A scientist has two acid solutions: Artesia, which is a $70\%$ acid solution, and Borromeo, which is a $20\%$ acid solution. (a) Find the amount of Artesia (in mL) that must be added to $500$ mL of Borromeo in order to produce a $60\%$ acid solution. (b) Find the amount of Artesia and Borromeo (in mL) that can be combined in order to form a $100$ mL solution that is $55\%$ acid. (c) Does there exist a combination of Artesia and Borromeo that is $80\%$ acid?

(a) 2000 mL

(b) 70 mL, 30 mL

(c) No

Step-by-step explanation:

Given that Artesia is a 70% acid solution, and Borromeo is a 20% acid solution.

(a) Let x mL of Artesia is added to 500  mL of Borromeo to produce a 60% acid solution.

So, 70% of x + 20% of 500 = 60% of (x+500)

mL.

(b) Let x mL of Artesia and y mL be combined in order to form a 100 mL solution having 55% acid solution.

So, x+y=100\Rightarrow y= 100-x

And, 70% of x + 20% of y = 55% of (100)\\\\

mL.

and y= 100-x= 100-70=30 mL.

(c)  Let x mL of Artesia and y mL of Borromeo be combined in order to make 80% acid solution.

And, 70% of x + 20% of y = 80% of (x+y)

As x and y are amount of Artesia and Borromeo, so x>0 and y>0

, so the negative value of x/y is not possible.

Ad the negative value of x/y is not so x/y=-6 is not possible, hence,  any combination of Artesia and Borromeo that is 80% acid solution is not possible.

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