This is the solution to a new algebra mixture problem as asked by Califax who asked, "A 50-lb solution of acid and water is 20% (by weight). How much pure acid must be added to this solution to make it 30% acid?"

1/5 = 20 % ..... 3/10 = 30 % x = the unknown amount in lbs we want to add of 100% or 1/1 or 1 or just x of pure acid.

50(1/5) + x = 3/10 (50 + x) multiply by 10...

100 + 10x = 3(50 + x)

100 + 10x = 150 + 3x

7x = 50 = 50/7 = 7 and 1/7th lbs of pure acid needs to be added to make the solution 30% acid.

## Sunday, October 9, 2011

### Mixture Problem 7 Solution

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Hey man, what's with this

ReplyDelete7x = 50 = 50/7 so how come

50 = 50/7 ????? Like wow man, if you get $50

suddenly it is worth $50/7 , like that's nasty!

If you could explain how and why this happened

some insight could be gained.