Question #c4c19

1 Answer
Mar 13, 2017

Answer:

The question is has an error: I am assuming that the last % written should be 60%. It makes sense as this is the target concentration of the blend (2nd % mentioned)

Explanation:

Percentage of a percentage is a little tricky to get your head round.

If you have all of one solution then you have none of the other. This is also true the other way round.

So the two are linked and can be represented by considering just one of them. This can be plotted against the resulting mixture concentration as you change the amount of the one in you are focusing on.

I elect to observe the change in concentration as you increase the proportion of the stronger solution. Thus we have:

RTony

The slope for all is the same as the slope for part.

#x/(60-50) =100/(75-50)#

#x=1000/25 = 40#

So the blend will be #40/100# of 75% solution

and #60/100# of 50% solution.
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Check:

#(40/100xx75%)+(60/100xx50%) = 30%+30%=60%#
#color(red)(" "uarr" "uarr)#
#color(red)(" ? fluid ounces 60 fluid onces")#

All as required
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So the proportion of #60/100# is the 60 fluid ounces she starts with.

Let the final volume of the blend be #v#

Thus the volume of the final blend is such that

#60/100v=60" fluid ounces"#

Thus #v=60xx100/60 = 100" fluid ounces"#

#color(blue)("I will let you take it from this point as your question's intent is not clear")#