📘 Introduction
The CBSE Class 12 Computer Science (083) Board Exam 2026 was successfully conducted today 🎯
In this post, you will find the complete solution of today’s board paper, including:
- ✅ Section-wise answers
- ✅ Detailed explanations
- ✅ Python & SQL solutions
- ✅ Answer key for objective questions
👉 This solution will help students analyze their performance and estimate their scores 📊
📢 Exam Overview
- 📅 Exam Date: 25-03-2026
- 🕒 Duration: 3 Hours
- 📝 Subject: Computer Science (083)
- 📊 Difficulty Level: Moderate (based on student feedback)
🎯 Paper Analysis (Quick Review)
- Section A (MCQs): Easy to Moderate
- Section B (Short Answer): Concept-based
- Section C (Long Answer): Slightly tricky
- Section D (Case Study): Moderate
👉 Overall, the paper was balanced and scoring. 💯
📂 Section-Wise Solutions
Class 12 Computer Science Code 083
CBSE Board Examination Question Paper 2026 – Solution
Date of Examination: 25-03-2026 | Series: Q3SPR | QP Code – 91 | SET – 4
Do you want Question Paper 2026 – Click Here
| SECTION – A [1 Mark Each] | |
|---|---|
| 1. | False |
| 2. | A. The truth |
| 3. | C. 3 > 1 and 3 > 2 |
| 4. | A. (‘War and Peace ‘, ‘by’, ‘Leo Tolstoy’) |
| 5. | mroPo |
| 6. | C. (‘t’, ‘u’, ‘p’, ‘l’, ‘e’, ‘p’) |
| 7. | B. A dictionary cannot have two elements with same key. |
| 8. | B. L.pop(6) |
| 9. | C. 3-None |
| 10. | B. rb |
| 11. | False |
| 12. | B. 1 |
| 13. | B. ALTER TABLE |
| 14. | D. ID and name of machines with quantity less than or equal to 100 from INVENTORY table. |
| 15. | C. 6 |
| 16. | A. MIN( ) |
| 17. | D. Registered Jack 45 |
| 18. | B. Gateway |
| 19. | Extensible Markup Language |
| 20. | A. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation for Assertion (A). |
| 21. | C. Assertion (A) is true, but Reason (R) is false. |
| SECTION – B [2 Marks Each] | |
|---|---|
| 22. | Difference between default parameters and positional parameters. * Default parameters have their default value, and positional parameters do not. * Default Parameters are optional, and positional parameters are required arguments. Example: def functionName( a, b = 10) : |
| 23. | (i) L1 = sorted(L) (ii) ch.isalnum( ) |
| 24. | (i) (a) ‘RNo’ in D1 OR ‘RNo’ in D1.keys( ) OR (b) 12 in D1.values( ) (ii) (a) D1.setdefault(‘RNo’, 12) OR (b) D1.clear( ) |
| 25. | D. 5-1-2-4- |
| 26 | Corrected Code: def CountVowels( s ): c = 0 for ch in s: if ch in ‘aeiouAEIOU’ : c += 1 return (c) |
| 27. | (i) (a) create table w_stock( w_code char(5) primary key, w_description varchar(20), b_qty int, u_price float(10,2)); OR (b) no, because it may contain duplicate value. (ii) (a) alter table w_stock add e_date date; OR (b) alter table w_stock drop column b_qty; |
| 28. | Bus Topology: Advantage: Less cable is required. Easy to expand and connect Disadvantage: Depending on main cable, if main cable is damaged then network also damaged. Difficult to find problems in cable. |
| SECTION – C [3 Marks Each] | |
|---|---|
| 29. | (a) def countDigit(): fobj = open(‘Space.txt’, ‘r’) data = fobj.read() count = 0 for ch in data: if ch.isdigit(): # if ch in ‘0123456789’ count += 1 return count countDigit() OR (b) def printWord(): fobj = open(‘Space.txt’, ‘r’) data = fobj.read() words = data.split() for w in words: if w.count(‘e’) >= 2: print(w, end=” “) printWord() |
| 30. | (a) FruitStack = [ ] (i) def push_fruit(FruitStack, fruit): if fruit[‘Price’] < 100: FruitStack.append(fruit) (ii) def pop_fruit(FruitStack): if FruitStack == [ ]: #if len(FruitStack) == 0: print(‘UNDERFLOW’) else: data = FruitStack.pop() return data (iii) def display(FruitStack): if FruitStack == [ ]: print(‘Empty Stack’) else: for f in FruitStack[::-1]: print(f) (b) EvenStack = [ ] for i in range(10): num = int(input(‘Enter No : ‘)) if num >= 100 and num <= 999: if num%2 == 0: EvenStack.append(num) print() # pop operation while EvenStack != [ ]: print(EvenStack.pop(), end= ‘ ‘) |
| 31. | Output: (a) [‘4’, ‘2’, ‘-‘, ‘d’, ‘l’, ‘o’] OR (b) 1-6-10-13-15-16- |
| SECTION – D [4 Marks Each] | |
|---|---|
| 32. | (a) (i) select type, max(price) from stock group by type; (ii) update stock set price = price + 0.5 where type = ‘F’; (iii) select sum(qty*price) as Total from stock; (iv) select * from stock where code like ‘A%’; OR (b) (i) Volume Qty Price 0.5 300 38.00 0.5 250 36.50 1.0 50 52.00 (ii) Code Qty AF0.5 300 MF0.5 250 PT0.5 78 (iii) DISTINCT Type F T D (iv) Volume Count(*) 0.5 3 1.0 3 |
| 33 | Method-1: import csv f1 = open(‘Stats.csv’, ‘r’, newline= ”) reader = csv.reader(f1) records=[ ] for rec in reader: records.append(rec) f1.close() f2 = open(‘More.csv’, ‘w’, newline=”) writer = csv.writer(f2) for i in range(1, len(records)): rec = records[i] if int(rec[2]) > 10000000: writer.writerow(rec) f2.close() Method-2 import csv f1 = open(‘Stats.csv’, ‘r’, newline= ”) f2 = open(‘More.csv’, ‘w’, newline=”) writer = csv.writer(f2) reader = csv.reader(f1) next(reader)#read one record for rec in reader: if int(rec[2]) > 10000000: writer.writerow(rec) f1.close() f2.close() |
| 34. | (i) SELECT COUNT(*) FROM LOANS WHERE ROI > 7.0; (ii) SELECT C_NAME FROM CUSTOMERS C, LOANS L WHERE C.C_ID = L.C_ID AND L_AMT >1000000; (iii) SELECT C.C_ID, C_NAME, TERMS FROM CUSTOMERS C, LOANS L WHERE C.C_ID = L.C_ID AND L_DATE > ‘2024-12-31’; (iv) (a) SELECT * FROM LOANS ORDER BY ROI DESC; OR (b) SELECT C_ID, AVG(TERMS) FROM LOANS GROUP BY C_ID; |
| 35. | import mysql.connector as sql conn = sql.connect(host = ‘localhost’ , user=’admin’, password = ‘root’,database = ‘SCHOOL’) cursor = conn.cursor( ) query = ‘select * from Account where fees < 5000’ cursor.execute(query) for rec in cursor: print(rec) conn.close() |
| SECTION – E [5 Marks Each] | |
|---|---|
| 36. | import pickle (i) def Append(): fobj = open(‘Resources.dat’, ‘ab’) r_id = int(input(‘Enter RID : ‘)) r_name = input(‘Enter Name : ‘) r_exp = input(‘Enter Expertise : ‘) charges = float(input(‘Enter Charges : ‘)) rec = (r_id, r_name, r_exp, charges) pickle.dump(rec, fobj) fobj.close() (ii) def Update(): fobj = open(‘Resources.dat’, ‘rb+’) try: while True: pos = fobj.tell( ) rec = pickle.load(fobj) rec[3] += 500 fobj.seek(pos) pickle.dump(rec, fobj) except EOFError: fobj.close() |
| 37. | (i) Server at ACADEMIC block, because it have maximum number of computers. (ii) Cable Layout – Answer given in bottom. (iii) Two wired media are – (a) Twisted Pair Cable , (b) Co-axial Cable or (c) Fiber Optics (iv) (A) Radio Waves (v) (a) Voice Over Internet Protocol (VoIP) OR (b) As per layout given in question number (ii) – Repeater is required between Academic and Sports block. As per distance given in the questions – Repeater is required between Admin and Hostel Academic and SportsHostel and Sports |
Cable Layout –
🎥 Video Solution (Watch Now)
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❓ FAQs
❓ Was the paper difficult?
👉 No, overall it was moderate and scoring.
❓ Were PYQs helpful?
👉 Yes, many questions were similar to previous years.
❓ Is this solution accurate?
👉 Yes, it is prepared based on expert analysis.
📢 Final Words
The CBSE Class 12 Computer Science 2026 paper was well-balanced and gave students a fair chance to score high 💯
👉 If you practiced PYQs and SQPs, this paper must have felt familiar 😊
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🔥 Hope you performed well! Best wishes for great results 🎯✨