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Jane gave Karen a 5 m head start in a 100 race and Jane was

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neoreaves
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Jane gave Karen a 5 m head start in a 100 race and Jane was [#permalink] New post   Updated on: 09 Jul 2013, 09:45
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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?

A 5m
B. 7m
C. 4.5m
D. 5.25m
E. 6m
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D

Originally posted by neoreaves on 28 Apr 2010, 21:06.
Last edited by Bunuel on 09 Jul 2013, 09:45, edited 1 time in total.
Renamed the topic, edited the question added the answer choices and OA.
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Re: Jane gave Karen a 5 m head start [#permalink] New post   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] New post   21 Oct 2010, 05:48
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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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Re: Jane gave Karen a 5 m head start [#permalink] New post   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?


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.


That is an excellent explanation
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Re: Jane gave Karen a 5 m head start [#permalink] New post   30 Apr 2010, 12:56
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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] New post   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] New post   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?


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.


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 in a 100 race and Jane was [#permalink] New post   04 Aug 2013, 19:27
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1. The initial gap between Jane and Karen was 5m. It was reduced to 0.25 m as Jane covers 99.75 m. Jane gains 4.75m.
2. So the gap of 0.25m would be reduced to 0 or Jane would gain 0.25m, if Jane further covers (99.75/4.75) * 0.25 = 5.25 m.
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Re: Jane gave Karen a 5 m head start in a 100 race and Jane was [#permalink] New post   05 Aug 2013, 12:00
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my approach is-

Jane covered 4.75m when she ran 99.75m
So for 1m gain 99.75/4.75 = 21
and so .25 meter gain .25*21 = 5.25
D
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Re: Jane gave Karen a 5 m head start in a 100 race and Jane was [#permalink] New post   23 Feb 2015, 02:09
How do you calculate 99.75/4.75 so easily ? I solved the problem quite quickly but got completely stuck in that calculation... Do you have a more straightforward way to solve that than doing the division as in primary school ?
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Re: Jane gave Karen a 5 m head start in a 100 race and Jane was [#permalink] New post   23 Feb 2015, 03:24
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Lapetiteflo wrote:
How do you calculate 99.75/4.75 so easily ? I solved the problem quite quickly but got completely stuck in that calculation... Do you have a more straightforward way to solve that than doing the division as in primary school ?


Even if you actually had 99.75/4.75, calculating this is also quite simple.

99.75/4.75 = 9975/475 (get rid of the decimals by multiplying and dividing by 100)

Now, you need to find common factors. Any number ending in 25/50/75/00 will be divisible by 25. Now the question is 9975 is which multiple of 25. Note that if 9975 had another 25, it would have been 10,000 and that is 25*400. So 9975 must be 25*399.
Similarly if 475 had another 25, it would have been 500 and 25*20 = 500. So 25*19 = 475

We get 9975/475 = 399/19 = 21
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Re: Jane gave Karen a 5 m head start in a 100 race and Jane was [#permalink] New post   30 Jun 2021, 01:53
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?

A 5m
B. 7m
C. 4.5m
D. 5.25m
E. 6m


Find my answer in the attached image.
Hope this helps.

The trick is to remember to add 0.25m after calculating the remaining distance.
Attachment:
Screenshot 2021-06-30 at 2.22.33 PM.png
Screenshot 2021-06-30 at 2.22.33 PM.png [ 1.85 MiB | Viewed 4709 times ]
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Re: Jane gave Karen a 5 m head start in a 100 race and Jane was [#permalink] New post   06 Jul 2023, 21:05
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Re: Jane gave Karen a 5 m head start in a 100 race and Jane was [#permalink]
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