Two vessels A and B contain alcohol and water in the ratio 7 : 5 and 17 : 7 respectively. In what ratio should the mixtures from two vessels A and B be mixed to get a new mixture containing alcohol and water in the ratio 5 : 3?
Solution
Correct Answer: Option B
When we form a new mixture by adding two mixtures in each other, then we will solve the question by considering the part of only one thing in the whole mixture –
First method:
Alcohol in vessel \(A = \frac{7}{{7 + 5}} = \frac{7}{{12}}\)
Alcohol in vessel \(B = \frac{{17}}{{17 + 7}} = \frac{{17}}{{24}}\)
Required alcohol in the new mixture \(= \frac{5}{{5 + 3}} = \frac{5}{8}\)
Now, required of the two mixtures would be –
\(\Rightarrow \frac{{first\ mixture}}{{Second\ mixture}} = \frac{{\frac{5}{8} \sim \frac{{17}}{{24}}}}{{\frac{5}{8} \sim \frac{{7}}{{12}}}} = \frac{{\frac{1}{{12}}}}{{\frac{1}{{24}}}} = \frac{2}{1}\)
Hence, required ratio of the two mixtures would be 2:1
Second method:
Water in vessel \(A = \frac{5}{{7 + 5}} = \frac{5}{{12}}\)
Water in vessel \(B = \frac{7}{{17 + 7}} = \frac{7}{{24}}\)
Required water in new mixture \(= \frac{3}{{5 + 3}} = \frac{3}{8}\)
Now required ratio of the two mixtures would be
\(\Rightarrow \frac{{first\ mixture}}{{Second\ mixture}} = \frac{{\frac{7}{{24}} \sim \frac{3}{8}}}{{\frac{5}{{12}} \sim \frac{3}{8}}} = \frac{{\frac{1}{{12}}}}{{\frac{1}{{24}}}} = 2:1\)
Hence required in the new mixture would be 2:1
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