Question

Soybean meal is 16% protein; commealis 8% protein. How many pounds of each should be mixed together in order to get 320-b mixture that is 14% protein. How many pounds of the cornmeal should be in the mixture?

Answer 1

`"Cornmeal " -> 80lb`

Explanation:

`color(blue)("Initial condition")`

Let the amount of Soybean be `s`
Let the amount of Cornmeal be `c`

Then `c+s=320lb" "=>" "color(red)(s=320-c)` .....Equation(1)

The target is 14% protein `-> 14/100xx320 = 44.8lb`

Equating the protein content we have:

`16/100s+8/100c=44.8lb...........Equation(2)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the value of c")`

Substitute for `s` in equation(2) using equation(1)

`color(green)(16/100color(red)(s)+8/100c=44.8" "->" "16/100(color(red)(320-c))+8/100c=44.8)`

`51.2-16/100c+8/100c=44.8`

`51.2-8/100c=44.8`

`8/100c=51.2-44.8 = 6.4`

`c=100/8xx6.4=80lb`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("As a check-Determine the value of s")`

Using equation(1)

`s=320-c" "->" "s=320-80 = 240lb`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Check")`

`16/100s+8/100c=44.8lb`

Consider the left hand side

`16/100s+8/100c" " ->" " (16/100xx240)+(8/100xx80)`

`LHS-> 38.4+6.4=44.8 -> RHS`

Answer 2

Another method

cornmeal`= 80lb`

Explanation:

`color(blue)("Introduction to the method")`

Given:

Soybean protein content = 16%
Cornmeal protein content = 8%
Target protein content =14%

Total weight of blend to be 320lb

There is a direct relationship in the blend between the two foods

If you have some weight of soybean the amount of cornmeal is 320 - weight of soybean. So by considering the weight of just one of the foods you are by default implying the weight of the other.

If you have no soybean that it is all cornmeal at 8% protein
If you have all soybean at 16% protein then there is no cornmeal
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determining the amount of soybean in the mix")`

The gradient of all the slope is the same as the gradient of part of the slope.

By ratio

`(xlb)/(14%-8%)=(320lb)/(16%-8%)`

`xlb=(320lbxx6cancel(color(white)(.)%))/(8cancel(color(white)(.)%)) = 240lb" soybean"`

So cornmeal `=320-240=80lb`