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 |
= | ❨ | 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 |
= | ❨ | 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, 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 |
Total distance covered |
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∴ Time taken |
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= 3 min. |
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.
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 = | ❨ | 240 | ❩m/sec = 10 m/sec. |
24 |
∴ Required time = | ❨ | 240 + 650 | ❩sec = 89 sec. |
10 |
(A's speed) : (B's speed) = √b : √a = √16 : √9 = 4 : 3.
= | ❨ | 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 |
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.
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