PodiumRound

What will be the output of the following code snippet?

my_list = [1, 2, 3, 4, 5]

my_list[2] = 10

print(my_list)

• a) [1, 2, 3, 4, 5]

• b) [1, 2, 10, 4, 5]

• c) [1, 10, 3, 4, 5]

• d) Error: Lists are immutable

What will be the output of the following code snippet?

my_tuple = (1, 2, [3, 4])

my_tuple[2][0] = 10

print(my_tuple)

• a) (1, 2, [3, 4])

• (1, 2, [10, 4])

• c) Error: Tuples are immutable

• d) (1, 10, [3, 4])

What will be the output of the following code snippet?

my_dict = {'a': 1, 'b': {'x': 10, 'y': 20}, 'c': 3}

my_dict['b']['z'] = 30

print(my_dict)

• a) {'a': 1, 'b': {'x': 10, 'y': 20}, 'c': 3}

• b) {'a': 1, 'b': {'x': 10, 'y': 20, 'z': 30}, 'c': 3}

• c) {'a': 1, 'b': {'z': 30}, 'c': 3}

• d) Error: Cannot add new keys to a dictionary

What will be the output of the following code snippet?

set_a = {1, 2, 3, 4}

set_b = {3, 4, 5, 6}

result = set_a.symmetric_difference(set_b)

print(sorted(result))

• a) [1, 2, 3, 4, 5, 6]

• c) [1, 2, 5, 6]

• c) {1, 2, 5, 6}

• d) Error: symmetric_difference() is not a method of sets

Which of the following set operations will return a set containing elements that are in either set, but not in both?

• A) union()

• B) intersection()

• C) difference()

• D) symmetric_difference()

Given the following list:

data = [[1, 5, 3], [10, 7, 2], [6, 9, 4]]

number1= data[1][2] )

number2= data[0][1]

print(number1, number2)

• A) 2, 5

• B) 3, 5

• C) 2, 3

• D) 2, 1

Given the following tuple:

t = (1, 2, 3, 4, 5)

Which of the following operations will not raise an error?

• A) t[2] = 10

• B) t.append(6)

• C) t = t + (6,)

• D) del t[0]

What will the following code print?

for i in range(1, 4):

for j in range(1, 4):

if i == j:

break

print(i, j)

• a) (2, 1), (3, 1), (3, 2)

• b) (1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2)

• c) (2, 1), (2, 3), (3, 1), (3, 2)
  

• d) (1, 2), (2, 3), (3, 1), (3, 2)

How many times will "Hello" be printed?

count = 0

while count < 5:

print("Hello")

if count % 2 == 0:

continue

count += 1

• a) 5 times

• b) Infinite times

• c) 2 times

• d) 0 times

What does the following list comprehension produce?

result = [x for x in range(5) if x % 2 == 0 for y in range(x)]

print(result)

• a) [0, 0, 2, 2, 4, 4]

• b) [0, 0, 2, 2, 2, 4, 4, 4, 4]

• c) [0, 2, 4]

• d) [0, 2, 4, 6, 8]