A jar contains a mixture of water and alcohol. Gopal takes out of the mixture and substitutes it by water of the same amount. The process is repeated once again. The percentage of water in the mixture is now
A jar contains a mixture of water and alcohol. Gopal takes out of the mixture and substitutes it by water of the same amount. The process is repeated once again. The percentage of water in the mixture is now
Solution
We understand what we have:
Water: 175 ml
Alcohol: 700 ml
Total mixture: 175 + 700 = 875 ml
Here's a crucial strategy for mixture problems: always track the component that's getting diluted (reduced).
Since Gopal is adding water each time, the alcohol content will decrease while water increases. It's easier to track alcohol and then find water at the end.
Initial alcohol fraction =
Fractions are more precise than decimals and easier to work with in calculations.
When Gopal removes 10% of the mixture:
He removes 10% of water AND 10% of alcohol
He replaces this with pure water
If 10% is removed, then 90% remains.
After removing 10%, the alcohol that remains =
We multiply by 9/10 because only 90% of the original alcohol remains after removing 10% of the mixture.
After 1st dilution:
Alcohol fraction =
After 2nd dilution:
Alcohol fraction =
Alcohol percentage = 0.648 × 100% = 64.8%
Therefore, Water percentage = 100% - 64.8% = 35.2%
For mixture dilution problems:
If you remove x% and replace with pure solvent, the concentration of solute becomes: Original × (1 - x/100)
For multiple dilutions, multiply this factor repeatedly
Final Answer: 35.2%