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Jack and Christina are standing 210 feet apart on a level [#permalink]
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17 Sep 2010, 05:44
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Question Stats:
83% (02:40) correct
17% (02:57) wrong based on 331 sessions
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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
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Re: Jack, Christina and Lindy [#permalink]
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17 Sep 2010, 06:00
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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....



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Re: Jack, Christina and Lindy [#permalink]
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17 Sep 2010, 06:20
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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.



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Re: Jack, Christina and Lindy [#permalink]
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17 Sep 2010, 06:37
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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



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Re: Jack, Christina and Lindy [#permalink]
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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 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



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Re: Jack, Christina and Lindy [#permalink]
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19 Sep 2010, 20:30
Good question (a KUDOS from me for that). And yes, I have seen such a question before.



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Re: Jack, Christina and Lindy [#permalink]
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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.



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Re: Jack, Christina and Lindy [#permalink]
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20 Sep 2010, 09:26
Nice question on relative speed. Easy 300 using relative concepts liek cano
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Re: Jack, Christina and Lindy [#permalink]
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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 9069 meters in 7 sec L covers 70 meters in 7 sec  3
So total dist covered 150+80+70=300 mtrs



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Re: Jack and Christina are standing 210 feet apart on a level [#permalink]
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20 Sep 2013, 00:09
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Re: Jack and Christina are standing 210 feet apart on a level [#permalink]
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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



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Re: Jack and Christina are standing 210 feet apart on a level [#permalink]
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24 Aug 2014, 06:52
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.
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Re: Jack and Christina are standing 210 feet apart on a level [#permalink]
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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. 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?



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Re: Jack and Christina are standing 210 feet apart on a level [#permalink]
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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. 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.
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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.
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Re: Jack and Christina are standing 210 feet apart on a level [#permalink]
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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



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Re: Jack and Christina are standing 210 feet apart on a level [#permalink]
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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.
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Re: Jack and Christina are standing 210 feet apart on a level
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