CareerCup

40 meters.

yt / y-8 = (y-10) t / y - 16

y is length of track
t is time in which A finished the race.

Forget about A.
When B goes 8 meters, C goes only 6 meters. Since their speeds are constant, C falls another 2 meters behind every 8 meters that B goes. Since B has left C 10 meters behind in the end, the track has to be (10/2)*8-40 meters long.

suppose d is distance and Va, Vb and Vc are their respective velocities. now when A finishes, time taken is d/Va. Also
d/Va = d/Vc - 16/Vc
d/Va = d/Vb - 8/Vb

this implies Vb/Vc = (d-8)/(d-16)
also, if you assume time taken by B to cover the remaining distance is t then
Vb = 8/t
and Vc = 6/t
this gives Vb/Bc = 8/6

now solve the two equations and u will get the value of d.

thank you very much for the detailed explanation @ anonymous

@Anon
would you like explaining your approach in detail for others sake?

i get 30 mts.
va vb vc be the velocites.
from the change in position betw two given instances
vb = 8/t
vc = 6/t
vb/vc = 4/3
take the velocities as 4v and 3v
take the track distance as D
since both have travelled for the same time. equating their time,
D/3v = (D-10)/4v
which gives D as 30 .
Tell me if the ans is write and also if there is a better approach

Please check the logic:

Speed of A = Sa, Speed of B = Sb, Speed of C = Sc
Let D be the track length, T is time taken by A to travel the track
D/T = Sa, D/(T+8) = Sb, D/(T+16) = Sc.

In 8 minutes, the diff between A and B is 10 minutes.
In 1 minute, the diff is 8/10 =4/5 mins, so Sb=5/4*Sc, Sa=5/2*Sc
Substituting from above
D/(T+8) = 5/4*(D/T+16) => T=24.

Since all speeds can be represented in Sc=D/(24+16), the smallest integer value that makes Sc=1, makes D=40.
The distance is 40 distance units.

S.M - Dude take a vacation. You need one!

No need to complicate things when it can be explained as simply as what "interviewtomorrow" added.

Suppose X is the length of the track.
When A covered X m, B covered X-8 m and C X-16 m.
When B covered X m, C covered X-10 m.
So When B covered 8 m,

Suppose X is the length of the track.
When A covered X m, B covered X-8 m and C X-16 m.
When B covered X m, C covered X-10 m.
So When B covered 8 m, C covered 6 m ( X-10 - X +16)
i.e x-8/x-16 = 8/6
x=40 m

kinda doubt does google really define a preliminary-school level prob as brainteaser?

what about it's a multi-lap run?

Answer is 40 meter,

Let total distance is D and Total time taken by A is T

S(a) = D/ T , S(b) = D-8 / T , S(c) = D-16 / T;

Now lets assume times taken by B to cover 8ms to finish line was t

S(b) = 8 / t ; S(c) = 6 / t ( Since C only covered 6 meters in that duration);

On Solving this we get S(c) = 3/4 S(b);

Replace the value of S(b) & S(c) inside equation 1

D-8 / T = 4/3 (( D - 16 ) / T );

On Solving we get D = 40

L = Va*T
L-8 = Vb*T
L=16 = Vc*T
==============
L= Vb*T1
L-10 = Vc*T1
L/(L-10) = Vb/Vc= (L-8)/(L-16)
2L = 80
L = 40 Meter.