Question

A petroleum company has two different sources of crude oil. The first source provides crude oil that is 65% hydrocarbons, and the second one provides crude oil that is 90% hydrocarbons. In order to obtain 80 gallons of crude oil that is 75% hydrocarbons, how many gallons of crude oil must be used from each of the two sources?

Answer 1

Define Terms. What terms? What does it want?

S = Gallons of 65%
N = Gallons of 90%

Now, add things you know!

Gallons: S + N = 80
HydroCarbons: S(0.65) + N(0.90) = 80(0.75)
Other Stuff: S(0.35) + N(0.10) = 80(0.25)

That's plenty of information.

Answer 2

Let \(f\) be the first source, and \(s\) be the second source.

\[f+s = 80\]
\(0.65f\) is the amount of hydrocarbons in 1 unit of \(f\)
\(0.90s\) is the amount of hydrocarbons in 1 unit of \(s\)

In order to produce 80 gallons of 75% hydrocarbons, the following must be true:
\[0.65f + 0.90s = 0.75*80\]

Answer 3

i dont get this still

Answer 4

Any trouble naming variables? I think that's the hardest part.

Answer 5

@jim_thompson5910

Answer 6

What exactly is confusing you? Do you agree that the first equation is right?
The two quantities added together total 80?

Answer 7

65+90=80? no

Answer 8

No, the amount of 65% mix and the amount of 90% mix add up to e total number of gallons of mixture you need.

Answer 9

If you don't agree, we have a problem...because that's all we are combining to make our 80 gallons of 75% hydrocarbon content mixture

Answer 10

Oh, yes I agree

Answer 11

Good. The other equation is a bit harder to decipher, but not too much. First, do you agree that if we have 80 gallons of 75% hydrocarbon content, that means the total hydrocarbon content is 0.75*80=60?

Answer 12

That's 60 gallons

Answer 13

So, that 60 gallons of hydrocarbon content comes from 65% of the first mixture and 90% of the second mixture. The number of gallons of the first mixture we have is \(f\) and the number of gallons of the second mixture is \(s\).

The hydrocarbon content of the first mixture is \(0.65f\)
The hydrocarbon content of the second mixture is \(0.90s\)

Together, they must add to 60 gallons:
\[0.65f + 0.90s = 60\]

That gives us two equations in two unknowns:
\[f+s = 80\]\[0.65f+0.90s = 60\]

We can solve this by substitution. Solve the first equation for one variable in terms of the other. \[f+s=80\]\[f=80-s\]

Now substitute \((80-s)\) wherever we see \(f\) in the second equation (this is why it is called substitution):

\[0.65(80-s) + 0.90s = 60\]\[52 - 0.65s + 0.90s = 60\]\[52+0.25s = 60\]\[0.25s = 8\]\[s=32\]

Now we plug that into the substitution equation and find the value of the other variable:\[f = 80-s\]\[f = 80-32\]\[f = 48\]

So, we have 48 gallons of the 65% mixture and 32 gallons of the 90% mixture. 48+32=80, so the total amount of fuel mixture produced is correct. Let's verify that we have the right hydrocarbon content:

\[48*0.65+32*0.9 = 60\]\[31.2+28.8 = 60\]\[60=60\]So the final mixture also contains the right amount of hydrocarbon content. Our work is done.

Any questions about anything we did here?