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.
Speed = | ❨ | 45 x | 5 | ❩m/sec | = | ❨ | 25 | ❩m/sec. |
18 | 2 |
Time = 30 sec.
Let the length of bridge be x metres.
Then, | 130 + x | = | 25 |
30 | 2 |
⟹ 2(130 + x) = 750
⟹ x = 245 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 = | ❨ | 60 x | 5 | ❩m/sec | = | ❨ | 50 | ❩m/sec. |
18 | 3 |
Length of the train = (Speed x Time).
∴ Length of the train = | ❨ | 50 | x 9 | ❩m = 150 m. |
3 |
Speed = | ❨ | 300 | ❩ | m/sec = | 50 | m/sec. |
18 | 3 |
Let the length of the platform be x metres.
Then, | ❨ | x + 300 | ❩ | = | 50 |
39 | 3 |
⟹ 3(x + 300) = 1950
⟹ x = 350 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.
= | ❨ | 66 x | 5 | ❩m/sec |
18 |
= | ❨ | 55 | ❩m/sec. |
3 |
∴ Time taken to pass the man = | ❨ | 110 x | 3 | ❩sec = 6 sec. |
55 |
Then, | x | = 8 ⟹ x = 8y |
y |
Now, | x + 264 | = y |
20 |
⟹ 8y + 264 = 20y
⟹ y = 22.
∴ Speed = 22 m/sec = | ❨ | 22 x | 18 | ❩ | km/hr = 79.2 km/hr. |
5 |
= | ❨ | 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.
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.
= | ❨ | 36 x | 5 | ❩m/sec |
18 |
= 10 m/sec.
Distance to be covered = (240 + 120) m = 360 m.
Comments
There are no comments.Copyright ©CuriousTab. All rights reserved.