A mixture contains ethanol and water in the ratio 4:3. If 5 litres of water is added to the mixture, the ratio becomes 4:5. Find the quantity of ethanol in the mixture.

Steps to Solve

1. Define the initial quantities Assume the initial quantity of ethanol is $E$ and the initial quantity of water is $W$.

2. Set up the initial ratio If the initial ratio of ethanol to water is given as - say 3:2 - then we can express it as: $$ \frac{E}{W} = \frac{3}{2} $$ From this ratio, we can express $W$ in terms of $E$: $$ W = \frac{2}{3} E $$

3. Add water to the mixture Let’s say we add a certain amount of water, denoted as $A$, to the mixture. The new quantity of water is: $$ W_{new} = W + A $$

4. Set up the new total ratio The new ratio of ethanol to water will now change. The new ratio is: $$ \frac{E}{W_{new}} = \frac{3}{5} $$ Plugging in the expression for $W_{new}$ we found earlier: $$ \frac{E}{W + A} = \frac{3}{5} $$

5. Solve for the new quantities Substitute $W = \frac{2}{3}E$ into the equation: $$ \frac{E}{\frac{2}{3}E + A} = \frac{3}{5} $$ Cross-multiply to solve for $A$: $$ 5E = 3\left(\frac{2}{3}E + A\right) $$ Distributing the 3 gives: $$ 5E = 2E + 3A $$ Now, isolate $A$: $$ 3A = 5E - 2E $$ Thus: $$ 3A = 3E $$ And finally, we can divide by 3: $$ A = E $$