Question

The owner of a 4000​-gal oil truck loads the truck with gasoline and kerosene. The profit on each gallon of gasoline is 16​¢ and on each gallon of kerosene it is 13​¢. How many gallons of each fuel did the owner load if the profit was ​$610​?

Answer 1

The owner load 3000 gallons of gasoline and 1000 gallons of kerosene.

Explanation:

Given:

Let the number of gallons of gasoline be 'x'.

Let the number of gallons of kerosene be 'y'.

Number of gallons of oil = 4000

Now we know that;

Number of gallons of oil is equal to sum of the number of gallons of gasoline and the number of gallons of kerosene.

framing in equation form we get;

`x+y=4000 \ \ \ \ equation\ 1`

Also Given:

Cost of each gallon of gasoline = 16¢

Cost of each gallon of kerosene = 13¢

Total Profit = $610

1¢ = $0.01

16¢ = $0.16

13¢ = $0.13

Now we know that;

Total Profit is equal to sum of the number of the number of gallons of gasoline multiplied by Cost of each gallon of gasoline and the number of gallons of kerosene multiplied by Cost of each gallon of kerosene.

framing in equation form we get;

`0.16x+0.13y=610 \ \ \ \ equation \ 2`

First we will multiply equation 1 with 0.13 we get;

`0.13(x+y)=4000* 0.13\\\\0.13x+0.13y= 520 \ \ \ \ equation \ 3`

Now we will subtract equation 3 from equation 2 we get;

`0.16x+0.13y-(0.13x+0.13y)=610-520\\\\0.16x+0.13y-0.13x-0.13y=90\\\\0.03x=90`

Now Dividing both side by 0.03 we get;

`(0.03x)/(0.03)=(90)/(0.03)\\\\x = 3000\ gallons`

Now we will substitute the value of 'x' in equation 1 we get;

`x+y=4000\\\\3000+y=4000\\\\y=4000-3000\\\\y=1000 \ gallons`

Hence The owner load 3000 gallons of gasoline and 1000 gallons of kerosene.