Answer:
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 = [tex]x[/tex] gallons
Now, we are given that total volume is 9 gallons, then
amount of gas with 78% ethanol used = 9 - [tex]x[/tex] gallons
As per the question statement:
[tex]x\times 15\% + (9-x)\times 78\%=9\times 50\%\\\Rightarrow x\times 15 + (9-x)\times 78=9\times 50\\\Rightarrow 9\times 78-9\times 50=63x\\\Rightarrow 9\times 28=63x\\\Rightarrow \bold{x = 4\ gallons}[/tex]
Therefore, the answers are:
4 gallons of 15% ethanol
9 - 4 = 5 gallons of 78% 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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