Boolean logic confused, please help me guys

Question

Please look this question and help me to solve it, because i'm so confused to what this app want. Or it just me who didn't understand after all?

I need to create a program that outputs whether a store is Open or Closed, based on the hour and the day of the week.

The store is open on all days from 10 to 21, except Saturday and Sunday.
I need to take the hour and the day of the week as input.

The day of the week is represented as an integer (1 for Monday, 2 for Tuesday, etc.)

Sample Input:
15
4

Sample Output:
Open

I try to answer those question with this:

hour = int(input())

day = int(input())
if (day==1 and hour>=10 and hour<=21):print("Open")
elif (day==2 and hour>=10 and hour<=21):print("Open")
elif (day==3 and hour>=10 and hour<=21):print("Open")
elif (day==4 and hour>=10 and hour<=21):print("Open")
elif (day==5 and hour>=10 and hour<=21):print("Open")
elif (day==6 or day==7):print("Closed")

#why_it_still_wrong_???

Answer 1

Novian , may be it's worth to re-think about the way you check the day and the time. There is some duplication in code that can be reduced.

Instead of using several if - elif conditions, you could use range() to define the valid time the shop is open, the same can be done for the day.
Doing so, you need only one if - else condition to check. Then the membership operator "in" can be used to check if day and time is in range.

keep in mind that using range, the end parameter is not included. range(1,5) does create numbers 1,2,3,4.

Happy coding!

Answer 2

This is what your completed code
hour = int(input())
day = int(input())
if (day==1 and hour>=10 and hour<=21):print("Open")
elif (day==2 and hour>=10 and hour<=21):print("Open")
elif (day==3 and hour>=10 and hour<=21):print("Open")
elif (day==4 and hour>=10 and hour<=21):print("Open")
elif (day==5 and hour>=10 and hour<=21):print("Open")
elif (day==6 or day==7):print("Closed")
elif hour<10 or hour>21:print("Closed")

But you're able to do it like this after learning about range() as Lothar said

hour = int(input())
day = int(input())

if hour in range(10,22) and day in range(1,6):
print("Open")
else:
print("Closed")

Answer 3

Fill in the blanks to output ”Closed” if the hour variable is larger than 20 or if the day variable is equal to ”Sunday”.

answer:-
if(hour > 20 or day == “Sunday”):
print(“Closed”)
else:
print(“Open”)

Answer 4

Simba thanks a lot mate, it works

Answer 5

Fill in the blanks to output ”Closed” if the hour variable is larger than 20 or if the day variable is equal to ”Sunday”.

if(hour > ______or______== “Sunday”):
print(“Closed”)
else:
print(“Open”)

can help me about this task please..

Answer 6

I try to answer like this, and still wrong:

hour = int(input())
day = int(input())
if (day==1 and day==2 and day==3 and day==4 and day==5):print("Open")
elif(hour>=10 and hour<=21):print("Open")
elif (hour<=10 or hour>=21):print("Closed")
elif(day==6 and day==7):print("Closed")
else:
print("Closed")
# your code goes here

Answer 7

Lothar it is work with boolean logic? I mean mu course still stuck on boolean logic number 2. I didn't even know what is range() and how to use it.

Answer 8

Also trying to figure this out. My code works for all test cases but one:
if day >= 1 and day <= 5:
if hour > 9 and hour < 21:
print("Open")
else:
print("Closed")

Answer 9

here's a short one
hour = int(input())
day = int(input())
if (10<=hour<=21) and (day<=5) :
print("Open")
else :
print("Closed")

Answer 10

This was my solution:

hour = int(input())
day = int(input())
# your code goes here

if hour in range(10,22):
if day in range(1,6):
print("Open")

else:
print("Closed")

but test case 4 was the only one that was wrong... Then I tried Simba's solution and it was correct. Can anyone tell me what the issue might have been in my code?

Answer 11

its ans is
20 or
day