= | ❨ | 150 x | 5 | ❩m/sec |
18 |
= | ❨ | 125 | ❩m/sec. |
3 |
Distance covered = (1.10 + 0.9) km = 2 km = 2000 m.
Required time = | ❨ | 2000 x | 3 | ❩sec = 48 sec. |
125 |
= | ❨ | 66 x | 5 | ❩m/sec |
18 |
= | ❨ | 55 | ❩m/sec. |
3 |
∴ Time taken to pass the man = | ❨ | 110 x | 3 | ❩sec = 6 sec. |
55 |
= | ❨ | 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.
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 |
= | ❨ | 36 x | 5 | ❩m/sec |
18 |
= 10 m/sec.
Distance to be covered = (240 + 120) m = 360 m.
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 = | ❨ | 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 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 |
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.
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 = | ❨ | 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 |
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