Let 1st distance be x km.
Then, 2nd is x-15 and third is x-18.
Similarly, time taken of 1st = y
time taken of 2nd = y+12
time taken of 3rd = y+15
Total distance = S km
Now as per the data given in the teaser, at a time T min
x = V1 * T ----> 1

x-15 = V2 * T ----> 2

x-18 = V3 * T ----> 3

At a Distance S Km.

S = V1 * y ----> 4

S = V2 * (y + 12) ----> 5

S = V3 * (y + 15) ----> 6

Thus, There are 6 equations and 7 unknown data that means it has infinite number of solutions.

Time taken by first biker, T1 = 60 Min.

Time taken by Second biker, T2 = 72 Min.

Time taken by first biker, T3 = 75 Min.

Also, we get

Speed of 1st biker, V1 = 90/y km/min

Speed of 2nd biker, V2 = (5/6)V1 = 75/y km/min

Speed of third biker, V3 = (4/5)V1 = 72/y km/min

Also, the length of the course, S = 5400/y km

So, algebra rules, By Graph we get respective values.

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Let distance = x
speed of first biker = v1
speed of second biker = v2
speed of third biker = v3

Now (x/v2) - (x/v1) = 12 .... (1)

(x/v3) - (x/v1) = 3 ... (2)

(x/v1) = (x-15)/v2 = (x-18)/v3 .... (3)

Now we can see from the third eqn that (x/v1) = (x-15)v2 which can be re-written as

(x/v2) - (x/v1) = 15/v2 .... (4)

Now Using eqn (1) and (4) we can find out value of v2 which on solving is 5/4

now we can solve for v3 similarly

hence x = 90
v1 = 3/2
v2 = 5/4
v3 = 6/5

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Let distance = x
speed of first biker = v1
speed of second biker = v2
speed of third biker = v3

Now (x/v2) - (x/v1) = 12 .... (1)

(x/v3) - (x/v2) = 3 ... (2)

(x/v1) = (x-15)/v2 = (x-18)/v3 .... (3)



Now we can see from the third eqn that (x/v1) = (x-15)/v2 which can be re-written as
(x/v2) - (x/v1) = 15/v2 .... (4)
Now Using eqn (1) and (4) we can find out value of v2 which on solving is 5/4
now we can solve for v3 similarly

hence x = 90
v1 = 3/2
v2 = 5/4
v3 = 6/5

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