4(3)/(7)+1(1)/(5)= 1f the sum is a mixed number, draw a road under the bus. If the sum is in fifths, draw the principal standing in the school doorway.

To solve this problem, we need to add the two mixed numbers.<br /><br />First, let's convert the mixed numbers to improper fractions:<br />$4\frac {3}{7} = \frac{4 \times 7 + 3}{7} = \frac{31}{7}$<br />$1\frac {1}{5} = \frac{1 \times 5 + 1}{5} = \frac{6}{5}$<br /><br />Now, we can add the two improper fractions:<br />$\frac{31}{7} + \frac{6}{5} = \frac{31 \times 5 + 6 \times 7}{7 \times 5} = \frac{155 + 42}{35} = \frac{197}{35}$<br /><br />Since the sum is not a mixed number, we cannot draw a road under the bus.<br /><br />Therefore, the answer is $\frac{197}{35}$.