Test: Problems On Trains- 2 - CUET Commerce MCQ

45*(5/18) = 12.5 
=>(130+x)/12.5 = 30 
=>(130+x) = 375 
=> x = 375 - 130 
=>x = 245

When a moving train with a velocity V₁ passes a man with velocity V₂, then,
Relative velocity of train w.r.t man V= V₁-V₂
Velocity of the man V2 = 2 km/h = 2000/(60x60) = (5/9) m/s
Thus, V = V₁ - (5/9)
Distance (d) = Time (t) x velocity (V)
d=150 m
t = 3 sec
V = V₁- (5/9)
Hence,
150 = 3 x [V₁- (5/9)]
50 = V₁- (5/9)
V₁ = 50+ (5/9)
On simplification,
V₁ = (455/9) m/s = (455/9)x(3600/1000) km/h= 182 km/h
Speed of the train= 182 km/h
 

Speed = 240/24 m/sec
= 10m/sec
∴ Requiredtime = (240+650)/10sec.
= 89sec.

Formula For Converting From Km/hr to m/s:
Xkm/hr = X × 5/18m/s
Therefore, Speed = 45 × 5/18m/sec
= 252m/sec
Total Distance To Be covered = (360+140)m=500m
Formula for finding
Time = Distance/Speed
∴ Requiredtime = (500×20/25sec
= 40sec.

Let the speeds of the two trains be x m/sec and y m/sec respectively.
Then, length of the first train = 27x metres,
Length of the second train = 17y metres.
[because distance = speed*time]
(27x+17y)/(x+y) = 23
=> 27x + 17y = 23x + 23y
=> 4x = 6y
=> x/y = 6/4

Speed of train relative to jogger = (45 - 9) km/hr = 36 km/hr.
= (36 x 5/18)m/sec
= 10 m/sec.
Distance to be covered = (240 + 120) m = 360 m.
Time taken = 360/10
= 36 sec.
 

Let the length of each train be x metres
Then, distance covered = 2x metres.
Relative speed =(46−36)km/hr
= [10 * 5/8 * 10]m/sec
= 25/9 m/sec
2x/36 = 25/9
x = 50
Length of each train is 50m.
 

Relative speed = (60+40)km/hr
= (100×5/18) m/sec
= (250/9) m/sec
Distance covered in crossing each other =(140+160)m
= 300m
Required time = Distance/Time 
=> 300/(250/9) 
= 300*(9/250)
= 54/5
= 10.8 sec

Relative speed (60+90)km/hr
= 150km/hr
= 150 × 5/18 m/sec
= 125/3 m/sec
Distance covered by the slower train to cross the faster train =(1.1+0.9)km
= 2km  = 2000m
The time taken by the slower train to cross the faster train in second = distance/speed = 2000/(125/3)
= 16 * 3
= 48 sec