A family has two cars. The first car has a fuel efficiency of 40 miles per gallon of gas and the second has a fuel efficiency of 30 miles per gallon of gas. During one particular week, the two cars went a combined total of 1250 miles, for a total gas consumption of 35 gallons. How many gallons were consumed by each of the two cars that week?
let distance covered by the 40 mpg car be x miles
let the distance covered by the 30 mpg car be 1250-x
miles
gas used by the 40 mpg car = x/40
gas used by the 30 mpg car = (1250-x)/30
x/40 + (1250-x)/30 = 35
times 120
3x + 4(1250-x) = 4200
-x = -800
x = 800
So the more efficient car went 800 miles and used 20 gallons
the other car went 1250-800 or 450 miles and used 15 gallons
To find out how many gallons were consumed by each car, we can use a system of equations.
Let's assume that the first car consumed x gallons of gas and the second car consumed y gallons of gas.
We know that the fuel efficiency of the first car is 40 miles per gallon, so the total miles driven by the first car can be calculated by dividing x by 40: x/40.
Similarly, the fuel efficiency of the second car is 30 miles per gallon, so the total miles driven by the second car can be calculated by dividing y by 30: y/30.
From the problem statement, we also know that the two cars went a combined total of 1250 miles, which can be expressed as x/40 + y/30 = 1250.
We also know that the total gas consumption was 35 gallons, so x + y = 35.
Now we have a system of equations:
x/40 + y/30 = 1250
x + y = 35
To solve this system of equations, we can use elimination or substitution method. Let's use the elimination method.
Multiply the first equation by 3 to eliminate the denominators:
3(x/40 + y/30) = 3(1250)
3x/40 + 3y/30 = 3750/3
3x/40 + y/10 = 1250
Multiply both sides of the second equation by 3:
3(x + y) = 3(35)
3x + 3y = 105
Now we have the equations:
3x/40 + y/10 = 1250
3x + 3y = 105
Multiply the first equation by 4 to eliminate the fraction:
4(3x/40 + y/10) = 4(1250)
3x + 4y/10 = 5000/4
3x + 2y/5 = 1250
Now we have the equations:
3x + 2y/5 = 1250
3x + 3y = 105
Subtract the second equation from the first equation:
(3x + 2y/5) - (3x + 3y) = 1250 - 105
(3x + 2y - 15x - 15y)/5 = 1145
-12y/5 = 1145
Multiply both sides by 5 to isolate y:
-12y = 1145*5
-12y = 5725
Divide both sides by -12 to solve for y:
y = 5725/-12
y = -476.0417
Since we can't have a negative number of gallons, we disregard this solution.
Now let's substitute y = -476.0417 back into one of the original equations:
x + (-476.0417) = 35
x - 476.0417 = 35
x = 35 + 476.0417
x = 511.0417
Again, we can't have a negative number of gallons, so we disregard this solution as well.
In this case, it appears that there is no valid solution where both x and y are positive integers.
Therefore, we cannot determine the exact number of gallons consumed by each car that week.
To determine the amount of gas consumed by each car during the week, we can use the given information about their fuel efficiencies and the total gas consumption.
Let's assume the number of gallons consumed by the first car is x, and the number of gallons consumed by the second car is y.
According to the problem, the first car's fuel efficiency is 40 miles per gallon. So, the first car traveled x miles and consumed x/40 gallons of gas.
Similarly, the second car's fuel efficiency is 30 miles per gallon. So, the second car traveled y miles and consumed y/30 gallons of gas.
Given that the combined total distance traveled by both cars is 1250 miles, we can write the equation:
x + y = 1250 (Equation 1)
And the total gas consumption is 35 gallons, which gives us another equation:
x/40 + y/30 = 35 (Equation 2)
To solve these equations and find the values of x and y, we can use various methods such as substitution or elimination. Let's solve them using elimination:
Multiply equation 2 by 120 to eliminate fractions:
3x + 4y = 4200 (Equation 3)
Now we have a system of two equations:
x + y = 1250 (Equation 1)
3x + 4y = 4200 (Equation 3)
Solving these equations simultaneously, we can find the values of x and y.