a lawnmower uses a mix of gas and oil. for each ounce of oil the mixture contains 12 ounces of gas. if the tank holds 65 OZ of mixture, how many OZ of oil and gas are required to fill the tank?

Answer 1

Hi Rafael,

This question boils down to being able to set up an equation based on the information given in the question. Let's use variables X and Y in our equation. Doing this would give us an equation:

X + Y = 65

where X is the number of ounces of oil and Y is the number of ounces of gas. However, we know that for each ounce of oil the mixture contains, it contains 12 ounces of gas. Thus, we can replace the Y term with 12X, and the equation becomes:

X + 12X = 65

Then I solve for X.

13X = 65 (simplifying like terms)

X = 5 (dividing by 13 on both sides)

Thus, we know that the number of ounces of oil used (X) is 5 ounces. The number of ounces of gas used (Y) again is equal to 12X, so Y= 12*5 = 60 ounces of gas.

Please let me know if you would like further explanation about this problem or any others from your class. I'd be happy to set up a lesson!

Regards,

Danny