tejal777 wrote:

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?

Track on alcohol; use water only to determine the percentage of alcohol in each solution.

Mixture A = .75 alcohol

\(\frac{3A}{1W}\): 4 parts total

Alcohol is 3 parts of 4, or

\(\frac{3}{4}\) = .75 of Mixture A

Mixture B = .40 alcohol

\(\frac{2A}{3W}\): 5 parts total

Alcohol is 2 parts of 5, or

\(\frac{2}{5}\) = .40 of Mixture B

Percent alcohol = concentration

\((Concen_{A})(Vol_{A}) + (Concen_{B})(Vol_{B})=\) \((Concen_{A+B})(Vol_{A+B})\)Mixture A has 4 times the volume of Mixture B. Let volume of A = 4 and volume of B = 1, where x = percentage of alcohol in final mixture

(.75)(4) + (.40)(1) = (x)(4+1)

3 + .4 = 5x

3.4 = 5x

x = \(\frac{3.4}{5} = \frac{34}{50} = \frac{68}{100} = x\)

x = 68 percent

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