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