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

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Jane gave Karen a 5 m head start in a 100 race and Jane was  [#permalink]

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New post Updated on: 09 Jul 2013, 09:45
5
1
55
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

43% (02:46) correct 57% (02:46) wrong based on 420 sessions

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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

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]

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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]

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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]

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New post 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]

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New post 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]

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New post 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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New post 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]

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New post 29 Apr 2010, 06:56
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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?


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]

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New post 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?


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.

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]

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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]

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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]

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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.
Attachment:
doc1.jpg
doc1.jpg [ 9.2 KiB | Viewed 14722 times ]

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

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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  [#permalink]

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New post 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?


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



Karishma's explanation is actually the same, just a little bit more grafical and therefore easier to visualize.
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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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Re: Jane gave Karen a 5 m head start  [#permalink]

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New post 04 Aug 2013, 16:21
This is a very good explanation but I feel that the questions wording is very ambiguous. I had trouble rationalizing that Jane and Karen traveled their respective distances in the same time!

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.
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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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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]

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New post 05 Aug 2013, 10:10
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 my problem is understanding that Jane and Karen started the race at the same time and Karen started from 5m ahead of Jane. If Jane and Karen started from the same spot and Jane only started running after Karen was 5m in front of her in which case Karen would have run for more time than Jane. How do I know to assume this for future problems?

If Jane and Karen start running at the same time with Karen starting 5m in front of Jane, and Karen beats Jane by .25m then over the same course of time Jane covers 99.75 m and gains 4.75m on Karen. I guess you have to assume that both instantaneously stop as soon as Karen hit's the finish line.

The problem asks about the distance Jane would have to run in addition to the 99.75m she did to overtake Karen. In other words, her rate is greater than Karen's and we have to find how many meters it takes for her to gain an additional .25m on Karen. If Jane gained 4.75m on Karen over 99.75m then Jane gains 99.75/4.75 = 1m on Karen every 21 meters she runs. If Jane gains 1m for every 21m she runs and we need to find how many meters she gains to cover just 1/4th of the distance (1/4th of 1m): 21/4 = 5.25.

Jane will pass Karen after another 5.25m of the race.

Given that Jane and Karen run at the same rates and that Karen is in front of Jane when the race starts we know that Jane has to be running at a faster rate than Karen. We can figure out how many meters Jane gained on Karen over the course of the race. If we know that Jane gains a certain amount on Karen over a fixed distance we can figure out how many more meters she needs to run to achieve a certain goal (i.e. passing Karen)

We are looking for Jane's gains on Karen.

ANSWER: D. 5.25m
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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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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
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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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New post 28 Jan 2014, 22:44
Time is equal so 95/x = 99.75/y. Solve for y = 1.05x.

Then set up another equation where time is equivalent again with new distances solving for z. xz = (x+.25)*1.05. Solve for Z and get 21/4 or 5.25.
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Re: Jane gave Karen a 5 m head start in a 100 race and Jane was   [#permalink] 28 Jan 2014, 22:44

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