Practicing Success
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? |
\(\frac{27}{5}\) ltr \(\frac{27}{6}\) ltr \(\frac{8}{9}\) ltr \(\frac{15}{4}\) ltr |
\(\frac{27}{5}\) ltr |
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 = \(\frac{27}{5}\)
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