Question:
Two trains, P and Q, pass a pole in 40 seconds and 2 minutes 20 seconds, respectively. If the length of train P is two-thirds that of train Q, what is the ratio of the speed of train P to that of train Q?
Two trains, P and Q, pass a pole in 40 seconds and 2 minutes 20 seconds, respectively. If the length of train P is two-thirds that of train Q, what is the ratio of the speed of train P to that of train Q?
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Always ensure that all units are consistent before performing calculations. In this problem, time was given in seconds and a mix of minutes and seconds. Converting everything to seconds is the first crucial step to avoid errors.
Updated On: Oct 2, 2025
- 9:4
- 5:2
- 7:3
- 12:5
- 14:3
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
When a train passes a pole (or a stationary point object), the distance it covers is equal to its own length. The relationship between speed, distance, and time is fundamental to solving this problem.
Step 2: Key Formula or Approach:
\[ \text{Speed} = \frac{\text{Distance}}{\text{Time}} \] Here, Distance = Length of the train.
Step 3: Detailed Explanation:
Let \(L_P\) and \(S_P\) be the length and speed of train P.
Let \(L_Q\) and \(S_Q\) be the length and speed of train Q.
For train P:
Time taken (\(T_P\)) = 40 seconds.
Distance covered = \(L_P\).
\[ S_P = \frac{L_P}{T_P} = \frac{L_P}{40} \quad \text{(Equation 1)} \] For train Q:
Time taken (\(T_Q\)) = 2 minutes 20 seconds = \(2 \times 60 + 20 = 120 + 20 = 140\) seconds.
Distance covered = \(L_Q\).
\[ S_Q = \frac{L_Q}{T_Q} = \frac{L_Q}{140} \quad \text{(Equation 2)} \] Given relationship between lengths:
The length of train P is two-thirds that of train Q.
\[ L_P = \frac{2}{3} L_Q \] Step 4: Final Answer:
We need to find the ratio of the speeds, \(S_P : S_Q\) or \( \frac{S_P}{S_Q} \).
Using Equation 1 and Equation 2:
\[ \frac{S_P}{S_Q} = \frac{L_P/40}{L_Q/140} = \frac{L_P}{L_Q} \times \frac{140}{40} \] Now, substitute the relationship \(L_P = \frac{2}{3} L_Q\):
\[ \frac{S_P}{S_Q} = \left(\frac{2/3 \cdot L_Q}{L_Q}\right) \times \frac{140}{40} \] \[ \frac{S_P}{S_Q} = \frac{2}{3} \times \frac{14}{4} = \frac{2}{3} \times \frac{7}{2} \] Cancel out the 2s:
\[ \frac{S_P}{S_Q} = \frac{7}{3} \] The ratio of the speed of train P to that of train Q is 7:3.
When a train passes a pole (or a stationary point object), the distance it covers is equal to its own length. The relationship between speed, distance, and time is fundamental to solving this problem.
Step 2: Key Formula or Approach:
\[ \text{Speed} = \frac{\text{Distance}}{\text{Time}} \] Here, Distance = Length of the train.
Step 3: Detailed Explanation:
Let \(L_P\) and \(S_P\) be the length and speed of train P.
Let \(L_Q\) and \(S_Q\) be the length and speed of train Q.
For train P:
Time taken (\(T_P\)) = 40 seconds.
Distance covered = \(L_P\).
\[ S_P = \frac{L_P}{T_P} = \frac{L_P}{40} \quad \text{(Equation 1)} \] For train Q:
Time taken (\(T_Q\)) = 2 minutes 20 seconds = \(2 \times 60 + 20 = 120 + 20 = 140\) seconds.
Distance covered = \(L_Q\).
\[ S_Q = \frac{L_Q}{T_Q} = \frac{L_Q}{140} \quad \text{(Equation 2)} \] Given relationship between lengths:
The length of train P is two-thirds that of train Q.
\[ L_P = \frac{2}{3} L_Q \] Step 4: Final Answer:
We need to find the ratio of the speeds, \(S_P : S_Q\) or \( \frac{S_P}{S_Q} \).
Using Equation 1 and Equation 2:
\[ \frac{S_P}{S_Q} = \frac{L_P/40}{L_Q/140} = \frac{L_P}{L_Q} \times \frac{140}{40} \] Now, substitute the relationship \(L_P = \frac{2}{3} L_Q\):
\[ \frac{S_P}{S_Q} = \left(\frac{2/3 \cdot L_Q}{L_Q}\right) \times \frac{140}{40} \] \[ \frac{S_P}{S_Q} = \frac{2}{3} \times \frac{14}{4} = \frac{2}{3} \times \frac{7}{2} \] Cancel out the 2s:
\[ \frac{S_P}{S_Q} = \frac{7}{3} \] The ratio of the speed of train P to that of train Q is 7:3.
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