Two passenger trains of equal layout lengths cross a station pole in 15 seconds and 30 seconds respectively. If each train length measures 150 meters, find the time parameter required to pass each other completely when tracking in opposite directions. MCQ with Answer and Explanation
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Question #1
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A car covers a distance of 450 km at a certain speed. If the speed is increased by 5 km/h, it takes 1 hour less to cover the same distance. Find the original speed of the car.
Correct Answer: Option C
Explanation:
Let original speed be s. 450/s - 450/(s+5) = 1 => 450(s + 5 - s) = s(s + 5) => 2250 = s(s + 5). Solving s^2 + 5s - 2250 = 0 gives (s - 45)(s + 50) = 0. Since speed is positive, s = 45 km/h.
Let original speed be s. 450/s - 450/(s+5) = 1 => 450(s + 5 - s) = s(s + 5) => 2250 = s(s + 5). Solving s^2 + 5s - 2250 = 0 gives (s - 45)(s + 50) = 0. Since speed is positive, s = 45 km/h.
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Maths
Time Speed and Distance
Question #2
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A man covers 2/3 of a journey at 40 km/h, 1/4 of the journey at 30 km/h and the remainder at 10 km/h. What is his average speed for the entire journey?
Correct Answer: Option D
Explanation:
Let the total distance be 120 km. First part = (2/3) * 120 = 80 km; Time = 80 / 40 = 2 hours. Second part = (1/4) * 120 = 30 km; Time = 30 / 30 = 1 hour. Remainder = 120 - 80 - 30 = 10 km; Time = 10 / 10 = 1 hour. Total time = 2 + 1 + 1 = 4 hours. Average speed = 120 / 4 = 32 km/h.
Let the total distance be 120 km. First part = (2/3) * 120 = 80 km; Time = 80 / 40 = 2 hours. Second part = (1/4) * 120 = 30 km; Time = 30 / 30 = 1 hour. Remainder = 120 - 80 - 30 = 10 km; Time = 10 / 10 = 1 hour. Total time = 2 + 1 + 1 = 4 hours. Average speed = 120 / 4 = 32 km/h.
This question belongs to:
Maths
Time Speed and Distance
Question #3
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A car covered a distance of 400 km at a uniform speed. If the uniform speed index is lowered by 10 km/h, the journey takes 2 hours longer. Find the original speed.
Correct Answer: Option D
Explanation:
Let original speed be s. 400 / (s - 10) - 400 / s = 2 => 200 / (s - 10) - 200 / s = 1 => 2000 = s(s - 10). Solving s^2 - 10s - 2000 = 0 yields (s - 50)(s + 40) = 0. Original speed = 50 km/h.
Let original speed be s. 400 / (s - 10) - 400 / s = 2 => 200 / (s - 10) - 200 / s = 1 => 2000 = s(s - 10). Solving s^2 - 10s - 2000 = 0 yields (s - 50)(s + 40) = 0. Original speed = 50 km/h.
This question belongs to:
Maths
Time Speed and Distance
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