General Questions
Practice and master this topic with our carefully crafted questions.
| Relative speed = | = (45 + 30) km/hr | |||||||
|
||||||||
|
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.
Required time = |
![]() |
500 x | 6 | ![]() |
= 24 sec. |
| 125 |
| Speed of the first train = | ![]() |
120 | ![]() |
m/sec = 12 m/sec. |
| 10 |
| Speed of the second train = | ![]() |
120 | ![]() |
m/sec = 8 m/sec. |
| 15 |
Relative speed = (12 + 8) = 20 m/sec.
Required time = |
![]() |
(120 + 120) | ![]() |
sec = 12 sec. |
| 20 |
Let the speed of the second train be x km/hr.
| Relative speed | = (x + 50) km/hr | |||||||
|
||||||||
|
Distance covered = (108 + 112) = 220 m.
|
220 | = 6 | ||
|
250 + 5x = 660
x = 82 km/hr.
Relative speed = (40 - 20) km/hr
| = | ![]() |
20 x | 5 | ![]() |
m/sec = | ![]() |
50 | ![]() |
m/sec. |
| 18 | 9 |
Length of faster train
| = | ![]() |
50 | x 5 | ![]() |
m = | 250 | m = 27 | 7 | m. |
| 9 | 9 | 9 |
4.5 km/hr :
| = | ![]() |
4.5 x | 5 | ![]() |
m/sec = | 5 | m/sec = 1.25 m/sec, and |
| 18 | 4 |
5.4 km/hr :
| = | ![]() |
5.4 x | 5 | ![]() |
m/sec = | 3 | m/sec = 1.5 m/sec. |
| 18 | 2 |
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 :
| = | ![]() |
22.5 x | 18 | ![]() |
km/hr = 81 km/hr. |
| 5 |
| 2 kmph = | ![]() |
2 x | 5 | ![]() |
m/sec = | 5 | m/sec. |
| 18 | 9 |
| 4 kmph = | ![]() |
4 x | 5 | ![]() |
m/sec = | 10 | m/sec. |
| 18 | 9 |
Let the length of the train be x metres and its speed by y m/sec.
| Then, | ![]() |
x | ![]() |
= 9 and | ![]() |
x | ![]() |
= 10. |
|
|
9y - 5 = x and 10(9y - 10) = 9x
9y - x = 5 and 90y - 9x = 100.
On solving, we get: x = 50.
Length of the train is 50 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.
20x + 25(x - 1) = 110
45x = 135
x = 3.
So, they meet at 10 a.m.
Let the length of the first train be x metres.
| Then, the length of the second train is | ![]() |
x | ![]() |
metres. |
| 2 |
Relative speed:
| = (48 + 42) kmph = | ![]() |
90 x | 5 | ![]() |
m/sec = 25 m/sec. |
| 18 |
|
[x + (x/2)] | = 12 or | 3x | = 300 or x = 200. |
| 25 | 2 |
Length of first train = 200 m.
Let the length of platform be y metres.
Speed of the first train :
| = | ![]() |
48 x | 5 | ![]() |
m/sec = | 40 | m/sec. |
| 18 | 3 |
(200 + y) x |
3 | = 45 |
| 40 |
600 + 3y = 1800
y = 400 m.
Let us name the trains as A and B. Then,
(A's speed) : (B's speed) = b : a = 16 : 9 = 4 : 3.
Relative speed = (120 + 80) km/hr
| = | ![]() |
200 x | 5 | m/sec |
| 18 |
| = | ![]() |
500 | m/sec. |
| 9 |
Let the length of the other train be x metres.
| Then, | x + 270 | = | 500 |
| 9 | 9 |
x + 270 = 500
x = 230.



![]](https://learnfrenzy.com/assets/images/aptitude/common/1-sym-cbracket-h1.gif)

