A sprinter starts running on a circular path of radius r metres. Her

The circumference of the path is 2πr meters and she covers (πr)/(1/2) meters (i.e. one-fourth of one round) in the first time period (30 seconds). It is obvious that, in each of the subsequent time periods, she will be covering the same distance (1/4th of one round) because the decrease in speed is offset by a proportionate increase in time (speed is halved and the time is doubled). So she will complete 1 round in 4 time periods.

Time taken for the 1st round = (1/2+1+2+4) mins = T1 (say)
Time taken for the 2nd round = (8+16+32+64) mins =16(1/2+1+2+4) = T2 (say)....> T2:T1=16:1

This pattern (i.e. the time taken to cover each round is 16 times the time taken to cover the previous round) is repeated for all subsequent rounds. Therefore:
Tn:T(n-1)=16:1. ANS: C

The circumference of the circular path (or the distance of one round) is 2πr meters. To find the ratio of the time taken for the nth round to that of the previous round, we can just compare the time taken for the second round and the time taken for the first round, assuming the ratio is always the same. So let’s determine these two times. Recalling that distance = rate x time, we have:

Distance traveled during the first ½ minute is: πr x ½ = πr/2 meters

Distance traveled during the first ½ + 1 = 1 ½ minutes is: πr/2 + πr/2 x 1 = πr meters

Distance traveled during the first 1 ½ + 2 = 3 ½ minutes is: πr + πr/4 x 2 = 3πr/2 meters

Distance traveled during the first 3 ½ + 4 = 7 ½ minutes is: 3πr/2 + πr/8 x 2 = 2πr meters

At this point, we see that it takes 7 ½ minutes to complete the first round of 2πr meters. Let’s continue keeping track of the distance traveled. However, we can see that when the time doubles, the rate is halved, so the increment of distance traveled is always πr/2 meters.

Distance traveled during the first 7 ½ + 8 = 15 ½ minutes is: 2πr + πr/2 = 5πr/2 meters

Distance traveled during the first 15 ½ + 16 = 31 ½ minutes is: 5πr/2 + πr/2 = 3πr meters

Distance traveled during the first 31 ½ + 32 = 63 ½ minutes is: 3πr + πr/2 = 7πr/2 meters

Distance traveled during the first 63 ½ + 64 = 127 ½ minutes is: 7πr/2 + πr/2 = 4πr meters

At this point, we see that it takes an extra 127 ½ - 7 ½ = 120 minutes to run the extra 4πr - 2πr = 2πr meters. In other words, it takes 120 minutes to run the second round of 2πr meters.

Therefore, the ratio between the two times is 120/7.5 = 240/15 = 16.

Answer: C

The circumference of the circle is \(2πr\). Note that every time the speed reduces to half, the time is becoming double. This means that at each speed, same distance is being covered.
Since speed \(πr\) is maintained for 30 secs i.e. 0.5 mins, distance covered at this speed is \(\frac{πr}{2}\) metres.
This means that a quarter of a circle is covered at a fixed speed. So first circle is covered at the given 4 speeds in total 0.5 + 1 + 2 + 4 = 7.5 mins.
The second circle will be covered in 8 + 16 + 32 + 64 = 120 mins
Required ratio = 120/7.5 = 64. Since it is a PS question which will have a single unique answer, we know that the ratio would be the same for each pair of rounds

Answer (E)

Alternatively, note that the time taken has become 16 times for each quarter. First quarter was covered in 0.5 mins in the first round but in 8 mins in the second round. Second quarter was covered in 1 min in the first round but in 16 mins in the second round and so on. So overall time becomes 16 times.

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