A, B, C, D, and E practice running at a stadium that has several tracks. Each one starts running from the same point Z. A runs around a path which is an equilateral triangle ZYX. B runs around a square path ZYWV. C runs on a regular pentagonal path ZYUTS. D runs on a regular hexagonal path ZYRQPO. E runs along the trapezium ZYPO and completes one round in 20 seconds. Among all the tracks, there is a common side ZY, having length 100 m. The ratio of speeds of B and C is 5 : 4 respectively whereas the ratio of the time taken by A and D to complete one round of their tracks is 3 : 4 respectively. E's speed is the same as B's speed whereas A's speed is half that of C's.Note: The athletes maintain the same speeds in any track.Q 110.   If E runs along the paths taken by A and B, then the time taken by him to complete 1 round each of the paths is approximately what percentage more or less than the total time taken by A and B?a)  46.24 % less timeb)  35.03 % more timec)  39.13 % less timed)  24.46 % less time

To solve this problem, we first need to calculate the perimeter of each shape, which represents the total distance each person has to run.

1. For A, the path is an equilateral triangle. The perimeter of an equilateral triangle is 3 times the length of one side, so the total distance A has to run is 3*100 = 300m.

2. For B, the path is a square. The perimeter of a square is 4 times the length of one side, so the total distance B has to run is 4*100 = 400m.

3. For E, the path is a trapezium. We don't have enough information to calculate the exact perimeter of the trapezium, but we know that E completes one round in 20 seconds and his speed is the same as B's speed. Therefore, we can calculate E's speed as 400m/20s = 20m/s.

Now, we can calculate the time it would take E to run along the paths taken by A and B.

1. For A's path, the time would be distance/speed = 300m/20m/s = 15 seconds.

2. For B's path, the time would be distance/speed = 400m/20m/s = 20 seconds.

So, the total time taken by E to complete 1 round each of the paths is 15s + 20s = 35s.

Next, we need to calculate the total time taken by A and B.

1. We know that A's speed is half that of C's, and C's speed is 4/5 that of B's. Therefore, A's speed is (1/2)*(4/5)*20m/s = 8m/s. So, the time taken by A is distance/speed = 300m/8m/s = 37.5s.

2. The time taken by B is distance/speed = 400m/20m/s = 20s.

So, the total time taken by A and B is 37.5s + 20s = 57.5s.

Finally, we can calculate the percentage difference between the time taken by E and the time taken by A and B. The formula for percentage difference is (difference/original)*100%.

The difference is 57.5s - 35s = 22.5s. The original time is 57.5s. So, the percentage difference is (22.5s/57.5s)*100% = 39.13%.

Therefore, the answer is c) 39.13% less time.