A person distributed two different amounts among his sons $A$, $B$, $C$ and $D$. First amount was distributed in the ratio of 4 : 3 : 2 : 1 and the second amount was distributed in the ratio of 5 : 6 : 7 : 8. If the first amount is half of the second amount, then which son will get the maximum amount?

### Concept & Weighted Ratios When comparing shares from two different distributions, define the totals of both distributions using a common variable based on their given relationship. $$\text{Total Share} = \text{Share from Amount 1} + \text{Share from Amount 2}$$ ### Step-by-Step Solution 1. **Defining the Amounts:** Let the first amount be $X$ and the second amount be $Y$. We are given that $X = \frac{Y}{2}$, which means $Y = 2X$. 2. **First Distribution:** The ratio is $4:3:2:1$. Sum of parts = $4+3+2+1 = 10$. - $A_1 = \frac{4}{10}X$ - $B_1 = \frac{3}{10}X$ - $C_1 = \frac{2}{10}X$ - $D_1 = \frac{1}{10}X$ 3. **Second Distribution:** The ratio is $5:6:7:8$. Sum of parts = $5+6+7+8 = 26$. Substitute $Y = 2X$. - $A_2 = \frac{5}{26}Y = \frac{5}{26}(2X) = \frac{10}{26}X = \frac{5}{13}X$ - $B_2 = \frac{6}{26}Y = \frac{6}{26}(2X) = \frac{12}{26}X = \frac{6}{13}X$ - $C_2 = \frac{7}{26}Y = \frac{7}{26}(2X) = \frac{14}{26}X = \frac{7}{13}X$ - $D_2 = \frac{8}{26}Y = \frac{8}{26}(2X) = \frac{16}{26}X = \frac{8}{13}X$ 4. **Total Shares:** To compare, find a common denominator for $10$ and $13$, which is $130$. - Total $A = \frac{4}{10}X + \frac{5}{13}X = (\frac{52}{130} + \frac{50}{130})X = \frac{102}{130}X$ - Total $B = \frac{3}{10}X + \frac{6}{13}X = (\frac{39}{130} + \frac{60}{130})X = \frac{99}{130}X$ - Total $C = \frac{2}{10}X + \frac{7}{13}X = (\frac{26}{130} + \frac{70}{130})X = \frac{96}{130}X$ - Total $D = \frac{1}{10}X + \frac{8}{13}X = (\frac{13}{130} + \frac{80}{130})X = \frac{93}{130}X$ 5. **Conclusion:** Comparing the numerators, $102 > 99 > 96 > 93$. Son $A$ receives the maximum amount. ### Exam Strategy & Shortcut Assume convenient values to eliminate fractions. Let $X = 130$ and $Y = 260$. $A = (\frac{4}{10} \times 130) + (\frac{5}{26} \times 260) = 52 + 50 = 102$. By observing the pattern (first part decreases by 13, second increases by 10, net change is -3), $A$ is clearly the highest. ### Common Pitfall Ignoring the relationship between the two total amounts (assuming $X = Y$) will lead to an incorrect calculation where $D$ might appear to get the most. ### Final Answer Therefore, the correct answer is **$A$**.