One type of liquid contains $20\%$ water and the second type of liquid contains $35\%$ of water. A glass is filled with $10$ parts of first liquid and $4$ parts of second liquid. The percentage of water in the new mixture in the glass is

### Concept & Formula This problem requires calculating the **Weighted Average** of two mixtures. Because the volumes (parts) of the two liquids are different, you cannot simply average their percentages. We must find the total absolute units of water and divide by the total units of mixture. $$ \text{Final Percentage} = \frac{(P_1 \times V_1) + (P_2 \times V_2)}{V_1 + V_2} $$ ### Step-by-Step Solution * **Given:** * Liquid 1: Volume = $10$ parts, Water = $20\%$ * Liquid 2: Volume = $4$ parts, Water = $35\%$ * **Calculation:** * Calculate absolute water units from Liquid 1: $20\%$ of $10 = 0.20 \times 10 = 2$ parts of water. * Calculate absolute water units from Liquid 2: $35\%$ of $4 = 0.35 \times 4 = 1.4$ parts of water. * Total absolute water in the glass = $2 + 1.4 = 3.4$ parts. * Total volume of liquid in the glass = $10 + 4 = 14$ parts. * Calculate the final percentage: $\left( \frac{3.4}{14} \right) \times 100$. * Simplify the fraction: $\frac{34}{140} \times 100 = \frac{17}{70} \times 100 = \frac{170}{7}\%$. * Convert to a mixed fraction: $170 \div 7 = 24$ with a remainder of $2$, giving $24 \frac{2}{7}\%$. ### Exam Strategy & Shortcut **Alligation Method:** The ratio of the parts is $10 : 4$, which simplifies to $5 : 2$. The total gap between the percentages ($20\%$ and $35\%$) is $15\%$. The final mixture will divide this $15\%$ gap in the *inverse* ratio of their volumes ($2 : 5$). Calculate the shift from the $20\%$ side (using the $2$ parts): $\frac{2}{7} \text{ of } 15\% = \frac{30}{7}\% = 4 \frac{2}{7}\%$. Add this shift to the lower percentage: $20\% + 4 \frac{2}{7}\% = 24 \frac{2}{7}\%$. Solved without decimals! ### Common Pitfall A very frequent mistake is calculating the simple, unweighted average: $\frac{20 + 35}{2} = 27.5\%$. This incorrectly assumes the glass was filled with equal amounts (e.g., $7$ parts and $7$ parts) of each liquid. Always factor in the specific volume ratios. ### Final Answer **Therefore, the correct answer is $24 \frac{2}{7}\%$.**