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BSchool Thread Master
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Jack and Christina are standing 210 feet apart on a level [#permalink]
 17 Sep 2010, 05:44
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Question Stats:
81% (02:18) correct
18% (02:16) wrong based on 44 sessions
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]
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]
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]
17 Sep 2010, 06:37
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
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Re: Jack, Christina and Lindy [#permalink]
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 
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Re: Jack, Christina and Lindy [#permalink]
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]
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]
20 Sep 2010, 09:26
Nice question on relative speed. Easy 300 using relative concepts liek cano
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If you like my post, consider giving me some KUDOS !!!!! Like you I need them
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Re: Jack, Christina and Lindy [#permalink]
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
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Re: Jack, Christina and Lindy
[#permalink]
23 Apr 2012, 10:20
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