Question
A cat, running at a speed of 54 km/h and positioned 240
meters ahead of a dog, can be caught by the dog in 6(2/3) seconds. Determine the time it takes for the dog to overtake a bus traveling at 21 m/s. The bus can pass a stationary man in 21(2/3) seconds, and there is no initial separation between the bus and the dog. (Note: The lengths of the cat and the dog are considered negligible.)Solution
ATQ, Speed of the Cat = 54 × (5/18) = 15 m/s Distance covered by the dog in 6(2/3) seconds = 240 + distance covered by the cat in 6(2/3) seconds = 240 + 15 × (20/3) = 240 + 100 = 340 metres So, speed of the dog = 340 ÷ (20/3) = 51 m/s Relative speed of the dog with respect to the bus = 51 – 21 = 30 m/s Length of the bus = 21(2/3) × 21 = (65/3) × 21 = 455 metres So, time taken by the dog to overtake the bus = 455/30 = 91/6 seconds
2 4 5 19 70...
6000 3002 1503 ? 378.75 191.375 97.6875
...If 204 196 223 x 284
Then, what is the average of the numbers of the above series?
...8 24 12 ? 18 54
3 ? 7 16 71 346
...104 106 110 113 ? 126
12, 18, 28, 42, 52, ?
18 29 51 84 128 182
5, 8, 17, ?, 37, 48
(32.03 + 111.98) ÷ 18.211 = 89.9 – 20.23% of ?
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