A person distributed two different amounts among his four sons A, B, C, and D. The first amount was distributed in the ratio of 4:3:2:1, while the second amount was distributed in the ratio of 5:6:7:8. If the first amount is half of the second amount, which son received the maximum total amount?

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Let's assume the first amount is ₹x and the second amount is ₹2x (since the first amount is half of the second).

**Calculate each son's share from the first amount:**  
   - A's share: $$\frac{4}{10} 	imes x = \frac{4x}{10}$$  
   - B's share: $$\frac{3}{10} 	imes x = \frac{3x}{10}$$  
   - C's share: $$\frac{2}{10} 	imes x = \frac{2x}{10}$$  
   - D's share: $$\frac{1}{10} 	imes x = \frac{x}{10}$$

**Calculate each son's share from the second amount:**  
   - A's share: $$\frac{5}{26} 	imes 2x = \frac{10x}{26} = \frac{5x}{13}$$  
   - B's share: $$\frac{6}{26} 	imes 2x = \frac{12x}{26} = \frac{6x}{13}$$  
   - C's share: $$\frac{7}{26} 	imes 2x = \frac{14x}{26} = \frac{7x}{13}$$  
   - D's share: $$\frac{8}{26} 	imes 2x = \frac{16x}{26} = \frac{8x}{13}$$

**Calculate the total share for each son:**  
   - A's total: $$\frac{4x}{10} + \frac{5x}{13} = \frac{52x}{130} + \frac{50x}{130} = \frac{102x}{130}$$  
   - B's total: $$\frac{3x}{10} + \frac{6x}{13} = \frac{39x}{130} + \frac{60x}{130} = \frac{99x}{130}$$  
   - C's total: $$\frac{2x}{10} + \frac{7x}{13} = \frac{26x}{130} + \frac{70x}{130} = \frac{96x}{130}$$  
   - D's total: $$\frac{x}{10} + \frac{8x}{13} = \frac{13x}{130} + \frac{80x}{130} = \frac{93x}{130}$$

**Compare the totals:**  
   - A: $$\frac{102x}{130}$$  
   - B: $$\frac{99x}{130}$$  
   - C: $$\frac{96x}{130}$$  
   - D: $$\frac{93x}{130}$$

Since A's total amount ($$\frac{102x}{130}$$) is the largest, son A received the maximum total amount.

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