1. **State the problem:** We need to find the number of adults at a funfair given relationships between adults, boys, and girls. 2. **Define variables:** Let $B$ = number of boys, $G$ = number of girls, and $A$ = number of adults. 3. **Given information:** - There were 120 more adults than boys: $$A = B + 120$$ - There were \(\frac{3}{4}\) as many boys as girls: $$B = \frac{3}{4}G$$ - The number of boys was 12% of the total number of people: $$B = 0.12 (A + B + G)$$ 4. **Express total people in terms of $B$ and $G$:** Total people = $$A + B + G$$ Using $A = B + 120$, total = $$B + 120 + B + G = 2B + G + 120$$ 5. **Substitute $B = \frac{3}{4}G$ into total:** $$2B + G + 120 = 2\left(\frac{3}{4}G\right) + G + 120 = \frac{3}{2}G + G + 120 = \frac{5}{2}G + 120$$ 6. **Use the percentage equation:** $$B = 0.12 (A + B + G)$$ Substitute $B = \frac{3}{4}G$ and $A = B + 120$: $$\frac{3}{4}G = 0.12 \left( (\frac{3}{4}G + 120) + \frac{3}{4}G + G \right)$$ Simplify inside parentheses: $$\frac{3}{4}G = 0.12 \left( \frac{3}{4}G + 120 + \frac{3}{4}G + G \right) = 0.12 \left( \frac{3}{4}G + \frac{3}{4}G + G + 120 \right)$$ $$= 0.12 \left( \frac{3}{4}G + \frac{3}{4}G + G + 120 \right) = 0.12 \left( \frac{3}{4}G + \frac{3}{4}G + G + 120 \right)$$ Sum the $G$ terms: $$\frac{3}{4}G + \frac{3}{4}G + G = \frac{3}{4}G + \frac{3}{4}G + \frac{4}{4}G = \frac{10}{4}G = \frac{5}{2}G$$ So: $$\frac{3}{4}G = 0.12 \left( \frac{5}{2}G + 120 \right)$$ 7. **Solve for $G$:** $$\frac{3}{4}G = 0.12 \times \frac{5}{2}G + 0.12 \times 120$$ $$\frac{3}{4}G = 0.3G + 14.4$$ Bring terms with $G$ to one side: $$\frac{3}{4}G - 0.3G = 14.4$$ Convert to decimals: $$0.75G - 0.3G = 14.4$$ $$0.45G = 14.4$$ $$G = \frac{14.4}{0.45} = 32$$ 8. **Find $B$ and $A$:** $$B = \frac{3}{4} \times 32 = 24$$ $$A = B + 120 = 24 + 120 = 144$$ **Final answer:** There were **144** adults at the funfair.