Quantitative Aptitude
TRAINS MCQs
Problems On Trains
Speed of the train relative to man
= (63 - 3) km/hr
= 60 km/hr
= \(\left(60\times\frac{5}{18}\right)m/sec\)
=\(\left(\frac{50}{3}\right)m/sec\)
Thairfor Time taken to pass the man = \(\left(500\times\frac{3}{50}\right)sec\)
= 30 sec.
Relative speed = \(\left(45+30\right)m/sec\)
= \(\left(75\times\frac{5}{18}\right)m/sec\)
=\(\left(\frac{125}{6}\right)m/sec.\)
We have to find the time taken by the slower train to pass the DRIVER of the faster train and not the complete train.
So, distance covered = Length of the slower train.
Therefore, Distance covered = 500 m.
Therefore. = \(\left(500\times\frac{6}{125}\right) = 24sec.\)
Let the speed of each train be x m/sec.
Then, relative speed of the two trains = 2x m/sec.
So,2x= \(\frac{(120+120)}{12}\)
2x = 20
x = 10.
Therefor Speed of each train = 10 m/sec = \(\left(10\times\frac{18}{5}\right)km/hr =36 km/hr\)
Speed of the first train = \(\left(\frac{120}{10}\right)m/sec = 12m/sec.\)
Speed of the second train =\(\left(\frac{120}{15}\right)m/sec = 8m/sec\)
Relative speed = (12 + 8) = 20 m/sec.
Therefore Required time = \(\left[\frac{(120+120)}{20}\right]sec = 12sec.\)
Let the speed of the second train be x km/hr.
Relative speed = \(\left(x+50\right)km/hr\)
= \(\left[(x+50)\times\frac{5}{18}\right]m/sec\)
=\(\left[\frac{250+5x}{18}\right]m/sec.\)
Distance covered = (108 + 112) = 220 m.
\(\therefore \frac{220}{\left[\frac{250+5x}{18}\right]}=6\)
250 + 5x = 660
x = 82 km/hr.
Relative speed = (40 - 20) km/hr = \(\left(20\times\frac{5}{18}\right)m/sec=\left(\frac{50}{9}\right)m/sec\)
Therefore Length of faster train = \(\left(\frac{50}{9}\times5\right)m = \frac{250}{9}m=27\frac{7}{9}\)
2 kmph = \(\left(2\times\frac{5}{18}\right)m/sec= \frac{5}{9}m/sec\)
4 kmph =\(\left(4\times\frac{5}{18}\right)m/sec = \frac{10}{9}m/sec\)
Let the length of the train be x metres and its speed by y m/sec.
Then, \(\left[\frac{x}{y-\frac{5}{9}}\right]=9 and
\left[\frac{x}{y-\frac{10}{9}}\right] = 10\)
Therefore 9y - 5 = x and 10(9y - 10) = 9x
9y - x = 5 and 90y - 9x = 100.
On solving, we get: x = 50.
Therefore Length of the train is 50 m.
A train overtakes two persons walking along a railway track. The first one walks at 4.5 km/hr. The other one walks at 5.4 km/hr. The train needs 8.4 and 8.5 seconds respectively to overtake them. What is the speed of the train if both the persons are walking in the same direction as the train?
4.5 km/hr =\(\left(4.5\times\frac{5}{18}\right)m/sec=\frac{5}{4}m/sec = 1.25m/sec, and\)
5.4 km/hr =\(\left(5.4\times\frac{5}{18}\right)m/sec=\frac{3}{2}m/sec = 1.5m/sec,\)
Let the speed of the train be x m/sec.
Then, (x - 1.25) x 8.4 = (x - 1.5) x 8.5
8.4x - 10.5 = 8.5x - 12.75
0.1x = 2.25
x = 22.5
Speed of the train =\(\left(22.5\times\frac{18}{5}\right)km/hr = 81km/hr.\)
Let the length of the first train be x metres.
Then, the length of the second train is\(\left(\frac{x}{2}\right)metres.\)
Relative speed = (48 + 42) kmph =\(\left(90\times\frac{5}{18}\right)m/sec = 25m/sec\)
\(\frac{[x+(\frac{x}{2})]}{25}=12 or \frac{3x}{2}=300 . or . x= 200\)
Therefore Length of first train = 200 m.
Let the length of platform be y metres.
Speed of the first train = \(\left(48\times\frac{5}{18}\right)m/sec = \frac{40}{3}m/sec\)
\(\therefore\left(200+y\right)\times\frac{3}{40}= 45\)
600 + 3y = 1800
y = 400 m.
Suppose they meet x hours after 7 a.m.
Distance covered by A in x hours = 20x km.
Distance covered by B in (x - 1) hours = 25(x - 1) km.
Therefore 20x + 25(x - 1) = 110
45x = 135
x = 3.
So, they meet at 10 a.m.