Amar, Akbar and Anthony ran on a racetrack, with Amar finishing 160 m.

Question

Amar, Akbar and Anthony ran on a racetrack, with Amar finishing 160 m. ahead of Akbar and 400 m ahead of Anthony. Akbar finished the race 300 m ahead of Anthony. The three of them ran the entire distance with their respective constant speeds. What was the length of the racetrack?

(A) 400m
(B) 600m
(C) 650m
(D) 700m
(E) 800m

(A) 400m 
(B) 600m 
(C) 650m 
(D) 700m
(E) 800m

Vivor · Accepted Answer

Tl;dr shortcut: 
 
If 3 racers are competing in a race where the first racer beats the second racer by A metres, the first racer beats the third racer by B metres and the second racer beats the third by C metres, then the length of the race-track is given by the relation,
 L=\frac{AC}{A+C-B}

Explanation: 
Let Amar's victory be at time  T_1  and Akbar's at time  T_2  from the start of the race
Let Length of Race track be L
A, B, C have meanings given above. In our case, when Amar won (at  T_1 ), Akbar was 'A' metres behind finishing line and Anthony was 'B' metres behind. When Akbar reached the finishing line (at  T_2 ), Anthony was 'C' metres behind.

At time  T_1 ,
Distance covered by Akbar = L-A
Distance covered by Anthony = L-B
Also, Akbar's speed =  \frac{L-A}{T_1} 
Anthony's speed =  \frac{L-B}{T_1}

At time  T_2 ,
Distance covered by Akbar = L
Distance covered by Anthony = L-C
Also, Akbar's speed =  \frac{L}{T_2} 
Anthony's speed =  \frac{L-C}{T_2}

The ratio of speeds of Akbar and Anthony will be a constant number at any point during the race.
Hence, 
 \frac{\frac{L-A}{T_1}}{\frac{L-B}{T_1}}=\frac{\frac{L}{T_2}}{\frac{L-C}{T_2}} 
or,  \frac{L-A}{L-B}=\frac{L}{L-C} 
Simplifying, we get  L=\frac{AC}{A+C-B}

Length of race track =  \frac{160*300}{160+300-400} = 800 
Ans is E IMO

ShaikhMoice · Answer

Twin Question https://gmatclub.com/forum/three-runner ... 20830.html Also the solution provided by VeritasKarishma is fairly intuitive.

Fdambro294 · Answer

There are 2 different time periods: when Amar finishes the race and then when Akbar finishes the race.

During each of these time periods, the Time spent running is constant for each runner - therefore the Ratio of Distances covered is directly proportional to the Ratio of Constant Speeds

Let Amar’s Speed = X

Let Akbar’s Speed = Y

Let Anthony’s Speed = Z

Let D = Distance of the track

(Time 1) Amar finishes the race:

-160 m ahead of Akbar

X/Y = D / (D - 160) (eq 1)

And

-400 m ahead of Anthony

X/Z = D / (D - 400) (eq 2)

(Time 2) Akbar finishes the race

-300 m ahead of Anthony

Y/Z = D / (D - 300) (eq 3)

(1st) Multiply the reciprocal of (eq 1) by (eq 2) to eliminate X - speed of Amar

(Y/X) = (D - 160) / D

*

(X/Z) = D / (D - 400)
_________________

(Y/Z) = (D - 160) / (D - 400) (eq 4)

(2nd) set (eq 3) equal to (eq 4)

Y/Z = D / (D - 300) = (D - 160) / (D - 400)

at this point, rather than solve for a nasty quadratic, you can plug in the values of the answer choices and see which value of D = length of track will satisfy the above equation

(E)800 m

(800 - 160) / (800 - 400) = 800 / (800 - 300)

640/400 = 800/500

64/40 = 8/5

8/5 = 8/5

Answer is E - 800 m

Posted from my mobile device

Crytiocanalyst · Answer

The key to solving the equation is getting all the equation and getting the answer through meare but powerful substitution 
1 method :
let a, b, c be the speeds of the sprinters then and d be the distance of the track then 
a/b =d/d-160
=>a/c=d/d-400
=>b/c=d/d-300

Through plugging and simplyfication we can eventually reach the requisite value 
Therefore IMO E

otherwise simple logic is that A covered 160 m B covered 240m when C covered 100m
one thing for sure we have to understand only their multiple can lead to the same spot 
in the above distance we have multiple of 8 and 10 and it should above 400 only option that stands out tobe 800 
People can comment on the reliability and the soundness of the argument

Therefore IMO E

bumpbot · Answer

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Amar, Akbar and Anthony ran on a racetrack, with Amar finishing 160 m.

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Amar, Akbar and Anthony ran on a racetrack, with Amar finishing 160 m.

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