A mixture contains alcohol and water in the ratio of 4:3. If 7 liters of water…

Concept

In a ratio mixture problem, the ratio terms are not the actual amounts — they are multiples of a common unit. If two quantities are in the ratio a : b, their real amounts are a·k and b·k for some positive multiplier k. Adding a fixed amount to one quantity changes only that quantity; the unchanged quantity keeps the same a·k value, and you solve for k by equating the new ratio.

Application

1. Let the common multiplier be k, so alcohol = 4k litres and water = 3k litres.

2. Adding 7 litres of water leaves alcohol unchanged at 4k and makes water 3k + 7. The new ratio of alcohol to water is 3 : 4, giving the equation 4k / (3k + 7) = 3 / 4.

3. Cross-multiply: 4 × 4k = 3 × (3k + 7), i.e. 16k = 9k + 21.

4. Subtract 9k from both sides: 7k = 21, so k = 3.

5. Quantity of alcohol = 4k = 4 × 3 = 12 litres.

Cross-check

With k = 3 the original amounts are alcohol 12 and water 9 (ratio 12 : 9 = 4 : 3, correct). After adding 7 litres of water, water becomes 16, and 12 : 16 simplifies to 3 : 4 — exactly the stated new ratio. The answer 12 litres is consistent.