Nested conditional execution in Python - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if a number is positive.

Python

num = 5
if num [1] 0:
    print("Positive number")

Drag options to blanks, or click blank then click option'

A==

B<

C>

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' instead of '>' which checks for negative numbers.

Using '==' which only checks if the number is exactly zero.

The condition num > 0 checks if the number is positive.

2fill in blank

medium

Complete the code to check if a number is positive and even.

Python

num = 4
if num > 0 and num [1] 2 == 0:
    print("Positive even number")

Drag options to blanks, or click blank then click option'

A%

B//

C**

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using '//' which does floor division instead of checking remainder.

Using '+' or '**' which are not related to divisibility.

The modulo operator % checks if the number is divisible by 2 with no remainder, meaning it is even.

3fill in blank

hard

Fix the error in the nested if statement to check if a number is positive and less than 10.

Python

num = 7
if num > 0:
    if num [1] 10:
        print("Number is positive and less than 10")

Drag options to blanks, or click blank then click option'

A>

B==

C!=

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' which checks if number is greater than 10.

Using '==' which checks equality, not less than.

The inner condition should check if num < 10 to confirm it is less than 10.

4fill in blank

hard

Fill both blanks to print if a number is positive and even.

Python

num = 8
if num [1] 0:
    if num [2] 2 == 0:
        print("Positive even number")

Drag options to blanks, or click blank then click option'

A>

B<

C%

D//

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' for positive check which is incorrect.

Using '//' instead of '%' for even check.

The first blank checks if the number is greater than zero. The second blank uses modulo '%' to check evenness.

5fill in blank

hard

Fill all three blanks to print if a number is positive, even, and less than 20.

Python

num = 16
if num [1] 0:
    if num [2] 2 == 0:
        if num [3] 20:
            print("Positive even number less than 20")

Drag options to blanks, or click blank then click option'

A>

B%

C<

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' instead of '<' for the last condition.

Using '//' instead of '%' for even check.

The first blank checks if number is greater than zero, the second uses modulo '%' to check evenness, and the third checks if number is less than 20.

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