SOLUTION: A grocer mixes two kind of nuts One kind costs $5.00/kg and the other $5.80/kg. How many kilograms of each type are needed to make 40 kg of a blend worth $5.50/kg

Question 713402: A grocer mixes two kind of nuts One kind costs $5.00/kg and the other $5.80/kg. How many kilograms of each type are needed to make 40 kg of a blend worth $5.50/kg

Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!
You have a general question, someone mixes two kinds of materials. One has a lower given strength and the other has a higher given strength. How much of each material must be mixed to get a certain amount of mixture of a particular intermediate strength.

This "strength" used in that general question may be cost, price, money, percent, some other kind of concentration unit.

Here deals with this grocer's problem:

ASSIGN VARIABLES TO ALL QUANTITIES
L, the lower concentration of available material, $5.00/kg
H, the higher concentration of available material, $5.80/kg
T, the target concentration of resulting mixture, $5.50/kg
x, the amount of lower concentrated material to use, unknown
y, the amount of higher concentrated materaial to use, unknown
M, amount of the resulting mixture, 40 kg.

That can be partly changed by multiplying both sides by M to get Lx+yH=TM.
The mass or material sum equation will also be used for completing the system.

This one easily allows us to substitute for either x or for y in the rational equation to solve for just one variable and get its value.

, and since the two differences here would both be negative,
we can multiply by (-1)/(-1),

In brief summary, this general two part mixture problem can be well solved all symbolically using the above process, leading to the use of these two formulas:

and .
Substitute the given values to find x, and then find the value for y, using these two final formulas.

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