1. **State the problem:** We have grades A, B, C, and D with frequencies 20, 5, 10, and 15 respectively. 2. **Find the modal grade:** The modal grade is the grade with the highest frequency. Looking at the frequencies: 20 (A), 5 (B), 10 (C), 15 (D). The highest frequency is 20, so the modal grade is **A**. 3. **Solve 2m + 5n = 8 on the frequency axis:** This is a linear equation with two variables $m$ and $n$. We can express $n$ in terms of $m$: $$ 2m + 5n = 8 \implies 5n = 8 - 2m \implies n = \frac{8 - 2m}{5} $$ This equation represents a line. For example, if $m=0$, then $n=\frac{8}{5}=1.6$. If $m=2$, then $n=\frac{8 - 4}{5} = \frac{4}{5} = 0.8$. 4. **Scale on frequency axis:** Since the scale is 2 cm = 5 units frequency, for the bar graph frequencies 20, 5, 10, 15 units correspond to heights: $$ \frac{20}{5} \times 2cm = 8cm,\quad \frac{5}{5} \times 2cm = 2cm,\quad \frac{10}{5} \times 2cm=4cm,\quad \frac{15}{5} \times 2cm=6cm $$ 5. **Summary:** - Modal grade is **A**. - Equation $2m + 5n = 8$ represents a line and can be graphed on the frequency axis scale. - Bar graph heights in cm are 8, 2, 4, and 6 for grades A, B, C, D respectively.