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A jar contains a mixture of 175ml175 \mathrm{ml} water and 700ml700 \mathrm{ml} alcohol. Gopal takes out 10%10 \% 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

✅ Correct Option: 4

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 = 700875=45\dfrac{700}{875} = \dfrac{4}{5}

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 = 45×910\dfrac{4}{5} \times \dfrac{9}{10}

We multiply by 9/10 because only 90% of the original alcohol remains after removing 10% of the mixture.


After 1st dilution:

Alcohol fraction = 45×910\dfrac{4}{5} \times \dfrac{9}{10}

After 2nd dilution:

Alcohol fraction = 45×910×910\dfrac{4}{5} \times \dfrac{9}{10} \times \dfrac{9}{10}

45×910×910=4×9×95×10×10\dfrac{4}{5} \times \dfrac{9}{10} \times \dfrac{9}{10} = \dfrac{4 \times 9 \times 9}{5 \times 10 \times 10}

=324500=0.648= \dfrac{324}{500} = 0.648


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%

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