Nested conditional execution in Python - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of nested if-else with numbers

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 A

C C

D No output

💡 Hint

Check the first condition, then the nested one inside it.

x is 10, which is greater than 5 and less than 15, so the inner if prints 'A'.

❓ Predict Output

intermediate

Nested if-else with strings

What will this code print?

Python

word = "hello"
if len(word) > 3:
    if word[0] == 'h':
        print("Starts with h")
    else:
        print("Long word")
else:
    print("Short word")

A Long word

B Short word

C Error

D Starts with h

💡 Hint

Check the length and first letter of the word.

The word 'hello' has length 5 (>3) and starts with 'h', so it prints 'Starts with h'.

❓ Predict Output

advanced

Nested if-else with multiple conditions

What is the output of this code?

Python

num = 7
if num % 2 == 0:
    if num > 10:
        print("Even and large")
    else:
        print("Even and small")
else:
    if num > 5:
        print("Odd and large")
    else:
        print("Odd and small")

A Odd and large

B Odd and small

C Even and large

D Even and small

💡 Hint

Check if the number is even or odd, then check its size.

7 is odd (not divisible by 2) and greater than 5, so it prints 'Odd and large'.

❓ Predict Output

advanced

Nested if-else with boolean logic

What will this code print?

Python

flag1 = True
flag2 = False
if flag1:
    if flag2:
        print("Both True")
    else:
        print("Only flag1 True")
else:
    print("flag1 is False")

A flag1 is False

B Both True

C Only flag1 True

D No output

💡 Hint

Check the values of flag1 and flag2 carefully.

flag1 is True but flag2 is False, so it prints 'Only flag1 True'.

❓ Predict Output

expert

Nested if-else with variable reassignment

What is the value of variable result after running this code?

Python

x = 3
if x > 0:
    if x % 2 == 0:
        result = "Positive even"
    else:
        result = "Positive odd"
else:
    if x == 0:
        result = "Zero"
    else:
        result = "Negative"

A "Positive odd"

B "Negative"

C "Zero"

D "Positive even"

💡 Hint

Check if x is positive and then if it is even or odd.

x is 3, which is positive and odd, so result is 'Positive odd'.

Practice

1. What does nested conditional execution mean in Python?

easy

A. An if or else inside another if or else

B. Using multiple if statements one after another

C. Writing if statements without indentation

D. Using only one if statement in a program

Solution

1. Step 1: Understand the meaning of nested conditionals

   Nested conditionals mean putting one conditional inside another, like an if inside an if.

2. Step 2: Compare options to definition

   An if or else inside another if or else correctly describes this as an if or else inside another if or else. Other options describe different or incorrect ideas.

3. Final Answer:

   An if or else inside another if or else -> Option A

4. Quick Check:

   Nested conditional = if inside if [OK]

Hint: Look for conditionals inside other conditionals [OK]

Common Mistakes:

• Thinking multiple separate ifs are nested
• Ignoring indentation importance
• Confusing nested with chained conditionals

2. Which of the following is the correct syntax for nested conditionals in Python?

easy

A. if x > 0:
if x < 10:
print('x is between 1 and 9')

B. if x > 0
if x < 10:
print('x is between 1 and 9')

C. if x > 0:
if x < 10:
print('x is between 1 and 9')

D. if x > 0:
print('x is positive')
else if x < 10:
print('x is less than 10')

Solution

1. Step 1: Check indentation and colons

   Python requires colons after if and proper indentation for nested blocks.

2. Step 2: Analyze each option

   if x > 0: if x < 10: print('x is between 1 and 9') uses colons and indents the inner if correctly. Options A and B miss colons or indentation. if x > 0: print('x is positive') else if x < 10: print('x is less than 10') uses invalid else if instead of elif.

3. Final Answer:

   if x > 0: if x < 10: print('x is between 1 and 9') -> Option C

4. Quick Check:

   Colons + indentation = correct syntax [OK]

Hint: Check colons and indentation for nested blocks [OK]

Common Mistakes:

• Missing colons after if statements
• Incorrect indentation of nested if
• Using else if instead of elif

3. What is the output of this code?

score = 85
if score >= 90:
    print('Grade A')
else:
    if score >= 80:
        print('Grade B')
    else:
        print('Grade C')

medium

A. Grade A

B. Grade B

C. Grade C

D. No output

Solution

1. Step 1: Check first condition

   score is 85, which is not >= 90, so skip first print.

2. Step 2: Check nested else condition

   Inside else, check if score >= 80. 85 >= 80 is True, so print 'Grade B'.

3. Final Answer:

   Grade B -> Option B

4. Quick Check:

   85 >= 80 triggers nested if [OK]

Hint: Follow conditions step-by-step inside else [OK]

Common Mistakes:

• Assuming first if is true for 85
• Ignoring nested else block
• Confusing indentation levels

4. Find the error in this nested conditional code:

num = 5
if num > 0:
if num < 10:
print('Number is between 1 and 9')

medium

A. Missing colon after first if

B. No error, code is correct

C. Using print without parentheses

D. Incorrect indentation of inner if and print

Solution

1. Step 1: Check colons

   Both if statements have colons, so no missing colon error.

2. Step 2: Check indentation

   Inner if and print are not indented under outer if, causing syntax error.

3. Final Answer:

   Incorrect indentation of inner if and print -> Option D

4. Quick Check:

   Nested blocks must be indented [OK]

Hint: Indent nested if and its code properly [OK]

Common Mistakes:

• Forgetting to indent nested blocks
• Confusing missing colon with indentation error
• Assuming print syntax error without checking Python version

5. You want to check if a number is positive, and if positive, check if it is even or odd. Which nested conditional code correctly prints 'Positive even' or 'Positive odd' accordingly?

hard

A. if num > 0:
if num % 2 == 0:
print('Positive even')
else:
print('Positive odd')

B. if num > 0 and num % 2 == 0:
print('Positive even')
else:
print('Positive odd')

C. if num > 0:
if num % 2 == 0:
print('Positive even')
else:
print('Positive odd')

D. if num > 0:
if num % 2 == 0:
print('Positive even')
else:
print('Positive odd')

Solution

1. Step 1: Understand the logic needed

   First check if number is positive, then inside that check if even or odd.

2. Step 2: Analyze options for correct nesting and indentation

   if num > 0: if num % 2 == 0: print('Positive even') else: print('Positive odd') correctly nests the even/odd check inside the positive check with proper indentation. if num > 0 and num % 2 == 0: print('Positive even') else: print('Positive odd') combines conditions but does not handle odd positive numbers correctly. Options C and D have indentation errors.

3. Final Answer:

   if num > 0: if num % 2 == 0: print('Positive even') else: print('Positive odd') -> Option A

4. Quick Check:

   Nested if with correct indentation prints right message [OK]

Hint: Indent inner if inside outer if for stepwise checks [OK]

Common Mistakes:

• Combining conditions incorrectly
• Indentation errors in nested blocks
• Missing else for odd case