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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 20 miles per gallon of gas. During one particular week, the two cars went a combined total of 2100 miles, for a total gas consumption of 65 gallons. How many gallons were consumed by each of the two cars that week?

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 20 Miles Per Gallon Of Gas Durin class=

Sagot :

Answers:

  • First car =   40    gallons
  • Second car =    25     gallons

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Explanation:

I'll denote the cars with the labels A and B.

Car A has a fuel rating of 40 mpg (miles per gallon) and car B gets 20 mpg.

Let,

  • x = number of gallons used by car A
  • y = number of gallons used by car B

Since car A gets 40 mpg, and uses up x gallons, then it travels 40x miles. Multiply the fuel rating with the number of gallons used to determine the distance traveled. Car B travels 20y miles since it gets 20 mpg.

Overall, the two cars travel a combined distance of 40x+20y miles. This is set equal to the 2100 miles stated to form the equation 40x+20y = 2100.

Let's divided every term by 20

40x+20y = 2100

40x/20+20y/20 = 2100/20

2x+y = 105

Then isolate y to get

y = 105-2x

We'll use this equation later.

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The two cars use up a total of 65 gallons, which means,

x+y = 65

Then we'll plug in y = 105-2x and solve for x.

x+y = 65

x+105-2x = 65

-x+105 = 65

-x = 65-105

-x = -40

x = 40

Car A (the 40 mpg car) used 40 gallons. It traveled 40x = 40*40 = 1600 miles.

Car B (the 20 mpg car) used y = 105-2x = 105-2*40 = 25 gallons of gas. Or you could note that if x = 40, then x+y = 65 must lead to y = 25. It traveled 20y = 20*25 = 500 miles.

The two cars combined distance is 1600+500 = 2100 miles to help verify the answers.

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