A passenger train departs from Delhi at 6 p.m. for Mumbai. At 9 p.m., an express train, whose average speed exceeds that of the passenger train by 15 km/hour leaves Mumbai for Delhi. Two trains met each other mid-route. At what time do they meet, given that the distance between the cities is 1080 km?
6 a.m.
Understanding the Train Meeting Problem
This problem involves two trains traveling towards each other from two different cities, Delhi and Mumbai. We are given the total distance between the cities, the departure times of both trains, and the relationship between their speeds. We need to find the exact time when these two trains meet.
Setting Up the Variables and Information
- Distance between Delhi and Mumbai = \(1080\) km.
- Passenger train departs from Delhi at 6 p.m.
- Express train departs from Mumbai at 9 p.m.
- The Express train's average speed is \(15\) km/hour more than the Passenger train's speed.
- Let the speed of the Passenger train be \(S_P\) km/hour.
- Then the speed of the Express train is \(S_E = S_P + 15\) km/hour.
- The trains meet "mid-route". This implies they meet exactly halfway between the two cities.
Calculating the Meeting Point Distance
Since the trains meet mid-route, the meeting point is exactly halfway between Delhi and Mumbai.
Midpoint distance from Delhi = Total distance / 2 = \(1080 \text{ km} / 2 = 540\) km.
Midpoint distance from Mumbai = Total distance / 2 = \(1080 \text{ km} / 2 = 540\) km.
So, the Passenger train travels 540 km and the Express train travels 540 km until they meet.
Calculating Time Taken for Each Train to Reach the Midpoint
The formula for time is Distance / Speed.
- Time taken by Passenger train to travel 540 km = \(T_P = \frac{540}{S_P}\) hours.
- Time taken by Express train to travel 540 km = \(T_E = \frac{540}{S_E} = \frac{540}{S_P + 15}\) hours.
Relating the Departure Times and Travel Times
The Passenger train leaves at 6 p.m. and the Express train leaves at 9 p.m. This means the Passenger train has a head start of 3 hours (9 p.m. - 6 p.m.).
Both trains travel until they meet at the same time. The total time the Passenger train travels until they meet is \(T_P\). The total time the Express train travels until they meet is \(T_E\). Since the Passenger train started 3 hours earlier, its total travel time \(T_P\) is 3 hours more than the Express train's total travel time \(T_E\).
So, we can write the equation:
\(T_P - T_E = 3\) hours
Substituting the time expressions:
\(\frac{540}{S_P} - \frac{540}{S_P + 15} = 3\)
Solving for the Speed of the Passenger Train
We need to solve the equation for \(S_P\).
Divide the entire equation by 3:
\(\frac{180}{S_P} - \frac{180}{S_P + 15} = 1\)
Find a common denominator, which is \(S_P (S_P + 15)\):
\(\frac{180(S_P + 15) - 180S_P}{S_P(S_P + 15)} = 1\)
Multiply both sides by \(S_P(S_P + 15)\):
\(180(S_P + 15) - 180S_P = S_P(S_P + 15)\)
Expand both sides:
\(180S_P + 180 \times 15 - 180S_P = S_P^2 + 15S_P\)
\(180S_P + 2700 - 180S_P = S_P^2 + 15S_P\)
Simplify:
\(2700 = S_P^2 + 15S_P\)
Rearrange into a quadratic equation:
\(S_P^2 + 15S_P - 2700 = 0\)
Use the quadratic formula \(S_P = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a=1\), \(b=15\), \(c=-2700\).
\(S_P = \frac{-15 \pm \sqrt{15^2 - 4(1)(-2700)}}{2(1)}\)
\(S_P = \frac{-15 \pm \sqrt{225 + 10800}}{2}\)
\(S_P = \frac{-15 \pm \sqrt{11025}}{2}\)
The square root of 11025 is 105.
\(S_P = \frac{-15 \pm 105}{2}\)
Since speed must be positive, we take the positive root:
\(S_P = \frac{-15 + 105}{2} = \frac{90}{2} = 45\)
So, the speed of the Passenger train \(S_P = 45\) km/hour.
Calculating the Speed of the Express Train
The speed of the Express train is \(S_E = S_P + 15 = 45 + 15 = 60\) km/hour.
Calculating the Time Until Meeting
We can use the time taken by either train to find the meeting time.
Using the Passenger train:
- Distance covered by Passenger train = 540 km
- Speed of Passenger train = 45 km/hour
- Time taken by Passenger train = \(\frac{540 \text{ km}}{45 \text{ km/hour}} = 12\) hours.
- The Passenger train departed at 6 p.m.
- Meeting time = 6 p.m. + 12 hours.
- 6 p.m. + 12 hours is 6 a.m. the next day.
Using the Express train:
- Distance covered by Express train = 540 km
- Speed of Express train = 60 km/hour
- Time taken by Express train = \(\frac{540 \text{ km}}{60 \text{ km/hour}} = 9\) hours.
- The Express train departed at 9 p.m.
- Meeting time = 9 p.m. + 9 hours.
- 9 p.m. + 9 hours is 6 a.m. the next day.
Both calculations give the same meeting time, confirming our speeds are correct and they meet at 6 a.m.
Final Answer
The two trains meet at 6 a.m.
Revision Table: Train Meeting Problem
| Detail | Passenger Train | Express Train |
|---|---|---|
| Origin | Delhi | Mumbai |
| Departure Time | 6 p.m. | 9 p.m. |
| Speed | 45 km/h (\(S_P\)) | 60 km/h (\(S_P + 15\)) |
| Distance to Meeting Point | 540 km | 540 km |
| Time Taken to Meet | 12 hours | 9 hours |
| Meeting Time (Departure + Time Taken) | 6 p.m. + 12 hours = 6 a.m. | 9 p.m. + 9 hours = 6 a.m. |
Additional Information: Speed, Distance, and Time Concepts
This problem is a classic example of relative speed when objects are moving towards each other. However, because the trains start at different times, we first accounted for the head start.
- Speed, Distance, Time Relation: The fundamental relationship is \(\text{Speed} = \frac{\text{Distance}}{\text{Time}}\), which can be rearranged to \(\text{Distance} = \text{Speed} \times \text{Time}\) or \(\text{Time} = \frac{\text{Distance}}{\text{Speed}}\).
- Relative Speed (Approaching): If two objects move towards each other with speeds \(v_1\) and \(v_2\), their relative speed is \(v_1 + v_2\). If they start at the same time, the time to meet is \(\frac{\text{Total Distance}}{v_1 + v_2}\). However, in this problem, the start times differ, so accounting for the head start is necessary before using relative speed on the remaining distance, or, as done here, setting up time equations based on equal total time elapsed from a common reference point (the meeting time).
- Meeting Mid-Route: This specific phrase simplified the problem significantly by fixing the distance each train travels (half the total distance). If they didn't meet mid-route, we would use the equation \(S_P \times T_P(\text{total}) + S_E \times T_E(\text{total}) = 1080\) and \(T_P(\text{total}) = T_E(\text{total}) + 3\).
Similar Questions
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Important Questions from Relative Speed
In a circular race of 2500 m, a man and a woman start from a point towards opposite directions with speeds of 37 km/h and 35 km/h, respectively. After how much time from the start of the race will they meet for the first time?
- X and Yrun a 3 km race along a circular course of length 300 m. Their speeds are in the ratio 3 : 2. If they start together in the same direction, how many times would the first one pass the other (the start-off is not counted as passing)?
A train of length 600 meters passes another train of length 1000 meters moving in the opposite direction in 96 seconds. If the speed of both the trains is the same, then what is the speed of each train?
A train, running at a speed of 36 kmph, takes 60 seconds to cross an electric pole. How much time it will take to cross a platform of length 250 meter?
A truck covered a distance of 780 km in 12 hours. What distance would a car cover in 11 hours travelling at a speed which is seven kmph more than that of the truck?
