A party mix is made by adding nuts that sell for ​$2.50per kg to a cereal mixture that sells for ​$11 per kg. How much of each should be added to obtain 30 kg of a mix that will sell for ​$2.10per​ kg?
If the mix has x kg of the $2.50 nuts, then the rest (30-x) is $11 nuts. Now just add up the value of each part, and they must add up to the total:
2.50x + 11(30-x) = 2.10*30
now solve for x ...
To solve this problem, we'll use a system of equations. Let's assume that x represents the amount of nuts in kg, and y represents the amount of cereal mixture in kg.
1. We can set up an equation based on the given information about the selling price of the mix:
2.10(x + y) = 30 * 2.10
2. The next equation can be created from the information about the cost per kg of the ingredients:
2.50x + 11y = 30 * 2.10
Now, let's solve the system of equations to find the values of x and y.
First Equation:
2.10(x + y) = 30 * 2.10
2.10x + 2.10y = 63
Second Equation:
2.50x + 11y = 30 * 2.10
2.50x + 11y = 63
We have the same value for the left side of both equations. Since the coefficients of y in both equations are the same, we can eliminate the variable y by subtracting the second equation from the first equation:
(2.10x + 2.10y) - (2.50x + 11y) = 63 - 63
-0.40x - 8.90y = 0
Simplifying, we get:
-0.40x - 8.90y = 0 -> -4x - 89y = 0
Now we can solve for one variable in terms of the other:
-4x - 89y = 0
-4x = 89y
x = -89y/4
x = -22.25y
Since x represents the amount of nuts in kg, it cannot be negative. Therefore, y must be 0. We can substitute y = 0 back into the equation to find x:
x = -22.25 * 0
x = 0
This implies that the amount of nuts and the amount of cereal mixture needed to achieve a $2.10 per kg mix is 0 kg of nuts and 0 kg of cereal mixture.
In summary, according to the given information, no nuts or cereal mixture is needed to obtain 30 kg of a mix that will sell for $2.10 per kg.