let the race be x meters long.
when A finished the race, B was (x-11) meters behind and C was (x-90) meters behind.
when B finished the race, C was (x-80) meters behind.
so, when B ran 11 meters, C ran 10 meters.
when B ran (x-11) meters, C ran = \(\frac{10*(x-11)}{11} = x-90\)
\(\frac{10x}{11} -10 = x-90\)
\(\frac{x}{11} = 80\)
x =880 meters
Ans D

Length of track = L

\(V_F : V_S : V_T = L : (L-11) : (L-90)\)

Also, \(V_S : V_T = L : (L-80)\)

\(\frac{(L-11)}{(L-90)} = \frac{L}{(L-80)}\)

(L-11)(L-80) = (L)(L-90)

\(L^2 -91L +880 = L^2-90L\)

L= 880

Let the distance be x
A beats B by 11 meter
So A/B= x/(x-11)
B beats C by 80 meter.
So B/C = x/(x-80)
A beats C by.
A/B*B/C=A/C
X/(X-11)*X/(X-80)=X/(X-90)
X^2/X^2-91X+880=X/(X-90)
X^3-90x^2= X^3-91x^2+880
X^2=880X
X=880
So OA is D

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Let the race length be a meter

When the 1st horse beats the 2nd and the 3rd by 11 meters and 90 meters, respectively, the distance traveled by 2nd and 3rd horse at that time is (x-11) meters and (x-90) meters, respectively.

===> Ratio of average speed of 2nd and 3rd horse = (x-11)/(x-90) ...(1)

When the 2nd horse beats the 3rd by 80 metres, the distance traveled by 2nd and 3rd at that time is x and (x-80), respectively.

==> Ratio of average speed of 2nd and 3rd horse = (x)/(x-80) ...(2)

(1) = (2)
(x-11)/(x-90) = x/(x-80)
(x-11)*(x-80) = x*(x-90)
x^2-91x+880 = x^2 - 90x
x = 880 meters


FINAL ANSWER IS (D)

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1st observation: x x-11 x-90
2nd : x x-80

so speed ratios of second and third horses must be same,
hence,
x-11/x-90 = x/x-80 (since time is same on that particular observation)..
on solving, x=880

Ans D

d/f=d-11/s=d-90/t
d/s=d-80/t

d/s-11/s=d-90/t
(d-80)/t-11/s=d-90/t
(d-80)s-11t/s=d-90
ds-80s-11t=ds-90s
10s=11t
s/t=11/10

d/s=d-80/t, d/d-80=s/t
d/d-80=(11/10), 10d=11d-880
-d=880, d=880

Ans (D)

Let the length of the track is x mts
So when FH travels =x mts
SH = x-11 mts
TH = x-90

Ratio of speed of SH to TH =
(x-11)/(x-90)

Now when SH travels =x mts
TH travels = x-80


Ratio of speed of SH to TH =
x/x-80


(x-11)/(x-99)= (x)/(x-80)

X=880
Option D is the answer

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Another nice observation to solve this sum
When SH travels 11m the gap between SH and TH increases by 1m
thatis 79 to 80, Now initially they all were at the same position and when the gap between SH and TH becomes 79 then SH must have travel 79×11 =869
And adding 11 to it you get 880
So option D is the answer.

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Let length of race = x meters
Let first horse =h1 ,second horse=h2 ,and third horse=h3
We know R•T=D
When time is constant (same time)
R=D

Now if h2 beat h3 by 80meters ,therefore distance traveled by h2= x meters and h3= (x-80)meters (time remains constant here)
Also h1 beat h2 by 11 metres and h3 by 90 metres, so distance traveled by h2= (x-11) and h3=(x-90)

Since time is constant
S2/S3 =D2/D3
(x-11)/(x-90) = x/(x-80)
(x-11)•(x-80) = x(x-90)
x = 880
D like disease

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_________________

In a race of three horses, the first horse beat the second horse by 11 metres and the third horse by 90 metres. If the second horse beat the third horse by 80 metres, what was the length, in metres, of the race?

A. 800
B. 860
C. 870
D. 880
E. 900

let d=length of race
assuming constant speeds for horses 2 and 3,
these two h2/h3 distance ratios will be equal:
(d-11)/(d-90)=d/(d-80)→
d=880 meters
D

Let the distance be 'd' meters and the speeds of the first, second and third horses be s1, s2 and s3 respectively.

s1/s2 = d/(d-11)..... (i)
s1/s3 = d(d-90)...... (ii)
Dividing (ii) by (i), we get:
s2/s3 = (d-11)/(d-90). But we know that s2/s3 is also equal to d/(d-80). Therefore:
(d-11)/(d-90) = d/(d-80)....> d = 880. ANS: D