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 |
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 |
Speed of the train relative to man | = (63 - 3) km/hr | |||||||
= 60 km/hr | ||||||||
|
||||||||
|
||||||||
∴ Time taken to pass the man |
|
|||||||
= 30 sec. |
Relative speed = (60 + 40) km/hr = | ❨ | 100 x | 5 | ❩m/sec | = | ❨ | 250 | ❩m/sec. |
18 | 9 |
Distance covered in crossing each other = (140 + 160) m = 300 m.
Required time = | ❨ | 300 x | 9 | ❩sec | = | 54 | sec = 10.8 sec. |
250 | 5 |
(A's speed) : (B's speed) = √b : √a = √16 : √9 = 4 : 3.
Then, | x | = 15 ⟹ y = | x | . |
y | 15 |
∴ | x + 100 | = | x |
25 | 15 |
⟹ 15(x + 100) = 25x
⟹ 15x + 1500 = 25x
⟹ 1500 = 10x
⟹ x = 150 m.
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.
Total distance covered |
|
|||||||||
|
∴ Time taken |
|
|||||||
|
||||||||
|
||||||||
= 3 min. |
Then, relative speed of the two trains = 2x m/sec.
So, 2x = | (120 + 120) |
12 |
⟹ 2x = 20
⟹ x = 10.
∴ Speed of each train = 10 m/sec = | ❨ | 10 x | 18 | ❩ | km/hr = 36 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.
Comments
There are no comments.Copyright ©CuriousTab. All rights reserved.