7.1. Masha came up with a number $A$, and Pasha came up with a number $B$. It turned out that $A+B=2020$, and the fraction $\frac{A}{B}$ is less than $\frac{1}{4}$. What is the greatest value that the fraction $\frac{A}{B}$ can take?
Answer: $\frac{403}{1617}$

Solution 1: The sum of the numerator and denominator is 2020. Therefore, the larger the numerator of the fraction, the smaller its denominator - and the larger the fraction itself (since both the numerator and the denominator are positive numbers). ${ }^{404} / 1616$ is exactly equal to $1 / 4$, while ${ }^{403} / 1617$ is less than $1 / 4$.

Solution 2: Let the fraction be $\frac{a}{b}$. Since $a+b=2020$, then $a=2020-b$.

By the condition,

$$
\frac{a}{b}=\frac{2020-b}{b}=\frac{2020}{b}-1<\frac{1}{4}
$$

from which $\frac{2020}{b}<\frac{5}{4}$. Since 2020 is divisible by 5, represent the second fraction as a fraction with the numerator 2020:

$$
\frac{2020}{b}<\frac{2020}{1616}
$$

From which the largest $b=1617$.

# Criteria: 

7 points - correct answer and complete correct solution;

partially correct solution - no more than 5 points;

answer only - 0 points.