Question

A coffee mixture has beans that sell for $0.20 a pound and beans that sell for $0.68. If 120 pounds of beans create a mixture worth $0.54 a pound, how much of each bean is used?

Answer 1

`$0.68` type `-> 85^("lb")`

`$0.20` type `-> 35^("lb")`

Explanation:

One of the better ways to tackle this problem type is to consider them as a set of fraction that sum to the value of 1. Where 1 represents the whole of the final blend.

`color(blue)("Determine the fractional proportion of each type of bean")`

Let the proportion of the $0.20 be `x`

Then the proportion of the $0.68 is `1-x`

Set the target at $0.54

Dropping the unit of measurement ($ ) for now we have

`0.20x+(1-x)0.68=1xx0.54`

`0.20x-0.68x+0.68=0.54`

`-0.48x+0.68=0.54`

Lets get rid of the decimals for now and multiply everything by 100

`-48x+68=54`

`48x=68-54`

`x=14/48`

`x=7/24` of the blend at $0.20

So the proportion of the $0.68 is `1-7/24=17/24`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the weight proportion of each bean type")`

`$0.68` type `->17/24xx120^("lb") = 85^("lb")`

`$0.20` type `->7/24xx120^("lb")= 35^("lb")`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Check:

`(35xx0.2)+(85xx0.68)=$64.8`

`120xx0.54 = $64.8`

Answer 2

Method 2 of 2

An amazing approach that is not commonly used. Very fast once used to it.

`85^("lb")" of $0.68 beans"`
`35^("lb")" of $0.20 beans"`

Explanation:

This uses the principle of a straight line graph.

Let the weight of the $0.20 bean be `w_(0.2)`
Let the weight of the $0.68 bean be `w_(0.68)`

Then by considering only say `w_0.68` the value of `w_(0.2)` is directly inferred by `w_(0.2)=120-w_(0.68)`

So it is possible to determine the part of the answer by considering just `w_(0.68)`

If all `w_(0.68)` total value is `................" "120xx$0.68 = $ 81.60`
If all `w_(0.2)` total value is `................." "120xx$0.20=$24.00`
If all the target blend total value is `." "120xx$0.54 = $64.80`

Plotting this on the graph we have:

The slope of all is the same as the slope of a part of it

`color(green)("Slope of all "->("change in y")/("change in x") ->(81.60-24.00)/(120)=57.6/120)`

`color(red)("Slope of part: "(64.8-24.00)/x=40.8/xcolor(green)(=57.6/120))`

`x=40.8xx120/57.6 = 85^("lb") = w_(0.68)`

So we have:
`85^("lb")" of $0.68 beans"`
`120-85=35^("lb")" of $0.20 beans"`