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Re: Jane gave Karen a 5 m head start [#permalink]
29 Apr 2010, 03:03

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

Re: Jane gave Karen a 5 m head start [#permalink]
29 Apr 2010, 05:56

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.

Re: Jane gave Karen a 5 m head start [#permalink]
29 Apr 2010, 09:20

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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 \(21*0.25=5.25m\).

Re: Jane gave Karen a 5 m head start [#permalink]
29 Apr 2010, 23:45

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

Re: Jane gave Karen a 5 m head start [#permalink]
29 Apr 2010, 23:51

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 \(21*0.25=5.25m\).

Re: Jane gave Karen a 5 m head start [#permalink]
04 Jan 2011, 12:20

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 _________________

Consider giving Kudos if you liked my post. Thanks!

Re: Jane gave Karen a 5 m head start [#permalink]
12 Feb 2011, 02:01

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.

Re: Jane gave Karen a 5 m head start [#permalink]
04 Aug 2013, 15:21

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 \(21*0.25=5.25m\).

Re: Jane gave Karen a 5 m head start in a 100 race and Jane was [#permalink]
04 Aug 2013, 18:27

1

This post received KUDOS

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. _________________

Re: Jane gave Karen a 5 m head start in a 100 race and Jane was [#permalink]
05 Aug 2013, 09:10

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)

Back to hometown after a short trip to New Delhi for my visa appointment. Whoever tells you that the toughest part gets over once you get an admit is...