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

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Two horses begin running on an oval course at the same time. [#permalink]

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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
[Reveal] Spoiler: OA

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New post 29 Apr 2012, 23:12
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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'
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New post 29 Apr 2012, 23:31
Thank you for the clarification Ravi. Your method makes sense to me.

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Re: Two horses begin running on an oval course at the same time. [#permalink]

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rakp wrote:
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


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.
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Re: Two horses begin running on an oval course at the same time. [#permalink]

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Re: Two horses begin running on an oval course at the same time. [#permalink]

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New post 14 Nov 2013, 18:19
The easiest way is to check with answers

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

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New post 16 Nov 2013, 06:29
Isn't the LCM of the two numbers the solution since the question is asking for mins?
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Re: Two horses begin running on an oval course at the same time. [#permalink]

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New post 20 Jan 2015, 20:52
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


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

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Re: Two horses begin running on an oval course at the same time. [#permalink]

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New post 28 Jan 2015, 20:29
gmatprav wrote:
Isn't the LCM of the two numbers the solution since the question is asking for mins?



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
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Two horses begin running on an oval course at the same time. [#permalink]

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.............. 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
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New post 09 Jul 2015, 12:35
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

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Re: Two horses begin running on an oval course at the same time. [#permalink]

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New post 13 Jul 2015, 22:55
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.
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Re: Two horses begin running on an oval course at the same time. [#permalink]

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New post 09 Jul 2016, 11:16
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.



Bunuel wrote:
rakp wrote:
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


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.

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Two horses begin running on an oval course at the same time. [#permalink]

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

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Re: Two horses begin running on an oval course at the same time. [#permalink]

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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

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Re: Two horses begin running on an oval course at the same time. [#permalink]

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New post 18 Aug 2017, 13:01
Hello from the GMAT Club BumpBot!

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Re: Two horses begin running on an oval course at the same time.   [#permalink] 18 Aug 2017, 13:01
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