A mixture contains wine and water in the ratio 3 : 2 and another mixture contains them in the ratio 4 : 5. How many litres of the latter must be mixed with 3 litres of the former so that the resulting mixture may contain equal quantities of wine and water = ?

Solution(By Examveda Team)

The first mixture contains wine and water in the ratio of 3 : 2
Wine in 3 liters of mixture = $$\frac{3}{5} \times 3 = \frac{9}{5}$$
Water in 3 liters of mixture = $$\frac{2}{5} \times 3 = \frac{6}{5}$$

Let us consider that the first mixture is mixed with the second mixture that has quantity as 9x liters (4x liters of wine and 5x liters of water).

After mixing,
The total quantity of wine = Total quantity of water
$$\eqalign{ & \Rightarrow \frac{9}{5} + 4x = \frac{6}{5} + 5x \cr & \Rightarrow x = \frac{9}{5} - \frac{6}{5} \cr & \Rightarrow x = \frac{3}{5} \cr} $$

Second Mixture required = 9x = $$9 \times \frac{3}{5}$$ = $$5\frac{2}{5}{\text{ litres}}$$