Jane gave Karen a 5 m head start

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Jane gave Karen a 5 m head start [#permalink]

28 Apr 2010, 21:06

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Jane gave Karen a 5 m head start in a 100 race and Jane was beaten by 0.25m. In how many meters more would Jane have overtaken Karen?

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Re: Jane gave Karen a 5 m head start [#permalink]

29 Apr 2010, 03:43

neoreaves wrote:

Jane gave Karen a 5 m head start in a 100 race and Jane was beaten by 0.25m. In how many meters more would Jane have overtaken Karen?

It took a lot of time.. Let me know if the answer is Correct..!!!

Jane traveled 100-0.25 in the same time in which Karen traveled 95.
So, 99.75 * J = 95 * K, where J and K are respective speeds of Jane and Karen.

so J = 1.05 K.

Now the distance between J and K, at the finishing line was 0.25 and difference in speeds of Jane and Karen is 0.05k.

Also, let say they both travel distance "m" from the finishing line..So distance traveled by karen is m and by Jane is 0.25 + m

and after time t they are at the same point.. so m/K = m+0.25/J
put the value of J=1.05K in the equation above we get m as 5.

so any value more than 5.00 meters will ensure that Jane wins..

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Re: Jane gave Karen a 5 m head start [#permalink]

29 Apr 2010, 03:47

hmmm nice one ....yes i got the same answer as 5 ..lets see if someone can prove that wrong but i thihnk that should be the OA ...

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Re: Jane gave Karen a 5 m head start [#permalink]

29 Apr 2010, 03:52

neoreaves wrote:

hmmm nice one ....yes i got the same answer as 5 ..lets see if someone can prove that wrong but i thihnk that should be the OA ...

What is the source of this question...?? And how much time u took to answer the Question..??

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Re: Jane gave Karen a 5 m head start [#permalink]

29 Apr 2010, 04:03

i took a looong time to solve this ....the source is unknown ...it was posted on one of the forums and i noted htis in my notes .....now revising my notes and going through them i found it again

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Re: Jane gave Karen a 5 m head start [#permalink]

29 Apr 2010, 06:56

neoreaves wrote:

Jane gave Karen a 5 m head start in a 100 race and Jane was beaten by 0.25m. In how many meters more would Jane have overtaken Karen?

Another way is to perhaps apply ratio & proportion.

99.75 is proportional to 4.75, hence what is proportional to 5?
We get 9975*5/475 = 9975/95 = 105
Now at 105 Jane and Karen will be neck to neck and the talks about Jane overtaking Karen. Hence according to me the answer must be > 105 - 100.
Hence my answer is greater than 5.

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Re: Jane gave Karen a 5 m head start [#permalink]

29 Apr 2010, 10:20

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neoreaves wrote:

Jane gave Karen a 5 m head start in a 100 race and Jane was beaten by 0.25m. In how many meters more would Jane have overtaken Karen?

We are asked: In how many meters more would Jane have overtaken Karen? So I think we are asked to calculate the distance Jane should cover to overtake Karen but all solutions above are calculating the distance Karen should cover. Can you please post answer choices to clear my doubts.

Anyway below is my solution:

Jane covered 100m-0.25m=99.75m and Karen covered 100-5=95m (in the same time interval).

Initial distance between them was 5m and final distance between Jane and Karen was 0.25m. Thus Jane gained 99.75m-95m=4.75m over Karen in 99.75m, hence Jane is gaining 1m over Karen in every \frac{99.75}{4.75}=21m.

Hence, Jane in order to gain remaining 0.25m over Karen should cover 21*0.25=5.25m.

Answer: 5.25.
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Re: Jane gave Karen a 5 m head start [#permalink]

30 Apr 2010, 00:45

Bunuel wrote:

neoreaves wrote:

Jane gave Karen a 5 m head start in a 100 race and Jane was beaten by 0.25m. In how many meters more would Jane have overtaken Karen?

I think you are right. I interpreted it this way- We now have a 100 meter track. How many more meters should the track have been if Jane were to overtake Karen;
hence my answer-greater than 5

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Re: Jane gave Karen a 5 m head start [#permalink]

30 Apr 2010, 00:51

Bunuel wrote:

neoreaves wrote:

Jane gave Karen a 5 m head start in a 100 race and Jane was beaten by 0.25m. In how many meters more would Jane have overtaken Karen?

That is an excellent explanation

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Re: Jane gave Karen a 5 m head start [#permalink]

30 Apr 2010, 12:56

ratio of distance covered = ratio of their speed in time t

hence \frac{95}{99.75} = \frac{vK}{vJ} --1

now Jane is 0.25 meter behind, suppose it overtakes him at x from 100m mark.

again using
ratio of distance covered = ratio of their speed in time t

\frac{(x+0.25)}{x} = \frac{vJ}{vK}---2

multiply equation 1 and 2
\frac{95}{99.75} = \frac{x}{(x+0.25)} => x =5

hence Jane needs to cover x+0.25 = 5.25
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Re: Jane gave Karen a 5 m head start [#permalink]

20 Oct 2010, 14:00

Bunuel,

art thou a kind of math GOD?

how can I raise unto this magnitude?

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Re: Jane gave Karen a 5 m head start [#permalink]

21 Oct 2010, 05:48

Since speeds of the two are constant, we can directly use variation here.

Since Jane covers a gap of 4.75m by running 99.75m, she will cover a gap of 0.25 m by running another (99.75) x 0.25/4.75 = 5.25 m.

In races questions, making a diagram can give you a clear picture.

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doc1.jpg [ 9.2 KiB | Viewed 1860 times ]

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Re: Jane gave Karen a 5 m head start [#permalink]

04 Jan 2011, 13:20

Bunuel wrote:

neoreaves wrote:

Jane gave Karen a 5 m head start in a 100 race and Jane was beaten by 0.25m. In how many meters more would Jane have overtaken Karen?

Excellent explanation!!! Much simpler than all the others i've read.

Thanks Bunuel
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Re: Jane gave Karen a 5 m head start [#permalink]

12 Feb 2011, 03:01

rahul321 wrote:

Bunuel wrote:

neoreaves wrote:

Jane gave Karen a 5 m head start in a 100 race and Jane was beaten by 0.25m. In how many meters more would Jane have overtaken Karen?

Excellent explanation!!! Much simpler than all the others i've read.

Thanks Bunuel

Karishma's explanation is actually the same, just a little bit more grafical and therefore easier to visualize.

Re: Jane gave Karen a 5 m head start [#permalink] 12 Feb 2011, 03:01

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Jane gave Karen a 5 m head start

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