# Thread: Mixture Problem

1. ## Mixture Problem

I encountered the following problem:

10 grams of water is added to a 30% solution of barium chloride to dilute the solution to 20%. How many grams of the original solution did we start with?

The answer is 20 grams calculated as follows:

Let x = the number of grams of the original solution.

(0.20)(10+x) = 0.30x
x = 20

I then asked myself this question: How would the solution to this problem be changed if, instead of adding 10 grams of water to dilute the solution to 20%, 10 grams of barium chloride were added to strengthen the solution to 40%?

It seems to me that the only thing that would change would be that 0.20 would be changed to 0.40. That results in a negative value for x. What am I missing?

Thanks for any input.

... doug

2. ## Re: Mixture Problem

Originally Posted by ddjolley
I encountered the following problem:

10 grams of water is added to a 30% solution of barium chloride to dilute the solution to 20%. How many grams of the original solution did we start with?

The answer is 20 grams calculated as follows:

Let x = the number of grams of the original solution.

(0.20)(10+x) = 0.30x
x = 20

I then asked myself this question: How would the solution to this problem be changed if, instead of adding 10 grams of water to dilute the solution to 20%, 10 grams of barium chloride were added to strengthen the solution to 40%?

It seems to me that the only thing that would change would be that 0.20 would be changed to 0.40. That results in a negative value for x. What am I missing?

Thanks for any input.

... doug
when you add water, it is a 0% concentration of the substance

10(0%) + x(30%) = (10 + x)(20%)

30x = 200 + 20x

x = 20 g

when you add 10 g of pure substance ...

10(100%) + x(30%) = (10 + x)(40%)

1000 + 30x = 400 + 40x

600 = 10x

x = 60 g

3. ## Re: Mixture Problem

Ah!!! I knew that it was something stupid like that! Thanks so much.

... doug