Respuesta :
The percentage of alcohol of a solution i is given by the quotient:
[tex]p_i=\frac{v_i}{V_i},_{}[/tex]where v_i is the volume of alcohol in the solution i and V_i is the volume of the solution i.
From the statement of the problem we know that:
1) Solution A has 10% of alcohol, i.e.
[tex]p_A=\frac{v_A_{}}{V_A}=0.1.\Rightarrow v_A=0.1\cdot V_A.[/tex]2) Solution B has 60% of alcohol, i.e.
[tex]p_B=\frac{v_B}{V_B}=0.6\Rightarrow v_B=0.6\cdot V_B.[/tex]3) The volume of solution A is V_A = 200ml.
4) The resulting mixture must have a percentage of 40% of alcohol, so we have that:
[tex]p_M=\frac{v_M}{V_M}=0.4.[/tex]5) The volume of the mixture v_M is equal to the sum of the volumes of alcohol in each solution:
[tex]v_M=v_A+v_{B\text{.}}_{}[/tex]6) The volume of the mixtureVv_M is equal to the sum of the volumes of each solution:
[tex]V_M=V_A+V_B\text{.}[/tex]7) Replacing 5) and 6) in 4) we have:
[tex]\frac{v_A+v_B}{V_A+V_B_{}}=0.4_{}\text{.}[/tex]8) Replacing 1) and 2) in 7) we have:
[tex]\frac{0.1\cdot V_B+0.6\cdot V_B}{V_A+V_B}=0.4_{}\text{.}[/tex]9) Replacing 3) in 8) we have:
[tex]\frac{0.1\cdot200ml_{}+0.6\cdot V_B}{200ml_{}+V_B}=0.4_{}\text{.}[/tex]Now we solve the last equation for V_B:
[tex]\begin{gathered} \frac{0.1\cdot200ml+0.6\cdot V_B}{200ml_{}+V_B}=0.4_{}, \\ \frac{20ml+0.6\cdot V_B}{200ml_{}+V_B}=0.4_{}, \\ 20ml+0.6\cdot V_B=0.4_{}\cdot(200ml+V_B), \\ 20ml+0.6\cdot V_B=80ml+0.4\cdot V_B, \\ 0.6\cdot V_B-0.4\cdot V_B=80ml-20ml, \\ 0.2\cdot V_B=60ml, \\ V_B=\frac{60}{0.2}\cdot ml=300ml. \end{gathered}[/tex]We must use 300ml of Solution B to have a 40% alcohol solution as the resulting mixture.
Answer: 300ml of Solution B.
