A train started with $540$ passengers. At the first stop $\frac{1}{9}$ of them got down and $24$ got up. On its second stop $\frac{1}{8}$ of the passengers then existing got down and $9$ got up. With how many passengers did it reach the third stop?

## Concept & Logic This is a sequential arithmetic problem. We need to track the number of passengers dynamically by applying the fractional changes and additions step-by-step as the train moves from one stop to the next. ## Step-by-step Solution * **Initial State:** * Number of passengers at start = $540$. * **First Stop:** * Passengers getting down = $\frac{1}{9} \times 540 = 60$. * Remaining passengers = $540 - 60 = 480$. * Passengers getting up = $24$. * Total passengers leaving the first stop = $480 + 24 = 504$. * **Second Stop:** * Passengers getting down = $\frac{1}{8} \times 504$. Let's calculate: $504 \div 8 = 63$. * Remaining passengers = $504 - 63 = 441$. * Passengers getting up = $9$. * Total passengers leaving the second stop = $441 + 9 = 450$. * **Third Stop:** * The train reaches the third stop with this final count of passengers. ## Exam Strategy & Shortcut To calculate remaining passengers after a fraction leaves, use the complement fraction directly. If $\frac{1}{9}$ leave, $\frac{8}{9}$ remain. Step 1: $(\frac{8}{9} \times 540) + 24 = 480 + 24 = 504$. If $\frac{1}{8}$ leave, $\frac{7}{8}$ remain. Step 2: $(\frac{7}{8} \times 504) + 9 = (7 \times 63) + 9 = 441 + 9 = 450$. This saves the subtraction step each time. ## Common Pitfall Applying the second fraction ($\frac{1}{8}$) to the original starting number ($540$) instead of the current running total ($504$). Always read "passengers *then existing*" carefully. ## Final Answer Therefore, the correct answer is **450**.