$6$ c.c. of a $20\%$ solution of alcohol in water is mixed with $4$ c.c. of a $60\%$ solution of alcohol in water. The alcoholic strength of the mixture is

### Concept & Logic This requires calculating the **Weighted Average** of two mixtures. To find the final strength, we must determine the total absolute amount of alcohol from both constituent solutions and divide it by their combined total volume. $$ \text{Final Percentage} = \frac{(P_1 \times V_1) + (P_2 \times V_2)}{V_1 + V_2} \times 100 $$ ### Step-by-Step Solution * **Given:** * Solution 1: Volume = $6$ c.c., Concentration = $20\%$ * Solution 2: Volume = $4$ c.c., Concentration = $60\%$ * **Calculation:** * Calculate absolute alcohol in Solution 1: $20\%$ of $6 = 0.20 \times 6 = 1.2$ c.c. * Calculate absolute alcohol in Solution 2: $60\%$ of $4 = 0.60 \times 4 = 2.4$ c.c. * Total absolute alcohol in the combined mixture = $1.2 + 2.4 = 3.6$ c.c. * Total combined volume of the mixture = $6 + 4 = 10$ c.c. * Calculate the final alcoholic strength: $\left(\frac{3.6}{10}\right) \times 100$. * $\frac{3.6}{10} = 0.36$. * $0.36 \times 100 = 36\%$. ### Exam Strategy & Shortcut **Alligation (Weighted Average Line):** The weights (volumes) of the two mixtures are $6$ c.c. and $4$ c.c., giving a weight ratio of $6:4$ or $3:2$. The concentrations are $20\%$ and $60\%$. The total distance between them is $60\% - 20\% = 40\%$. The final mixture concentration will divide this $40\%$ gap in the *inverse* ratio of their volumes ($2:3$). Calculate the $2$ parts (pulling away from the $20\%$ side): $\frac{2}{5} \text{ of } 40\% = 16\%$. Add this to the lower concentration: $20\% + 16\% = 36\%$. ### Common Pitfall A very common trap is taking a simple unweighted average of the two percentages: $\frac{20 + 60}{2} = 40\%$. This completely ignores the fact that there is more of the $20\%$ solution ($6$ c.c.) than the $60\%$ solution ($4$ c.c.), meaning the final answer must heavily skew closer to $20\%$. Trap option (d) catches students who make this mistake. ### Final Answer **Therefore, the correct answer is $36\%$.**