Mixture problem for college algebra?
An octane number of 89 on gasoline means that it will fight engine "knock" as effectively as a reference fuel that is 89% isooctane, a type of gas. Suppose you want to fill an empty 20-gallon tank with some 89-octane gasoline and some 94-octane fuel to produce a mixture that is 92-octane. How much of each type of gasoline must you use?
If you could show how you did this it would be greatly appreciate. Thank you!
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Answers & Comments
you'll put in x gallons of 89 octane
and since there are 20 gallons total, the remaining 20 - x will be 94 octane
you know that this mixture will be 20 gallong of 92 octane
treating the octane as percentages...
the x-gallons of 89 octane will contain .89x gallons of isooctane
the 20 - x gallons of 94 octane will contain .94(20 - x) gallons of isooctane
together, the 20 gallons of 92 octane mixture will contain .92(20) gallons of isooctane
these must balance, so
.89x + .94(20 - x) = .92(20)
.89x + 18.8 - .94x = 18.4
.4 = .05x
x = .4 / .05 = 8
so 8 gallons of 89 octane (with .89(8) = 7.12 gallons isooctane
mixed with 20- 8 = 12 gallons of 94 octane (with .94(12) = 11.28
gives a 20 gallon mixture with .92 octane (.92(20) = 18.4)
and yes, 7.12 + 11.28 = 18.4
89x +94y =92*20, so
89x + 94y=1840
So if you mix 8 gallons at 89% (712) and 12 gallons at 94% (1128), you get 20 gallons at 92%
one million) 2 equations and a pair of unknowns equation one million) (.3*x + .5*y)/800 = .40 two equation 2) x + y = 800 remedy equation 2 for x or y and sub right into one million x = 800 - y .3*(800 -y) + .5y = .40 two*800 240 - .3y + .5y = 336 .2y = ninety six y = 480 sub into equation 2 x + 480 = 800 x = 320 2) Use the comparable approach as one million) equation one million) .40*x + 18*.24 = .28*y equation 2) x + 18 = y 3) comparable approach returned equation one million) 5*40 5 + 2.60*x = 3.40*y equation 2) 40 5 + x = y