Difficulties with python exercises

Question

So, I'm doing exercises from the Practical Programming book on Python, but I am stuck on the 9th exercise in Chapter 2, which is related to the 7th:

7.: In the United States, a car’s fuel efficiency is measured in miles per gallon. In the metric system, it is usually measured in liters per 100 kilometers. Write a function called convert_mileage that converts from miles per gallon to liters per 100 kilometers.

I wrote the program like this:

def convert_mileage(miles_per_gallon):
    liters_per_gallon = 3.785411784
    kilometers_per_mile = 1.609344
    liters_per_100 = (100*liters_per_gallon)/(kilometers_per_mile*miles_per_gallon)
    print miles_per_gallon,'miles per gallon are',liters_per_100,'liters per 100 kilometers.'

convert_mileage(40)
convert_mileage(20)

Now, the 9th exercise is the following:

9.: Define a function called liters_needed that takes a value representing a distance in kilometers and a value representing gas mileage for a vehicle and returns the amount of gas needed in liters to travel that distance. Your definition should call the function convert_mileage that you defined as part of a previous exercise.

I have no clue how to link the first function into the second one... and I am having difficulties understanding the whole mileage thing compared to liter to travel. If anyone could help me out, that'd be great! Thanks :)

Answer 1

This excercise asks you to reuse your convert_mileage function. Therefore, instead of just printing the calculated value, you have to return it. Change your function to something like this:

LITERS_PER_GALLON = 3.785411784
KILOMETERS_PER_MILES = 1.609344

def convert_mileage(miles_per_gallon):
    """convert miles-per-gallon to liters per 100 kilometers"""
    return (100*LITERS_PER_GALLON)/(KILOMETERS_PER_MILES*miles_per_gallon)

Now you can call this function and reuse its result in another calculation:

def liters_needed(distance_km, miles_per_gallon):
    """determine liters needed for distance with given miles per gallon"""
    liters_per_100km = convert_mileage(miles_per_gallon)
    return liters_per_100km * distance_km / 100

Now you have to print the results when you call the functions:

print "Liters needed for 200km with 15mpg:", liters_needed(200, 15)

Answer 2

#onverting l/100km into mpg
def liters_100km_to_miles_gallon(liters): 
    kms_per_mile=1.609344
    liters_per_gallon=3.785411784
    kms_per_liter=100/liters
    kms_per_gallon=kms_per_liter*liters_per_gallon
    miles_per_gallon=kms_per_gallon/kms_per_mile
    return miles_per_gallon

#onverting mpg into 1/100km
def miles_gallon_to_liters_100km(miles):
    kms_per_mile=1.609344
    liters_per_gallon=3.785411784
    gallons_per_100miles=100/miles
    gallons_per_100kms=gallons_per_100miles/kms_per_mile
    liters_per_100kms=gallons_per_100kms*liters_per_gallon
    return liters_per_100kms

print(liters_100km_to_miles_gallon(3.9))
print(liters_100km_to_miles_gallon(7.5))
print(liters_100km_to_miles_gallon(10.))
print(miles_gallon_to_liters_100km(60.3))
print(miles_gallon_to_liters_100km(31.4))
print(miles_gallon_to_liters_100km(23.5))

# output
60.31143162393162
31.36194444444444
23.52145833333333
3.9007393587617467
7.490910297239916
10.009131205673757