If–else execution flow in Python - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of nested if–else blocks

What is the output of this Python code?

Python

x = 10
if x > 5:
    if x > 15:
        print("A")
    else:
        print("B")
else:
    print("C")

A B

B C

C A

D No output

💡 Hint

Check the conditions step by step, starting from the outer if.

The variable x is 10, which is greater than 5, so the outer if is true.

Then the inner if checks if x is greater than 15, which is false.

So the else of the inner block runs, printing "B".

❓ Predict Output

intermediate

If–else with multiple conditions

What will be printed by this code?

Python

score = 75
if score >= 90:
    print("Excellent")
elif score >= 70:
    print("Good")
else:
    print("Try again")

A Excellent

B No output

C Try again

D Good

💡 Hint

Check which condition matches the score first.

The score is 75.

The first condition score >= 90 is false.

The second condition score >= 70 is true, so it prints "Good".

❓ Predict Output

advanced

If–else with assignment expressions

What is the output of this code using the walrus operator?

Python

if (n := 4) > 3:
    print(f"Number {n} is greater than 3")
else:
    print(f"Number {n} is not greater than 3")

A Number 4 is greater than 3

B Number 4 is not greater than 3

C No output

D SyntaxError

💡 Hint

The walrus operator assigns and returns the value in the condition.

The walrus operator n := 4 assigns 4 to n and returns 4.

Since 4 > 3 is true, it prints "Number 4 is greater than 3".

❓ Predict Output

advanced

If–else with match-case and else

What will this code print?

Python

value = 2
match value:
    case 1:
        print("One")
    case 2:
        print("Two")
    case _:
        print("Other")

A One

B Two

C Other

D SyntaxError

💡 Hint

The match-case checks the value against each case.

The variable value is 2.

The case 2 matches, so it prints "Two".

❓ Predict Output

expert

Complex if–else with multiple conditions and logical operators

What is the output of this code?

Python

a = 5
b = 10
if a > 0 and b < 5:
    print("X")
elif a > 0 or b < 5:
    print("Y")
else:
    print("Z")

A Z

B X

C Y

D No output

💡 Hint

Evaluate each condition carefully with the values of a and b.

a is 5 (greater than 0), b is 10 (not less than 5).

The first condition a > 0 and b < 5 is false because b < 5 is false.

The second condition a > 0 or b < 5 is true because a > 0 is true.

So it prints "Y".

Practice

1. What does an if-else statement do in Python?

easy

A. It chooses between two paths based on a condition.

B. It repeats code multiple times.

C. It defines a function.

D. It creates a list.

Solution

1. Step 1: Understand the purpose of if-else

   An if-else statement lets the program decide which code to run based on a condition being true or false.

2. Step 2: Compare with other options

   Repeating code is done by loops, functions define reusable code blocks, and lists store multiple items, so these are not correct.

3. Final Answer:

   It chooses between two paths based on a condition. -> Option A

4. Quick Check:

   If-else = choose path [OK]

Hint: If-else picks one of two paths based on condition [OK]

Common Mistakes:

• Confusing if-else with loops
• Thinking if-else creates data structures
• Mixing if-else with function definitions

2. Which of the following is the correct syntax for an if-else statement in Python?

easy

A. if x > 0:
print('Positive')
else:
print('Non-positive')

B. if x > 0 then print('Positive') else print('Non-positive')

C. if (x > 0) { print('Positive'); } else { print('Non-positive'); }

D. if x > 0
print('Positive')
else
print('Non-positive')

Solution

1. Step 1: Recall Python if-else syntax

   Python uses a colon after the condition and indentation for the code blocks.

2. Step 2: Check each option

   if x > 0: print('Positive') else: print('Non-positive') uses colons and indentation correctly. if x > 0 then print('Positive') else print('Non-positive') uses 'then' which is not Python syntax. if (x > 0) { print('Positive'); } else { print('Non-positive'); } uses braces and semicolons, which are for other languages. if x > 0 print('Positive') else print('Non-positive') misses colons and indentation.

3. Final Answer:

   if x > 0: print('Positive') else: print('Non-positive') -> Option A

4. Quick Check:

   Colon + indent = correct if-else [OK]

Hint: Remember colons and indentation for if-else in Python [OK]

Common Mistakes:

• Using 'then' keyword
• Forgetting colons after if and else
• Not indenting code blocks

3. What will be the output of this code?

age = 18
if age >= 18:
    print('Adult')
else:
    print('Minor')

medium

A. Minor

B. Adult

C. 18

D. No output

Solution

1. Step 1: Evaluate the condition age >= 18

   Since age is 18, the condition age >= 18 is true.

2. Step 2: Determine which block runs

   Because the condition is true, the code inside the if block runs, printing 'Adult'.

3. Final Answer:

   Adult -> Option B

4. Quick Check:

   Condition true -> print 'Adult' [OK]

Hint: Check if condition is true or false to pick output [OK]

Common Mistakes:

• Assuming >= means less than
• Printing else block by mistake
• Confusing output with variable value

4. Find the error in this code:

num = 5
if num > 0
    print('Positive')
else:
    print('Non-positive')

medium

A. Variable 'num' is not defined

B. Wrong indentation on print statements

C. Missing colon ':' after 'if num > 0' line

D. 'else' should be 'elif' here

Solution

1. Step 1: Check syntax of if statement

   The if statement must end with a colon ':' to mark the start of the block.

2. Step 2: Identify the missing colon

   The code misses the colon after 'if num > 0', causing a syntax error.

3. Final Answer:

   Missing colon ':' after 'if num > 0' line -> Option C

4. Quick Check:

   Colon needed after if condition [OK]

Hint: Always put colon after if condition [OK]

Common Mistakes:

• Forgetting colon after if
• Confusing else with elif
• Incorrect indentation

5. You want to write a program that prints 'Even' if a number is even, and 'Odd' if it is odd. Which code correctly uses if-else to do this?

hard

A. num = 4
if num % 2:
print('Even')
else:
print('Odd')

B. num = 4
if num / 2 == 0:
print('Even')
else:
print('Odd')

C. num = 4
if num % 2 != 0:
print('Even')
else:
print('Odd')

D. num = 4
if num % 2 == 0:
print('Even')
else:
print('Odd')

Solution

1. Step 1: Understand how to check even numbers

   A number is even if dividing by 2 leaves no remainder, so num % 2 == 0 is true for even numbers.

2. Step 2: Check each option's condition

   num = 4 if num % 2 == 0: print('Even') else: print('Odd') correctly uses num % 2 == 0. num = 4 if num / 2 == 0: print('Even') else: print('Odd') uses division instead of modulo, which is wrong. num = 4 if num % 2: print('Even') else: print('Odd') treats nonzero remainder as even, which is incorrect. num = 4 if num % 2 != 0: print('Even') else: print('Odd') reverses the logic.

3. Final Answer:

   num = 4 if num % 2 == 0: print('Even') else: print('Odd') -> Option D

4. Quick Check:

   Modulo equals zero means even [OK]

Hint: Use 'num % 2 == 0' to check even numbers [OK]

Common Mistakes:

• Using division instead of modulo
• Reversing even and odd logic
• Checking truthiness of modulo without comparison