Train Passengers B81213

1. **State the problem:** We have a train with adults and children at Station X, Y, and Z with given ratios and changes. We want to find: (a) The total number of passengers at the end. (b) The number of children who alighted at Station Y. 2. **Define variables:** Let $A$ = number of adults at Station X. Let $C$ = number of children at Station X. 3. **From Station X:** Given $C = \frac{3}{5}A$. 4. **At Station Y:** - $\frac{2}{9}$ of adults alight, so adults remaining = $A - \frac{2}{9}A = \frac{7}{9}A$. - Let $x$ = number of children alighting at Station Y. - Children remaining = $C - x$. - New ratio: $\frac{C - x}{\frac{7}{9}A} = \frac{3}{7}$. 5. **Express $x$ in terms of $A$:** $$\frac{C - x}{\frac{7}{9}A} = \frac{3}{7} \implies C - x = \frac{3}{7} \times \frac{7}{9} A = \frac{3}{9} A = \frac{1}{3} A$$ Substitute $C = \frac{3}{5} A$: $$\frac{3}{5} A - x = \frac{1}{3} A \implies x = \frac{3}{5} A - \frac{1}{3} A = \frac{9}{15} A - \frac{5}{15} A = \frac{4}{15} A$$ 6. **At Station Z:** - 358 children and 98 adults board. - Adults after Station Z: $\frac{7}{9} A + 98$. - Children after Station Z: $C - x + 358 = \frac{1}{3} A + 358$ (from step 5). - Given adults = children after Station Z: $$\frac{7}{9} A + 98 = \frac{1}{3} A + 358$$ 7. **Solve for $A$:** $$\frac{7}{9} A - \frac{1}{3} A = 358 - 98$$ $$\frac{7}{9} A - \frac{3}{9} A = 260$$ $$\frac{4}{9} A = 260 \implies A = 260 \times \frac{9}{4} = 585$$ 8. **Find $C$ and $x$:** $$C = \frac{3}{5} \times 585 = 351$$ $$x = \frac{4}{15} \times 585 = 156$$ 9. **Find total passengers at the end:** Adults after Station Z: $$\frac{7}{9} \times 585 + 98 = 455 + 98 = 553$$ Children after Station Z: $$\frac{1}{3} \times 585 + 358 = 195 + 358 = 553$$ Total passengers: $$553 + 553 = 1106$$ **Final answers:** (a) Total passengers at the end = **1106** (b) Children alighted at Station Y = **156**