SOLUTION: A family has two cars. The first car has a fuel efficiency of 15 miles per gallon of gas and the second has a fuel efficiency of 35 miles per gallon of gas. During one particular w

Question 1111708: A family has two cars. The first car has a fuel efficiency of 15 miles per gallon of gas and the second has a fuel efficiency of 35 miles per gallon of gas. During one particular week, the two cars went a combined total of 825 miles, for a total gas consumption of 35 gallons. How many gallons were consumed by each of the two cars that week?
First car: ______ gallons
Second car: ______ gallons
Found 3 solutions by mananth, josgarithmetic, greenestamps:
Answer by mananth(16946)   (Show Source): You can put this solution on YOUR website!
Car I x gallons for the for the week
Car II y gallons for the for the week
Equation for tickets total distance

1 x + 1 y = 35 .............1

Equation for amount spent
15 x + 35 y = 825 .............2
Eliminate y
multiply (1)by -35.00
Multiply (2) by 1
-35.00 x -35.00 y = -1225.00
15.00 35.00 y = 825.00
Add the two equations
-20.00 x = -400.00
/ -20.00
x = 20.00
plug value of x in (1)
1.00 x + 1.00 y = 35.00
20.00 + 1.00 y = 35.00
1.00 y = 15.00
y= 15.00

Car I 20.00 gallons for the for the week
Car II 15.00 gallons for the for the week
m.ananth@hotmail.ca

Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!

-

-
Fuel efficiency rate R, volume of fuel V, distance traveled D

CAR RATE VOLUME DISTANCE First Car 15 v 15v Second Car 35 35-v 35(35-v) Totals 35 825

Volume is gallons and distance is miles; fuel efficiency in MILES per GALLON

v is the gallons used by First Car.
Solve the equation .
Use this to evaluate gallons used by the Second car.
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

The algebraic solutions shown by the other tutors are of course fine. If you don't need to use algebra, there is an easy way to solve a problem like this using basic arithmetic and a bit of logical reasoning.

Here is how you could solve this problem that way.

If all 35 gallons of gas were used by the car that gets only 15 miles per gallon, the total number of miles would be 35*15 = 525 -- 300 miles short of the actual 825 miles.
The more efficient car gets 20 more miles per gallon than the other. To get 300 "extra" miles with the 35 gallons of gas, the number of gallons used by the more efficient car must be 300/20 = 15.
That means the gas guzzler used 35-15=20 gallons.

CHECK: 20*15 + 15*35 = 300+525 = 825
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