Train and Platform Aptitude Questions:

Answer: (c) \(130 \ km/hr\)

Solution: Given, speed of the train x \((s_1) = 65 \ km/hr\)
time taken by the train x to reach the station B \((T_1) = 20 \ hours\)
time taken by the train y to reach the station A \((T_2) = 5 \ hours\), then $$ \frac{s_1}{s_2} = \sqrt{\frac{T_2}{T_1}} $$ $$ \frac{65}{s_2} = \sqrt{\frac{5}{20}} $$ $$ \frac{65}{s_2} = \sqrt{\frac{1}{4}} $$ $$ \frac{65}{s_2} = \frac{1}{2} $$ $$ s_2 = 130 \ km/hr $$

Answer: (b) \(26 \ m/sec\)

Solution: Given, length of the train \((l_m) = 260 \ meter\)
time taken by the train to cross the train \((T) = 10 \ seconds\), then $$ time \ taken \ (T) = \frac{l_m}{s_m} $$ $$ 10 = \frac{260}{s_m} $$ $$ s_m = 26 \ m/sec $$

Answer: (d) \(480 \ km\)

Solution: Given, speed of the man from station A to B = \(40 \ km/hr\)
speed of the man from station B to A = \(60 \ km/hr\)
time taken by the man on traveling = \(20 \ hours\)
Let the distance between between the two stations = \(d \ km\), then $$ \frac{d}{40} + \frac{d}{60} = 20 $$ $$ d \ \left(\frac{1}{40} + \frac{1}{60}\right) = 20 $$ $$ d \ \left(\frac{3 + 2}{120}\right) = 20 $$ $$ d = \frac{120 \times 20}{5} = 480 \ km $$

Answer: (a) \(6.4 \ seconds\)

Solution: Given, speed of the train \((s_m) = 45 \ km/hr = 45 \times \frac{5}{18} \ m/sec\)
length of the train \((l_m) = 80 \ meter\), then $$ time \ taken \ (T) = \frac{l_m}{s_m} $$ $$ = \frac{80}{45 \times \frac{5}{18}} $$ $$ = \frac{80 \times 18}{45 \times 5} $$ $$ = \frac{1440}{225} = 6.4 \ sec $$

Answer: (b) \(225 \ km\)

Solution: Given, distance between two stations = \(500 \ km\)
speed of the train x = \(45 \ km/hr\)
speed of the train y = \(55 \ km/hr\)
time taken by the trains to meet each other = \(\frac{500}{45 + 55}\) = \(5 \ hours\)
distance from station A when both the trains will cross each other $$ = 5 \times 45 = 225 \ km $$

Answer: (c) \(1.38 \ m/sec\)

Solution: Given, length of the bridge = \(250 \ meter\)
time taken by the man to cross the bridge \(= 3 \times 60 = 180 \ seconds\), then $$ speed = \frac{distance \ covered}{time \ taken} $$ $$ = \frac{250}{180} = 1.38 \ m/sec $$

Answer: (a) \(57.35 \ km/hr\)

Solution: Given, speed of the car from city A to B \((s_1) = 75 \ km/hr\)
Average speed of the car = \(65 \ km/hr\)
Let the distance between city A and city B = \(1 \ km\), then $$ Average \ speed = \frac{distance \ covered}{time \ taken} $$ $$ Average \ speed = \frac{d_1 + d_2}{\frac{d_1}{s_1} + \frac{d_2}{s_2}} $$ $$ 65 = \frac{1 + 1}{\frac{1}{75} + \frac{1}{s_2}} $$ $$ 65 = \frac{2 \times 75 \times s_2}{(s_2 + 75)} $$ $$ 65 \ s_2 + 4875 = 150 \ s_2 $$ $$ s_2 = \frac{4875}{85} = 57.35 \ km/hr $$

Answer: (b) \(55.5 \ meter\)

Solution: Given, length of the train = \(125 \ meter\)
speed of the train = \(65 \ km/hr\) = \(65 \times \frac{5}{18} = 18.05 \ m/sec\)
time taken by the train to cross the bridge = \(10 \ seconds\)

distance covered by the train in \(10 \ seconds\) with the speed of \(18.05 \ m/sec\) = \(18.05 \times 10\) = \(180.5 \ meter\), then length of the bridge $$ = 180.5 - length \ of \ the \ train $$ $$ = 180.5 - 125 = 55.5 \ meter $$

Answer: (a) \(333.36 \ meter\)

Solution: Given, speed of the train \((s_m) = 150 \ km/hr\) = \(150 \times \frac{5}{18}\) = \(41.67 \ m/sec\)
time taken by the train to cross the pole = \(8 \ seconds\), then $$ time \ taken = \frac{l_m}{s_m} $$ $$ 8 = \frac{l_m}{41.67} $$ $$ l_m = 333.36 \ meter $$

Answer: (c) \(22.49 \ seconds\)

Solution: Given, length of the train \((l_m) = 1000 / meter\)
length of the bridge \((l_s) = 250 \ meter\)
speed of the train \((s_m) = 200 \ km/hr\) = \(200 \times \frac{5}{18} = 55.56 \ m/sec\), then $$ time \ taken = \frac{l_m + l_s}{s_m} $$ $$ = \frac{1000 + 250}{55.56} = 22.49 \ seconds $$