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 secondsÂ
What value should come in the place of (?) in the following questions?
√? + 10 * 18 + 16 * 4 = 280
[{70 + (40 - 22) ÷ 3} ÷ 4] = ?
What will come in place of (?) in the given expression.
{(60% of 250) + √144} ÷ (3² - 4) = ?- What will come in place of (?) in the given expression.
(0.2 × 0.5) + (0.3 × 0.4) = ? 654.056 + 28.9015 × 44.851 – 43.129 =?
800 + 900 X (3)-2 - ? = 25 X 60 ÷ 2Â
What will come in the place of question mark (?) in the given expression?
8 × 12 + 110 ÷ 5 = 72 + ?
Simplify: (1 ÷ 0.08)
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