On her way home from the school board meeting, Kelly fills up her car. She likes the idea of using gasoline with ethanol, but thinks her car can only handle 50% ethanol. At the gas station, she can use regular gas with 15% ethanol or E78 fuel with 78% ethanol. How many gallons of each type of fuel should Kelly use if she wants to fill up her car with 9 gallons of fuel containing 50% ethanol?

The number of gallons required of both kind would be as follows:

4 gallons having 15% ethanol

5 gallons having 78% ethanol

Given that,

The 2 kinds of gases are:

One with 15% ethanol

Other with 78% ethanol

Given that,

The gases(ethanol and gasoline) are diluted in a such a manner that it involves a total of only 50% ethanol.

Number of gallons she wants to use = 9

This shows that only 50% of 9 gallons i.e. 4.5 gallons of ethanol can be used.

Assuming x being the gasoline's measure that is diluted with 15% ethanol.

Now,

The gasoline's measure that is diluted with 78% ethanol would be = 9 - x gallons

Therefore,

x × 15% + (9-x) × 78% = 4.5

On solving, we get

x = 4 gallons

Thus,

The measure of gallons having 15% ethanol = 4

The measure of gallons having 78% ethanol = (9 - x)

= (9 - 4)

= 5 gallons

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4 gallons of 15% ethanol

5 gallons of 78% ethanol

Step-by-step explanation:

Given that:

Two types of gases:

1st gas with 15% ethanol

2nd gas with 78% ethanol

The gases are to be mixed in such a way that the mixture of gases comes out to be with 50% ethanol with volume of 9 gallons.

Let amount of gas with 15% ethanol used = gallons

Now, we are given that total volume is 9 gallons, then

amount of gas with 78% ethanol used = 9 - gallons

As per the question statement:

Therefore, the answers are:

4 gallons of 15% ethanol

9 - 4 = 5 gallons of 78% ethanol

the probability is 0

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