A person walks from his home to his office at 4.5 km/h and reaches 12 minutes late. If he walks at 6 km/h, he reaches 3 minutes early. Find the distance. MCQ with Answer and Explanation
A person walks from his home to his office at 4.5 km/h and reaches 12 minutes late. If he walks at 6 km/h, he reaches 3 minutes early. Find the distance.
A. 4.5 km
B. 5.0 km
C. 5.5 km
D. 4.0 km
Answer: Option A
Solution (By JKExamLibrary)
Time difference = 12 - (-3) = 15 minutes = 15/60 = 0.25 hours. Let distance be d. d / 4.5 - d / 6 = 0.25 => (4d - 3d) / 18 = 0.25 => d / 18 = 0.25 => d = 4.5 km.
A person sets off from location M at 6:00 AM tracking at 30 km/h. Another starts from M at 8:00 AM at 50 km/h along the same direction pathway. At what distance from M do they meet?
Explanation:
By 8:00 AM, the first person tracks for 2 hours, covering 30 * 2 = 60 km. Relative tracking speed = 50 - 30 = 20 km/h. Time taken to meet = 60 / 20 = 3 hours. Distance covered from point M = Speed of second traveler * Meeting Time = 50 * 3 = 150 km.
Two standard vehicles depart a warehouse at the same instant at a right angle. Their speeds are 45 km/h and 60 km/h. Find the direct distance separating them after 20 minutes.
Explanation:
Time = 20 minutes = 1/3 hours. Distance covered by first vehicle = 45 * (1/3) = 15 km. Distance covered by second vehicle = 60 * (1/3) = 20 km. Direct line distance = sqrt(15^2 + 20^2) = sqrt(225 + 400) = sqrt(625) = 25 km.
Excluding operational corridor stops, the speed of an inter-provincial bus is 60 km/h, and including stops it tracks at 48 km/h. How many minutes does it stop per hour?
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