November 15, 2018 November 15, 2018 10:00 PM MST 11:00 PM MST EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth $100 with the 3 Month Pack ($299) November 17, 2018 November 17, 2018 07:00 AM PST 09:00 AM PST Nov. 17, 7 AM PST. Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.
Author 
Message 
TAGS:

Hide Tags

BSchool Thread Master
Joined: 19 Feb 2010
Posts: 350

Jack and Christina are standing 210 feet apart on a level
[#permalink]
Show Tags
17 Sep 2010, 04:44
Question Stats:
81% (02:08) correct 19% (02:47) wrong based on 354 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
Official Answer and Stats are available only to registered users. Register/ Login.




Intern
Joined: 18 Jul 2010
Posts: 44

Re: Jack, Christina and Lindy
[#permalink]
Show Tags
17 Sep 2010, 05:00
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: 154
Schools: Wharton/Lauder  Class of 2013
WE 1: Project Management, Telecommunications

Re: Jack, Christina and Lindy
[#permalink]
Show Tags
17 Sep 2010, 05:20
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.



BSchool Thread Master
Joined: 19 Feb 2010
Posts: 350

Re: Jack, Christina and Lindy
[#permalink]
Show Tags
17 Sep 2010, 05:37
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: 44

Re: Jack, Christina and Lindy
[#permalink]
Show Tags
17 Sep 2010, 05:44
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 Uturn, 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!
Affiliations: IIT Madras
Joined: 31 Jul 2010
Posts: 196
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, 19:30
Good question (a KUDOS from me for that). And yes, I have seen such a question before.



Manager
Joined: 16 Sep 2010
Posts: 198
Location: United States
Concentration: Finance, Real Estate

Re: Jack, Christina and Lindy
[#permalink]
Show Tags
19 Sep 2010, 19:48
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.



Manager
Joined: 20 Jul 2010
Posts: 215

Re: Jack, Christina and Lindy
[#permalink]
Show Tags
20 Sep 2010, 08:26
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: 15
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, 09:20
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 9069 meters in 7 sec L covers 70 meters in 7 sec  3
So total dist covered 150+80+70=300 mtrs



Manager
Joined: 28 May 2014
Posts: 54

Re: Jack and Christina are standing 210 feet apart on a level
[#permalink]
Show Tags
24 Aug 2014, 02:58
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: 50585

Re: Jack and Christina are standing 210 feet apart on a level
[#permalink]
Show Tags
24 Aug 2014, 05:52
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. Answer: D. Similar question to practice: 12easypiecesornot126366.html#p1033924Hope it helps.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 28 Mar 2014
Posts: 4

Re: Jack and Christina are standing 210 feet apart on a level
[#permalink]
Show Tags
02 Sep 2014, 01:23
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. Answer: D. Similar question to practice: 12easypiecesornot126366.html#p1033924Hope 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: 50585

Re: Jack and Christina are standing 210 feet apart on a level
[#permalink]
Show Tags
02 Sep 2014, 02:28
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. Answer: D. Similar question to practice: 12easypiecesornot126366.html#p1033924Hope 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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 20 Aug 2014
Posts: 7
Location: United States
GPA: 3.27

Re: Jack and Christina are standing 210 feet apart on a level
[#permalink]
Show Tags
19 Jun 2015, 06:51
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



NonHuman User
Joined: 09 Sep 2013
Posts: 8774

Re: Jack and Christina are standing 210 feet apart on a level
[#permalink]
Show Tags
29 Dec 2017, 19:25
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.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: Jack and Christina are standing 210 feet apart on a level &nbs
[#permalink]
29 Dec 2017, 19:25






