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Jack and Christina are standing 210 feet apart on a level [#permalink]
17 Sep 2010, 04:44

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This post received KUDOS

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A

B

C

D

E

Difficulty:

25% (medium)

Question Stats:

83% (02:31) correct
17% (02:39) wrong based on 108 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?

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

1

This post received 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.

Re: Jack, Christina and Lindy [#permalink]
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

Re: Jack, Christina and Lindy [#permalink]
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 U-turn, it will take time, etc etc.... imagine the dog is running as much as time as J and C need to meet.

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

Re: Jack and Christina are standing 210 feet apart on a level [#permalink]
19 Sep 2013, 23:09

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