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29 Jun 2017, 22:16
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Let the distance be \(D\), and the original speed of the train before the accident be \(S\) km/h.
\(\frac{50}{S} + \frac{4(D-50)}{3S} = \frac{35}{60}\) hours late
Had the accident occurred 24 km farther - had the train traveled 74 km while the accident took place.
\(\frac{74}{S} + \frac{4(D-74)}{3S} = \frac{25}{60}\) hours late
(i) - (ii):
\((\frac{50}{S} + \frac{4D-200}{3S}) - (\frac{74}{S} + \frac{4D-296}{3S}) = \frac{1}{6}\)
\((\frac{150 + 4D - 200}{3S}) - (\frac{222 + 4D - 296}{3S}) = \frac{1}{6}\)
\(\frac{150 + 4D - 200 - 222 - 4D + 296}{3S} = \frac{1}{6}\)
\(\frac{24}{3S} = \frac{1}{6}\)
\(3S = 24*6\)
\(S = 48 km/h\).
\(\frac{50}{S} + \frac{4(D-50)}{3S} = \frac{35}{60}\) hours late
Had the accident occurred 24 km farther - had the train traveled 74 km while the accident took place.
\(\frac{74}{S} + \frac{4(D-74)}{3S} = \frac{25}{60}\) hours late
(i) - (ii):
\((\frac{50}{S} + \frac{4D-200}{3S}) - (\frac{74}{S} + \frac{4D-296}{3S}) = \frac{1}{6}\)
\((\frac{150 + 4D - 200}{3S}) - (\frac{222 + 4D - 296}{3S}) = \frac{1}{6}\)
\(\frac{150 + 4D - 200 - 222 - 4D + 296}{3S} = \frac{1}{6}\)
\(\frac{24}{3S} = \frac{1}{6}\)
\(3S = 24*6\)
\(S = 48 km/h\).
29 Jun 2017, 22:43
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Blackbox
1. Let y be the speed of the train
2. The train can save 10 minutes by traveling the 24 km at its original speed instead of the reduced speed
3. 24/(3y/4)-24/y=10/60, So y=48km/hr
22 Jul 2018, 18:35
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Blackbox
We can let the speed of the train be r kmph and the distance to its destination be d km. We can create the following equations for the time:
50/r + (d - 50)/(3r/4) = d/r + 35/60
and
74/r + (d - 74)/(3r/4) = d/r + 25/60
Multiplying both equation by 60r, we have:
3000 + 80(d - 50) = 60d + 35r
and
4440 + 80(d - 74) = 60d + 25r
Subtracting these two equations, we have:
-1440 + 1920 = 10r
480 = 10r
48 = r
Answer: C
a bit long 
