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20 Apr 2007, 17:21
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Heres a problem,
Aribica coffee costs .4 per ounce, robusta .3 per ounce. if the blend of both aribica and robusta costs .33, what is the share of arabica in this blend.
The solution gives the equations,
let A= ounces Arib, R= ounces Robusta
then,
(A*.4+R*.3)/(A+R)= .33
my question is why do you set up this type of equation, whats the principle? I just figured out combinations that would work, which I'm sure is the long way. So, I'm sure this eqn is more efficient, but I'm not sure why its being set up.
TIA.
Aribica coffee costs .4 per ounce, robusta .3 per ounce. if the blend of both aribica and robusta costs .33, what is the share of arabica in this blend.
The solution gives the equations,
let A= ounces Arib, R= ounces Robusta
then,
(A*.4+R*.3)/(A+R)= .33
my question is why do you set up this type of equation, whats the principle? I just figured out combinations that would work, which I'm sure is the long way. So, I'm sure this eqn is more efficient, but I'm not sure why its being set up.
TIA.
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20 Apr 2007, 18:03
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ayl989
Let the amt of Arabica coffee in 1 ounce of Blend be x
Let the amt of Arabica coffee in 1 ounce of Blend be y
->x+y =1 .....(1)
Cost of 1 ounce of arabica coffee is .4
Therefore cost of x ounce of arabica cofee is .4x
Cost of 1 ounce of Robustica coffee is .3
Therefore cost of y ounce of arabica cofee is .4y
Given :cost of blend/ounce = .33
->.4x+.3y = .33 ....(2)
Now solve equstions 1 and 2 to get values of x and y
x = .3
y =.7
20 Apr 2007, 18:27
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goalsnr
But it is not specified that cost of 1 ounce of blend is 0.33
Is it ok to assume that its per ounce price?
