Question:
Only a single rail track exists between stations A and B on a railway line. One hour after the northbound super fast train N leaves station A for station B, a south-bound passenger train S reaches station A from station B. The speed of the super fast train is twice that of a normal express train E, while the speed of a passenger train S is half that of E. On a particular day, N leaves for B from A, 20 min behind the normal schedule. In order to maintain the schedule, both N and S increased their speeds. If the super fast train doubles its speed, what should be the ratio (approximately) of the speeds of passenger train to that of the super fast train so that the passenger train S reaches exactly at the scheduled time at A on that day?
Only a single rail track exists between stations A and B on a railway line. One hour after the northbound super fast train N leaves station A for station B, a south-bound passenger train S reaches station A from station B. The speed of the super fast train is twice that of a normal express train E, while the speed of a passenger train S is half that of E. On a particular day, N leaves for B from A, 20 min behind the normal schedule. In order to maintain the schedule, both N and S increased their speeds. If the super fast train doubles its speed, what should be the ratio (approximately) of the speeds of passenger train to that of the super fast train so that the passenger train S reaches exactly at the scheduled time at A on that day?
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On a single track the two trains must cross at one fixed point. Find where and when they cross on the normal schedule, then see how the delayed N changes the time at which S can get through that point.
Updated On: Nov 4, 2026
- 1 : 3
- 1 : 4
- 1 : 6
- 1 : 5
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The Correct Option is C
Solution and Explanation
Step 1: Speeds.
Let the express train have speed \(e\). Then N has speed \(2e\) and S has speed \(e/2\). Take \(e = 1\) and measure time in hours from the moment N normally leaves A.
Step 2: Normal crossing.
S reaches A at \(t = 1\). If \(AB = D\), S leaves B at \(t = 1 - 2D\). The trains cross when \(2t = D - \tfrac12\big(t - (1 - 2D)\big)\), i.e. \(2.5t = 0.5\), so \(t = 0.2\) h and the crossing point is \(2 \times 0.2 = 0.4\) from A (independent of \(D\)). Check: S then needs \(0.4 \div 0.5 = 0.8\) h to reach A, arriving at \(t = 1\).
Step 3: Delayed day.
N leaves \(20\) min \(= \tfrac13\) h late and now runs at \(2 \times 2 = 4\). It reaches the same crossing point at \(t = \tfrac13 + \tfrac{0.4}{4} = \tfrac13 + \tfrac1{10} = \tfrac{13}{30}\) h.
Step 4: Speed of S.
S passes the crossing point at that moment and must still reach A at \(t = 1\), i.e. cover \(0.4\) in \(1 - \tfrac{13}{30} = \tfrac{17}{30}\) h:
\[ v_S = \frac{0.4}{17/30} = \frac{12}{17} \approx 0.706. \]
Step 5: Ratio.
\[ \frac{v_S}{v_N} = \frac{12/17}{4} = \frac{3}{17} \approx \frac{1}{5.67}, \] which is nearest to \(1 : 6\).
Final Answer:
The correct option is (3): 1 : 6
Let the express train have speed \(e\). Then N has speed \(2e\) and S has speed \(e/2\). Take \(e = 1\) and measure time in hours from the moment N normally leaves A.
Step 2: Normal crossing.
S reaches A at \(t = 1\). If \(AB = D\), S leaves B at \(t = 1 - 2D\). The trains cross when \(2t = D - \tfrac12\big(t - (1 - 2D)\big)\), i.e. \(2.5t = 0.5\), so \(t = 0.2\) h and the crossing point is \(2 \times 0.2 = 0.4\) from A (independent of \(D\)). Check: S then needs \(0.4 \div 0.5 = 0.8\) h to reach A, arriving at \(t = 1\).
Step 3: Delayed day.
N leaves \(20\) min \(= \tfrac13\) h late and now runs at \(2 \times 2 = 4\). It reaches the same crossing point at \(t = \tfrac13 + \tfrac{0.4}{4} = \tfrac13 + \tfrac1{10} = \tfrac{13}{30}\) h.
Step 4: Speed of S.
S passes the crossing point at that moment and must still reach A at \(t = 1\), i.e. cover \(0.4\) in \(1 - \tfrac{13}{30} = \tfrac{17}{30}\) h:
\[ v_S = \frac{0.4}{17/30} = \frac{12}{17} \approx 0.706. \]
Step 5: Ratio.
\[ \frac{v_S}{v_N} = \frac{12/17}{4} = \frac{3}{17} \approx \frac{1}{5.67}, \] which is nearest to \(1 : 6\).
Final Answer:
The correct option is (3): 1 : 6
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