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# If my car gets 24mpg highway and 16mpg city, how much did it cost me for the highway portion of the trip, and how much for the city portion of the trip?

I drove 280 miles, made up of both highway and city driving.  I used 15 gallons of gas that cost \$3.15 per gallon.  If my car gets 24mpg highway and 16mpg city, how much did it cost me for the highway portion of the trip, and how much for the city portion of the trip?  Explain work.

### 6 Answers by Expert Tutors

Arthur D. | Effective Mathematics TutorEffective Mathematics Tutor
5.0 5.0 (9 lesson ratings) (9)
1
let x= # of gallons used driving on the highway
let y= # of gallons used driving in the city
x+y=15
24x+16y=280
if x+y=15, then y=15-x
substitute y=15-x in the second equation
24x+16(15-x)=280
24x+240-16x=280
8x+240=280
8x=40
x=5 gallons used on the highway
5+y=15 and y=10
y=10 gallons used in the city
check: (5x24)+(10x16)=280
120+160=280 miles
5x\$3.15=\$15.75 for the highway cost
10x\$3.15=\$31.50 for the city cost
check: 15x\$3.15=47.25 and \$15.75+\$31.50=\$47.25
By the way, if you want to determine your average gas mileage and you drive the same number of miles in the city as on the highway, you can use a special case of what is called the harmonic mean., For example, if you get 10 mpg in the city and 15 mpg on the highway and you drive 30 mile in the city and 30 miles on the highway, you use 3 gallons of gas in the city and 2 gallons of gas on the highway. Therefore you use 5 gallons altogether(3+2) and you drive 60 miles altogether(30+30).
Therefore 60/5=12 mpg. Using the special case of the harmonic mean, we have (2xrate#1xrate#2)/(rate#1+rate#2)=(2x10x15)/(10+15)=
300/25=12 mpg
Using your problem, we have (2x24x16)/(24+16)=768/40=19.2 mpg.
This is your average miles per gallon if you drive the same distance in the city as on the highway. Another example:20 mpg city and 30 mpg highway-(2x20x30)/(20+30)=1200/50=24 mpg average
Notice the average is always closer to the city mpg !

Vivian L. | Microsoft Word/Excel/Outlook, essay composition, math; I LOVE TO TEACHMicrosoft Word/Excel/Outlook, essay comp...
3.0 3.0 (1 lesson ratings) (1)
1
Hi Ryan;
24 m/g highway
x=dollars highway
16 m/g city
280 miles
15 g
\$3.15/g

First...

(\$3.15/g)(15g)
Let's cancel our units...
Gallons is a denominator and numerator...
(\$3.15/g)(15g)
(\$3.15)(15)=\$47.25

Next...

\$47.25[(280 m)/(15 g)]=[(24 m/g)(x)]+[(16 m/g)(\$47.25-x)]
First, let cancel our units...
On both sides, we have miles/gallon...
\$47.25[(280 m)/(15 g)]=[(24 m/g)(x)]+[(16 m/g)(\$47.25-x)]
\$47.25[(280)/(15)]=[(24)(x)]+[(16)(\$47.25-x)]
The only unit remaining is \$, which is what we are solving for...
\$882.00=24x+\$756.00-16x
Let's combine like units...
\$882.00=8x+\$756.00
Let's subtract \$756.00 from both sides...
\$882.00-\$756.00=8x+\$756.00-\$756.00
\$126.00=8x
Let's divide both sides by 8...
\$126.00/8=(8x)/8
\$15.75=x

\$15.75 spent on highway
\$47.25-\$15.75=\$31.50 spent on city

Let's check our work...
[(\$15.75)(24 m/g)]+[(\$31.50)(16 m/g)]=(\$3.15/g)(280 m)
Our units align in that each side is \$-m/g...
[(15.75)(24)]+[(31.50)(16)]=(3.15)(280)
378+504=882
882=882

Jason S. | My goal is the success of my students. Knowledge-Patience-HonestyMy goal is the success of my students. K...
4.9 4.9 (115 lesson ratings) (115)
1

x = miles of hwy driving

280-x = miles of city driving

Gallons for hwy + Gallons for city  = Total gallons for the trip
(x miles)/(24mpg) + (280-x)/16 = 15

2x/48 + 3(280-x)/48 = 15

(2x + 840 - 3x)/48 = 15

(-x + 840)/48 = 15

-x + 840 = 48 * 15

-x + 840 = 720
x = 120 miles hwy

x/24 = 120/24 = 5 gallons for the hwy

(280-x)/16 = (280 - 120) / 16 = 160/16 = 10 gallons city

5 gallons hwy * (3.15) = \$15.75 cost of hwy
10 gallons city * (3.15) = \$31.50 cost of city
----------------------------------------------------

15 gallons * (3.15) = 47.25
Vandana D. | Tutor for elementary and maths (up to 5 grade)Tutor for elementary and maths (up to 5...
0
Hi Ryan,
Assumptions:
x = no. of gallons used on highway
y = no. of gallons used in city
x+y = 15 ................................................................equation 1
24x+16y = 280 .......................................................equation 2
Multiply Equation 1 by 16, u will get
16x+16y = 240 ........................................................equation 3

Now subtraction equation 3 from 2, u will get
8x = 40
therefore x= 5
putting the value of x in equation 1
5+y = 15, y=10

Check: putting the value of x and y in equation 2
24*5+16*10 =280

Price :
5*3.15 = 15.75 cost on highway
10*3.15 = 31.5 cost in the city
Total cost = 15.75+31.5 = 47.25
Pamela E. | Math Tutor for Middle School, High School, and College LevelsMath Tutor for Middle School, High Schoo...
-1
Let H represent the number of gallons of gas used for Highway travel.
Let C represent the number of gallons of gas used for City travel.

15 total gallons of gas are used, so one equation, I'll call the "gallons" equation is:

H gal + C gal = 15 gal         or more simply:          H + C = 15
We have 2 unknowns, but only one equation so far. So the second equation comes from the MPG information (I will write this equation including the units, so you can see how it "works").

This one I'll call the "miles" equation:

24 mi / gal * H gal + 16 mi / gal * C gal = 280 mi

Notice that for the left side of this equation, in both terms the "gallon" units will "cancel," since one of them is in the denominator (from the MPG part -- miles per gallon, or mi / gal), and the other is in the numerator (from the unknowns H and C, which represent "gallons").

After the "gallon" units cancel, the left side of the equation is only expressed in terms of "miles," which matches the "miles" on the right side of the equation.

Here is the updated "miles" equation:

24 mi * H + 16 mi * C = 280 mi               or more simply:               24 H + 16 C = 280

When we combine these 2 equations (the "gallons" equation with the "miles" equation, we get the following system of equations:

H +      C =   15         ("gallon" equation)
24 H + 16 C = 280         ("miles" equation)

You can now solve this system in any way you choose, depending on what your teacher requires.
I suspect that the solution method (substitution, elimination, graphing, Cramer's rule, or matrix methods) is not what was giving you trouble; it was setting up the system of equations in the first place.

If you are familiar with matrix methods on the graphing calculator, then that is probably the fastest way to go. You simply enter this system of equations as a 2x3 augmented matrix, and then find the solution using the "rref" (reduced-row echelon form) method.
(feel free to message me if you would like the detailed, step-by-step instructions)

To solve using the substitution method (probably the fastest "by-hand" way)
Solve "gallons" equation for either variable (I'll choose H):
H = 15 - C
Now, substitute (15 - C) for H in the second, "miles" equation of   24 H + 16 C = 280, like so:
24 ( 15 - C ) + 16 C = 280                Now, you have an equation only in terms of one variable, C.
So, simplify and solve this equation:
360 - 24 C + 16 C = 280
- 8 C = 280 - 360
- 8 C = - 80
So C = 10 gallons.   But the question is not asking how many gallons were used for City travel; it is asking how much it COST for the City travel. All you do is multiply the cost per gallon from the given information:

Cost of City driving = 10 gal * \$3.15 / gal = \$31.50 total cost for City driving.

(Note that this problem did NOT ask about the Highway driving, but this is easy enough to get.)
You already know that C = 10, so substitute this back into the "gallons" equation of   H + C = 15:

H + 10 = 15         so    H = 5 gallons,
and cost  of Highway driving = 5 gal * \$3.15 / gal = \$15.75  total cost for Highway driving.

Joseph A. | SAT, Math 6-12SAT, Math 6-12
-2
Total cost of gas = \$3.15 per gallon * 15 gallons = \$47.25.

distance of highway travel is unknown; call it x.
distance of city travel is 280-x.

distance = rate * time.

x = 24 *time1
280-x = 16 * time 2

Without knowing how long you traveled on each type of road, you cannot solve this question.

in other words, time 1 + time 2 has not been specified, and without it, I cannot solve.

Have a suspicion that there is way to solve this, but I only spent a minute or so looking at the question.  At the least, this gives you a framework to start you off.