Walking at 5/7 of his usual speed, a student reaches his exam center 12 minutes late. What is his exact usual time to reach the center? MCQ with Answer and Explanation
Walking at 5/7 of his usual speed, a student reaches his exam center 12 minutes late. What is his exact usual time to reach the center?
A. 30 minutes
B. 32 minutes
C. 26 minutes
D. 35 minutes
Answer: Option A
Solution (By JKSSB Mock Tests)
New speed = 5/7 of usual speed => New time = 7/5 of usual time. Difference = 2/5 of usual time = 12 minutes. 1/5 of usual time = 6 minutes => Usual time = 6 * 5 = 30 minutes.
Explanation:
Let each third segment distance be 40 km. Total sector distance = 120 km. Total time = 40/20 + 40/40 + 40/40 = 2 + 1 + 1 = 4 hours. Average speed across route = 120 / 4 = 30 km/h.
Two trains 140 meters and 160 meters long are running at the speeds of 60 km/h and 40 km/h respectively in opposite directions on parallel tracks. What is the time taken by them to cross each other completely?
A boat goes 6 km upstream and returns back to the starting point in 2 hours. If the speed of the stream is 4 km/h, find the speed of the boat in still water.
Explanation:
Let speed of boat in still water be v. 6 / (v - 4) + 6 / (v + 4) = 2 => 3 / (v - 4) + 3 / (v + 4) = 1 => 3(v + 4 + v - 4) = v^2 - 16 => 6v = v^2 - 16 => v^2 - 6v - 16 = 0. Solving the quadratic equation gives (v - 8)(v + 2) = 0. Since speed must be positive, v = 8 km/h.
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