Find the 3 equations and solve

cjebens

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A coffee wholesaler blends together three types of coffee that sell for $8.80, $9.20, and$10.40 per kilogram respectively. When the blending process is complete, the wholesaler has100 kilograms of coffee worth $9.60 per kilogram. If the wholesaler uses the same amount ofthe two higher-priced coffees, how much of each type must be used in the blend?
 
A coffee wholesaler blends together three types of coffee that sell for $8.80, $9.20, and $10.40 per kilogram, respectively. When the blending process is complete, the wholesaler has 100 kilograms of coffee worth $9.60 per kilogram. If the wholesaler uses the same amount of the two higher-priced coffees, how much of each type must be used in the blend?
I've moved this question out of your original choice of "Differential Equations". You may be in a differential-equations class now, having completed trigonometry and calculus, but this exercise does not require anything past what you'd learned back in pre-calc algebra. So just go ahead and use that. ;)
 
Coffe!

Let the amount used for each coffee be a,b,c resp.

Then a + b + c = 100 Equation #1
But we are told

a = b ............ Eqn #2

So 2a +c = 100 is our new Eqn #1

The total cost for 100 Kg is
(I will use * to mean multipy)

8.80*a + 9.20*b + 10.40*c

But a = b

Total cost = (8.80+ 9.20)*a + 10.40*c

But we are told the total cost per Kg so we must divided by 100

Total Cost per Kg

9.60 = (8.80+ 9.20)*a + 10.40*c Eqn #3

IN summary we have two equations and two unknowns a,c

2a +c = 100 Eqn #1
(8.80+ 9.20)*a + 10.40*c = 9.60 Eqn #3


From Eqn #1 c = (100 - 2a)

So substitute for c in eqn #3 and solve for a

(8.80+ 9.20)*a + 10.40*(100 – 2a) = 9.60

But b = a so we now have b also

And from Eqn #1 we can now get c

Check my maths. The method is sound.
 
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