Question

How many pounds of gourmet candy selling for $2.20 per pound should be mixed with 6 pounds of gourmet candy selling for $1.20 per pound to obtain a mixture selling for $1.60 per pound?

Answer 1

`color(green)("$1.20 candy amount is "6"lb"`
`color(green)("$2.20 candy amount is "4"lb"`

Explanation:

There are two ways of solving this.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Method 1")`

`color(Brown)("Straight line graph approach")`

Standard form equation`" "->y=mx+c`

In this case:`" "c=1.2"; "m=(2.20-1.20)/100 = 1/100"; "y=1.60`

`1.6=x/100+1.2`

`x=(1.6-1.2)xx100" "color(purple)(-> x=0.4xx100)`

`x=40-> 40%`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Method 2")`

`color(Brown)("Ratios, which is the same thing as method 1 just that this fact is in disguise.")`

The approach is based on the principle that the gradient is constant.

`100/(2.2-1.2)=x/(1.6-1.2)`

Basically this is saying that the gradient of the whole is the same gradient as part of it!

`100/1=x/0.4`

`color(purple)(0.4xx100=x)`

`x=40" " -> 40%`

`color(green)('"NOTE THAT IS 40% OF THE $2.20 CANDY")`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the weights of each constituent")`

so `100%-40% =60%` as the proportion of the $1.20 candy

Let the whole weight be `w` then

`60/100 w=6`

`w=(6xx100)/60" "=" "100/10" "=" "10lb`

So

`color(green)("$1.20 candy amount is "6"lb"`
`color(green)("$2.20 candy amount is "10-6 = 4"lb"`