concentrate percentage

Question

.3 ounces of a liquid at 40% concentration is mixed with 8 ounces of a different solution that is 100% pure, what does the concentrate level of the .3 ounce liquid become after mixing? i need to know what the concentrate level is of the original 40% solution. please and thank you!

Answer 1

This is a classic mixture problem where the formula that needs used is

 

Amount*Cost=Value       (here cost is replaced by percentage)

 

So, for the first liquid, it's value is .3*.40=.12      (.40=40%)

 

the value of the second liquid is 8*1=8        (1=100%)

 

Now, the sum of these values must equal the value of the mixture.

 

For the mixture, we know that the amount is .3+8=8.3  (since you combined 2 liquids)

However, we do not know the percent concentration of the mixture. Call it "x". Then, since we know that the sum of the values of the individual liquids equals the value of the mixture, we have

 

.12+8=8.3x

8.12=8.3x

 

x=.978=97.8%

 

This intuitively makes sense since you only had .3 oz of 40% solution that you are mixing a relatively huge amount (8 oz) of 100% solution.