Question : One type of liquid contains 20% water, and the second type contains 35% water. A glass is filled with 10 parts of the first liquid and 4 parts of the second liquid. What is the percentage of water in the new mixture in the glass?
Option 1: $37$%
Option 2: $46$%
Option 3: $12\tfrac{1}{7}$%
Option 4: $24\tfrac{2}{7}$%
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Correct Answer: $24\tfrac{2}{7}$%
Solution :
Given: The first liquid contains 20% water and the second liquid contains 35% of water.
A glass is filled with 10 parts of the first liquid and 4 parts of the second liquid.
Amount of water in first liquid $=\frac{20}{100}×10=2$ parts
Amount of water in second liquid $=\frac{35}{100}×4=\frac{7}{5}$ parts
Total water $=2+\frac{7}{5}=\frac{17}{5}$parts
Total liquid $=10+4=14$ parts
So, the required percentage of water $=\frac{\frac{17}{5}}{14}×100=\frac{170}{7}=24\tfrac{2}{7}$%
Hence, the correct answer is $24\tfrac{2}{7}$%.
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