Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 29 May 2017, 23:26

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Manager
Joined: 09 Apr 2010
Posts: 79
Followers: 1

Kudos [?]: 68 [4] , given: 3

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

### Show Tags

28 Apr 2010, 21:06
4
KUDOS
27
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

42% (03:31) correct 58% (07:26) wrong based on 489 sessions

### HideShow timer Statistics

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
[Reveal] Spoiler: OA

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.
Senior Manager
Joined: 24 Jul 2009
Posts: 293
Followers: 3

Kudos [?]: 149 [0], given: 0

### Show Tags

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..
Manager
Joined: 09 Apr 2010
Posts: 79
Followers: 1

Kudos [?]: 68 [0], given: 3

### Show Tags

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 ...
Senior Manager
Joined: 24 Jul 2009
Posts: 293
Followers: 3

Kudos [?]: 149 [0], given: 0

### Show Tags

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..??
Manager
Joined: 09 Apr 2010
Posts: 79
Followers: 1

Kudos [?]: 68 [0], given: 3

### Show Tags

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
Intern
Joined: 03 Jan 2010
Posts: 7
Followers: 0

Kudos [?]: 2 [1] , given: 1

### Show Tags

29 Apr 2010, 06:56
1
KUDOS
1
This post was
BOOKMARKED
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.
Math Expert
Joined: 02 Sep 2009
Posts: 39066
Followers: 7759

Kudos [?]: 106591 [16] , given: 11630

### Show Tags

29 Apr 2010, 10:20
16
KUDOS
Expert's post
14
This post was
BOOKMARKED
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 $$21*0.25=5.25m$$.

_________________
Intern
Joined: 03 Jan 2010
Posts: 7
Followers: 0

Kudos [?]: 2 [0], given: 1

### Show Tags

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;
Manager
Joined: 09 Apr 2010
Posts: 79
Followers: 1

Kudos [?]: 68 [0], given: 3

### Show Tags

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 $$21*0.25=5.25m$$.

That is an excellent explanation
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2786
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 238

Kudos [?]: 1731 [2] , given: 235

### Show Tags

30 Apr 2010, 12:56
2
KUDOS
1
This post was
BOOKMARKED
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
_________________

Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned

Jo Bole So Nihaal , Sat Shri Akaal

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html

Intern
Affiliations: BYU M S
Joined: 01 Sep 2010
Posts: 8
Schools: Harvard, BYU
Followers: 0

Kudos [?]: 1 [0], given: 31

### Show Tags

20 Oct 2010, 14:00
Bunuel,

art thou a kind of math GOD?

how can I raise unto this magnitude?
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7381
Location: Pune, India
Followers: 2292

Kudos [?]: 15170 [11] , given: 224

### Show Tags

21 Oct 2010, 05:48
11
KUDOS
Expert's post
3
This post was
BOOKMARKED
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 [ 9.2 KiB | Viewed 9820 times ]

_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Manager
Joined: 23 Oct 2009
Posts: 72
Followers: 1

Kudos [?]: 31 [0], given: 14

### Show Tags

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 $$21*0.25=5.25m$$.

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

Thanks Bunuel
_________________

Consider giving Kudos if you liked my post. Thanks!

Manager
Joined: 19 Aug 2010
Posts: 76
Followers: 3

Kudos [?]: 27 [0], given: 2

### Show Tags

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 $$21*0.25=5.25m$$.

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.
Math Expert
Joined: 02 Sep 2009
Posts: 39066
Followers: 7759

Kudos [?]: 106591 [0], given: 11630

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

### Show Tags

09 Jul 2013, 09:45
Bumping for review and further discussion.
_________________
Senior Manager
Joined: 13 May 2013
Posts: 468
Followers: 3

Kudos [?]: 172 [0], given: 134

### Show Tags

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 $$21*0.25=5.25m$$.

Senior Manager
Joined: 17 Dec 2012
Posts: 481
Location: India
Followers: 27

Kudos [?]: 427 [1] , given: 15

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

### Show Tags

04 Aug 2013, 19:27
1
KUDOS
Expert's post
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.
_________________

Srinivasan Vaidyaraman
Sravna
http://www.sravnatestprep.com

Classroom and Online Coaching

Senior Manager
Joined: 13 May 2013
Posts: 468
Followers: 3

Kudos [?]: 172 [0], given: 134

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

### Show Tags

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.

Manager
Joined: 06 Jul 2013
Posts: 108
GMAT 1: 620 Q48 V28
GMAT 2: 700 Q50 V33
Followers: 0

Kudos [?]: 26 [1] , given: 42

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

### Show Tags

05 Aug 2013, 12:00
1
KUDOS
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
Current Student
Joined: 05 Dec 2013
Posts: 32
Concentration: Technology
GMAT Date: 02-01-2014
GPA: 3.95
WE: Information Technology (Venture Capital)
Followers: 0

Kudos [?]: 6 [0], given: 3

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

### Show Tags

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.
Re: Jane gave Karen a 5 m head start in a 100 race and Jane was   [#permalink] 28 Jan 2014, 22:44

Go to page    1   2    Next  [ 31 posts ]

Similar topics Replies Last post
Similar
Topics:
21 In a 100 metre race, A can allow B to start the race either 10 metres 6 18 Jan 2017, 10:13
9 John and Jane went out for a dinner and they ordered the 4 17 Apr 2016, 15:50
17 Jane started baby-sitting when she was 18 years old. 20 01 Jan 2017, 14:45
24 John and Jane started solving a quadratic equation. John mad 11 21 Jan 2017, 13:47
38 In a 100m race, Sam beats John by 4 seconds. On the contrary 16 20 May 2017, 18:40
Display posts from previous: Sort by