Question

The distance from B to C is thrice that from A to B. Two trains travel from A to C via B. The speed of train 2 is double that of train 1 while traveling from A to B and their speeds are nterchanged while traveling from B to C. The ratio of the time taken by train 1 to that taken by train 2 in travelling from A to C is

• A. 1: 4
• B. 7: 5
• C. 5:7
• D. 4:1

Answer 1

The Correct Option is C

Let the speed of Train \(1\) from point \(A\) to \(B\) be \(s\).
Then the speed of Train \(2\) from \(A\) to \(B\) is \(2s\).

Time taken by Train 1 to go from \(A\) to \(C\):
\[ \frac{D}{s} + \frac{3D}{2s} = \frac{5D}{2s} \]

Time taken by Train 2 to go from \(A\) to \(C\):
\[ \frac{D}{2s} + \frac{3D}{s} = \frac{7D}{2s} \]

Required time ratio (Train 1 : Train 2):
\[ \frac{5D}{2s} : \frac{7D}{2s} = 5 : 7 \]

✅ Final Answer: Option (C): \(5 : 7\)

Answer 2

Assumptions:

• Distance from A to B = \( x \)
• Distance from B to C = \( 3x \)
• Train 1: Speed from A to B = \( v \), from B to C = \( 2v \)
• Train 2: Speed from A to B = \( 2v \), from B to C = \( v \)

Time taken by Train 1:

• A to B: \( \frac{x}{v} \)
• B to C: \( \frac{3x}{2v} \)
• Total: \[ \frac{x}{v} + \frac{3x}{2v} = \frac{2x + 3x}{2v} = \frac{5x}{2v} \]

Time taken by Train 2:

• A to B: \( \frac{x}{2v} \)
• B to C: \( \frac{3x}{v} \)
• Total: \[ \frac{x}{2v} + \frac{3x}{v} = \frac{x + 6x}{2v} = \frac{7x}{2v} \]

Ratio of Time Taken (Train 1 : Train 2):

\[ \frac{\frac{5x}{2v}}{\frac{7x}{2v}} = \frac{5}{7} \]

✅ Final Answer: Option (C): 5 : 7