two different solutions of alcohol with respective proportions of alcohol to water of 3:1 and 2:3 were combined. what is the concentration of alcohol in the new solution if the first solution was 4 times the amount of the second solution?

a. 17%
b. 25%
c. 34%
d. 50%
e. 68%

Question

two different solutions of alcohol with respective proportions of alcohol to water of 3:1 and 2:3 were combined. what is the concentration of alcohol in the new solution if the first solution was 4 times the amount of the second solution?

a. 17%
b. 25%
c. 34%
d. 50%
e. 68%

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [], "new_concepts": [ "Mixture Problems", "Ratio and Proportion" ], "current_concepts": [ "Mixture Problems", "Ratio and Proportion" ] } ## Determine alcohol concentrations of both solutions $$ \begin{aligned} &\text{Solution 1 ratio (alcohol:water)} = 3:1 \implies C_1 = \frac{3}{3+1} = \frac{3}{4} = 0.75\ &\text{Solution 2 ratio (alcohol:water)} = 2:3 \implies C_2 = \frac{2}{2+3} = \frac{2}{5} = 0.40 \end{aligned} $$ ## Set up weighted average concentration $$ \begin{aligned} &V_1 = 4V_2\ &C_{\text{new}} = \frac{C_1 V_1 + C_2 V_2}{V_1 + V_2} = \frac{0.75(4V_2) + 0.40(V_2)}{4V_2 + V_2} \end{aligned} $$ ## Calculate final concentration $$ \begin{aligned} &C_{\text{new}} = \frac{3.0V_2 + 0.4V_2}{5V_2} = \frac{3.4}{5} = 0.68 = 68\% \end{aligned} $$

A. 17%

B. 25%

C. 34%

D. 50%

E. 68%

{ "subject": "Mathematics", "question_type": "Multiple Choice", "knowledge_point": [ "Mathematics", "Algebra", "Mixture Problems" ] }

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QUESTION IMAGE

Question

Explanation:

Determine alcohol concentrations of both solutions

Set up weighted average concentration

Calculate final concentration

Answer: