Two horses begin running on an oval course at the same time.

Question

Two horses begin running on an oval course at the same time. One runs each lap in 9 minutes; the other takes 12 minutes to run each lap. How Many minutes after the start will the faster horse have a one lap lead?

A. 36
B. 12
C. 9
D. 4
E. 3

A. 36
B. 12
C. 9
D. 4
E. 3

Bunuel · Accepted Answer

The rate of the faster horse is 1/9 lap/minute;
The rate of the slower horse is 1/12 lap/minute;

Their relative rate is 1/9-1/12=1/36 lap/minute;

The faster horse to gain one full lap will need time=distance/rate=1/(1/36)=36 minutes.

Answer: A.

Hope it's clear.

vsrsankar · Answer

The best approach in my view is like this:
 Check from the answer options .
As one horse completes the lap in 9 minutes and the other in 12 minutes,  the answer is definitely greater than 12 , because by 12 minutes, faster on completes 12/9 = 1.33 laps and the slower one completes only 1 lap, which results in a gap of 1.33 - 1 = 0.33 laps, which is less than 1 lap.
The only option looks feasible is option-A, ie., 36 min.
With out checking we can go for that. But let us check:
After 36 min, faster one completes 36/9 = 4 laps and 
slower one completes 36/12 = 3 laps.
The faster one covered exactly one more lap than the slower one did.
Answer  'A'

rakp · Answer

Thank you for the clarification Ravi. Your method makes sense to me.

adityapagadala · Answer

The easiest way is to check with answers

A) 36 min i.e horse A 4 laps and horse B 3 laps 
Hence A

gmatprav · Answer

Isn't the LCM of the two numbers the solution since the question is asking for mins?

thorinoakenshield · Answer

Two horses begin running on an oval course at the same time. One runs each lap in 9 minutes; the other takes 12 minutes to run each lap. How Many minutes after the start will the faster horse have a one lap lead?

A. 36
B. 12
C. 9
D. 4
E. 3

A. 36
B. 12
C. 9
D. 4
E. 3

SOLUTION:

Here is how I thought about it with minimal math:

The problem asks after how many minutes H2 will lead by 1 lap. In other words, both horses are running for the same amount of time until H2 leads by 1 lap. Sp check the LCM of 9 and 12, which is 36 = ANSWER A

PareshGmat · Answer

That's correct in this case.

LCM of 9 & 12 = 36

In 36 minutes, the faster would be 1 LAP ahead of the slower one

Answer = A

PareshGmat · Answer

.............. Rate .............. Time ...................... Total Laps

Slow horse ........  \frac{1}{12}  ............. x ...................  \frac{x}{12}  (Let "x" is the time taken by both horses)

Fast horse .........  \frac{1}{9}  ................ x ....................  \frac{x}{12} + 1  (Given that fast horse is 1 lap ahead compared to slow horse)

\frac{1}{9} * x = \frac{x}{12} + 1

x = 36

Answer = A

dishamaitra · Answer

let length of 1 lap be 'd'.
2nd horse travels 'd' distance in 12 mins . then in 9 mins it travels = (3d)/4.

So 1st horse leads by d/4 .
hence 1st horse takes 9 mins to give a lead of d/4 , 
then ??? to give a lead of d
=> 9 * d * (1/(d/4)) => 9 * d * 4/d = 36min

mahakmalik · Answer

it can be done like:-

let faster horse take X laps in t time:so total time = 9X
and in same time slower one will take X-1 laps so total time =12(X-1)

so 9x=12(x-1)
as time is same for both.

12x-9x=12
x=4

total time for faster horse =36

I hope it helps.
Give kudos

jazyb · Answer

Hi Bunuel,

Can you please check my approach?

Let's assume distance is 36miles.
Speeds are in this ratio: 4:3, so ratio of distance will also be in the same ratio because time is constant.
ratio of distance is 4:3 and difference is 36, so 4 parts are equal to 4x36. Then time is : 4x36 divided by 4 equals 36.

gracie · Answer

faster horse gains 1/9-1/12=1/36 lpm
1 lap/(1/36) lpm=36 minutes

hsbinfy · Answer

or a simple method

1)9 min 12 min(+3 min lead)
2)9 min 12 min(+6 min lead)
3)9 min 12 min(+9 min lead)

after third total lead time is +9 min(time taken for faster horse to complete on lap or that means the faster horse is now exactly one lap ahead of the slower)
So after 36 min the father horse will be one lap ahead

(9+9+9 +9(leadtime))=12+12+12=36

stne · Answer

Let T be the time , at which the lead of 1 lap occurs 
So in T minutes faster horse makes  \frac{T}{9}  laps , slower horse in T minutes makes  \frac{T}{12}  laps

Eqn can be written as  \frac{T}{9}  - \frac{T}{12}  =1
 \frac{12T- 9T} {108}  =1 -> 3T =108 -> T =36, Ans = A. Hope this helps.

Gmat2Cracker · Answer

s=d/t
d is same
s= 1/9 and 1/12
now take the LCM of 9 and 12 to check where they meets
answer= 36 (A)

Carlosfernando25 · Answer

Horse 1= 1/9

Horse 2= 1/12

Relative speed= 1/9 - 1/12

Relative speed= 1/36

T= D/R

T= 1/(1/36)= 36 = A.

bumpbot · Answer

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Two horses begin running on an oval course at the same time.

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