Quantitative Aptitude
TRAINS MCQs
Problems On Trains
Relative speed = (120 + 80) km/hr
=\(\left(200\times\frac{5}{18}\right)m/sec\)
=\(\left(\frac{500}{9}\right)m/sec\)
Let the length of the other train be x metres.
Then,\(\frac{x+270}{9}=\frac{500}{9}\)
x + 270 = 500
x = 230.
Speed =\(\left(72\times\frac{5}{18}\right)m/sec=20m/sec\)
Time = 26 sec.
Let the length of the train be x metres.
Then, \(\frac{x+250}{26} = 20\)
x + 250 = 520
x = 270.
Let the speed of the slower train be x m/sec.
Then, speed of the faster train = 2x m/sec.
Relative speed = (x + 2x) m/sec = 3x m/sec.
\(\left(\frac{100+100}{8}\right) = 3x\)
24x = 200
x=\(\frac{25}{3}.\)
So, speed of the faster train = \(\frac{50}{3}m/sec\)
= \(\left(\frac{50}{3}\times\frac{18}{5}\right)km /hr\)
= 60 km/hr.
Relative speed = (60 + 40) km/hr = \(\left(100\times\frac{5}{18}\right)m/sec=\left(\frac{250}{9}\right)m/sec\)
Distance covered in crossing each other = (140 + 160) m = 300 m.
Required time = \(\left(300\times\frac{9}{250}\right)sec= \frac{54}{5}sec=10.8sec.\)
Speed of train relative to man = (60 + 6) km/hr = 66 km/hr.
= \(\left(66\times\frac{5}{18}\right)m/sec\)
=\(\left(\frac{55}{3}\right)m/sec.\)
Thairfor Time taken to pass the man = \(\left(100\times\frac{3}{55}\right)sec = 6sec.\)
Total distance covered = \(\left(\frac{7}{2}+\frac{1}{4}\right)miles\)
=\(\frac{15}{4}miles.\)
Thairfor Time taken = \(\left(\frac{15}{4\times75}\right)hrs\)
= \(\frac{1}{20}hrs\)
=\(\left(\frac{1}{20}\times60\right)min.\)
= 3 min.
Speed =\(\left(78\times\frac{5}{18}\right)m/sec = \left(\frac{65}{3}\right)m/sec.\)
Time = 1 minute = 60 seconds.
Let the length of the tunnel be x metres.
Then,\(\left(\frac{800+x}{60}\right) = \frac{65}{3}\)
3(800 + x) = 3900
x = 500.
Speed = \(\left(\frac{300}{18}\right)m/sec = \frac{50}{3}m/sec.\)
Let the length of the platform be x metres.
Then, \(\left(\frac{x+300}{39}\right) = \frac{50}{3}\)
3(x + 300) = 1950
x = 350 m.
Let the length of the train be x metres and its speed by y m/sec.
Then,\(\frac{x}{y} = 15 \Rightarrow y = \frac{x}{15}\)
Thairfor \(\frac{x+100}{25} = \frac{x}{15}\)
15(x + 100) = 25x
15x + 1500 = 25x
1500 = 10x
x = 150 m.
Let the length of the train be x metres and its speed by y m/sec.
Then, \(\frac{x}{y} = 8 \Rightarrow x = 8y\)
Now,\(\frac{x+264}{20} = y\)
8y + 264 = 20y
y = 22.
Speed = \(22m/sec = \left(22\times\frac{18}{5}\right)km/hr = 79.2 km/hr\)