Python Conditions Practice

Practise if, elif and else, combined and nested conditions, and controlling loops with break, continue and pass.

Beginner8 problemsUpdated October 23, 2026

Review first: if, if else, elif, Nested Conditions, break, continue, pass

1.Marks to grades
Beginner
For each mark in marks = [95, 82, 67, 45, 74], print the mark and its grade:
- 90 and above: A
- 75 to 89: B
- 60 to 74: C
- below 60: F
Show hint
Use if / elif / else and check from the highest grade down, so each check only needs a lower limit.
Show solution
Python
marks = [95, 82, 67, 45, 74] for mark in marks: if mark >= 90: grade = "A" elif mark >= 75: grade = "B" elif mark >= 60: grade = "C" else: grade = "F" print(mark, grade)
Output
95 A 82 B 67 C 45 F 74 C
Because elif branches are checked in order and only the first true one runs, elif mark >= 75 already knows the mark is below 90.
2.Leap years
Beginner
A year is a leap year if it is divisible by 4, except years divisible by 100, which are leap years only if they are also divisible by 400. For each year in [1900, 2000, 2024, 2026], print whether it is a leap year.
Show hint
Combine the rules with and and or: (year % 4 == 0 and year % 100 != 0) or year % 400 == 0.
Show solution
Python
for year in [1900, 2000, 2024, 2026]: is_leap = (year % 4 == 0 and year % 100 != 0) or year % 400 == 0 if is_leap: print(year, "is a leap year") else: print(year, "is not a leap year")
Output
1900 is not a leap year 2000 is a leap year 2024 is a leap year 2026 is not a leap year
The standard library already knows this rule, import calendar then calendar.isleap(year), which is what you should use in real code.
3.Cinema ticket price
Beginner
Tickets cost 250. Children under 12 pay 120. Everyone else aged under 25 pays 180 if they are a student. Write a function ticket_price(age, is_student) and print the price for (8, False), (19, True), (19, False) and (40, True).
Show hint
Check the child case first. Inside the else branch, check the student discount.
Show solution
Python
def ticket_price(age, is_student): if age < 12: return 120 else: if age < 25 and is_student: return 180 return 250 for age, student in [(8, False), (19, True), (19, False), (40, True)]: print(age, student, ticket_price(age, student))
Output
8 False 120 19 True 180 19 False 250 40 True 250
The nested if can be flattened with elif age < 25 and is_student:. Both versions are correct; flatter code is usually easier to read.
4.Find the first negative number
Beginner
Loop over readings = [12, 7, 0, -3, 8, -1] and print the first negative reading and its position, then stop. If there is no negative reading, print All readings are fine.
Show hint
Use break to stop the loop. A for ... else block runs only when the loop was not stopped by break.
Show solution
Python
readings = [12, 7, 0, -3, 8, -1] for i, value in enumerate(readings): if value < 0: print(f"First negative reading: {value} at position {i}") break else: print("All readings are fine")
Output
First negative reading: -3 at position 3
5.Skip the blanks
Beginner
Print a greeting for every name in names = ["Asha", "", "Ravi", " ", "Meera"], skipping names that are empty or only spaces.
Show hint
name.strip() removes spaces; an empty string is falsy. Use continue to skip to the next name.
Show solution
Python
names = ["Asha", "", "Ravi", " ", "Meera"] for name in names: if not name.strip(): continue print(f"Hello, {name}!")
Output
Hello, Asha! Hello, Ravi! Hello, Meera!

6.Check a password

Beginner

A password is strong when it has at least 8 characters, at least one digit, and at least one uppercase letter. For each password in ["sunshine", "Sunshine1", "Ab1"], print strong or the first rule it breaks.

Show hint

any(ch.isdigit() for ch in password) is True if at least one character is a digit.

Show solution

Python

def check(password):
    if len(password) < 8:
        return "too short"
    elif not any(ch.isdigit() for ch in password):
        return "needs a digit"
    elif not any(ch.isupper() for ch in password):
        return "needs an uppercase letter"
    else:
        return "strong"


for pw in ["sunshine", "Sunshine1", "Ab1"]:
    print(pw, "->", check(pw))

Output

sunshine -> needs a digit
Sunshine1 -> strong
Ab1 -> too short

7.FizzBuzz

Beginner

Print the numbers 1 to 15, but print Fizz for multiples of 3, Buzz for multiples of 5, and FizzBuzz for multiples of both. Print them on one line separated by spaces.

Show hint

Check "multiple of both" (divisible by 15) first, otherwise the multiple-of-3 branch catches 15.

Show solution

Python

results = []

for n in range(1, 16):
    if n % 15 == 0:
        results.append("FizzBuzz")
    elif n % 3 == 0:
        results.append("Fizz")
    elif n % 5 == 0:
        results.append("Buzz")
    else:
        results.append(str(n))

print(" ".join(results))

Output

1 2 Fizz 4 Buzz Fizz 7 8 Fizz Buzz 11 Fizz 13 14 FizzBuzz

8.Plan first with pass
Beginner
Write a function process(order) that will later handle three statuses. For now, only "new" orders should print Processing new order. Leave branches for "paid" and "shipped" that do nothing yet, then call it with "new", "paid" and "shipped".
Show hint
A block cannot be empty in Python. pass is a statement that does nothing.
Show solution
Python
def process(order): if order == "new": print("Processing new order") elif order == "paid": pass # TODO: send receipt elif order == "shipped": pass # TODO: send tracking link for status in ["new", "paid", "shipped"]: process(status) print("Done")
Output
Processing new order Done
Without pass, Python raises an IndentationError because the elif blocks have no statements. A # TODO comment makes the unfinished work easy to find later.

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