Help me to solve this problem.

Question

https://sololearn.com/compiler-playground/cS3Yu0qNW44b/?ref=app

Answer 1

Thanks bro😁

Answer 2

# first you need to declare a variable
student_marks = 70
# the if statement in python begin with lower case if
if student_marks >60:
#note that for your code to be in if block it needs
#to be indedted by 4 spaces or 1 tab
print("passed")
#the print statement should be enclosed in bracket
#this is the else if sector example
elif student_marks >90:
print("passed with an A")

else:
print("failed")
#if you have another condition you put in the else if sector

Answer 3

Anirudh Kavediya

"If" must be in all lowercase

Python is a case sensitive language

Answer 4

Anirudh Kavediya And, use "elif", not "else". else always goes after "if" and "elif" (elif stands for else if)

Else is a last resort if all conditions are false, it runs.

Answer 5

Anirudh Kavediya Sorry for another ping

there were more issues, like, you need to convert the user input to a integer as python saves them as strings. and the correct syntax for print is print("string")

here is the corrected code:

student_marks = int(input())

if student_marks > 60:
print("pass")
elif student_marks < 60:
print("failed")

print(student_marks)

Answer 6

np

Answer 7

If you enter exactly 60 there will be no result.

this is correct and more compact:

student_marks = int(input())
print(student_marks, "pass" if student_marks>=60 else "failed")

Answer 8

https://sololearn.com/compiler-playground/cefL6WriOC3t/?ref=app

Answer 9

https://sololearn.com/compiler-playground/cJ66R3MnGFS4/?ref=app