Quantitative Aptitude
TRAINS MCQs
Problems On Trains
Total Questions : 842
| Page 6 of 85 pages
Answer: Option B. -> 4 : 3
Let us name the trains as A and B. Then,
(A's speed) : (B's speed) = b : a = 16 : 9 = 4 : 3.
Answer: Option A. -> 48 sec
 - Relative sped = (60 + 90) km/hr
= [150 x 5/18] m/sec = [125 / 3] m/sec.
Distance covered = (1.10 + 0.9) km = 2 km = 2000 m.
Required time = [2000 x 3/125] sec = 48 sec.
 - Relative sped = (60 + 90) km/hr
= [150 x 5/18] m/sec = [125 / 3] m/sec.
Distance covered = (1.10 + 0.9) km = 2 km = 2000 m.
Required time = [2000 x 3/125] sec = 48 sec.
Answer: Option C. -> 10.8 sec
 - Relative speed = (60 + 40) km/hr = [100 x 5/18] m/sec
= [250/9] m/sec.
Distance covered in crossing each other = (140 + 160) m = 300 m.
Required time = [300 x 9/250] sec = 54/5 sec
= 10.8 sec.
 - Relative speed = (60 + 40) km/hr = [100 x 5/18] m/sec
= [250/9] m/sec.
Distance covered in crossing each other = (140 + 160) m = 300 m.
Required time = [300 x 9/250] sec = 54/5 sec
= 10.8 sec.
Answer: Option D. -> 230
 - Relative speed = (120 + 80) km/hr
[200 x 5/18] m/sec = [500/9] m/sec.
Let the length of the other tain be x metres.
Then, x + 270 / 9 = 500/9
⇔ x + 270 = 500
⇔ x = 230.
 - Relative speed = (120 + 80) km/hr
[200 x 5/18] m/sec = [500/9] m/sec.
Let the length of the other tain be x metres.
Then, x + 270 / 9 = 500/9
⇔ x + 270 = 500
⇔ x = 230.
Answer: Option B. -> 50
 - Let the length of each tain be x metres.
Then, distance covered = 2x metres.
Relaive speed = (46 - 36) km/hr = [10 x 5/18] m/sec = [25/9] m/sec.
∴ 2x/36 = 25/9
⇔ 2x = 100
⇔ x = 50.
 - Let the length of each tain be x metres.
Then, distance covered = 2x metres.
Relaive speed = (46 - 36) km/hr = [10 x 5/18] m/sec = [25/9] m/sec.
∴ 2x/36 = 25/9
⇔ 2x = 100
⇔ x = 50.
Answer: Option D. -> 50 km/hr
 - Speed of the train relative to man = [125/10] m/sec = [25/2] m/sec.
= [25/2 x 18/5] km/hr = 45 km/hr.
Let the speed of the train be x kmph.
Then, relative speed = (x - 5) kmph.
∴ x - 5 = 45 or x = 50 kmph.
 - Speed of the train relative to man = [125/10] m/sec = [25/2] m/sec.
= [25/2 x 18/5] km/hr = 45 km/hr.
Let the speed of the train be x kmph.
Then, relative speed = (x - 5) kmph.
∴ x - 5 = 45 or x = 50 kmph.
Answer: Option C. -> 36 sec
 - Speed of train relative to jogger = (45 - 9) km/hr = 36 km/hr
= 36 x 5 m/sec 18
= 10 m/sec.
Distance to be covered = (240 + 120) m = 360 m
Time taken = 360 sec = 36 sec. 10
 - Speed of train relative to jogger = (45 - 9) km/hr = 36 km/hr
= 36 x 5 m/sec 18
= 10 m/sec.
Distance to be covered = (240 + 120) m = 360 m
Time taken = 360 sec = 36 sec. 10
Answer: Option A. -> 230 m
 - Relative speed = (120 + 80) km/hr
= 200 x 5 m/sec 18 = 500 m/sec. 9
 - Relative speed = (120 + 80) km/hr
= 200 x 5 m/sec 18 = 500 m/sec. 9
Answer: Option D. -> 270 m
 - Speed = 72 x 5 m/sec = 20 m/sec. 18
Time = 26 sec.
Let the length of the train be x metres.
Then, x + 250 = 20 26
x + 250 = 520
x = 270.
 - Speed = 72 x 5 m/sec = 20 m/sec. 18
Time = 26 sec.
Let the length of the train be x metres.
Then, x + 250 = 20 26
x + 250 = 520
x = 270.
Answer: Option C. -> 60 km/hr
 - 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.
(100 + 100) = 3x 8
24x = 200
x = 25 . 3 So, speed of the faster train = 50 m/sec 3 = 50 x 18 km/hr 3 5
= 60 km/hr.
 - 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.
(100 + 100) = 3x 8
24x = 200
x = 25 . 3 So, speed of the faster train = 50 m/sec 3 = 50 x 18 km/hr 3 5
= 60 km/hr.