In a 100-meter race, A can beat B by 20 meters and B can beat C by 10 meters. By how many meters can A beat C in the same race? MCQ with Answer and Explanation
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Question #1
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A person travels from P to Q at a speed of 30 km/h and returns from Q to P at a speed of 40 km/h. What is the ratio of time taken by him in the forward journey to that in the return journey?
Correct Answer: Option B
Explanation:
Since distance is constant, time is inversely proportional to speed. Ratio of times = Speed2 : Speed1 = 40 : 30 = 4:3.
Since distance is constant, time is inversely proportional to speed. Ratio of times = Speed2 : Speed1 = 40 : 30 = 4:3.
This question belongs to:
Maths
Time Speed and Distance
Question #2
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An athlete loops a regular hexagonal field of side 50 meters. If his speed is 15 km/h, find the time taken to complete one full lap.
Correct Answer: Option C
Explanation:
Perimeter of the hexagonal field = 6 * 50 = 300 meters. Speed = 15 km/h = 15 * (5/18) = 25/6 m/s. Time = Distance / Speed = 300 / (25/6) = 300 * 6 / 25 = 12 * 6 = 72 seconds.
Perimeter of the hexagonal field = 6 * 50 = 300 meters. Speed = 15 km/h = 15 * (5/18) = 25/6 m/s. Time = Distance / Speed = 300 / (25/6) = 300 * 6 / 25 = 12 * 6 = 72 seconds.
This question belongs to:
Maths
Time Speed and Distance
Question #3
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Walking at 3/5 of his typical pace, a technician reaches his field assignment 16 minutes late. Find his typical baseline time.
Correct Answer: Option A
Explanation:
New speed = 3/5 of normal speed => New time = 5/3 of normal time. Difference = 2/3 of normal time = 16 minutes. Normal baseline time = (16 * 3) / 2 = 24 minutes.
New speed = 3/5 of normal speed => New time = 5/3 of normal time. Difference = 2/3 of normal time = 16 minutes. Normal baseline time = (16 * 3) / 2 = 24 minutes.
This question belongs to:
Maths
Time Speed and Distance
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