Formula for converting from km/hr to m/s: X km/hr = | ❨ | X x | 5 | ❩ | m/s. |
18 |
Therefore, Speed = | ❨ | 45 x | 5 | ❩m/sec | = | 25 | m/sec. |
18 | 2 |
Total distance to be covered = (360 + 140) m = 500 m.
Formula for finding Time = | ❨ | Distance | ❩ |
Speed |
∴ Required time = | ❨ | 500 x 2 | ❩sec | = 40 sec. |
25 |
= | ❨ | 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.
Then, | x | = 15 ⟹ y = | x | . |
y | 15 |
∴ | x + 100 | = | x |
25 | 15 |
⟹ 15(x + 100) = 25x
⟹ 15x + 1500 = 25x
⟹ 1500 = 10x
⟹ x = 150 m.
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 |
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.
= | ❨ | 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, length of the first train = 27x metres,
and length of the second train = 17y metres.
∴ | 27x + 17y | = 23 |
x+ y |
⟹ 27x + 17y = 23x + 23y
⟹ 4x = 6y
⟹ | x | = | 3 | . |
y | 2 |
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 |
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.
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