How many ounces of iodine worth 30 cents an ounce must be mixed with 50 ounces of iodine worth 18 cents an ounce so that the mixture can be sold for 20 cents an ounce?

If we sell #30# cents an ounce for #20# cents an ounce, we will be losing #10# cents every ounce sold.

If we sell #18# cents an ounce for #20# cents an ounce, we will be making #2# cents every ounce sold.

In order for us to not lose or gain any money, we would need to sell #5# times more #18# cents iodine than #30# cents iodine in order to balance it out.

So, if there are #50# ounces of #18# cents iodine,

#50/5=10#

There will be #10# ounces of #30# cents iodine.

Check:

#10# ounces of #30# cents iodine #rarr 300# cents
#50# ounces of #18# cents iodine #rarr 900# cents

#60# ounces of #20# cents iodine #rarr 1200# cents

#10+50=60#
#300+900=1200#

Rather than write '30 cents iodine' or '18 cents iodine' each time do the following:

Let the iodine worth 30 cents be represented by #I_30 #
Let the iodine worth 18 cents be represented by #I_18#

It is given that we start with #50 " ounces of "I_18 -> 50I_18#

Let the amount (in ounces) of #color(red)(I_30" be "x ->xI_30)#

We calculate that the value of #50I_18 = 50xx18=900# cents

So the final value of the blend is such that we have a total cost of:

#("ounces "xx" cost") +("ounces "xx" cost")=("ounces"xx"cost")#

#ubrace((50I_18color(white)(".d")xxcolor(white)("dd")18))color(white)("dd")+color(white)("dd") ubrace((color(red)(xI_30)xx30))color(white)("ddd") =color(white)("ddd")ubrace((50+x)xx20) #

#"Total cost of "I_18color(white)("dd")"Total cost of "I_30color(white)("ddd")"Total cost of blend"#

At the moment all we are interested is #x# so lets drop the #I_18 and I_30# designation giving:

#(50xx18)+(30x)=1000+20x#

#900+30x=1000+20x#

#10x=100#

#x=10#

So #x= 10" ounces of "I_30#