1. **State the problem:** There are initially more boys than girls in a hall. Specifically, there are 5 more boys than girls. 2. **Define variables:** Let the number of girls be $g$. Then the number of boys is $g + 5$. 3. **After some leave:** $\frac{2}{5}$ of the girls leave, so girls remaining = $g - \frac{2}{5}g = \frac{3}{5}g$. $\frac{3}{7}$ of the boys leave, so boys remaining = $(g + 5) - \frac{3}{7}(g + 5) = \frac{4}{7}(g + 5)$. 4. **Given condition:** The number of boys and girls remaining are equal: $$\frac{3}{5}g = \frac{4}{7}(g + 5)$$ 5. **Solve the equation:** Multiply both sides by 35 (LCM of 5 and 7) to clear denominators: $$35 \times \frac{3}{5}g = 35 \times \frac{4}{7}(g + 5)$$ $$7 \times 3g = 5 \times 4 (g + 5)$$ $$21g = 20(g + 5)$$ 6. **Expand and simplify:** $$21g = 20g + 100$$ Subtract $20g$ from both sides: $$21g - \cancel{20g} = \cancel{20g} + 100 - 20g$$ $$g = 100$$ 7. **Find number of boys:** $$g + 5 = 100 + 5 = 105$$ 8. **Find total pupils initially:** $$g + (g + 5) = 100 + 105 = 205$$ **Final answer:** The total number of pupils in the hall at first is **205**.