Question 21
1 pts
How many milliliters of a $5\mathrm{\%}$ acid solution and how many milliliters of a $17\mathrm{\%}$ acid solution must be mixed to obtain $60\mathrm{mL}$ of a $13\mathrm{\%}$ acid solution?
$40\mathrm{mL}$ of $5\mathrm{\%}$ solution and $20\mathrm{mL}$ of $17\mathrm{\%}$ solution
$15\mathrm{mL}$ of $5\mathrm{\%}$ solution and $45\mathrm{mL}$ of $17\mathrm{\%}$ solution
$20\mathrm{mL}$ of $5\mathrm{\%}$ solution and $40\mathrm{mL}$ of $17\mathrm{\%}$ solution
$30\mathrm{mL}$ of $5\mathrm{\%}$ solution and $30\mathrm{mL}$ of $17\mathrm{\%}$ solution
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Step 1 ：Let's denote the volume of the 5% solution as $x$ (in mL) and the volume of the 17% solution as $y$ (in mL).

Step 2 ：We know that the total volume of the solution is 60 mL, so we can write the first equation as: $x+y=60$.

Step 3 ：The total amount of acid in the final solution is 13% of the total volume, or $0.13\times 60=7.8$ mL. The amount of acid in the 5% solution is $0.05x$ and in the 17% solution is $0.17y$. So we can write the second equation as: $0.05x+0.17y=7.8$.

Step 4 ：Let's multiply the first equation by 0.05: $0.05x+0.05y=3$.

Step 5 ：Subtract this equation from the second equation: $0.12y=4.8$.

Step 6 ：Solving for $y$ gives: $y=\frac{4.8}{0.12}=40$ mL.

Step 7 ：Substituting $y=40$ into the first equation gives: $x+40=60$.

Step 8 ：Solving for $x$ gives: $x=60-40=20$ mL.

Step 9 ：$\boxed{{\displaystyle 20}}$ mL of a 5% acid solution and $\boxed{{\displaystyle 40}}$ mL of a 17% acid solution must be mixed to obtain 60 mL of a 13% acid solution.