Answer:
Fuel efficiencies of two cars in that week are 20 miles per gallon and 40 miles per gallon.
Solution:
Let the two cars of family be A and B.
Gas consumed by car A = 15 gallons
Gas consumed by car B = 35 gallons
Let’s assume number of miles car A drove in that week = x miles
Assume number of miles car B drove in that week = y miles
Given that two cars drove a combine of 1700 miles in that week.
So we can say,
x + y = 1700 --- eqn (1)
Fuel efficiency of any car = [tex]\frac{\text { Number of miles covered }}{\text { gallons of gas consumed }}[/tex]
So fuel efficiency of car A = [tex]\frac{x \text { miles }}{15 \text { gallons }}[/tex]
And fuel efficiency of car B = [tex]\frac{y \text { miles }}{35 \text { gallons }}[/tex]
Given that sum of efficiencies of two cars = 60 miles per gallons.
[tex]\frac{x}{15}+\frac{y}{35}=60[/tex]
[tex]\frac{35 x+15 y}{35 \times 15}=60[/tex]
On cross-multiplication we get,
5(7x +3y) = [tex]15 \times 35 \times 60[/tex]
7x +3y = 6300 --------- eqn (2)
Now we have following two equations and two variable to be determine.
x + y = 1700 -------- (1)
7x +3y = 6300 -------- (2)
On modifying equation (1) we get
y = 1700 - x --------(3)
On substituting value of y from equation (3) in equation (2) we get
7x + 3(1700-x) = 6300
7x + 5100 – 3x = 6300
4x = 6300 – 5100
4x = 1200
x = 300
Substituting value of x in equation 3 to get value of y
y = 1700 – 300 = 1400
Fuel efficiency of car A = [tex]\frac{x}{15}[/tex] = [tex]\frac{300}{15}[/tex] = 20 miles per gallon
Fuel efficiency of car B = [tex]\frac{y}{35}[/tex] = [tex]\frac{1400}{35}[/tex] = 40 miles per gallon.
Hence fuel efficiencies of two cars in that week are 20 miles per gallon and 40 miles per gallon.
