In a race of three horses, the first horse beat the second horse by 11

Question

Competition Mode Question

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

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

Are You Up For the Challenge: 700 Level Questions

ArunSharma12 · Accepted Answer

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

freedom128 · Answer

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)

Posted from my mobile device

nick1816 · Answer

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

yashikaaggarwal · Answer

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

Posted from my mobile device

monikakumar · Answer

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

exc4libur · Answer

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)

gurmukh · Answer

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

Posted from my mobile device

gurmukh · Answer

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.

Posted from my mobile device

Staphyk · Answer

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

Posted from my mobile device

gracie · Answer

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

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

effatara · Answer

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

sachin_27 · Answer

can we solve this question using lcm?
randomly tried and the answer was correct.
lcm (11,90,80) = 7920
using the options 
option D, divides 7920 completely. getting an integer. 
7920/880=9

can someone tell if this can be another method to solve this question?

HamP95 · Answer

observe that at x distance travelled, the gap between 2nd and 3rd was 80m

at x-11 distance travelled, the gap was 90-11 so 79m, so every 11m travelled, 2nd gained 1m

so at x - (11*80) the gap will be 0m, the start of the race, so x - (11*80) = 0

x - 880 = 0 so x = 880m

Vivek1707 · Answer

10 second approach to solving this question

Lets first horse, 2nd horse and 3rd horse A, B and C respectively

Given

C ___79___B ___11___A

C_______80________B

So notice that when B is covering 11 meters, B its taking a lead of 1 meter over C

Now since B finishes the race with an 80 meter lead over C and since they all started at the same time (since its a race), to get a lead of 80, B will have to travel 11 x 80 i.e 880 meters. So the distance HAS to be 880 meters.

egmat · Answer

Hi! I understand your curiosity about the LCM method - it's interesting that it gave you the right answer! However,  this is purely coincidental  and not a valid approach for race problems.

Why the LCM method doesn't work:

The LCM of  11 ,  90 , and  80  gives you  7920 , and yes,  880  divides it perfectly. But there's  no mathematical relationship  between the LCM of the given distances and the race length.

Think about it this way: If the problem had different numbers (say the horses were separated by  13 ,  87 , and  75  meters), the LCM method wouldn't give you the correct race length.

The correct approach:

This is a  speed ratio problem . Here's the key insight:

When Horse 1 finishes  L  meters:
- Horse 2 has run  (L-11)  meters
- Horse 3 has run  (L-90)  meters

When Horse 2 finishes  L  meters:
- Horse 3 has run  (L-80)  meters

Since speed ratios must be consistent:
 Speed of Horse 3 ÷ Speed of Horse 2 = (L-90)/(L-11) = (L-80)/L

Cross-multiplying and solving:
L(L-90) = (L-11)(L-80)
L2 - 90L = L2 - 91L + 880
L =  880

Answer: 880 meters (Option D)

In a race of three horses, the first horse beat the second horse by 11

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

In a race of three horses, the first horse beat the second horse by 11

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards