Answer:
A.Rate of Mrs. Allan's car=28 mile/gallon
Rate of Mrs. Owen's car=35 mile/gallon
B.Mrs. Owen's car
C.Mrs.Allen's car used gas for 560 miles=20 gallons
Mrs.Owen's car used gas for 560 miles=16 gallons
Step-by-step explanation:
We are given that
Mrs. Allan's car uses 8 gallons of gas for 224 mile tripe.
Mrs. Owens car uses 6 gallons of a gas for 210 mile tripe.
A.Rate=[tex]\frac{Distance\;traveled}{Gas;used}[/tex]
Rate of Mrs. Allan's car=[tex]\frac{224}{8}=28 mile/gallon[/tex]
Rate of Mrs. Owen's car=[tex]\frac{210}{6}=35 mile/gallon[/tex]
B.Mrs.Allan's car travel in 1 gallon of gas =28 miles
Mrs. Owen' s car travel in one gallon of gas=35 miles
Therefore, Mrs. Owen's car travel large distance in one gallon of gas.Hence, it is better.
C.If distance traveled by the cars=560 miles
Mrs. Allen's car used gas for 28 mile=1 gallon
Mrs. Allen's car used gas for 1 mile=[tex]\frac{1}{28}[/tex]
Mrs.Allen's car used gas for 560 mile=[tex]\frac{560}{28}=20 gallons[/tex]
Mrs.Allen's car used gas for 560 miles=20 gallons
Mrs.Owen's car used gas for 35 miles=1 gallon
Mrs. Owen's car used gas for 1 mile=[tex]\frac{1}{35}[/tex]
Mrs. Owen's car used gas for 560 miles=[tex]\frac{560}{35}=16 gallons[/tex]
Mrs.Owen's car used gas for 560 miles=16 gallons