Mixture...3

Suppose a car can run on ethanol and gas and you have a 15 gallons tank to fill. You can buy fuel that is either 30 percent ethanol or 80 percent ethanol. How much of each type of fuel should you mix so that the mixture is 40 percent ethanol?

A method that I found online tells me to do the following.

Let A = amount

0.30A = ethanol amount

0.80(15 - A) = gas

Equation:

0.30A + 0.80(15 - A) = 0.40(15)

1. What does (15 - A) represent?

2. Why must 0.40 be multiplied by 15?

I don't get it.

 

Let x represent number of gallons of gas that contain 30% ethanol.
Let 15 - x be number of gallons of gas that contain 80% ethanol.

Since the mixture contains 40% ethanol, only 40% of the 15 gallons will be ethanol.

40% of 15 = (40/100) *15 = 600 / 100 = 6

In order for x gallons of gas to contain 30% ethanol, we must take 30% of x or 30% times x

In order for 15 - x gallons of gas to contain 80% ethanol, we must take 80% of 15 - x or 80% times 15 - x

0.30 × x + 0.80 × (15 - x) = 6

0.30x + 0.80 × 15 - 0.80 × x = 6

0.30x + 12 - 0.80x = 6

0.30x - 0.80x + 12 = 6

-0.50x + 12 = 6

-0.50x = -6

x = 12

So 12 gallons of gas contain 30% ethanol and 15 - 12 = 3 gallons contain 80% ethanol.

Therefore, mix 12 gallons of a 30% ethanol with 3 gallons of an 80% ethanol.

Indeed 30% of 12 = 0.30 × 12= 3.6 and 80% of 3 = 0.80 × 3 = 2.4

3.6 + 2.4 = 6

 

Here I am MR. PRECALCULUS struggling with basic high school word problems. Sad. Sad. Sad.