It is currently 25 Jun 2017, 14:17

### 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

# Jack and Christina are standing 210 feet apart on a level

Author Message
TAGS:

### Hide Tags

Joined: 19 Feb 2010
Posts: 394
Jack and Christina are standing 210 feet apart on a level [#permalink]

### Show Tags

17 Sep 2010, 05:44
5
KUDOS
8
This post was
BOOKMARKED
00:00

Difficulty:

15% (low)

Question Stats:

83% (02:40) correct 17% (02:57) wrong based on 331 sessions

### HideShow timer Statistics

Jack and Christina are standing 210 feet apart on a level surface. Their dog, Lindy, is standing next to Christina. At the same time, they all begin moving toward each other. Jack walks in a straight line toward Christina at a constant speed of 3 feet per second and Christina walks in a straight line toward Jack at a constant speed of 4 feet per second. Lindy runs at a constant speed of 10 feet per second from Christina to Jack, back to Christina, back to Jack, and so forth. What is the total distance, in feet, that Lindy has traveled when the three meet at one place?

A. 105
B. 210
C. 280
D. 300
E. 420
[Reveal] Spoiler: OA
Intern
Joined: 18 Jul 2010
Posts: 47
Re: Jack, Christina and Lindy [#permalink]

### Show Tags

17 Sep 2010, 06:00
4
KUDOS
In 30 seconds, J and C meet
J has done 90 feet and C has done 120 feet (210 - 120 = 90)
So in 30 seconds, the dog has done 30 * 10 = 300 feet

ANS: D.

Hope it's clear enough....
Manager
Joined: 29 Aug 2010
Posts: 156
Schools: Wharton/Lauder - Class of 2013
WE 1: Project Management, Telecommunications
Re: Jack, Christina and Lindy [#permalink]

### Show Tags

17 Sep 2010, 06:20
1
KUDOS
Let J represent the distance Jack travels
Let C represent the distance Christina travels
Let t represent the time it takes for Jack and Christina to meet in the middle

For them to meet in the middle, J + C = 210 ft and Speed * Time = Distance

J + C = 210
3t + 4t = 210
t = 30 seconds

In 30 seconds, Lindy will have traveled 10ft/s * 30 seconds = 300 ft. Answer is D.
Joined: 19 Feb 2010
Posts: 394
Re: Jack, Christina and Lindy [#permalink]

### Show Tags

17 Sep 2010, 06:37
1
KUDOS
Well done, guys.
I see that you are advanced already in this kind of exercises. Somehow you kind of leap steps, at least just for me, a beginner.
What I have learned is that is "easier" to combine the speeds of Jack and Christina to find out the time:
3 feet per second + 4 feet per second = 7 feet per second
Then, divide: 210/7 = 30 seconds.
Then as you did, multiply for the speed of Lindy: 30*10 = 300
Intern
Joined: 18 Jul 2010
Posts: 47
Re: Jack, Christina and Lindy [#permalink]

### Show Tags

17 Sep 2010, 06:44
Hey Cano!
I would like to tell you something!
First, the solution of redjam is much better that mine to understand how it works. But I did mine because I did some questions like this one and I am telling you, the answer is never too complicated! That's why on those kind of questions, instead of thinkig of the equation I should ind, I am looking for an easy to solve.

In this particular case, it seems really complicated! So first, when do they meet.
Okay, how much feet in 30 seconds for J and C? Did they meet? Not yeat ! So how about in 1 minute ? etc etc... usually it works pretty fastly...
And then, you know when they meet, you just have to compute he the distance the dog run!! And don't think about: the dog has to do a U-turn, it will take time, etc etc.... imagine the dog is running as much as time as J and C need to meet.

Hope I get myself clear
Manager
Status: Keep fighting!
Joined: 31 Jul 2010
Posts: 229
WE 1: 2+ years - Programming
WE 2: 3+ years - Product developement,
WE 3: 2+ years - Program management
Re: Jack, Christina and Lindy [#permalink]

### Show Tags

19 Sep 2010, 20:30
Good question (a KUDOS from me for that). And yes, I have seen such a question before.
Manager
Joined: 16 Sep 2010
Posts: 220
Location: United States
Concentration: Finance, Real Estate
GMAT 1: 740 Q48 V42
Re: Jack, Christina and Lindy [#permalink]

### Show Tags

19 Sep 2010, 20:48
Thought it was a good question as well. The math isn't hard its more in the framing of the question. The reader could easily be confused and start to try and calculate each there and back by the dog instead of quickly determining the time the two people would meet and then multiplying out the time the dog would be constantly running.
Senior Manager
Joined: 20 Jul 2010
Posts: 263
Re: Jack, Christina and Lindy [#permalink]

### Show Tags

20 Sep 2010, 09:26
Nice question on relative speed. Easy 300 using relative concepts liek cano
_________________

If you like my post, consider giving me some KUDOS !!!!! Like you I need them

Intern
Affiliations: IEEE
Joined: 27 Jul 2010
Posts: 19
Location: Playa Del Rey,CA
WE 1: 2.5 yrs - Medicaid
WE 2: 2 yrs - Higher Ed
Re: Jack, Christina and Lindy [#permalink]

### Show Tags

23 Apr 2012, 10:20
Initially I had a different approach!

we know J & C will meet in 210/7 = 30 sec, J will travel 120 mtrs and C will travel 90 mtrs

At start L moves towards J and they have relative speed of 10+4 and they meet after 210/14=15 sec

In 15 sec L traveled 150 mtrs --- 1

Now in 15 sec C traveled 45 meters and J traveled 60 meters So now distance between J & C is 105 mtrs

when L moves towards C they have relative speed 10+3=13

so now L & C meet in 105/13= 8 sec aprrox and in this time L traveled 80 meters --- 2

C now is at 23*3=69 meters she covers remaining 90-69 meters in 7 sec
L covers 70 meters in 7 sec -- 3

So total dist covered 150+80+70=300 mtrs
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15978
Re: Jack and Christina are standing 210 feet apart on a level [#permalink]

### Show Tags

20 Sep 2013, 00:09
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Manager
Joined: 28 May 2014
Posts: 61
Schools: NTU '16
GMAT 1: 620 Q49 V27
Re: Jack and Christina are standing 210 feet apart on a level [#permalink]

### Show Tags

24 Aug 2014, 03:58
redjam,

Shouldn't we consider relative speed of Lindy with Christina and Jack? Sometimes it gets added up and sometimes wen Lindy returns back to Christina we may have to subtract the speeds of Lindy and Christina? Please explain
Math Expert
Joined: 02 Sep 2009
Posts: 39673
Re: Jack and Christina are standing 210 feet apart on a level [#permalink]

### Show Tags

24 Aug 2014, 06:52
Expert's post
4
This post was
BOOKMARKED
sri30kanth wrote:
redjam,

Shouldn't we consider relative speed of Lindy with Christina and Jack? Sometimes it gets added up and sometimes wen Lindy returns back to Christina we may have to subtract the speeds of Lindy and Christina? Please explain

Jack and Christina are standing 210 feet apart on a level surface. Their dog, Lindy, is standing next to Christina. At the same time, they all begin moving toward each other. Jack walks in a straight line toward Christina at a constant speed of 3 feet per second and Christina walks in a straight line toward Jack at a constant speed of 4 feet per second. Lindy runs at a constant speed of 10 feet per second from Christina to Jack, back to Christina, back to Jack, and so forth. What is the total distance, in feet, that Lindy has traveled when the three meet at one place?

A. 105
B. 210
C. 280
D. 300
E. 420

The relative speed of Jack and Christina is 3 + 4 = 7 feet per second.
The distance between them is 210 feet, hence they will meet in (time) = (distance)/(relative speed) = 210/7 = 30 seconds.

For all this time Lindy was running back and forth, so it covered (distance) = (speed)*(time) = 10*30 = 300 feet.

Similar question to practice: 12-easy-pieces-or-not-126366.html#p1033924

Hope it helps.
_________________
Intern
Joined: 28 Mar 2014
Posts: 4
Schools: Rotman '17
Re: Jack and Christina are standing 210 feet apart on a level [#permalink]

### Show Tags

02 Sep 2014, 02:23
Bunuel wrote:
sri30kanth wrote:
redjam,

Shouldn't we consider relative speed of Lindy with Christina and Jack? Sometimes it gets added up and sometimes wen Lindy returns back to Christina we may have to subtract the speeds of Lindy and Christina? Please explain

Jack and Christina are standing 210 feet apart on a level surface. Their dog, Lindy, is standing next to Christina. At the same time, they all begin moving toward each other. Jack walks in a straight line toward Christina at a constant speed of 3 feet per second and Christina walks in a straight line toward Jack at a constant speed of 4 feet per second. Lindy runs at a constant speed of 10 feet per second from Christina to Jack, back to Christina, back to Jack, and so forth. What is the total distance, in feet, that Lindy has traveled when the three meet at one place?

A. 105
B. 210
C. 280
D. 300
E. 420

The relative speed of Jack and Christina is 3 + 4 = 7 feet per second.
The distance between them is 210 feet, hence they will meet in (time) = (distance)/(relative speed) = 210/7 = 30 seconds.

For all this time Lindy was running back and forth, so it covered (distance) = (speed)*(time) = 10*30 = 300 feet.

Similar question to practice: 12-easy-pieces-or-not-126366.html#p1033924

Hope it helps.

I get the question, and I solved it exactly like you did. However, when you think about it a little more, I feel there is a little more to this question. As Jack and Christina walk towards each other, the distance between them is reducing as well, and the question says Lindy is running back and forth between them. Shouldn't the fact that they are waling towards each other and reducing the distance between them affect the total distance Lindy runs back and forth?
Math Expert
Joined: 02 Sep 2009
Posts: 39673
Re: Jack and Christina are standing 210 feet apart on a level [#permalink]

### Show Tags

02 Sep 2014, 03:28
nyoebic wrote:
Bunuel wrote:
sri30kanth wrote:
redjam,

Shouldn't we consider relative speed of Lindy with Christina and Jack? Sometimes it gets added up and sometimes wen Lindy returns back to Christina we may have to subtract the speeds of Lindy and Christina? Please explain

Jack and Christina are standing 210 feet apart on a level surface. Their dog, Lindy, is standing next to Christina. At the same time, they all begin moving toward each other. Jack walks in a straight line toward Christina at a constant speed of 3 feet per second and Christina walks in a straight line toward Jack at a constant speed of 4 feet per second. Lindy runs at a constant speed of 10 feet per second from Christina to Jack, back to Christina, back to Jack, and so forth. What is the total distance, in feet, that Lindy has traveled when the three meet at one place?

A. 105
B. 210
C. 280
D. 300
E. 420

The relative speed of Jack and Christina is 3 + 4 = 7 feet per second.
The distance between them is 210 feet, hence they will meet in (time) = (distance)/(relative speed) = 210/7 = 30 seconds.

For all this time Lindy was running back and forth, so it covered (distance) = (speed)*(time) = 10*30 = 300 feet.

Similar question to practice: 12-easy-pieces-or-not-126366.html#p1033924

Hope it helps.

I get the question, and I solved it exactly like you did. However, when you think about it a little more, I feel there is a little more to this question. As Jack and Christina walk towards each other, the distance between them is reducing as well, and the question says Lindy is running back and forth between them. Shouldn't the fact that they are waling towards each other and reducing the distance between them affect the total distance Lindy runs back and forth?

(distance) = (speed)*(time).

We know the speed of the dog and we know the time it runs. We need nothing more.
_________________
Intern
Joined: 20 Aug 2014
Posts: 7
Location: United States
GMAT 1: 690 Q49 V34
GPA: 3.27
Re: Jack and Christina are standing 210 feet apart on a level [#permalink]

### Show Tags

19 Jun 2015, 07:51
3 m/s + 4 m/s = 7 m/s * rate at which J & C close the gap between them

210 m / (7 m/s) = 30 s * time period it takes for J & C to meet

30 s x 10 m/s = 300 m * distance that Lindy Covers over a 30s time period
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15978
Re: Jack and Christina are standing 210 feet apart on a level [#permalink]

### Show Tags

25 Aug 2016, 01:14
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: Jack and Christina are standing 210 feet apart on a level   [#permalink] 25 Aug 2016, 01:14
Similar topics Replies Last post
Similar
Topics:
16 A certain store sells only jacks and marbles. A single jack costs 19 6 01 Jun 2017, 07:36
2 Fred and Sam are standing 45 miles apart and they start walking in a 4 07 Jan 2016, 21:51
3 Jack and Jill are marathon runners. Jack can finish a marathon (42 km) 6 24 Jun 2017, 08:14
7 A cubical tank is filled with water to a level of 2 feet. If 13 13 Dec 2015, 19:24
10 Jack is now 14 years older than Bill. If in 10 years Jack 9 23 Jun 2016, 09:36
Display posts from previous: Sort by

# Jack and Christina are standing 210 feet apart on a level

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.