Problem 1item = “quesadilla”meat = “steak”queso = Falseguacamole =
False
double_meat = False
#———————————————————–
#You may modify the lines of code above, but don’t move them!
#When you Submit your code, we’ll change these lines to
#assign different values to the variables.
#
#Let’s further expand our previous program to cover a broader
#menu variety. Instead of just burritoes, now the program
#should cover three menu items: quesadillas, burritoes, and
#nachos. Instead of separate booleans for steak and pork,
#we instead have a string that could be “steak”, “pork”,
#”chicken”, “tofu”, and “beef”. We still have booleans for
#queso and guacamole, but we also have a boolean for double
#meat.
#
#Your code should calculate the price as follows:
#
# – The base price for a quesadilla is 4.00, for nachos is
# 4.50, and for burritoes is 5.00.
# – If meat is steak or pork, add 0.50. Any other meat adds
# no money to the price.
# – guacamole always adds 1.00 to the price.
# – queso adds 1.00 to the price UNLESS the item is nachos,
# in which case it adds nothing.
# – double_meat adds 1.50 if the meat is steak or pork, or
# 1.00 otherwise.
base_price = 4.5
if item == “quesadilla”:
base_price = 4.0
elif item == “burrito”:
base_price = 5.0
if meat == “steak”:
base_price += 0.50
if meat == “pork”:
base_price += 0.50
elif meat == “steak” and double_meat:
base_price += 1.50
elif meat == “pork” and double_meat:
base_price += 1.50
elif double_meat:
base_price += 1.0
if guacamole:
base_price += 1.0
if queso and not “nachos”:
base_price += 1.0
print(base_price)
Problem 2
steak = False
pork = True
guacamole = False
queso = False
#You may modify the lines of code above, but don’t move them!
#When you Submit your code, we’ll change these lines to
#assign different values to the variables.
#
#Imagine you’re writing code for a cash register at a
#restaurant. This restaurant serves burritoes. The base price
#of a burrito is $5. If the customer wants steak or pork, it
#adds $0.50. If they want quacamole, it adds $1.00. If they
#want queso, it adds $1.00. The customer may only select one
#meat, but they may have both queso and guacamole, neither,
#or just one.
#
#Write some code below that will print the cost of the
#burrito based on the variables above. You do not need to
#print the dollar sign or extra 0s. Remember, your final answer
#should only print out the price: comment out any debug
#statements once you have the right answer.
price = 5.00
if steak or pork:
print(“Adding 0.50 for steak or pork!”)
price += 0.50
elif guacamole:
print(“Adding 1.00 for guacamole!”)
price += 1.00
else:
print(“Adding 1.00 for queso!”)
price += 1.00
print(price)
Problem 3
mystery_value = 87
#You may modify the lines of code above, but don’t move them!
#When you Submit your code, we’ll change these lines to
#assign different values to the variables.
#Writ a while loop that continues to add 9 to mystery_value
#until mystery_value is greater than 100. Each time 9 is
#added, print the *new* value of mystery_value. For example,
#with mystery_value = 87, your code should print 96 and 105.
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