Part A. In Europe, gasoline efficiency is measured in km/L. If your car\'s gas m
ID: 998044 • Letter: P
Question
Part A.
In Europe, gasoline efficiency is measured in km/L. If your car's gas mileage is 20.0 mi/gal , how many liters of gasoline would you need to buy to complete a 142-km trip in Europe? Use the following conversions: 1km=0.6214mi and 1gal=3.78L.
Express your answer numerically in liters.
Part B
While in Europe, if you drive 119 km per day, how much money would you spend on gas in one week if gas costs 1.10 euros per liter and your car's gas mileage is 30.0 mi/gal ? Assume that 1euro=1.26dollars.
Express your answer numerically in dollars.
Explanation / Answer
Part A:- Given car's gas mileage is 20.0 mi/gal
Distance covered , d = 142-km = 142 x 0.6214 mi since 1km=0.6214mi
= 88.24 mi
From the mileage of the car 1 gal of fuel will give 20.0 mi
M gal of fuel gave 88.24 mi
M = ( 1x88.24) / 20.0
= 4.41 gal
= 4.41 x 3.78 L Since 1gal=3.78L
= 16.68 L
Part B :- Given distance travelled by one day = 119 km
So distance travelled by one week is = 119 x 7 km
= 833 km
Given cars mileage is 30.0 mi/gal
= 30.0 mi /(3.78 L) since 1 gal = 3.78 L
= 7.94 mi/L
= 7.94(km/0.6214)/L
= 12.8 km/L
So for 1 L the distance travelled is 12.8 km
For N L the distance travelled is 833 km
N = ( 1x833) / 12.8
= 65.2 L
The gas costs 1.10 euros per liter
= 1.10 x 1.26 dollars since 1euro=1.26dollars
= 0.26 dollars
So for 65.2 L the cost of fuel is 65.2x0.26 dollars
= 16.9 dollars