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Word Problem - The figure above shows the railway tracks between three 22 Jan 2017, 01:35
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The figure above shows the railway tracks between three cities A, B and C. The distance between city A and city B is 100 kilometres and city A and city C is 240 kilometres. A train can travel from city A to city B via city C only and vice versa. Two trains are scheduled to start from city A and city B respectively at 9 AM and travel to city  B and city A respectively. The speeds of the train starting from city A and city B are 20 kilometres per hour and 25 kilometres per hour respectively. If the train from city A starts half an hour early and the train from city B starts half an hour late, at what time will the two trains meet?

    A. 7:10 PM
    B. 8:10 PM
    C. 8:23 PM
    D. 8:40 PM
    E. 8:53 PM

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Word Problem - The figure above shows the railway tracks between three Updated on: 22 Jan 2017, 11:41
Originally posted by rohit8865 on 22 Jan 2017, 11:01.
Last edited by rohit8865 on 22 Jan 2017, 11:41, edited 2 times in total.
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EgmatQuantExpert



The figure above shows the railway tracks between three cities A, B and C. The distance between city A and city B is 100 kilometres and city A and city C is 240 kilometres. A train can travel from city A to city B via city C only and vice versa. Two trains are scheduled to start from city A and city B respectively at 9 AM and travel to city  B and city A respectively. The speeds of the train starting from city A and city B are 20 kilometres per hour and 25 kilometres per hour respectively. If the train from city A starts half an hour early and the train from city B starts half an hour late, at what time will the two trains meet?

    A. 7:10 PM
    B. 8:10 PM
    C. 8:23 PM
    D. 8:40 PM
    E. 8:53 PM

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length of BC =260(right triangle properties)
distance between A to B via c =240+260= 500km
Since a train leaves 1/2hr early and another 1/2hr late ,we can combined say first train leave 1 hr early ie 8:00 AM
distance travelled by first train in this 1 hr =40km
thus they meet on time =distance between them /relative speed
=(500-20)/(20+25)= 10.66
thus they meet after 10hrs 40 mints after starting
09:30 AM +10hrs 40 mints = 8:10PM

Ans B
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Word Problem - The figure above shows the railway tracks between three Updated on: 28 Jan 2017, 00:43
Originally posted by EgmatQuantExpert on 22 Jan 2017, 01:36.
Last edited by EgmatQuantExpert on 28 Jan 2017, 00:43, edited 1 time in total.
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The official solution has been posted. Looking forward to a healthy discussion..:)
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Word Problem - The figure above shows the railway tracks between three 22 Jan 2017, 11:26
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Ans is B.. As we need to subtract 20 km from 500km as in one hour train A travelled 20 kms.
So 480/45= 10 2/3 ie 10 hrs and 40 minutes.
9:30+ 10 hrs 40 minutes= 8:10 PM
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Word Problem - The figure above shows the railway tracks between three 22 Jan 2017, 11:45
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Distance from B to C = \(\sqrt{100^2+240^2}\) = 260.

So total distance between B and A via C = 260+240 = 500

Both trains are supposed to start at 9:00 AM. However, Train from A starts 0.5 hours early(8:30 AM) and Train from B starts 0.5 hrs late(9:30AM).

So at 9:30 AM when Train from B is about to start, Train from A has already travelled for 1 hour. Distance covered by train A in hour = 20 km (20km/h * 1 hour).

At 9:30 AM, distance between Train from A and Train from B = 500-20 = 480.

So time required to meet = 480/(20+25) = 480/45 = 32/3 = 10 2/3 hours or 10 hours 40 mins.

So time at which they meet = 9:30 AM + 10hours 40mins = 8:10PM

Ans B
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Word Problem - The figure above shows the railway tracks between three 28 Jan 2017, 00:41
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Hey,

PFB the official solution.

Given:

    • Distance between City A and City B = \(100\)kilometers
    • Distance between City A and City C = \(240\)kilometers
      o So, distance between city B and city C = \(√(100^2 + 240^2) = 260\)kilometers



    • Time at which train-1 starts from city A = \(8:30\) AM
      o (Speed of train -1)= \(20\) kilometres per hour
    • Time at which train-2 starts from city B =\(9 :30\) AM
    o (Speed of train - 2)= \(25\) kilometres per hour

    To Find:

      • The time at which the two trains would meet?

    Approach:

      1. Time at which the two trains would meet = 8: 30 AM + t hours, where t is the time for which train-1 traveled before it met Train - 2
      2. As train-2 started 1 hour late compared to train-1, train-2 would have traveled (t-1) hours before it met Train - 1.
      3. The total distance traveled by both the trains = (Distance traveled by train-1 from A to the meeting point) + (Distance traveled by train-2 from point B to the meeting point)
        o Irrespective of the point at which the two trains would meet, the total distance travelled by both the trains would be the total distance between A and B. As the trains can travel from A to B via point C only, the total distance between A and B can be written as sum of AC and BC
        o So, total distance travelled by both the trains = AC + BC.
      4. As we know the speeds of both the trains and the distance covered by both the trains, we can find the value of t using Distance = Speed * Time

    Working Out:

      1. (Distance covered by train-1 in t hours) = \(20 * t\)
      2. (Distance covered by train-2 in (t-1) hours) = \(25 *(t-1)\)
      3. Total distance covered by the trains = AC + BC = \(240 + 260 = 500\) kilometres. So, we can write
        • \(500 = 20t + 25(t - 1)\)
      Solving this gives us \(t = (11 + \frac{2}{3})\)hours = \(11\) hours + \(40\) minutes
      4. Thus, the two trains would meet at \(8:30\) AM \(+ 11\) hours \(+ 40\) minutes \(= 8:10\) PM

    Answer: B


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