Question 60

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

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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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