A person walks from his villa to a local station at 4 km/h and arrives 10 minutes late. If he routes at 5 km/h, he arrives 2 minutes early. Find the distance. MCQ with Answer and Explanation
A person walks from his villa to a local station at 4 km/h and arrives 10 minutes late. If he routes at 5 km/h, he arrives 2 minutes early. Find the distance.
A. 3.5 km
B. 5.0 km
C. 4.0 km
D. 4.5 km
Answer: Option C
Solution (By JKSSB Mock Tests)
Time difference = 10 - (-2) = 12 minutes = 12/60 = 0.2 hours. Let distance be d. d/4 - d/5 = 0.2 => d/20 = 0.2 => d = 4 km.
Two cars start from the same spot and travel along perpendicular roads. One travels at 15 km/h and the other at 20 km/h. Find the distance between them after 2 hours.
Explanation:
In 2 hours, the first car travels 30 km and the second car travels 40 km. Direct distance = sqrt(30^2 + 40^2) = sqrt(900 + 1600) = sqrt(2500) = 50 km.
A thief is spotted by a policeman from a distance of 150 meters. The thief runs at 11 km/h and the policeman chases him at 12 km/h. Find the distance covered by the thief before he is caught.
Explanation:
Relative speed = 12 - 11 = 1 km/h. Time to catch the thief = 0.15 km / 1 km/h = 0.15 hours. Distance covered by thief = Speed * Time = 11 km/h * 0.15 hours = 1.65 km = 1650 meters.
Explanation:
Total distance = 30 + 40 = 70 km. Time for first part = 30 / 10 = 3 hours. Time for second part = 40 / 20 = 2 hours. Total time = 5 hours. Average speed = 70 / 5 = 14 km/h.
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