Respuesta :
Answer: 40 ml of 25% solution & 60 ml of 15% solution
Step-by-step explanation:
[tex]\left\begin{array}{c|c|c|l}&Quantity&Percent&Total\ Solution\\Solution\ 1&x&25\%=0.25&0.25x\\\underline{Solution\ 2}&\underline{\ 100-x\ }&\underline{15\%=0.15}&\underline{\qquad \quad 0.15(100-x)}\\Mixture&100&19\%=0.19&0.25x+0.15(100-x)\end{array}\right\\\\\\Mixture: 100(0.19)=0.25x+0.15(100-x)\\.\qquad \qquad \qquad 19\ \quad =0.25x+15-0.15x\\.\qquad \qquad \qquad 4\qquad =0.10x\\\\.\qquad \qquad \quad \dfrac{4}{0.10}\qquad =\dfrac{0.10}{0.10}x\\\\.\qquad \qquad \qquad 40\qquad =x[/tex]
Solution 1: x = 40
Solution 2: 100 - x --> 100 - 40 = 60
We need to find how each of each solution did the chemist mixed, and we will find that with a system of equations. We can conclude that the chemist used 40 mm^3 of the 25% solution and 60 mm^3 of the 15% solution.
Writing the system of equations.
First, we need to define the variables that we will use.
- x = mm^3 of the 25% acid solution.
- y = mm^3 of te 15% acid solution.
We know that at the end there are 100 mm^3 of the mixture then we must have:
x + y = 100.
We also know that the concentration of the final 100 mm^3 is 19%, then the concentration at the left must be the same than at the right, then we have:
x*0.25 + y*0.15 = 100*0.19 = 19
Where I multiplied each term by the correspondent concentration in decimal form.
Now we can see that our system of equations is:
x + y = 100
x*0.25 + y*0.15 = 19
To solve it we need to isolate one variable in one of the equations, I will isolate x on the first one:
x = 100 - y
Now we replace this on the other equation:
(100 - y)*0.25 + y*0.15 = 19
25 - y*0.10 = 19
-y*0.10 = 19 - 25 = -6
y = -6/-0.10 = 60
and:
x = 100 - y = 100 - 60 = 40
Then we can conclude that the chemist used 40 mm^3 of the 25% solution and 60 mm^3 of the 15% solution.
If you want to learn more about systems of equations, you can read:
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