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Puzzle: Speed, distance & time

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In a Road Race, one of the three bikers was doing 15km less than the first and 3km more than the third. He also finished the race 12 minutes after the first and 3 minutes before the third.Can you find out the speed of each biker, the time taken by each biker to finish the race and the length of the course?Assume that there were no stops in the race and also they were driving with constant speeds through out.
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RAHUL KUMAR RAHUL KUMAR   5 years ago

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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Sandeep Seth Sandeep Seth   5 years ago

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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Sandeep Seth Sandeep Seth   5 years ago

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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