dave13 wrote:

VeritasPrepKarishma wrote:

sameerkamath21 wrote:

VeritasPrepKarishma Hello Karishma. I recently read your blog posts on weighted averages and found the w1/w2 method and the scale method to be very useful, fast and accurate. So i've just started solving mixture problems using your techniques as much as possible. Will take sometime for me to master it. Could you please explain the how you can solve this sum using your method? Thanks

Right now the concentration of water in the mix is 4/8. We want to add pure water (8/8) to it and make the concentration of water in the final solution (5/8).

w1/w2 = (8/8 - 5/8)/(5/8 - 4/8) = 3/1

So for every 3 parts of the original mix, we should add 1 part of pure water.

The original mix is 8 quarts. So we should add 8/3 quarts of pure water.

Answer (D)

Hi

VeritasPrepKarishma,

thanks for the link you shared in the previous post

https://www.veritasprep.com/blog/2017/1 ... -averages/that confusion regarding nnomenclature is clear

now after reading the explanation which is followed by an example i have a question:

regarding this : w1/w2 = (8/8 - 5/8)/(5/8 - 4/8) = 3/1

where does it say that we need to add 8/8 pure water, it says 3 parts of alcohol and only 5 parts of water and not 8

please explain

Now another question, when you get this 3/1

you say " So for every 3 parts of the original mix, we should add 1 part of pure water.

The original mix is 8 quarts. So we should add 8/3 quarts of pure water. "

let me break it down:

"So for every 3 parts of the original mix" doesn't "3" in numerator refer to water?

"we should add 1 part of pure water" doesn`t "1" in denominator refer to alcohol ?

thank you:-)

Note that 8/8 is the concentration of water in water i.e. the water is fully water only and hence its concentration is 1. How many parts is the weight, not the concentration.

The concentration of water in the mix is 4/8 and final concentration of water is 5/8 so to ease our calculations, we write 1 as 8/8 (= 1).

The concentration of an ingredient (C1, C2 and Cavg) is expressed as Ingredient/Total (not Ingredient1/Ingredient2).

w1 and w2 are the weights of total solution (including both ingredients) in the two cases

w1/w2 = 3/1

So for every 3 parts of mix (water + alcohol), we put 1 part of water (pure water). When we do this, we will get water concentration of 5/8.

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