Question

• c.Imagine that a company produces bottled lemonade. For this company, an order consists of:
  • the number of required bottles and
  • the size of the bottles, with a choice from: ‘small’, ‘medium’ and ‘large’.

You can assume that the bottles in a single order are all of the same size.

Given an order, the problem is to calculate the number of litres of lemonade required to fill all the bottles for that order.

The number of litres required for a single bottle is calculated as follows: Multiply 0.9560 with 1, 2 or 3 depending on whether the bottle size is ‘small’, ‘medium’ and ‘large’, respectively.

• i.Include your initial decomposition of the problem in your solution document using the chevron notation (> and >>) from the module materials.
• ii.Include the algorithm for solving the problem in your solution document.

Part iii of this question involves writing one Python function definition.

iii.Provide a single Python function that implements the algorithm. Follow the instructions above for submitting code.

Your answer must be a translation of your algorithm from part c (ii), otherwise no marks will be awarded.

Answer 1

Solution:

C(i)

Take the total number of bottles required for the user

Take the size of the bottle required i.e small,medium,large

calculate the number of litres for each size

   if size is "small" then

>> output =number of bottles required *0.9560*1

if size is "medium" then

>> output =number of bottles required *0.9560*2

if size is "large" then

>> output =number of bottles required *0.9560*3

C(ii)

Step1:

Take the total number of bottles required for the user

Step2:

Take the size of the bottle required i.e small,medium,large

Step3:

if size is "small" then

output =number of bottles required *0.9560*1

Step4:

if size is "medium" then

output =number of bottles required *0.9560*2

Step5:

if size is "large" then

output =number of bottles required *0.9560*3

C(iii)

def solution(cou,sie):
    value=0.9560
    result=cou*sie*value
    return result
B_count=float(input("enter the number of required bottles:"))
size_string=input("enter the size small,medium or large:")
if(size_string=="small"):
    size=1.0
elif(size_string=="medium"):
    size=2.0
elif(size_string=="large"):
    size=3.0
print(size)
res=solution(B_count,size)
print(res)