A mixture contains wine and water in the ratio 3:2 and another mixture contain them in the ratio 4:5. How many liters of latter must be mixed with 3 liters of the former so that the resulting mixture may contain equal quantities of wine and water?
Answer & explanation
Correct answer: option 1
Wine = 3 × \(\frac{3}{5}\) + n × \(\frac{4}{9}\) = \(\frac{9}{5}\) + \(\frac{4n}{9}\)
Water = 3 × \(\frac{2}{5}\) + n × \(\frac{5}{9}\) = \(\frac{6}{5}\) + \(\frac{5n}{9}\)
\(\frac{\frac{9}{5} + \frac{4n}{9}}{\frac{6}{5} +\frac{5n}{9}}\) = \(\frac{1}{1}\)
\(\frac{9}{5}\) + \(\frac{4n}{9}\) = \(\frac{6}{5}\) + \(\frac{5n}{9}\)
\(\frac{5n}{9}\) - \(\frac{4n}{9}\) = \(\frac{9}{5}\) - \(\frac{6}{5}\)
\(\frac{n}{9}\) = \(\frac{3}{5}\)
⇒ n = \(\frac{27}{5}\)