Two cars travel the same distance starting at 10:00 am and 11:00 am, respectively, on the same day. They reach their common destination at the same point of time. If the first car travelled for at least 6 hours, then the highest possible value of the percentage by which the speed of the second car could exceed that of the first car is?
Question
Two cars travel the same distance starting at 10:00 am and 11:00 am, respectively, on the same day. They reach their common destination at the same point of time. If the first car travelled for at least 6 hours, then the highest possible value of the percentage by which the speed of the second car could exceed that of the first car is?
Solution
To solve this problem, we need to first understand the given conditions.
- Both cars travel the same distance.
- The first car starts at 10:00 am and travels for at least 6 hours.
- The second car starts at 11:00 am and reaches the destination at the same time as the first car.
Given these conditions, we can infer that the second car travels for one hour less than the first car, since it starts one hour later but arrives at the same time.
Let's denote the speed of the first car as v1, the speed of the second car as v2, and the time the first car travels as t (which is at least 6 hours).
Since both cars travel the same distance, we can write the equation: v1t = v2(t-1).
We want to find the highest possible value of the percentage by which the speed of the second car could exceed that of the first car, which is (v2-v1)/v1 * 100%.
Substituting v2 from the equation above, we get ((v1*(t-1))/v1 - v1)/v1 * 100% = ((t-1)-1)*100% = (t-2)*100%.
Since t is at least 6 hours, the highest possible value of the percentage by which the speed of the second car could exceed that of the first car is (6-2)*100% = 400%.
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