Guidelines to crack CBSE Examination
XII
(Computer science-083)
For Academically Challenged Students (Slow Learners)
In-Depth Analysis of Questions & Suggestions
By:
Mr. Arun Kumar, PGT CS
K.V. No.-2, Binnaguri Cantt
Slow learners should not be treated as burden or mentally weak students. They belong to those group of students who are not able to get proper guidance & timing from parents as well as teachers from very beginning. They need only more time to understand the concepts. They need different teaching techniques rather than traditional. If they were told some simple ways to do, definitely they can do better.
Directions for Academically challenged students:
1. First know your weak points Chapter-wise, take the help of subject teacher in this regard as he/she is the evaluator of your performance.
2. Have a glance on your answer scripts of all term examinations, Half yearly, Monthly test & Pre-boards.
3. Note down the types of mistakes you committed Question-wise. Also note down your strong points.
4. Have a New Blue-print of CBSE examination. If, recent is not available follow the blue print of previous year after consultation with your teacher.
5. As per blue print, start self evaluation of your knowledge. Target the scoring & easy chapter.
6. Practice min 5-6 questions of each type from previous board paper, clear your all doubts from with help of your subject teacher.
7. Ask the subject teacher to give you remedial class in extra time.
8. Be honest with yourself & with your teacher.
Chapter-wise analysis of score is given, which may be followed:
Class & Object 4+2=6 marks
Inheritance, constructor/Destructor 4+2= 6 marks
Data File Handling 1+2+3=6 marks
Data structure (Array) 3+3+2=8 marks
Stack, Queue 2+4=6 marks
Database concepts, SQL 2+6=8 marks
Boolean algebra 2+2+1+3=8 marks
Network Concept 1+1+1+1+4+1+1=10 mark
Revision of XI 2+1+2+3+2+2= 12 marks
UNIT WISE ANALYSIS FOR SLOW LEARNERS/WEAK STUDENTS
Unit-1(Review of C++ covered in Class XI)
The total weightage of this unit is 12 marks. Weak students can easily score as per following criterion & suggestions:
Question.-1 (i)- Theoretical question ( 2 marks) : Students are advised to prepare 6-7 questions from previous board paper with example. If not full marks they can easily get 11/2 marks.
(ii) Related to header files (1 marks): Students are advised to remember header files of 10-15 library function from previous board papers. 1 mark can be scored easily.
(iii) Error finding (2 marks): students are advised to rewrite the code after removing the errors & indicate the errors with arrow. 2 marks can be scored easily. Practice 5-6 questions from previous board exams.
(iv) Finding output (3 marks): It may be a simple or little bit complex. Students are advised not to waste their time in this question. It should be taken care at end, as it may take more time. For weak students, I exclude this mark.
(v) Output- Random function (2 marks): It can be done by weak students. Just they have to grasp the concept of random function. Practice 5-6 different question from board paper. Full 2 marks can be scored.
(vi) Finding Output (2 marks): Again like bit (iv), I exclude this marks. It depends on complexity of question. It should be taken care at end.
Hence, in this unit min. 6 marks can be scored easily.
Unit-2 (Object Oriented programming)
The total weightage of this unit is 12 marks. It comprises class & object, Constructor/Destructor & Inheritance. Weak students can score as per following criterion & suggestions:
Question-2 (i)- Theoretical question (2 marks): Students are advised to prepare 6-7 questions from previous board paper with example. If not full marks they can easily get 11/2 marks. Always support your theory with small c++ programming code.
(ii) Programming based question of 4 marks. A class has to be defined with some given criterion. It is a scoring question with little efforts. Students are advised to prepare 6-7 questions from previous board papers & rectified it by teacher. Min. 21/2- 4 marks can be scored by each student depending on complexity. Take care of minor syntax errors.
(iii) Question from constructor/destructor of 2 marks. Students are advised to prepare 6-7 questions from previous board papers & rectified it from teacher. Have a clear concept of related questions. 2 marks can be scored easily.
(iv) Question from Inheritance of 4 marks. Students can do it easily by grasping few concepts of inheritance. Students are advised to prepare 6-7 questions from previous board papers & rectified them from teacher. Easily 3- 4 marks can be scored depends on complexity.
Hence, each weak student can easily score 6- 8 marks.
Unit-3 (Data Structure & Pointers)
The total weightage of this unit is 14 marks. It is not a scoring unit for weak students. But still they can score some marks as per following analysis & suggestions.
Question-3 (i) Application of 1-D array: It carries 3 marks. I exclude these marks. But still they can get ½ or 1 marks if they will partially attempt it.
(ii) Address calculation in 2’D array: It carries 3 marks also a scoring one. Students can easily score 3 marks if they will take care of few things. Writing correct formula. Placing the correct values & doing mathematical calculation correctly will certainly give them 3 marks. If mathematical calculation is wrong then for just writing the correct formula & placing the values, 1 marks is certain. Students are advised to prepare 6-7 questions from previous board papers & rectified them from teacher.
(iii) From stack & queue programming part: It carries 4 marks. I completely exclude these marks. So, student don’t try it right now. See it in last half hr of examination, if time is there.
(iv) Application of 2’D array: Application of 1-D array: It carries 3 marks. I exclude these marks. But still they can get ½ or 1 marks if they will partially attempt it.
(v) Postfix notation: This question carries 2 marks. Marks can be scored easily just by understanding the concept. Students are advised to prepare 6-7 questions from previous board papers & rectified them from teacher. 1-2 marks can be scored easily.
Hence, 3-7 marks can be scored by each slow learner.
Unit-4 (Data File Handling)
This unit carries 6 marks. Though, it is difficult unit for slow learners yet few marks can be get, if planned & work upon as per analysis & suggestions:
Question-4 (i) It is one mark question. Only two statements has to be written. ½ marks can be scored if basic concept of seekg(). Seekp(), tellg(), tellp() understood.
(ii) Operation on text file: It carries two marks & easily can be scored if 5-6 questions are practiced by student. The basic & mandatory part of program has to be understood. If partially written 1 marks can be get.
(iii) Operation on Binary file: It carries three marks. I exclude this mark but some mark can be scored by partially writing the answer. Mainly, read(), write() function, Opening the file with correct stream with surely give some mark. Atleast 1 mark can be scored by each student.
Hence, 2- 3 marks can be scored.
Unit-5 (Database & SQL)
This unit carries 8 marks. It is very easy & scoring unit. Full marks can be scored on basis of following analysis & suggestions:
Question-5 (i) Simple theoretical question: Students are suggested to mug-up all keys with example. 2 marks can be scored easily.
(ii) SQL statements based on table: It carries 4 marks for writing queries & 2 marks for writing output. Students are advised to prepare 6-7 questions from previous board papers & rectified them from teacher. This type of questions need more practice. Still 2-4 marks can be scored by each slow learner.
Hence 3-6 marks can be scored by each slow learner.
Unit-6 (Boolean Algebra)
This unit carries 8 marks. It is bless for slow learners. It can make them pass. Very easy concepts are there. Each student can ensure full marks if planned as per following analysis & suggestions:
Questions- 6 (i) Statement of any law (2 marks): Students are suggested to mug-up all the Boolean laws with their truth table verification. 2 marks can be scored
(ii) Boolean expression of logic circuit (2 marks): 2 marks can be scored easily just by understanding the logic circuit.
(iii) SOP/POS form of Boolean function in a table: (1 mark). Full mark can be achieved just by understanding basic concept of SOP & POS.
(iv) K-Map (3 marks): scoring & easing one. Full mark be scored. Students are advised to prepare 6-7 questions from previous board papers & rectified them from teacher.
Hence, 5-8 marks can be scored by each slow learner.
Unit-7 (Communication & Networking concepts)
Total weightage of this unit is 10 marks. Again, one more boon for slow learners. More theoretical less practical. Full marks can be scored if planned as per following analysis & suggestions:
Question-7 Generally, 6 bit of question are general questions on networking, communication & open source concepts. Students are suggested to go through solved questions in their text book & mug-up full form of all abbreviations. One question is related with layout design of network. It carried four bits of 1 mark each. Students can easily score full marks just by remembering few concepts. Students are advised to prepare 6-7 questions from previous board papers & rectified them from teacher.
Hence 6-8 marks can be scored easily.
Unit-wise maximum or minimum marks which can be scored by slow learners based on Analysis
UNIT Max/Min. Out of
Unit-1 8/6 12
Unit-2 8/6 12
Unit-3 7/3 14
Unit-4 3/2 6
Unit-5 6/3 8
Unit-6 8/5 8
Unit-7 8/6 10
Maximum marks scored by slow learners: 48/70
Maximum marks scored by slow learners: 31/70
So, this analysis & suggestions given above ensure 100% pass percentage in computer science. It reduces the last minute burden of preparation from the mind of slow learners.
Mr. Arun Kumar, PGT CS
K.V. No.-2, Binnaguri Cantt,