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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]  29 Apr 2012, 21:23
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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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Re: Word Problem [#permalink]  29 Apr 2012, 22:12
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The best approach in my view is like this:
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.
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Re: Word Problem [#permalink]  29 Apr 2012, 22:31
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]  29 Apr 2012, 22:51
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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.

Hope it's clear.
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Re: Two horses begin running on an oval course at the same time. [#permalink]  14 Nov 2013, 08:50
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Re: Two horses begin running on an oval course at the same time. [#permalink]  14 Nov 2013, 17:19
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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Re: Two horses begin running on an oval course at the same time. [#permalink]  16 Nov 2013, 05:29
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]  20 Dec 2014, 00:22
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Re: Two horses begin running on an oval course at the same time. [#permalink]  20 Jan 2015, 19:52
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]  28 Jan 2015, 19:29
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

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Two horses begin running on an oval course at the same time. [#permalink]  28 Jan 2015, 19:46
.............. 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

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Re: Two horses begin running on an oval course at the same time. [#permalink]  09 Jul 2015, 11:35
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]  13 Jul 2015, 21:55
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] 13 Jul 2015, 21:55
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