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Page: 1/11 
MARKING SCHEME 
Class: XII Session: 2024-25 
Computer Science (083) 
 
Time allowed: 3 Hours           Maximum Marks: 70 
Q No. SECTION A (21X1=21) Marks 
1.  False 
(1 mark for correct answer) 
(1) 
 
2.  (A) #THONPROGRAM       
(1 mark for correct answer)           
(1) 
3.  (A) not (True) and False 
(1 mark for correct answer) 
(1) 
4.  (B) ['I', 'ter', 'atio', 'al']  
(1 mark for correct answer) 
(1) 
5.  ce lo 
(1 mark for correct answer) 
(1) 
6.  (B) False  
(1 mark for correct answer)  
(1) 
7.  (B) print(my_dict['apple', 'banana']) 
(1 mark for correct answer) 
(1) 
8.  (B) Removes the first occurrence of value x from the list 
(1 mark for correct answer) 
(1) 
9.  (C) 3 
(1 mark for correct answer) 
(1) 
10.  file.seek(0) ( OR file.seek(0,0) )  
(1 mark for correct answer) 
(1) 
11.  False 
(1 mark for correct answer) 
(1) 
12.  (C) 12#15% 
(1 mark for correct answer) 
(1) 
13.  Alter (or Alter Table) 
(1 mark for correct answer) 
(1) 
 
 
Page 2


Page: 1/11 
MARKING SCHEME 
Class: XII Session: 2024-25 
Computer Science (083) 
 
Time allowed: 3 Hours           Maximum Marks: 70 
Q No. SECTION A (21X1=21) Marks 
1.  False 
(1 mark for correct answer) 
(1) 
 
2.  (A) #THONPROGRAM       
(1 mark for correct answer)           
(1) 
3.  (A) not (True) and False 
(1 mark for correct answer) 
(1) 
4.  (B) ['I', 'ter', 'atio', 'al']  
(1 mark for correct answer) 
(1) 
5.  ce lo 
(1 mark for correct answer) 
(1) 
6.  (B) False  
(1 mark for correct answer)  
(1) 
7.  (B) print(my_dict['apple', 'banana']) 
(1 mark for correct answer) 
(1) 
8.  (B) Removes the first occurrence of value x from the list 
(1 mark for correct answer) 
(1) 
9.  (C) 3 
(1 mark for correct answer) 
(1) 
10.  file.seek(0) ( OR file.seek(0,0) )  
(1 mark for correct answer) 
(1) 
11.  False 
(1 mark for correct answer) 
(1) 
12.  (C) 12#15% 
(1 mark for correct answer) 
(1) 
13.  Alter (or Alter Table) 
(1 mark for correct answer) 
(1) 
 
 
Page: 2/11 
14.  (A) Details of all products whose names start with 'App' 
(1 mark for correct answer) 
(1) 
15.  (D) CHAR 
(1 mark for correct answer) 
(1) 
16.  (B) count() 
(1 mark for correct answer) 
(1) 
17.  (B) FTP 
(1 mark for correct answer) 
(1) 
18.  (B) Gateway 
(1 mark for correct answer) 
(1) 
19.  (B) Packet Switching 
(1 mark for correct answer) 
(1) 
20.  (C) A is True but R is False. 
(1 mark for correct answer) 
(1) 
21.  (C) A is True but R is False. 
(1 mark for correct answer) 
(1) 
 
Q No. SECTION B (7 X 2 =14) Marks 
22.  A mutable object can be updated whereas an immutable object cannot be 
updated. 
Mutable object: [1,2] or  {1:1,2:2}  (Any one) 
Immutable object: (1,2) or ‘123’ (Any one) 
(1 mark for correct difference) 
(½ x 2 = 1 Mark for selecting correct objects) 
(2) 
23.  
(I) Arithmetic operators: +,-    
(II) Relational operators: >, >= 
(½ x 4 = 2 Marks for each correct operator) 
(2) 
24.  (I) 
     A) L1.count(4) 
OR 
     B) L1.sort() 
(1 mark for correct answer) 
 
(2) 
Page 3


Page: 1/11 
MARKING SCHEME 
Class: XII Session: 2024-25 
Computer Science (083) 
 
Time allowed: 3 Hours           Maximum Marks: 70 
Q No. SECTION A (21X1=21) Marks 
1.  False 
(1 mark for correct answer) 
(1) 
 
2.  (A) #THONPROGRAM       
(1 mark for correct answer)           
(1) 
3.  (A) not (True) and False 
(1 mark for correct answer) 
(1) 
4.  (B) ['I', 'ter', 'atio', 'al']  
(1 mark for correct answer) 
(1) 
5.  ce lo 
(1 mark for correct answer) 
(1) 
6.  (B) False  
(1 mark for correct answer)  
(1) 
7.  (B) print(my_dict['apple', 'banana']) 
(1 mark for correct answer) 
(1) 
8.  (B) Removes the first occurrence of value x from the list 
(1 mark for correct answer) 
(1) 
9.  (C) 3 
(1 mark for correct answer) 
(1) 
10.  file.seek(0) ( OR file.seek(0,0) )  
(1 mark for correct answer) 
(1) 
11.  False 
(1 mark for correct answer) 
(1) 
12.  (C) 12#15% 
(1 mark for correct answer) 
(1) 
13.  Alter (or Alter Table) 
(1 mark for correct answer) 
(1) 
 
 
Page: 2/11 
14.  (A) Details of all products whose names start with 'App' 
(1 mark for correct answer) 
(1) 
15.  (D) CHAR 
(1 mark for correct answer) 
(1) 
16.  (B) count() 
(1 mark for correct answer) 
(1) 
17.  (B) FTP 
(1 mark for correct answer) 
(1) 
18.  (B) Gateway 
(1 mark for correct answer) 
(1) 
19.  (B) Packet Switching 
(1 mark for correct answer) 
(1) 
20.  (C) A is True but R is False. 
(1 mark for correct answer) 
(1) 
21.  (C) A is True but R is False. 
(1 mark for correct answer) 
(1) 
 
Q No. SECTION B (7 X 2 =14) Marks 
22.  A mutable object can be updated whereas an immutable object cannot be 
updated. 
Mutable object: [1,2] or  {1:1,2:2}  (Any one) 
Immutable object: (1,2) or ‘123’ (Any one) 
(1 mark for correct difference) 
(½ x 2 = 1 Mark for selecting correct objects) 
(2) 
23.  
(I) Arithmetic operators: +,-    
(II) Relational operators: >, >= 
(½ x 4 = 2 Marks for each correct operator) 
(2) 
24.  (I) 
     A) L1.count(4) 
OR 
     B) L1.sort() 
(1 mark for correct answer) 
 
(2) 
Page: 3/11 
 (II)  
     A) L1.extend(L2) 
OR 
      B) L2.reverse() 
(1 mark for correct answer) 
25.  
(A), (C)   
 (½ x 2 = 1 Mark) 
Minimum and maximum possible values of the variable b: 1,6   
(½ x 2 = 1 Mark) 
(2) 
26.  def swap_first_last(tup): 
    if len(tup) < 2: 
        return tup 
    new_tup = (tup[-1],) + tup[1:-1] + (tup[0],) 
    return new_tup 
 
result = swap_first_last((1, 2, 3, 4)) 
print("Swapped tuple:", result) 
 
(½ mark each for correcting 4 mistakes) 
(2) 
27.  
(I)  
A) UNIQUE 
OR 
B) NOT NULL 
(1 mark for correct answer) 
(II)  
A) ALTER TABLE MOBILE DROP PRIMARY KEY; 
OR 
B) ALTER TABLE MOBILE ADD PRIMARY KEY (M_ID); 
(1 mark for correct answer) 
(2) 
28.  
A) Advantage: Network extension is easy. 
Disadvantage: Failure of switch/hub results in failure of the network. 
(1 mark for correct Advantage) 
(1 mark for correct Disadvantage) 
OR 
 
(2) 
Page 4


Page: 1/11 
MARKING SCHEME 
Class: XII Session: 2024-25 
Computer Science (083) 
 
Time allowed: 3 Hours           Maximum Marks: 70 
Q No. SECTION A (21X1=21) Marks 
1.  False 
(1 mark for correct answer) 
(1) 
 
2.  (A) #THONPROGRAM       
(1 mark for correct answer)           
(1) 
3.  (A) not (True) and False 
(1 mark for correct answer) 
(1) 
4.  (B) ['I', 'ter', 'atio', 'al']  
(1 mark for correct answer) 
(1) 
5.  ce lo 
(1 mark for correct answer) 
(1) 
6.  (B) False  
(1 mark for correct answer)  
(1) 
7.  (B) print(my_dict['apple', 'banana']) 
(1 mark for correct answer) 
(1) 
8.  (B) Removes the first occurrence of value x from the list 
(1 mark for correct answer) 
(1) 
9.  (C) 3 
(1 mark for correct answer) 
(1) 
10.  file.seek(0) ( OR file.seek(0,0) )  
(1 mark for correct answer) 
(1) 
11.  False 
(1 mark for correct answer) 
(1) 
12.  (C) 12#15% 
(1 mark for correct answer) 
(1) 
13.  Alter (or Alter Table) 
(1 mark for correct answer) 
(1) 
 
 
Page: 2/11 
14.  (A) Details of all products whose names start with 'App' 
(1 mark for correct answer) 
(1) 
15.  (D) CHAR 
(1 mark for correct answer) 
(1) 
16.  (B) count() 
(1 mark for correct answer) 
(1) 
17.  (B) FTP 
(1 mark for correct answer) 
(1) 
18.  (B) Gateway 
(1 mark for correct answer) 
(1) 
19.  (B) Packet Switching 
(1 mark for correct answer) 
(1) 
20.  (C) A is True but R is False. 
(1 mark for correct answer) 
(1) 
21.  (C) A is True but R is False. 
(1 mark for correct answer) 
(1) 
 
Q No. SECTION B (7 X 2 =14) Marks 
22.  A mutable object can be updated whereas an immutable object cannot be 
updated. 
Mutable object: [1,2] or  {1:1,2:2}  (Any one) 
Immutable object: (1,2) or ‘123’ (Any one) 
(1 mark for correct difference) 
(½ x 2 = 1 Mark for selecting correct objects) 
(2) 
23.  
(I) Arithmetic operators: +,-    
(II) Relational operators: >, >= 
(½ x 4 = 2 Marks for each correct operator) 
(2) 
24.  (I) 
     A) L1.count(4) 
OR 
     B) L1.sort() 
(1 mark for correct answer) 
 
(2) 
Page: 3/11 
 (II)  
     A) L1.extend(L2) 
OR 
      B) L2.reverse() 
(1 mark for correct answer) 
25.  
(A), (C)   
 (½ x 2 = 1 Mark) 
Minimum and maximum possible values of the variable b: 1,6   
(½ x 2 = 1 Mark) 
(2) 
26.  def swap_first_last(tup): 
    if len(tup) < 2: 
        return tup 
    new_tup = (tup[-1],) + tup[1:-1] + (tup[0],) 
    return new_tup 
 
result = swap_first_last((1, 2, 3, 4)) 
print("Swapped tuple:", result) 
 
(½ mark each for correcting 4 mistakes) 
(2) 
27.  
(I)  
A) UNIQUE 
OR 
B) NOT NULL 
(1 mark for correct answer) 
(II)  
A) ALTER TABLE MOBILE DROP PRIMARY KEY; 
OR 
B) ALTER TABLE MOBILE ADD PRIMARY KEY (M_ID); 
(1 mark for correct answer) 
(2) 
28.  
A) Advantage: Network extension is easy. 
Disadvantage: Failure of switch/hub results in failure of the network. 
(1 mark for correct Advantage) 
(1 mark for correct Disadvantage) 
OR 
 
(2) 
Page: 4/11 
B) SMTP: Simple Mail Transfer Protocol. 
    SMTP is used for sending e-mails from client to server. 
(1 mark for correct expansion) 
(1 mark for correct usage) 
 
 
Q No. SECTION C (3 X 3 = 9) Marks 
29.  
(A)  
def show(): 
    f=open("Email.txt",'r') 
    data=f.read() 
    words=data.split() 
    for word in words: 
        if '@cmail' in word: 
            print(word,end=' ') 
    f.close() 
(½ mark for correct function header)  
(½ mark for correctly opening the file) 
(½ mark for correctly reading from the file)  
(½ mark for splitting the text into words)  
(1 mark for correctly displaying the desired words) 
OR 
(B)  
def display_long_words(): 
    with open("Words.txt", 'r') as file: 
        data=file.read() 
        words=data.split() 
        for word in words: 
           if len(word)>5: 
               print(word,end=' ') 
(½ mark for correct function header)  
(½ mark for correctly opening the file) 
(½ mark for correctly reading from the file)  
( ½  mark for splitting the text into words)  
(1 mark for correctly displaying the desired words) 
(3) 
Page 5


Page: 1/11 
MARKING SCHEME 
Class: XII Session: 2024-25 
Computer Science (083) 
 
Time allowed: 3 Hours           Maximum Marks: 70 
Q No. SECTION A (21X1=21) Marks 
1.  False 
(1 mark for correct answer) 
(1) 
 
2.  (A) #THONPROGRAM       
(1 mark for correct answer)           
(1) 
3.  (A) not (True) and False 
(1 mark for correct answer) 
(1) 
4.  (B) ['I', 'ter', 'atio', 'al']  
(1 mark for correct answer) 
(1) 
5.  ce lo 
(1 mark for correct answer) 
(1) 
6.  (B) False  
(1 mark for correct answer)  
(1) 
7.  (B) print(my_dict['apple', 'banana']) 
(1 mark for correct answer) 
(1) 
8.  (B) Removes the first occurrence of value x from the list 
(1 mark for correct answer) 
(1) 
9.  (C) 3 
(1 mark for correct answer) 
(1) 
10.  file.seek(0) ( OR file.seek(0,0) )  
(1 mark for correct answer) 
(1) 
11.  False 
(1 mark for correct answer) 
(1) 
12.  (C) 12#15% 
(1 mark for correct answer) 
(1) 
13.  Alter (or Alter Table) 
(1 mark for correct answer) 
(1) 
 
 
Page: 2/11 
14.  (A) Details of all products whose names start with 'App' 
(1 mark for correct answer) 
(1) 
15.  (D) CHAR 
(1 mark for correct answer) 
(1) 
16.  (B) count() 
(1 mark for correct answer) 
(1) 
17.  (B) FTP 
(1 mark for correct answer) 
(1) 
18.  (B) Gateway 
(1 mark for correct answer) 
(1) 
19.  (B) Packet Switching 
(1 mark for correct answer) 
(1) 
20.  (C) A is True but R is False. 
(1 mark for correct answer) 
(1) 
21.  (C) A is True but R is False. 
(1 mark for correct answer) 
(1) 
 
Q No. SECTION B (7 X 2 =14) Marks 
22.  A mutable object can be updated whereas an immutable object cannot be 
updated. 
Mutable object: [1,2] or  {1:1,2:2}  (Any one) 
Immutable object: (1,2) or ‘123’ (Any one) 
(1 mark for correct difference) 
(½ x 2 = 1 Mark for selecting correct objects) 
(2) 
23.  
(I) Arithmetic operators: +,-    
(II) Relational operators: >, >= 
(½ x 4 = 2 Marks for each correct operator) 
(2) 
24.  (I) 
     A) L1.count(4) 
OR 
     B) L1.sort() 
(1 mark for correct answer) 
 
(2) 
Page: 3/11 
 (II)  
     A) L1.extend(L2) 
OR 
      B) L2.reverse() 
(1 mark for correct answer) 
25.  
(A), (C)   
 (½ x 2 = 1 Mark) 
Minimum and maximum possible values of the variable b: 1,6   
(½ x 2 = 1 Mark) 
(2) 
26.  def swap_first_last(tup): 
    if len(tup) < 2: 
        return tup 
    new_tup = (tup[-1],) + tup[1:-1] + (tup[0],) 
    return new_tup 
 
result = swap_first_last((1, 2, 3, 4)) 
print("Swapped tuple:", result) 
 
(½ mark each for correcting 4 mistakes) 
(2) 
27.  
(I)  
A) UNIQUE 
OR 
B) NOT NULL 
(1 mark for correct answer) 
(II)  
A) ALTER TABLE MOBILE DROP PRIMARY KEY; 
OR 
B) ALTER TABLE MOBILE ADD PRIMARY KEY (M_ID); 
(1 mark for correct answer) 
(2) 
28.  
A) Advantage: Network extension is easy. 
Disadvantage: Failure of switch/hub results in failure of the network. 
(1 mark for correct Advantage) 
(1 mark for correct Disadvantage) 
OR 
 
(2) 
Page: 4/11 
B) SMTP: Simple Mail Transfer Protocol. 
    SMTP is used for sending e-mails from client to server. 
(1 mark for correct expansion) 
(1 mark for correct usage) 
 
 
Q No. SECTION C (3 X 3 = 9) Marks 
29.  
(A)  
def show(): 
    f=open("Email.txt",'r') 
    data=f.read() 
    words=data.split() 
    for word in words: 
        if '@cmail' in word: 
            print(word,end=' ') 
    f.close() 
(½ mark for correct function header)  
(½ mark for correctly opening the file) 
(½ mark for correctly reading from the file)  
(½ mark for splitting the text into words)  
(1 mark for correctly displaying the desired words) 
OR 
(B)  
def display_long_words(): 
    with open("Words.txt", 'r') as file: 
        data=file.read() 
        words=data.split() 
        for word in words: 
           if len(word)>5: 
               print(word,end=' ') 
(½ mark for correct function header)  
(½ mark for correctly opening the file) 
(½ mark for correctly reading from the file)  
( ½  mark for splitting the text into words)  
(1 mark for correctly displaying the desired words) 
(3) 
Page: 5/11 
30.  (A) 
(I)  
def push_book(BooksStack, new_book): 
  BooksStack.append(new_book) 
(II)  
def pop_book(BooksStack): 
    if not BooksStack: 
        print("Underflow") 
    else: 
        return(BookStack.pop()) 
(III)  
def peep(BooksStack): 
    if not BooksStack: 
        print("None") 
    else: 
        print(BookStack[-1]) 
(3x1 mark for correct function body; No marks for any function header as it 
was a part of the question) 
OR 
(B)  
def push_even_numbers(N): 
           EvenNumbers = [] 
           for num in N: 
                  if num % 2 == 0: 
                     EvenNumbers.append(num) 
            return EvenNumbers 
 
 
VALUES = [] 
 
for i in range(5): 
    VALUES.append(int(input("Enter an integer: "))) 
 
EvenNumbers = push_even_numbers(VALUES) 
 
def pop_even(): 
   if not EvenNumbers: 
        print("Underflow") 
          else: 
               print(EvenNumbers.pop()) 
 
pop_even() 
 
 
 
(3) 
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