In a 40 litre mixture of alcohol and water, the ratio of alcohol to water is 5 : 3. If 20% of this mixture is removed and then the same volume is replaced with pure water, what will be the ratio of alcohol to water in the final mixture?

Introduction:
This mixture problem involves an initial alcohol water mixture, from which a percentage of the mixture is removed and then replaced with pure water. The aim is to find the final ratio of alcohol to water after this operation. It tests understanding of proportional removal and replacement and basic ratio manipulation.

Given Data / Assumptions:

• Initial volume of mixture = 40 litres.
• Initial ratio of alcohol to water = 5 : 3.
• 20% of the mixture is removed.
• The same volume is replaced with pure water.
• We must determine the final ratio of alcohol to water.

Concept / Approach:
First, we compute the initial amounts of alcohol and water using the ratio 5 : 3. Then we find the volume removed (20% of 40 litres) and calculate how much alcohol and water are removed in the same ratio. After removal, we add back the removed volume as pure water, which increases water but does not add alcohol. Finally, we find the ratio of alcohol to water in the new mixture.

Step-by-Step Solution:
Step 1: Initial total mixture = 40 litres, with alcohol : water = 5 : 3. Step 2: Total parts = 5 + 3 = 8. Step 3: Alcohol initially = 40 * 5 / 8 = 25 litres. Step 4: Water initially = 40 * 3 / 8 = 15 litres. Step 5: 20% of mixture is removed, so removed volume = 20 / 100 * 40 = 8 litres. Step 6: The removed 8 litres have alcohol and water in ratio 5 : 3. Step 7: Alcohol removed = 8 * 5 / 8 = 5 litres, water removed = 8 * 3 / 8 = 3 litres. Step 8: After removal, alcohol left = 25 - 5 = 20 litres. Step 9: After removal, water left = 15 - 3 = 12 litres. Step 10: Now 8 litres of pure water are added back to restore the volume to 40 litres. Step 11: Alcohol remains 20 litres; water becomes 12 + 8 = 20 litres. Step 12: Final ratio of alcohol to water = 20 : 20 = 1 : 1.

Verification / Alternative check:
We can verify the logic quickly: the total volume stays 40 litres. The fraction of alcohol after removal is still 25 / 40 = 5 / 8, and 8 litres out of 40 are removed, so the multiplicative factor for alcohol is (1 - 8 / 40) = 4 / 5, giving final alcohol 25 * 4 / 5 = 20 litres. The final water must then be 40 - 20 = 20 litres, again giving a 1 : 1 ratio. This confirms the earlier calculation.

Why Other Options Are Wrong:
2 : 1 and 3 : 1 would mean alcohol is still much higher than water, which contradicts the effect of adding pure water after removal.
1 : 2 would indicate water dominates, but our calculation shows equal amounts.
5 : 3 is the original ratio and does not reflect the change caused by the replacement operation.

Common Pitfalls:
Many students mistakenly remove 20% only from alcohol or only from water instead of from the entire mixture. Others forget to account for the new water added, or they miscalculate the removed quantities by applying the ratio incorrectly. Another frequent error is to assume that the ratio stays fixed after removal, which is not true when pure water is added back.

Final Answer:
The ratio of alcohol to water in the final mixture is 1 : 1.