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ISC Class XII Board Exam 2026 : Computer Science

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Pratima Patel
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ISC ANALYSIS OF PUPIL PERFORMANCE October 2025 COMPUTER SCIENCE Research Development and Curriculum Division Council for the Indian School Certificate Examinations New Delhi OCTOBER 2025 ____________________________________________________________________________________________ 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. Council for the Indian School Certificate Examinations (CISCE) MISSION STATEMENT The Council for the Indian School Certificate Examinations is committed to serving the nation's children, through high quality educational endeavours, empowering them to contribute towards a humane, just and pluralistic society, promoting introspective living, by creating exciting learning opportunities, with a commitment to excellence. ETHOS OF CISCE Trust and fair play. Minimum monitoring. Allowing schools to evolve their own niche. Catering to the needs of the children. Giving freedom to experiment with new ideas and practices. Diversity and plurality - the basic strength for evolution of ideas. Schools to motivate pupils towards the cultivation of: Excellence - The Indian and Global experience. Values - Spiritual and cultural - to be the bedrock of the educational experience. Schools to have an 'Indian Ethos', strong roots in the national psyche and be sensitive to national aspirations. FOREWORD The Council for the Indian School Certificate Examinations (CISCE) has endeavoured to sincerely take forward the vision of NEP 2020 in enhancing pedagogy and assessments through systemic reforms. In alignment with this vision, the CISCE also remains committed to ensuring that learning is meaningful, and that assessment practices are utilised as tools for growth, reflection, and continuous improvement rather than as an exercise in mere evaluation. The ISC and ICSE examinations have long served as the cornerstone of the CISCE s academic framework, especially in shaping the educational journeys and futures of countless CISCE students. In this context, the Analysis of Pupils Performance documents have played a significant role in strengthening the CISCE s assessment culture. These reports have, over the years, supported students in understanding subject-wise expectations and in preparing effectively for the Board examinations. Our heartfelt appreciation is extended to the Research Development and Curriculum Division (RDCD) of the CISCE for their dedicated efforts in preparing this detailed document for the benefit of students. We also express our gratitude to the examiners for their meticulous evaluation and insightful feedback on candidates performance, as well as for recommending pedagogical strategies to help teachers address common areas of difficulty. We are confident that students, teachers, and parents will make the best use of this document by engaging with its various sections in depth and by implementing the learnings to ensure enhanced and successful performance in the forthcoming examinations. October 2025 Dr. Joseph Emmanuel Chief Executive & Secretary CISCE PREFACE The Analysis of Pupil Performance document is a standing legacy of the Council for Indian School Certificate Examinations (CISCE) that aims at capturing performance trends of CISCE candidates in the annually conducted ICSE and ISC examinations. Over the years, these user-friendly and accessible reports have helped CISCE schools analyse patterns in candidates responses to board examination papers, while also providing valuable pedagogical insights and suggestions for teachers. Each year, CISCE undertakes this exercise of a detailed analysis of the performance of candidates for selected subjects at both the ICSE and ISC examinations. Each subject document includes the questions from the board examination along with the marking scheme for each question, enabling teachers and students to better understand the scope of the question and the appropriate approach to answering it. It also presents an infographic presentation of statistical analysis and contextualised commentaries on candidate performance with detailed insights and remedies on the errors. Towards the end of the document, a comprehensive summary highlights the topics revealing gaps in conceptual understanding, areas that candidates found confusing, and tailored suggestions to help students approach examinations more effectively. The uniqueness of this document lies in its micro-analysis of items in board examination papers where suggested pedagogical strategies are tied to each common errors identified from candidates answers, a strategy that CISCE believes is rudimentary in driving improvements rather than merely sharing a broad stroke macro analysis of candidate performance. While these reports are grounded in real candidate data, they also lend themselves to seamlessly to classroom action. The Analysis of Pupil Performance document for ICSE for the Examination Year 2025 covers the following subjects - English Language, Literature in English, Hindi, History and Civics, Geography, Mathematics, Physics, Chemistry, Biology, Commercial Studies, Economics, Computer Applications, Economic Applications, Commercial Applications, Environmental Science and Home Science. The subjects covered in the ISC Analysis of Pupil Performance document for the Year 2025 are Accounts, English Language, Literature in English, Hindi, Economics, Commerce, Business Studies, Mathematics, Physics, Chemistry, Biology, Elective English, History, Political Science, Geography, Psychology, Sociology, Computer Science, Environmental Science, Home Science and Legal Studies. I extend my heartfelt gratitude to all the ICSE and ISC examiners who have shared their valuable comments on each question. I also acknowledge the efforts of the RDCD team of Dr. Manika Sharma, Ms. Parul Kohli, Ms. Lyimee Saikia, Ms. Mansi Guleria, Ms. Gunjan Khurana, Ms. Madhusree Chatterjee and Ms. Aakriti Agrawal for their commitment to ensuring the compilation of the comments and data into its current form. We hope that the document is successful in its intended mission of better assisting our teachers in identifying areas that require focused intervention and allowing them to better support CISCE students. At the same time, we also hope that it guides our students in developing a roadmap of effective study methods for their examinations. October 2025 Dr. Bhawna Taragi Deputy Head RDCD, CISCE QUANTITATIVE ANALYSIS Computer Science Number of Candidates Appeared Maximum Marks 100 26067 Highest Marks Boys Girls 16460 9607 Lowest Marks Average Marks 100 83 Overall Performance (Mean Marks) 2 82.6 Performance by Gender 82.3 83 Region-wise Performance (Mean Marks) 81.6 81.5 82.5 81.8 81.8 Eastern 82.1 85 Northern Girls 83 0.14 Boys 82.3 0.12 t-value 3.8* Top 5 Perfoming States/UTs (Mean Marks) 83.5 85.8 82.7 SE *Significant at 0.05 level Western 82.3 Mean Gender Dubai (U.A.E) 93.2 New Delhi 92.6 Goa 91.8 Chandigarh 91.5 Haryana 91.2 87.3 Southern 89.9 89.3 92 Foreign 1 Performance State-wise and Foreign (Mean Marks) BOYS GIRLS OVERALL 85.2 85.0 85.2 85.8 ANDHRA PRADESH ASSAM 94.7 89.0 79.4 79.8 79.6 BIHAR CHANDIGARH 93.6 67.0 CHHATTISGARH 63.5 91.5 71.1 67.4 92.8 94.7 93.2 DUBAI (U.A.E.) 88.9 GOA 91.8 95.2 76.9 79.2 77.7 GUJARAT 90.8 92.7 91.2 HARYANA HIMACHAL PRADESH 81.3 73.2 78.9 74.6 76.6 75.4 79.7 80.6 80.1 JAMMU AND KASHMIR JHARKHAND 86.2 87.3 86.6 87.1 89.6 88.0 KARNATAKA KERALA 77.9 78.0 77.9 MADHYA PRADESH 86.6 87.2 86.8 MAHARASHTRA MEGHALAYA 81.5 81.5 92.8 92.4 92.6 NEW DELHI 75.2 76.7 75.8 77.1 73.4 76.1 ODISHA PUDUCHERRY 81.8 84.6 83.3 85.8 89.5 87.8 PUNJAB RAJASTHAN SHARJAH 70.0 67.3 SIKKIM 69.0 76.3 75.0 73.4 82.1 SINGAPORE 86.9 85.0 86.7 85.5 TAMIL NADU 72.5 TELANGANA 76.4 98.0 83.0 71.5 73.3 71.9 TRIPURA 81.8 82.8 82.2 85.3 85.2 85.3 82.9 83.3 83.0 UTTAR PRADESH UTTARAKHAND WEST BENGAL 0 20 40 2 60 80 100 QUALITATIVE ANALYSIS PART I 20 MARKS Answer all questions. While answering questions in this Part, indicate briefly your working and reasoning, wherever required. Question 1 (i) (ii) The complement of the Boolean expression (A B ) + (B C) is: (a) (A+ B ) (B + C) (b) (A B) + (B C ) (c) (A +B) (B+ C ) (d) (A B ) + (B C ) Given below are two statements marked, Assertion and Reason. Read the two statements carefully and choose the correct option. [1] [1] Assertion: The expression ~ ( X Y) is logically equivalent to ( ~X ~Y ) Reason: The commutative property of logical operators states that the order of the operands does not change the result of a binary operation. (iii) (a) Both Assertion and Reason are true and Reason is the correct explanation for Assertion. (b) Both Assertion and Reason are true but Reason is not the correct explanation for Assertion. (c) Assertion is true and Reason is false. (d) Both Assertion and Reason are false. According to the Principle of Duality, the Boolean equation [1] (1 + Y) (X + Y) = Y + X will be equivalent to: (iv) (a) (1 + Y ) (X + Y ) = Y + X (b) (0 Y) + (X Y) = Y X (c) (0 + Y) (X + Y) = Y + X (d) (1 Y) + (X Y) = Y X [1] The Associative Law states that: (a) A B = B A (b) A+B=B+A (c) A ( B + C ) = A B + A C (d) A+(B+C)=(A+B)+C 3 2025 (v) ISC- Computer Science [1] Consider the following code statement: public class Person { int age; public Person (int age) { this.age = age; } } Which of the following statements are valid for the given code? (vi) I. The keyword this in the constructor refers to the current instance of the class. II. The keyword this differentiates between the instance variable age and the parameter age. III. The keyword this can be used only in constructors. (a) Only I and II (b) Only II and III (c) Only I and III (d) Only III Given below are two statements marked, Assertion and Reason. Read the two statements carefully and choose the correct option. [1] Assertion: The break statement prevents fall through effect in switch case construct. Reason: The break statement enables unnatural exit from the loop. (vii) (a) Both Assertion and Reason are true and Reason is the correct explanation for Assertion. (b) Both Assertion and Reason are true but Reason is not the correct explanation for Assertion. (c) Assertion is true and Reason is false. (d) Both Assertion and Reason are false. The canonical expression of F( P, Q, R ) = (2, 5, 7) is: (a) ( P + Q + R ) ( P + Q + R ) ( P + Q + R ) (b) ( P Q R ) + ( P Q R )+( P Q R ) (c) ( P + Q + R ) ( P + Q + R ) ( P + Q + R ) (d) ( P Q R )+( P Q R )+( P Q R ) 4 [1] 2025 ISC- Computer Science (viii) Study the given propositions and the statements marked, Assertion and Reason that follow it. Choose the correct option on the basis of your analysis. [1] P It is a holiday Q It is a Sunday Assertion: If it is not a Sunday, then it is not a holiday. (Q => P ) Reason: Inverse is formed when antecedent and consequent are interchanged. (ix) (a) Both Assertion and Reason are true and Reason is the correct explanation for Assertion. (b) Both Assertion and Reason are true but Reason is not the correct explanation for Assertion. (c) Assertion is true and Reason is false. (d) Both Assertion and Reason are false. For the given code segment, write Big O notation for worst case complexity. [1] for ( int i=1; i<=P; i++) { Statements } for (int j=1; j<=P; ++j) for (int k=1; k<=Q; k++) { Statements } (x) Write the minterms in canonical form for the Boolean Function X (A, B), from the truth table given below: A B X 0 0 1 0 1 0 1 0 0 1 1 1 5 [1] 2025 ISC- Computer Science MARKING SCHEME (i) (c) or (A + B) (B+ C ) (ii) (b) or Both Assertion and Reason are true but Reason is not the correct explanation for Assertion. (iii) (b) or (0 Y) + (X Y) = Y X (iv) (d) or A + ( B + C ) = ( A + B ) + C (v) (a) or Only I and II (vi) (vii) (b) or Both Assertion and Reason are true but Reason is not the correct explanation for Assertion. (a) or ( P + Q + R ) ( P + Q + R ) ( P + Q + R ) (viii) (c) or Assertion is true and Reason is false. (ix) O(P) + O(PQ) or O(P + PQ) (x) 0 = A B and 3 = A B or (0,3) 6 2025 ISC- Computer Science Comments of Examiners (i) Most of the candidates were confused between dual and complement of Boolean expression . Some candidates could not understand the concept of complementation and use of De-Morgan s law and thus answered it incorrectly. (ii) Most of the candidates answered it correctly. Some candidates could not translate the Boolean laws into proposition and were confused. (iii) Many candidates found the concepts of complement and dual confusing, which led to incorrect answers. Some candidates altered the complements, while a few were misled by the nearly identical options provided in the question. (iv) Many candidates answered this question correctly. (v) A large number of candidates were able to answer it correctly. However, some candidates were confused by the various combinations and chose incorrect options. (vi) Most of the candidates were confused between option (a) and option (b) because of the term loop in reasoning. (vii) A few candidates answered it incorrectly because of lack of clarity to convert cardinal number to maxterm. (viii) Most of the candidates answered this question correctly. However, some candidates were unsure about the difference between antecedent and consequent as a result failed to mark the correct option. (ix) Most candidates could answer this question correctly. Many candidates answered directly by considering the dominant term instead of explaining the complexities of non-nested and nested loop . Some candidates gave examples to illustrate the answers while some presented graphs to illustrate the complexities. (x) Most candidates gave the correct answer to this question. Some gave vague answers despite the correct working. A few candidates could not understand the concept of the term canonical and wrote the incorrect answers. 7 Suggestions for teachers Provide adequate practice of finding the compliment of Boolean expression to make the students understand the application of De Morgan s law. Explain the difference between principle of duality and De Morgan s theorem and simultaneously give practice of both kind of sums to avoid confusion. Explain explicitly the conversion of the Boolean laws into its proposition with examples. Teach the analysis of the Assertion and Reason and interpretation of the statements to identify the correct option. Teach the difference between complement and dual with sufficient examples. Conduct regular practice sessions to reduce incidences of error. Advise students to write all the laws as small handouts in one sheet of paper along with the names, and discuss the same in class. Teach all the properties and their laws, along with their respective names. Elucidate the concept of canonical and cardinal form of expression. Explain conversion from canonical to cardinal form and vice versa with examples. Discuss all the terminologies like converse, inverse, contrapositive with suitable examples. Provide worksheets to clarify concepts on different terminologies. Teach all three cases of complexities with examples and factors that influence it. Clearly highlight the various programs where complexity differs- example loops, nested loops, conditional, recursion etc. Teach students about minterm and maxterm with or without the help of truth table. Explain the difference between M and m along with symbols and with respect to minterms, maxterms, SOP and POS expression. Acquaint students to obtain the SOP and POS expression using the truth table. 2025 ISC- Computer Science Question 2 (i) (ii) (iii) Convert the following infix notation to postfix form. (A B/C)+(D*E/F)*G A matrix M[ 1 10, 4 .13] is stored in the memory with each element requiring 2 bytes of storage. If the base address is 1200, find the address of M[2][7] when the matrix is stored Row Major Wise. [2] The following function int solve( ) is a part of some class. Assume m and n are positive integers. Answer the questions given below with dry run / working. (a) (b) (iv) [2] int solve(int m, int n) { int k=1; if(m<0) return k; else if(m==0) return m; else return k+(solve(m n, n+2)); } What will the functions solve ( ) return if: (1) m = 16, n = 1 (2) m = 9, n = 1 What is the function solve( ) performing a part from recursion? [2] [1] The following function duck( ) is a part of some class which is used to check if a given number is a duck number or not. There are some places in the code marked by ?1?, ?2?, ?3? which may be replaced by a statement /expression so that the function works properly. A number is said to be Duck if the digit zero (0) is present in it. boolean duck(int a) { int f= 1; if(a==0) return true; for(int i=a; i!=0;?1?) { int c = i%10; if(c==?2?) { f=1; break; } } return (f ==?3?)? false: true; } (a) What is the expression or statement at ?1? [1] (b) What is the expression or statement at ?2? [1] (c) What is the expression or statement at ?3? [1] 8 2025 ISC- Computer Science MARKING SCHEME (i) (A B/C)+(D*E/F)*G ( A B C/ ) + ( D E * / F) * G (ABC/ )+(DE*F/)*G (ABC/ )+(DE*F/G*) =ABC/ DE*F/G*+ (ii) Address of M[2][7] = B + W * ( ( I Lr) * COL + (J Lc)) = 1200 + 2 ((2+1) 10 +(7 - 4) ) = 1200 + 2(30 +3) = 1200 + 2 * 33 = 1200 + 66 = 1266 (iii) (iv) (a) = solve(16,1) 1 + solve(15,3) 1+1 + solve (12,5) 1+1+1 + solve (7,7) =1+1+1 + 1 = 4 (i) 4 = solve(9,1) 1+ solve(8,3) 1+1+solve(5,5) = 1+1+1 = 3 (ii) 3 (b) Returning Square root of a perfect square only (a) i /=10 (b) 0 (c) -1 or i = i/10 9 2025 ISC- Computer Science Comments of Examiners Most candidates were able to solve this Suggestions for teachers problem correctly. However, some of the Provide practice of examples involving candidates wrote the correct answer without conversion of Infix to Postfix notation, showing the working while some other the order of precedence. candidates applied the postfix correctly but Explain the operator s precedence could not derive the final answer due to rule/priority of operators, and concept of wrong operator precedence. left-to-right while evaluating an (ii) This part was answered well by most of the expression. candidates. Some of the errors noted were Give more practice to calculate addresses Directly writing the answer without using Row major and Column major wise. showing the working/formula. Explain the different terms used in address calculations. Calculation mistakes. Discuss in detail how stack concept is Doing column major instead of row used while executing recursive methods. major. Give sufficient practice to find the Not calculated the size of the matrix output/return value of recursive methods properly. with proper working. (iii) (a) Many candidates were unclear with the Ensure to provide adequate practice to the concept of stack and could not answer it students on Recursive technique correctly. Most of the candidates could (including identification of base case) and do the dry run well but made mistakes in how the values changes and stored in returning the final correct value. Some fresh memory locations and their of the candidates did wrong working working. Explain Memory blocks/tables to show while some other candidates incorrectly the concept of working of Stack and LIFO changed the function parameters during process in case of recursion. recursive call. A few candidates showed Emphasise that the students must obtain the working but were unable to find the the output along with dry run / working exact value. because it is a necessary component. (b) Since this was an interrelated question, Familiarise students with dry run of each many candidates who answered the first code so that they understand the logic part incorrectly also got the second part clearly. Provide sufficient practice of wrong. Some candidates carried out the such questions along with practical working correctly but failed to mention classes. the correct output. Practice should be given more on (iv) (a) Majority of the candidates answered this program using conditions / looping and part quite well, however, some other output related programs. candidates were confused in identifying Show the dry run/ working of program and emphasise that working is necessary the actual value for ?1? and wrote i = to get full credit. a/10 or only i/10 instead of i=i/10. (b) Most of the candidates answered it correctly. A few candidates gave vague answer. (c) Most of the candidates answered this sub-part correctly. (i) 10 2025 ISC- Computer Science 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 3 (i) A superhero is allowed access to a secure Avengers facility if he/she meets any of the following criteria: The superhero has Avengers membership and possesses a high-security clearance badge OR The superhero does not have Avengers membership but holds a special permit issued by S.H.I.E.L.D. along with a high-security clearance badge OR The superhero is not a recognised ally but holds a special permit issued by S.H.I.E.L.D. along with a high-security clearance badge The inputs are: INPUTS A S C L [5] Superhero has Avengers membership. Superhero holds a special permit issued by S.H.I.E.L.D. Superhero possesses a high-security clearance badge Superhero is a recognised ally (In all the above cases, 1 indicates YES and 0 indicates NO) Output: X Denotes allowed access [1 indicates YES and 0 indicates NO in all cases] Draw the truth table for the inputs and outputs given above. Write the POS expression for X (A, S, C, L). (ii) Reduce the above expression X (A, S, C, L) by using 4-variable Karnaugh map, showing the various groups (i.e., octal, quads and pairs). Draw the logic gate diagram using NOR gates only for the reduced expression. Assume that the variables and their complements are available as inputs. 11 [5] 2025 ISC- Computer Science MARKING SCHEME (i) A 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 S 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 L 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 X 0 0 0 0 0 0 1 1 0 0 1 1 0 0 1 1 X (A,S,C,L) = (0,1,2,3,4,5,8,9,12,13) X (A, S, C, L) = (A+S+C+L) . (A+S+C+L ) . (A+S+C +L) . (A+S+C +L ) . (A+S +C+L) .(A+S +C+L ) . (A +S+C+L) . (A +S+C+L ) . (A +S +C+L) . (A +S +C+L ) (ii) C+L A+S A + S A + S A + S 0 4 12 8 0 0 0 0 C + L 1 5 13 9 0 0 0 0 There is one Octal and one Quad: Octal (M0.M1.M4,M5.M8.M9.M12.M13) = C Quad 1( M0.M1.M2.M3) = A + S Hence X(A,S,C,L) = C (A + S ) 12 C + L 3 7 15 11 0 1 1 1 C + L 2 6 14 10 0 1 1 1 2025 ISC- Computer Science Comments of Examiners (i) Most of the candidates answered this question well and were able to score full credit. A few common errors observed were: Not mentioning the final simplified expression. Confusion between Sum of Products (SOP) and Product of Sums (POS) forms. Incorrectly used the output values with 0 s instead of 1 s for SOP. Used minterms while dealing with 0 s. Incorrect variable representation or incomplete solutions. (ii) Most candidates could answer this part correctly. However, some recurring mistakes were observed A few candidates were unable to draw the K-Map for the POS (Product of Sums) expression correctly, often confusing it with the SOP form. Incorrect variables were used to construct the K-Map . Additionally, some candidates included redundant groups in the final expression, which were unnecessary. Suggestions for teachers 13 Lay stress on the accurate interpretation of what is being asked, particularly the questions related to logical expressions. Provide more practice in deriving both SOP (Sum of Products) and POS (Product of Sums) expressions from given truth tables, including correct identification and use of minterms and maxterms. Remind students about the importance of clearly stating the final simplified expression in order to get full credit. Train students to reduce both SOP and POS expressions using Karnaugh Maps simultaneously. Emphasis must be laid on avoiding redundant groupings. Instruct students to include only essential groups in the final simplified expression. Ensure that students practice all the steps of working with K-Maps including drawing the map accurately, correctly filling it with 0 s and 1 s, identifying and marking valid groups, and reducing those groups to derive the most optimized Boolean expression. Insist upon reading the question carefully and respond precisely. 2025 ISC- Computer Science Question 4 (i) (ii) Reduce the Boolean function F(P, Q, R, S) = (0,1,2,5,7,8,9,10,13,15) by using 4-variable Karnaugh map , showing the various groups (i.e., octal, quads and pairs). (b) Draw the logic gate diagram using NAND gates only for the reduced expression. Assume that the variables and their complements are available as inputs. From the logic gate diagram given below: (a) [1] (a) Derive Boolean expression for (1), (2) and R. Reduce the derived expression. [4] (b) Name the logic gate that represents the reduced expression. [1] MARKING SCHEME (i) [4] (a) R S P Q P Q PQ PQ 0 4 12 8 R S 1 1 5 0 13 0 9 1 1 1 1 1 RS 3 7 15 11 0 1 1 0 There are 3 Quads. Quad 1 : m5+m7+m13+m15 = QS Quad 2: m0+m2+m8+m10 = Q S Quad 3: m0+m1+m8+m9 = Q R OR m1+m5+m9+m13 F(P,Q,R,S) = QS+Q'S'+Q'R' OR QS+Q'S'+R'S 14 RS 2 6 14 10 = R S 1 0 0 1 2025 ISC- Computer Science (b) (ii) (a) (1) = (2) = A+B R= ( )( + ) = ( + )( + ) = ( + ) + ( + ) = [( ) . ( ) ] + ( . ) = [( + ). ( + )] + ( . ) = + + = + = + = (A.B) (b) NAND gate 15 2025 ISC- Computer Science Comments of Examiners (i) (a) Most candidates answered this question correctly, However, a few common mistakes were observed Some candidates made errors in determining the place value and in correctly placing variables within the K-Map . Reduced the groups using Boolean laws instead of simplifying directly through the K-Map method . Some K-Maps were drawn incorrectly. Included redundant groups in the final expression, resulting in nonoptimized solutions. (b) Most of the candidates answered the question correctly. Some candidates drew the logic circuit using basic gates (such as AND, OR, and NOT) instead of using only NAND gates. Additionally, a few responses included vague or unclear circuit diagrams. (ii) (a) Several candidates answered this part of the question correctly. However, some candidates were unable to reduce the expressions properly, which affected the final result. In some cases, a single incorrect output led to an entirely incorrect reduction. (b) Most candidates represented the logic gate accurately. However, some reduced the Boolean expression incorrectly, which led to inaccurate gate representations. Additionally, a few candidates, despite correctly reducing the expression, overlooked the requirement to draw the corresponding logic circuit, resulting in incomplete answers. Suggestions for teachers Emphasis should be made on arranging 16 variables in the correct order when constructing Karnaugh Maps. Ensure that students understand the formation of groups and how these groups are then systematically reduced to derive the simplified Boolean expression. Lay stress on not including redundant groups in the final expression, as it may lead to unnecessarily complex results. Provide extensive practice to the students in drawing logic circuits using both basic gates (AND, OR, NOT) and universal gates (NAND and NOR). Reinforce the understanding of logical equivalence and gate substitution by regular exposure to designing circuits with different combinations of gates. Insist on clarity, accuracy, and neatness in circuit diagrams. Give adequate practice to the students on variety of circuit diagrams, covering both simple and complex logic gate combinations. Provide rigorous practice on the interconversion between logic circuits, Boolean expressions, and truth tables. Teach step-by-step methods for translating one form into another in detail, supported by diverse examples to build accuracy and confidence. Provide detailed examples to clearly demonstrate the representation of Boolean expressions using various logic gates. Provide detailed examples to clearly demonstrate the representation of Boolean expressions for various logic gates, including AND, OR, NOT, NAND, NOR, XOR, and XNOR. Teach the representation of each gate algebraically and interpretation of these expressions in circuit form. Explain both SOP (Sum of Products) and POS (Product of Sums) expressions for all seven basic gates, with emphasis on how to derive and simplify them. 2025 ISC- Computer Science Question 5 (i) What is an encoder? Draw the logic gate diagram for an octal to binary encoder. State one application of a decoder. [5] (ii) By using truth table, verify if the following proposition is valid or not. [3] (~X => Y) X = (X ~Y) (X Y) (iii) Study the logic gate diagram given below and answer the questions that follow: What will be the output of the above gate when: (a) A=1, B=0 [1] (b) A=1, B= 1 [1] 17 2025 ISC- Computer Science MARKING SCHEME (i) An encoder is a combinational circuit which inputs 2n or fewer lines and outputs n lines, where n represents bit code for the input. Application of a decoder Address Decoding Barcode Scanner Television Remote Control Traffic Light Control Making of Multiplexers Conversion of Decimal, Octal, Hexadecimal to Binary (ii) X Y ~X ~Y ~X=>Y (~X=>Y) X X ~Y 0 0 1 0 1 1 1 1 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 1 1 1 0 1 1 0 0 1 1 0 1 1 The given expression is VALID (iii) (a) 1 or True (b) 0 or False 18 X Y (X ~Y) (X Y) 2025 ISC- Computer Science Comments of Examiners Suggestions for teachers (i) Most candidates were able to draw the logic gate diagram correctly. However, a few Provide regular practice in drawing common errors were observed: encoders along with their corresponding Boolean expressions, truth tables, and Some candidates incorrectly drew a logic circuits. decimal or hexadecimal encoder instead of Ensure sufficient practice on the circuit the required octal encoder. diagrams, truth tables, logical A few candidates mistakenly drew an AND expressions, definitions, and applications gate in place of an OR gate. of key Boolean algebra components such Some vague or generic examples were also as half adders, full adders, encoders, and provided. decoders. (ii) Most of the candidates answered this question Emphasis should be laid on helping well. However, some candidates were students to understand the purpose and confused by the logical symbols such as ~ functioning of logic gates used in these applications. (NOT), (AND), (OR), and (implication), Encourage students to practice all the which led to errors in interpretation. Some fundamental laws of Boolean Algebra and errors made by the candidates were: Propositional Logic, with special Used a truth table for three variables emphasis on proving each law through instead of the required two. appropriate methods. The column representing the implication Lay stress on the correct use and ( ) was incorrect, resulting in an interpretation of logical symbols such as inaccurate final conclusion. (AND), (OR), ~ (NOT), Opted to prove the statement using logical (implication), and (bi-implication) laws instead of constructing a truth table. Give ample practice in identifying and analyzing the nature of propositions Missed mentioning whether the given whether they represent a tautology, proposition was a contradiction. contingency, or contradiction. (iii) (a)This part of the question was answered well Provide exercises involving randomly by most of the candidates. Several selected gates with assigned input values candidates drew the truth table to support to help students develop confidence and their answers. However, a few candidates accuracy in determining the correct were confused with the type of logic gate output. involved and consequently provided the Provide rigorous practice to the students incorrect answers. by framing questions in different twisted (b) Majority of the candidates answered the forms for various conditions. question correctly. However, a few candidates interchanged the answers of part (a) with part (b) leading to the incorrect answers. 19 2025 ISC- Computer Science SECTION B Question 6 Answer any two questions. A class Perni has been defined to accept a positive integer in binary number system from [10] the user and display if it is a Pernicious number or not. [A pernicious number is a binary number that has minimum of two digits and has prime number of 1 s in it.] Examples: 101 is a pernicious number as the number of 1 s in 101 = 2 and 2 is prime number. 10110 is a pernicious number as the number of 1 s in 10110 = 3 and 3 is prime number. 1111 is a NOT a pernicious number as the number of 1 s in 1111 = 4 and 4 is NOT a prime number. The details of the members of the class are given below: Class name : Perni : to store a binary number Perni ( ) : void accept( ) : int countOne(int k) : void check ( ) : constructor to initialise the data member with 0 to accept a binary number (containing 0 s and 1 s only) to count and return the number of 1 s in k using recursive technique to check whether the given number is a pernicious number by invoking the function countOne( ) and to display an appropriate message Data variable: member/instance num Methods / Member functions: Specify the class Perni giving the details of the constructor( ), void accept( ),int countOne(int) and void check( ). Define a main( ) function to create an object and call the functions accordingly to enable the task. 20 2025 ISC- Computer Science MARKING SCHEME import java.util.*; class Perni { int num; static Scanner S = new Scanner(System.in); Perni() { num = 0; } void accept() { System.out.println("Enter a number"); num = S.nextInt(); } int countOne(int k) int countOne(int k) { { if(k<10) if(k==0) return k; OR return 0; else if(k%10 ==1) else return k%10 + countOne(k/10); return 1+countOne(k/10); } else return 0+countOne(k/10); } void check() { int m = countOne(num); int k=0; for(int i=2;i<m;i++) { if (m%i==0) { k++; break; } } if(k==1 || m==0 ||m==1) System.out.println(num+" is not a Pernicious number"); else System.out.println(num+" is a Pernicious number"); } static void main() { Perni obj=new Perni(); obj.accept(); obj.check(); } } 21 2025 ISC- Computer Science Comments of Examiners Most candidates answered it correctly. However, several recurring issues were observed: Incorrectly used instance variables. Conceptual gaps in recursion : Several candidates did not understand the concept of recursion . Common mistakes included: Missed or wrong base cases. Improperly structured recursive cases. Used iteration (loops) to solve the recursive problem. Misunderstood parameters in countOne( ) method which affected the logical flow and correctness of the method. Issues in main( ) method: Object instantiation and method invocation were incorrectly done. Omitted the main( ) method entirely. Lack of documentation resulted in reducing code readability and clarity of logic. Suggestions for teachers 22 Ensure that students understand recursive techniques with examples. Lay special emphasis on teaching recursion with clear, step-by-step examples, illustrating how the recursive flow unfolds. Reinforce the identification and correct implementation of both the base case and the recursive case through recursive program. Train students to distinguish when a problem requires recursion and how to structure the logic accordingly. Provide more practice on invoking one method from another, particularly in recursive contexts, to build clarity on program control flow. Clearly explain the conceptual difference between iteration and recursion with examples. Advise students to read the question carefully and focus on addressing the exact requirements, rather than applying unrelated logic or modifying the problem structure. Reinforce the concepts of recursion, emphasizing proper structuring with base and recursive cases. Conduct focused practice on method parameter passing and object-oriented programming principles. Encourage students to follow the problem statement strictly without unnecessary additions. Emphasise the importance of writing a main method and commenting the code properly. 2025 ISC- Computer Science Question 7 [10] Design a class Colsum to check if the sum of elements in each corresponding column of two matrices is equal or not. Assume that the two matrices have the same dimensions. Example: Input: MATRIX A MATRIX B 2 3 1 7 4 2 7 5 6 1 3 1 1 4 2 2 5 6 Output: Sum of corresponding columns is equal. The details of the members of the class are given below: Class name : Colsum mat[ ] [ ] : to store the integer array elements m : to store the number of rows n : to store the number of columns Colsum(int mm, int nn) : void readArray( ) boolean check(Colsum A, Colsum B) : : parameterised constructor to initialise the data members m = mm and n = nn to accept the elements into the array to check if the sum of elements in each column of the objects A and B is equal and return true otherwise, return false void print( ) : Data members/instance variables: Member functions/methods: to display the array elements Specify the class Colsum giving details of the constructor(int, int), void readArray( ), boolean check(Colsum, Colsum), and void print( ). Define the main() function to create objects and call the functions accordingly to enable the task. 23 2025 ISC- Computer Science MARKING SCHEME import java.util.Scanner; class Colsum { int[][] mat; int m, n; static Scanner sc = new Scanner(System.in); public Colsum(int mm, int nn) { m = mm; n = nn; mat = new int[m][n]; } public void readArray() { System.out.println("Enter elements for the matrix"); for (int i = 0; i< m; i++) for (int j = 0; j < n; j++) mat[i][j] = sc.nextInt(); } public boolean check(Colsum A, Colsum B) { for (int j = 0; j <A.n; j++) { int csumA = 0; int csumB = 0; for (int i = 0; i<A.m; i++) { csumA += A.mat[i][j]; csumB += B.mat[i][j]; } if (csumA != csumB) return false; } return true; } public void print() { System.out.println("Matrix:"); for (int i = 0; i< m; i++) { System.out.println(); for (int j = 0; j < n; j++) System.out.print(mat[i][j] + "\t"); }} public static void main( ) { Colsum matA = new Colsum(3,4); Colsum matB = new Colsum(3,4); matA.readArray(); matB.readArray(); matA.print(); matB.print(); if (matA.check(matA, matB)) System.out.println("\nThe sum of elements in each column of both matrices is equal."); else System.out.println("\nThe sum of elements in each column of both matrices is NOT equal."); } } 24 2025 ISC- Computer Science Comments of Examiners Majority of the candidates answered it correctly. Some common errors made by most of the candidates were: Failed to allocate memory for the array using the new operator. Incorrectly handled the passing of objects between methods or constructors. Some candidates created a new array, deviating from the intended solution approach. Errors observed in object creation within the main() method. Unnecessarily created three objects instead of the required two. The invocation of the check() method was either incomplete or incorrectly structured. Lacked understanding of the return type of check() method, leading to logical errors or compilation issues. Suggestions for teachers Provide more exercises to students on how objects are passed to functions via parameters. Explain the concepts of onedimensional and two-dimensional arrays with multiple practical examples. Emphasis should be laid on array declaration, memory allocation, iteration, and manipulation. Clearly explain the difference between pass by value and pass by reference, using diagrams and traceable examples to demonstrate how data is transferred and modified in each case. Instruct students to read the question and respond strictly according to the given rubric. Elucidate the concept of dynamic binding (runtime polymorphism) with suitable examples involving method overriding and late binding. Explain the usage of the dot (.) operator for accessing members and methods and demonstrate it through examples. Highlight the purpose and usage of flag variables to help students manage control flow, especially in conditional checks and loop-based logic. Question 8 [10] A class Flipgram has been defined to flip the letters of the left and right halves of a nonheterogram word. If the word has odd number of characters, then the middle letter remains at its own position. A heterogram is a word where no letter appears more than once. Example 1: INPUT : BETTER OUTPUT: TERBET Example 2: INPUT : NEVER OUTPUT: ERVNE Example 3: INPUT : THAN OUTPUT: HETEROGRAM The details of the members of the class are given below: Class name : Flipgram Data member/instance variable: 25 2025 ISC- Computer Science : to store a word Flipgram(String s) : parameterised constructor to assign word = s boolean ishetero( ) : to return true if word is a heterogram else return false String flip( ) : void display( ) : to interchange the left and right sides of a non-heterogram word and return the resultant word to print the flipped word for a nonheterogram word by invoking the method flip( ).An appropriate message should be printed for a heterogram word word Methods/Member functions: Specify the class Flipgram giving the details of the constructor(String), boolean ishetero( ), String flip( ) and void display( ). Define a main( )function to create an object and call the functions accordingly to enable the task. MARKING SCHEME import java.util.Scanner; class Flipgram { String word; Flipgram(String s) { word =s; } boolean ishetero() { String w1=word; int ctr=0; for(int i=0;i<w1.length();i++) { char m=w1.charAt(i); if(w1.indexOf(m)==w1.lastIndexOf(m)) ctr++; } if(ctr==w1.length()) return true; else return false; } String flip() { String w2=word; String left =w2.substring(0,w2.length()/2); String right; if(w2.length()%2==0) { right= w2.substring(w2.length()/2); return right+left; } else { right= w2.substring(w2.length()/2+1); char c=w2.charAt(w2.length()/2); 26 2025 ISC- Computer Science return right+c+left; } } } void display() { if(ishetero()) System.out.println("HETEROGRAM"); else System.out.println(flip()); } public static void main( ) { Scanner sc=new Scanner (System.in); System.out.println("Enter a word"); String s=sc.next(); Flipgram fl=new Flipgram(s); OR Flipgram fl=new Flipgram( NEVER ); fl.display(); } Comments of Examiners Most candidates answered this question well. Several candidates used a variety of valid and logical approaches to count the frequency of characters/alphabet. Different logics were also applied to split and concatenate substrings, and most of the solutions met the expected outcomes. Some candidates correctly accepted the input word and converted it into uppercase, though this was not required in all cases. A few candidates, however, produced incorrect or logically inconsistent implementations of the flip() method. Suggestions for teachers 27 Provide ample practice to the students in extracting characters from words, words from sentences, and sentences from paragraphs to strengthen. Teach varied methods and logic for solving string-related problem and ensure that students are exposed to a wider range of approaches. Lay emphasis on using constructors to initialize strings and other data members to reinforce the fundamentals of object-oriented programming. Give additional practice to students in converting strings into character arrays and vice versa. Include concatenation of strings using different techniques (e.g., using +, concat(), StringBuilder, etc.) in exercises for clarity and proficiency. 2025 ISC- Computer Science SECTION C Answer any two questions. Question 9 A circular queue is a linear data structure that allows data insertion at the rear and removal from the front, with the rear end connected to the front end forming a circular arrangement. The details of the members of the class are given below: Class name : CirQueue Q[ ] : array to hold integer values cap : maximum capacity of the circular queue front : to point the index of the front rear : to point the index of the rear CirQueue(int n) : void push(int v) : constructor to initialise cap = n, front = 0 and rear = 0 to add integers from the rear index if possible else display the message "QUEUE IS FULL" int remove( ) : void print( ) : Data variables: members/instance Methods/Member functions: to remove and return the integer from front if any, else return 999 to display the elements of the circular queue in the order of front to rear (i) Specify the class CirQueue giving the details of the functions void push(int) and int [4] remove( ). Assume that the other functions have been defined. The main( ) function and algorithm need NOT be written. (ii) State one application of a circular queue. [1] MARKING SCHEME (i) class CirQueue { public void push(int v) { if ((rear + 1) % cap == front) System.out.println("QUEUE IS FULL"); else { rear = (rear + 1) % cap; Q[rear] = v; } } public int remove() { if (front == rear) return -999; 28 2025 ISC- Computer Science else { front = (front + 1) % cap; return Q[front]; } (ii) } Application of Circular queues is CPU scheduling / memory management / network traffic management, etc. Comments of Examiners (i) A majority of the candidates could not understand the concept of circular queues. Several common errors and misconceptions were observed: Incorrectly implemented the basic logic for underflow and overflow condition. Mishandled the incrementing and decrementing of start and end indices. Failed to manage the wrap-around logic required for circular queues. Implemented a linear queue instead of a circular queue. Difficulty with push() and remove() methods (ii) This part was answered well by most of the candidates. However, some candidates were unable to state the application clearly and wrote vague responses. A few candidates overlooked this part altogether, while others incorrectly mentioned the principle FIFO as an entity. Suggestions for teachers 29 Give extensive practice to students on data structure implementations including Stacks, Queues, Circular Queues, and Dequeues. Provide regular exercises and assignments focusing on both conceptual understanding and practical implementation. Demonstrate step-by-step workings of stacks and queues to the students by using visual aids or real-time dry runs. Use supporting examples and diagrams to show how insertion, deletion, overflow, and underflow operations occur. Teach the fundamental principles of LIFO (Last-In-First-Out) for stacks and FIFO (First-In-First-Out) for queues using real-world analogies (e.g., stack of plates, ticket queues). Teach the implementation of stacks, queues, circular queues, and dequeues using arrays as a base. Emphasise the role of logical pointers such as front, rear, or top to simulate the behaviour of these data structures accurately. Lay emphasis on the conceptual structure and flow of data within these models. Encourage students to understand the logical architecture of how data is managed in constrained environments (e.g., fixed size, no dynamic resizing). Give sufficient practice on stacks and queues along with its applications and uses. 2025 ISC- Computer Science Question 10 [5] A superclass Flight has been defined to store the details of a flight. Define a subclass Passenger to calculate the fare for a passenger. The details of the members of both the classes are given below: Class name : Flight flightno : to store the flight number in string dep_time : to store the departure time in string arr_time : to store the arrival time in string basefare : to store the base fare in decimal Flight( ) : parameterised constructor to assign values to the data members void show( ) : to display the flight details Data variables: members/instance Methods/Member functions: Class name Data variables: Passenger members/instance id : to store the ID of the passenger name : to store the name of the passenger tax : to store the tax to be paid in decimal tot : to store the total amount to be paid in decimal Passenger( ) : void cal( ) : void show( ) : parameterised constructor to assign values to the data members of both the classes to calculate the tax as 5% of base fare and total amount(base fare + tax) to display the flight details along with the passenger details and total amount to be paid Methods/Member functions: Assume that the super class Flight has been defined. Using the concepts of Inheritance, specify the class Passenger giving the details of constructor( ),void cal( ) and void show( ). The super class, main function and algorithm need NOT be written. 30 2025 ISC- Computer Science MARKING SCHEME class Passenger extends Flight { String id, name; double tax,tot; Passenger(String a,String b,String c,double d,String i,String n) { super(a,b,c,d); id=i; name=n; tax=0.0; tot=0.0; } void cal() { tax= 5.0/100 * basefare; tot= tax + basefare; } void show() { super.show(); System.out.println("ID: "+id); System.out.println("NAME: "+name); System.out.println("TAX: "+tax); System.out.println("TOTAL AMOUNT: "+tot); } } 31 2025 ISC- Computer Science Comments of Examiners The following errors were commonly observed: Lack of clarity in the concept of inheritance. Omission or misuse of essential keywords such as extends (for defining the subclass) and super (for accessing superclass members or constructors). Difficulty in handling constructors in the context of inheritance, particularly failing to invoke the superclass constructor using super(). Inadequate understanding of how a derived class accesses members of its superclass, especially methods and data members. Unnecessary re-declaration or redefinition of the base class, even when it was already provided or implied in the question. Improper declaration of data members, leading to incomplete or functionally incorrect programs. Failure to invoke the show() method from the superclass within the subclass. Submission of only algorithms or pseudocode instead of a fully functional program. Suggestions for teachers 32 Provide extensive practice on writing programs involving inheritance, covering both single and multilevel inheritance examples. Explain the purpose and correct use of the extends keyword (to establish inheritance) and the super keyword (to access superclass constructors and members in detail through code demonstrations. Reinforce the concept of invoking base class constructors using super() especially in cases involving parameterized constructors. Highlight the different visibility modes (private, protected, and public) along with how they affect the accessibility of members in subclasses. Teach how to call methods of the superclass to the students from the subclass using the appropriate syntax and logic. Remind students to carefully read and interpret the question, especially when the base class is already provided and does not need to be re-declared. Discuss at least one complete program to explain the concept of inheritance (extend keywords), and method overriding. 2025 ISC- Computer Science Question 11 (i) A linked list is formed from the objects of the class Cell. The class structure of the Cell is given below: [2] class Cell { char m; Cell right; } Write an Algorithm OR a Method to print the sum of the ASCII values of the lower case alphabets present in the linked list. The method declaration is as follows: void lowercase (Cell str) (ii) Answer the following questions based on the Binary Tree given below: (a) Write the in-order traversal of the right subtree. [1] (b) State the depth of the entire binary tree and depth of node E. [1] (c) Name the external nodes of the left subtree and internal nodes of the right subtree. [1] 33 2025 ISC- Computer Science MARKING SCHEME (i) (ii) 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. Step 4. Check for lowercase, if found, add ASCII value to sum. Step 5. Move pointer to the next node. Step 6: Display sum Step 7. End OR METHOD: void lowercase(Cell str) { int sum=0; Cell C = str; while(C !=null) { if(Character.isLowerrCase(C.m)= = true) OR (C.m>= a &&C.m<= z ) OR (C.m>=97&&C.m<=122) sum+=C.m ; C=C.right; } System.out.println(sum); } (a) KFNLCGM (b) Depth of the tree: 4 (c) External Nodes of Left subtree: E I J Depth of Node E: 2 Internal Nodes of Right subtree: F G L 34 2025 ISC- Computer Science Comments of Examiners (i) The question was well answered by most of Suggestions for teachers the candidates, with many scoring full credit. Provide rigorous practice of various Several candidates wrote the algorithm in methods and algorithms involving linked simple English and covered all the main steps lists and binary tree data structures, as effectively, others missed creating a these are foundational topics in data temporary pointer. However, some structure programming. candidates had difficulty in moving the Emphasise on the use of diagrams to pointer to the next node and in checking for visually illustrate how linked lists and the null condition. Additionally, in some binary trees are structured and responses, the loop construct was missing, manipulated for the conceptual clarity. leading to incomplete or incorrect logic flow. Lay stress on the use of temporary (ii) (a) Most candidates answered this part of the pointers, the importance of checking for null conditions, and the correct way of question correctly. However, a few moving the pointer to the next node common mistakes were observed: during transversal or insertion. Some candidates wrote traversal orders Highlight the key concepts related to other than in-order, indicating binary trees- such as root, height, depth, confusion between different types of size, degree, siblings internal and external tree traversals. nodes, levels, and tree traversals and Some performed the in-order traversal explain it using illustrative binary tree for the entire tree instead of focusing diagrams. only on the right sub-tree, as required. A few responses contained minor placement errors, where one or two nodes were incorrectly positioned in the traversal sequence. (b) The subpart was well answered by most of the candidates. Some candidates considered the level of the root as 1, which is acceptable and consistent with certain conventions, and they answered the question correctly based on that assumption. (c) Most of the candidates answered this subpart correctly. However, a few candidates included the root of the right sub-tree in their answer, which was not required as per the question. 35 2025 ISC- Computer Science GENERAL COMMENTS Topics found difficult by candidates Concepts in which candidates got confused Propositional statements for proving Converse and Contrapositive. Assertion and Reasoning type of questions. Complexity and computing O(P) and O(PQ). Output using recursive function. Identifying the missing statements/expressions. K-MAPS (grouping, redundant groups, map-rolling, place value) Drawing logic gate diagram using NAND gates and NOR gates. Recursive technique. Invoking methods in another method (nested methods). Passing objects to functions. Circular Queue operations for push() and remove() operations. Depth of a binary tree and depth of a particular node Market supply curve and individual supply curve. Propositional statements. Concept of this operator. Complexity and computing O(P) and O(PQ). Output using recursion. Passing objects to functions. Use of single instance variable for multiple operations in various functions. Constructor inheritance and method overriding (use of super keyword). Incrementing/Decrementing front and rear indexes in a circular queue. Link list and Circular Queue. Depth and internal nodes of a tree. 36 2025 ISC- Computer Science Download proper definitions, output programs, algorithms, and programming questions from a reliable source. Prepare summaries for each chapter and use highlighters or annotations to mark important terms, definitions, and key points for easy revision. Engage in regular hands-on practice on the system to reinforce theoretical knowledge, understand syntax, and learn to identify and correct coding errors effectively. Practice K-maps thoroughly, using the correct values for both SOP and POS forms. Include constructors as a standard part of programming exercises. Suggestions for candidates Focus on understanding the underlying logic. Practice specimen papers, previous years Board papers, and competency-focused questions. Ensure that answers and definitions are short, precise, and aligned with the marks allotted. Underline key terms and important words. Show working alongside each question, especially in logic and Boolean algebra. Read the question paper carefully. Write all answers in legible handwriting. Include documentation with each program, as it is compulsory. Keep answers precise and support them with examples whenever needed. 37

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