SOLUTION: A salesperson purchased an automobile that was advertised as averaging 27 mi/gal in the city and 38 mi/gal on the highway. A recent sales trip that covered 1192 miles required 36 g

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Question 899640: A salesperson purchased an automobile that was advertised as averaging 27 mi/gal in the city and 38 mi/gal on the highway. A recent sales trip that covered 1192 miles required 36 gallons of gasoline. Assuming that the advertised mileage estimates were correct, how many miles were driven in the city?
Found 3 solutions by josgarithmetic, richwmiller, MathTherapy:
Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!
This is Uniform Rates for fuel efficiency.
R = fuel efficiency
V = volume of fuel in gallons
D = distance in miles
RV=D

A data table might help.

___________________rate_________volume__________distance
CITY________________27
HIWAY_______________38
TOTALS___________________________36______________1192

Use the basic equation to help fill the missing volume and distance data; assign variables, either for the volumes, or the distances. In fact, you could assign variables just to either the City or to the Highway.


.(---Portion Cut Out and Moved Below, bottom of page field---)


ASSIGNING VARIABLES DIFFERENTLY
-
___________________rate_________volume__________distance
CITY________________27___________c/27_______________c
HIWAY_______________38___________h/38_______________h
TOTALS___________________________36______________1192

Form two equations.
-

LCD=2*2*3*3*3*19

-

-

Multiply the c+h equation by 54, subtract from the 76c+54h equation.



, miles of city driving.



-----------THIS ATTEMPT HAS UNKNOWN MISTAKE--------
This attempt will be to JUST assign c = distance done for the city.
The reason is that the question asks specifically for this.

___________________rate_________volume____________________distance
CITY________________27___________c/27_______________________c
HIWAY_______________38__________(1192-c)/38_________________1192-c
TOTALS___________________________36_________________________1192

That single variable assigning appears to be workable for a solution. The equation to solve will be .

The denominators are a bit inconvenient.
3*3*3, 2*19, 2*2*3*3
Lowest Common Denominator is , and most easily used in this factored form. Multiply left and right members by this.



22c---------------------------------AND THE SOLUTION IS FAILING BECAUSE MAKES A NEGATIVE VALUE FOR c
Answer by richwmiller(17219)   (Show Source): You can put this solution on YOUR website!
c+h=1192,
c/27+h/38=36
c = 432, h = 760
Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!
A salesperson purchased an automobile that was advertised as averaging 27 mi/gal in the city and 38 mi/gal on the highway. A recent sales trip that covered 1192 miles required 36 gallons of gasoline. Assuming that the advertised mileage estimates were correct, how many miles were driven in the city?


Let amount of miles driven in the city be D

Then amount of miles driven on highway = 1,192 – D

Equation to be formed:

Miles driven in city, divided by advertised mpg in city, plus miles driven on highway, 

divided by advertised mpg on highway, equals total gallons used. This is:



38D + 27(1,192 – D) = 36(27)(38) ------ Multiplying by LCD, 27(38)

38D + 32,184 – 27D = 36,936

38D – 27D = 36,936 – 32,184

11D = 4,752

D, or distance driven in the city = , or  miles 

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