Bike Riding Puzzle

Eight persons, A through H, were standing along a circular track, not necessarily in the same order, such that the distance along the track between any two persons standing adjacent to each other was exactly 1 km. At exactly 9:00 AM, all of them started riding along the track in the clockwise direction. However, four of them started riding at a speed of 1 km/h, while the other four started riding at a speed of 3 km/h.

The following information is known about their relative positions along the track at exactly 10:00 AM, as they were riding:
No two persons were at the same position.
G was immediately behind D, while E was diametrically opposite C.
B was not immediately behind H, while F was diametrically opposite A.
The following information is known about their relative positions along the track at exactly 11:00 AM, as they were riding:
Neither F nor B was either immediately ahead of or immediately behind H.
E was either immediately ahead of or immediately behind D.


PS: It can be solved completely. Find the arrangement at 9 AM,10 AM and 11 AM ?

The solution I found is:

The arrangement and speed (k/hr) of people at 9:00am are
D G A E B H F C : 3 1 3 1 3 1 3 1

Note: Bold writing indicates information that is given or has been deduced earlier.

I solved this by making the following inferences from the given information:

At 10:00am

No two persons were at the same position.
The order of 3k/hr and 1k/hr riders alternates (i.e. it is 3,1,3,1,3,1,3,1)

G was immediately behind D
G and D travel at different speeds

E was diametrically opposite C
E and C travel at the same speed

F was diametrically opposite A
F and A travel at the same speed

We also know 4 riders ride at each speed. For the above statements to all be true, E & C must travel at a different speed to F & A

At 11:00am

E was either immediately ahead of or immediately behind D
E and D travel at different speeds
D, F and A travel at the same speed
G, E and C travel at the same speed
Thus B and H travel at different speeds

Something to realize at this point is that any two riders who are 'adjacent' (and thus travelling at different speeds) cannot be 'adjacent' in +/-2 hours.

Also, any rider (Alice) who is not adjacent to another rider (Bob), who rides at a different speed must then be adjacent to the rider (Charlie) who is diametrically opposite of the other rider (Bob). Put simply; if Alice rides at a different speed to both Bob and Charlie, Alice is adjacent to exclusively Bob or Charlie if Bob and Charlie form a diametrically opposite pair.

At 11:00am

E and D cannot be adjacent at 9:00am
E was diametrically opposite C
E and D travel at different speeds
D and C must be adjacent at 9:00am

Neither F nor B was either immediately ahead of or immediately behind (adjacent to) H
B and H travel at different speeds
B and H must be adjacent at 9:00am

.

At this point, some visual representations are helpful. Consider 9:00am
E and C are diametrically opposite

09:00 _ _ _ E _ _ _ C

09:00 _ _ _ E _ _ D C
09:00 D _ _ E _ _ _ C

We can deduce that the latter must be correct and thus
C must be immediately behind D at 9:00am because
G is immediately behind D (at 10:00am)

09:00 D G _ E _ _ _ C (if G rides 1k/hr)
09:00 D _ _ E _ G _ C (if G rides 3k/hr)

Since A and F are diametrically opposite and travelling at a different speed to E and C, they are either each immediately behind E and C or each immediately ahead of E and C. In this case D is already immediately ahead of C so A and F must each be immediately behind E and C. However, we do not know which is A or F

09:00 D G X E _ _ X C (if G rides 1k/hr)
09:00 D _ X E _ G X C (if G rides 3k/hr)

B and H must be adjacent at 9:00am
B was not immediately behind H at 10:00am

09:00 D G X E B H X C (G rides 1k/hr)

B, D, F and A travel at the same speed
H, G, E and C travel at the same speed
F was not immediately ahead of or immediately behind H at 11:00am
F and H travel at different speeds

F and H must be adjacent at 9:00am

09:00 D G A E B H F C (G rides 1k/hr)

• Your first solution is correct. Sequence at 9 AM is DGAEBHFC. – sam Nov 5 '18 at 13:27
• Thanks. I realized why the 2nd solution doesn't hold. Edited answer accordingly – SockPastaRock Nov 5 '18 at 13:53

It can be deduced that:

The length of the track is 8 km.

At 10:00

As there weren't any people occupying the same position, then:

The only possible combination of speeds in kmph is 3 1 3 1 3 1 3 1 or 1 3 1 3 1 3 1 3. (I am still not sure of this, but any combination that fits here must have each 1 preceded by a 3 by a distance not equal to 2 km).

Since E was diametrically opposite to C, then:

E and C are moving at the same speed.

Since F was diametrically opposite to A, then:

F and A are also moving at the same speed.

Since G was immediately behind D, then:

G and D are not moving at the same speed.

At 11:00

The combination of speeds is similar to that at 10:00, with the positions of the 3's and 1's swapped.

And:

E and C are still on the same diameter. F and A are also on the same diameter.

Since E was either immediately ahead of or immediately behind D, then:

E and D are not moving at the same speed.

• You are on the right track. Good going! – sam Nov 4 '18 at 16:41

9am - DGFBHECA
10am - CAGDBFEH
11am - FCHGABDE

- the distance along the track between any two persons standing adjacent to each other was exactly 1 km

This indicates that the track is exactly 8km long, since the distance between the last and first person is also 1km (taking from any starting person). This also indicates that no one is at the same position at 9am.

Moving on to 10am:
- No two persons were at the same position
Taking a sample position grid:
Position

i,ii,iii,iv,v,vi,vii,viii

9am

1,2,3,4,5,6,7,8

10am

assuming that Person i is travelling at 3km, it would be at position 4 now. Since no person were at the same position at 10am, it means that person iii cannot be travelling at 1km. Repeating the same logic, it seems that alternate person at travelling at 1km/h or 3km/h respectively.
Hence 10am position will be:
10am > 4,3,6,5,8,7,2,1
11am > 7,4,1,6,3,8,5,2

- G was immediately behind D

Assume case 1: i = D, ii = G or case 2: i = G, iv = D

Lets further see a possible arrangement of the positions with the following statements:
- ... E was diametrically opposite C.
- ... F was diametrically opposite A.

123
8 4
765

so 1 is diametrically opposite 7, 2/6 etc .....

For both cases above, since 4 & 3/5 is occupied, only 1/7 & 2/6 is left for the E/C & F/A pair.

Lets visualize the 4 main cases possible (noting that E/C & F/A could switch):

DGFBHEAC
DGEBHFCA
GBFDHEAC
GBEDHFCA

• B was not immediately behind H

B/H is either in iv/v or ii/V, which is position 3,5,8 respectively. Hence this condition is fulfilled.

Moving on to 11am:
- Neither F nor B was either immediately ahead of or immediately behind H

Position ii & v is adjacent to each other (4/3), hence the last 2 cases are invalid.
If H is iv/v (6/3 at 11am), F cannot be vii/viii (5/2 at 11am) respectively.

- E was either immediately ahead of or immediately behind D

Since D is at i, it is at position 7 at 11am. D has to be at 6/8, making position vi (8) the only case. Final answer is therefore starting position of DGFBHEAC.

• Pls have a recheck of your solution. It is not correct. – sam Nov 4 '18 at 15:04