Question

A train approaches a tunnel AB. Inside the tunnel is a cat located at a point that is \( \frac{3}{8} \) of the distance AB measured from the entrance A. When the train whistles the cat runs. If the cat moves to the entrance of the tunnel A, the train catches the cat exactly at the entrance. If the cat moves to the exit B, the train catches the cat at exactly the exit. The speed of the train is greater than the speed of the cat by what order?

• A. 3 : 1
• B. 4 : 1
• C. 5 : 1
• D. None of these

Hint:

When comparing speeds, use time equations based on distances traveled by both objects.

Answer 1

The Correct Option is B

Let the total distance \( AB = L \). The cat is located at \( \frac{3}{8}L \) from A. - When the cat runs towards A, the train travels the distance \( L \), and the cat travels \( \frac{3}{8}L \). - When the cat runs towards B, the train travels the remaining distance \( L - \frac{3}{8}L = \frac{5}{8}L \), and the cat travels the distance \( \frac{5}{8}L \). Let the speed of the train be \( v_t \) and the speed of the cat be \( v_c \). Using the time to travel for each case, we have: \[ \frac{L}{v_t} = \frac{\frac{3}{8}L}{v_c} \quad \text{and} \quad \frac{\frac{5}{8}L}{v_t} = \frac{\frac{5}{8}L}{v_c}. \] This implies: \[ \frac{v_t}{v_c} = 4. \] Thus, the speed of the train is 4 times that of the cat.