwrd; int len; static Scanner x=new Scanner(System.in); SwapSort() { swapwrd=""; sortwrd=""; } void readword() { System.out.println("Enter word in Upper case"); wrd=x.next(); len=wrd.length(); } void swapchar() { swapwrd=wrd.charAt(len-1) + wrd.substring(1,len-1) + wrd.charAt(0); } void sortword() { char c; for(int i=65;i<=90;i++) { for(int j=0;j<len;j++) { c=wrd.charAt(j); if(c==i) sortwrd += c; } } } void display() { System.out.println("Original word = " + wrd); System.out.println("Swapped word = " + swapwrd); System.out.println("Sorted word = " + sortwrd); } static void main() { SwapSort x=new SwapSort(); x.readword(); x.swapchar(); x.sortword(); x.display(); } } 29 SECTION C Answer any two questions. Question 10 A super class Product has been defined to store the details of a product sold by a wholesaler to a retailer. Define a sub class Sales to compute the total amount paid by the retailer with or without fine along with service tax. Some of the members of both the classes are given below: Class name Data member/instance variable: name code amount Member functions/methods: Product(String n, int c, double p) void show( ) Class name: Data member/instance variable: day tax totamt Member functions/methods: Sales( ) : Product : : : stores the name of the product integer to store the product code stores the total sale amount of the product (in decimals) : parameterized constructor to assign data members name=n, code=c and amount = p displays the details of the data members : Sales : stores number of days taken to pay the sale amount to store the service tax (in decimals) to store the total amount (in decimals) : : : void compute( ) : void show( ) : parameterized constructor to assign values to data members of both the classes calculates the service tax @ 12 4% of the actual sale amount calculates the fine @ 2 5% of the actual sale amount only if the amount paid by the retailer to the wholesaler exceeds 30 days calculates the total amount paid by the retailer as (actual sale amount + service tax + fine) displays the data members of super class and the total amount Assume that the super class Product has been defined. Using the concept of inheritance, specify the class Sales giving the details of the constructor( ),void compute( ) and void show( ). The super class, main function and algorithm need NOT be written. 30 [5] Comments of Examiners The concept of Inheritance was not clear to many candidates. The keywords extends and super were missing in some cases. Constructor with inheritance was not answered correctly. Accessing the members of the super class by the derived class was not clear to a number of candidates. Some candidates declared the base class also, which was not required. Data members were not declared properly by some candidates. The function compute() in the derived class was not answered properly. In some cases, algorithm was written instead of a program. The rest of the function were well answered. Suggestions for teachers Practice should be given to students on inheritance. The importance of the keywords extends and super in inheritance must be explained properly. Use of constructor of the base class should be made clear. Explain the different visibility modes and their accessing capability. Calling the member function from the super class to the derived class must be made clear. Instruct students to read the question properly (base class not required) and answer accordingly. The concept of overriding in inheritance must be explained with examples. MARKING SCHEME Question 10 public class Sales extends Product { int day; double tax,totamt; Sales( String n,int a, double b, int d) { super(n,a,b); day=d; } void compute() { double f=0.0; tax= (12.4 /100) * amount; if(day>27) f=(2.5/100)* amount; totamt= amount+tax+f; } void show() { super.show(); System.out.println("No of days=" + day); System.out.println("Sales Tax=" + tax); System.out.println("Total Amount=" + totamt ); } } 31 Question 11 Queue is an entity which can hold a maximum of 100 integers. The queue enables the user to add integers from the rear and remove integers from the front. Define a class Queue with the following details: Class name : Queue Data Members / instance variables: Que[ ] : array to hold the integer elements size : stores the size of the array front : : to point the index of the front rear Member functions: Queue (int mm) to point the index of the rear constructor to initialize the data size = mm, front = 0, rear = 0 void addele(int v ) : to add integer from the rear if possible else display the message Overflow int delele( ) : returns elements from front if present, otherwise displays the message Underflow and return -9999 void display ( ) : displays the array elements Specify the class Queue giving details of ONLY the functions void addele(int) and int delele( ). Assume that the other functions have been defined. The main function and algorithm need NOT be written. 32 [5] Comments of Examiners The concept of queue was not clear to most of the candidates. Common errors made by candidates were as follows: (i) the condition / logic for underflow and overflow was not answered correctly; (ii) increment / decrement of front and rear index was not done properly. The methods addele ( ) and delete( ) were found to be difficult by some of the candidates. A few candidates also defined the constructor which was not required. In some cases, the class was not defined, only the functions addele( ) and delete( ) were defined. Suggestions for teachers More practice should be given in data structure programs like the stacks, queues, de queues, etc. Working must be shown as to how the stack or a queue performs (examples can be supportive). The concept of LIFO and FIFO must be explained to students with lively examples related to real world. Implementation of stacks, queues and de queues using arrays should be emphasized. Only the concept has to be explained taking the base as an array. It should be made clear to the students that it is not an array related program which can be manipulated by shifting / inserting or initializing by any value since these data structures require pointers and pointers are not supported in java. MARKING SCHEME Question 11 public class Queue { int Que[]=new int[100]; int max,f,r; void addele(int v) { if(r<max-1) Que[++r]=v; else System.out.println("Overflow"); } int delele() { if(f!=r) return Que[++f]; else return -9999; } } 33 Question 12 (a) A linked list is formed from the objects of the class Node. The class structure of the Node is given below: class { [2] Node int num; Node next; } Write an Algorithm OR a Method to count the nodes that contain only odd integers from an existing linked list and returns the count. The method declaration is as follows: int CountOdd( Node startPtr ) (b) Answer the following questions from the diagram of a Binary Tree given below: M N W G Y Z D F R (i) Write the postorder traversal of the above tree structure. [1] (ii) State the level numbers of the nodes N and R if the root is at 0 (zero) level. [1] (iii) List the internal nodes of the right sub-tree. [1] Comments of Examiners (a) Many candidates attempted this part well. Some candidates had problems in moving the pointer to the next node and checking for null. Some wrote the algorithm in simple English language, covering all the main steps. In some cases, the temporary pointer was not created. (b) (i) Several candidates wrote the preorder instead of post order of the tree. In some cases, one or two nodes were not placed correctly. (ii) This part was largely answered well. (iii) Most candidates attempted this part well. A few wrote the internal nodes of the entire tree. 34 Suggestions for teachers More methods / algorithms should be practiced with link list data structure. Use of diagrams to illustrate the link list must be practiced. Knowledge of temporary pointer, checking for null condition and moving pointer to the next node must be given. Root, height, depth, size, degree, siblings, nodes (internal and external), levels, tree traversals, etc. must be explained using a binary tree diagram. MARKING SCHEME Question 12 (a) 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. Return count Step 4. Check for odd and increment the counter. Step 5. Move pointer to the next node Step 6. End METHOD: int CountOdd(Node startPtr) { int c=0; Node temp=new Node(startPtr); while(temp != null) { if (temp.num % 2 != 0) c++; temp=temp.next; } return c; } (b) (i) (ii) (iii) WFYNRZDGM Level of N=1 and Level of R=3 G and Z 35 Topics found difficult by candidates Concepts in which candidates got confused Suggestions for candidates The symbols => , and v from propositional logic ( Inverse and Converse ) Interfaces and Classes Complexity Returning value of the base case in recursive output K-MAPS (Grouping, map-rolling, place value) Complement properties Recursive technique Passing objects to functions Queue operations for adding and removing elements The symbols in a proposition The terms complexity and interface Output using recursive technique Passing objects to functions Sorting and swapping character in a word Passing of objects Use of Single instance variable for multiple operations in various functions Link list and Queues Prepare summary for each chapter or use high lighters to recognize the important terms and definitions. Practical work on the system on a regular basis is necessary to understand the syntax and to correct errors. Answers and definitions should be short and precise and according to marks intended. Working should be shown at the side of each question wherever required. Laws must be mentioned while reducing a Boolean Expression. Practice one form of K-Map with proper place value for both SOP and POS. In programming, documentation is compulsory 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. Do not memorize the program, try to understand the logic. Practice constructors with every program. Treat each function of a class as separate program. 36

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Additional Info : ISC Class XII Analysis Of Pupil Performance 2017 : Computer Science
Tags : ISC Board, Class 11th, Class 12th, NDA/NA Entrance Examination  

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