An alcohol brand BE contains alcohol and water in ratio 3:5, and brand RE contains alcohol and water in ratio 4:1. In what ratio these two should be mixed such that the mixture has equal quantity of water and alcohol?
Solution
Let the volume of BE be 8m and volume of RE be 5n.
The concentration of alcohol in the mixture is $$\frac{\left(3m+4n\right)}{\left(8m+5n\right)}=\frac{1}{2}$$
6m+8n = 8m+5n
3n = 2m
m:n = 3:2
We need 8m:5n = $$\frac{8}{5}\cdot\frac{3}{2}\ =\frac{12}{5}$$
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