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The minimum time travelled by Car 1 = 6 hours.
Minimum time travelled by Car 2 = 5 hours
Now, the time taken can be of any value -> (6,5) (7,6) ...... (1000,999)
However, since we want to calculate the highest possible percentage by which speed of car 2 could exceed car 1, we consider the lowest time taken - 6 hours and 5 hours, respectively.
Let the distance between the start and finish be 30 Kms (LCM of 6 and 5)
So, Car 1 travels at 5 Kmph and Car 2 travels at 6 Kmph
Percentage increase in speed by Car 2 = x 100 = x 100 = 20%
Circumference of A and B are in the ratio 3 : 4
So, Ratio of Distance travelled in one revolution by A and B = 3: 4
Since they travel the same distance,
Ratio of number of revolutions of A and B = 4 : 3 ...(1)
We know that each wheel of A requires 5000 more revolutions than B
So, the Ratio of number of revolutions of A and B = (n + 5000) : n ...(2)
So, comparing (1) and (2)
Number of revolutions of A and B are 20000 and 15000 respectively
So, we know Bike B does 15000 revolutions in 45 minutes
Distance travelled = 2 x π x r x Number of revolutions
Speed of Bike B = mph
Speed of Bike B = 2 x π x 2/5 x 5 x 4 Kmph
Speed of Bike B = 16 π Kmph
 It is given that the cyclist starts at 10:00 am from A and reaches B at 11:00 am
Now, Motorcyclists start every minute from 10:01 am, and 45 such motorcyclists reach B before 11:00 am
If they leave one by one every minute, the 45th motorcyclist would have left by 10:45 am to reach B at 11:00 am.
Thus, time taken by one motorcyclist to reach B from A = 15 minutes.
Now, the cyclist doubles his speed. This means, he reaches B at 10:30 am
So, the last motorcyclist should have left A by 10:15 am.
Thus, 15 motorcyclists would have reached B by the time the cyclist reaches B.
It is given that the cyclist starts at 10:00 am from A and reaches B at 11:00 am
 By the time A and B meet for the first time, A covers 60% of the distance, while B covers 40% of the distance.
So, the speeds of A and b are in the ratio 60:40 or 3:2
Hence, the time they take to cover a particular distance will be in the ratio 2:3
We know that A covers 60% of the distance at 10:00 AM and covers 100% of the distance at 10:12 AM.
That means A takes 12 minutes to cover 40% of the track. So to cover the entire track he must have taken 12+12+6 = 30 minutes.
(because 40% + 40% + 20% = 100%)
Since the time taken by A and B to complete the track are in the ratio 2:3, the time taken by B to complete the track will be 45 minutes.
At 10:00 AM, B has covered 40% of the track. If we can find out what time does B take to complete the remaining 60% of the track, we can find the finish time of B.
Time required to complete 60% of the track = 60% of 45 = 27 minutes.
Hence, B complete one single round at 10:27 AM.
By the time A and B meet for the first time, A covers 60% of the distance, while B covers 40% of the distance.
Let the distance be = 60 Kms.
So, Bimal travels a distance of 20Kms in each mode and Amal travels 10, 20 and 30 Kms respectively in 1 hour each
Time taken by Bimal = Time taken to travel 20 kms in 10 Kmph + Time taken to travel 20 kms in 20 Kmph + Time taken to travel 20 kms in 30 Kmph
Time taken by Bimal = 2 + 1 + hours = 3 + hours
Extra time taken by Bimal = hour
Percentage increase in time = x 100 = x 100 = 22
So, Percentage increase in time = 22