Problem 1. Three brothers divided a certain amount of money such that the first brother received $\frac{1}{5}$ of the total amount, the second brother received $\frac{5}{8}$ of the total amount, and the third brother received the remainder of the money. Then, the third brother gave $\frac{3}{4}$ of his share to the first brother, and the remainder of his share to the second brother. What fraction of the total amount did the first brother receive?
Solution. Since the first and second brothers received $\frac{1}{5}$ and $\frac{5}{8}$ of the sum respectively, we get that the third brother received $1-\frac{1}{5}-\frac{5}{8}=\frac{40-8-25}{40}=\frac{7}{40}$ of the total sum. He gave the first brother $\frac{3}{4} \cdot \frac{7}{40}=\frac{21}{160}$ of the total sum, so the first brother in total received $\frac{1}{5}+\frac{21}{160}=\frac{53}{160}$ of the total sum.