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# Sam and Chris leave City A for City B simultaneously at 6 A.

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Joined: 24 Jan 2014
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Sam and Chris leave City A for City B simultaneously at 6 A.  [#permalink]

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Updated on: 08 Apr 2014, 03:35
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Difficulty:

55% (hard)

Question Stats:

73% (02:57) correct 27% (03:16) wrong based on 218 sessions

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Sam and Chris leave City A for City B simultaneously at 6 A.M in the morning driving in two cars at speeds of 60 mph and 80 mph respectively. As soon as Chris reaches City B, he returns back to City A along the same route and meets Sam on the way back. If the distance between the two cities is 210 miles, how far from City A did Sam and Chris meet?
A.60 miles
B.120 miles
C.30 miles
D.180 miles
E.150 miles

Originally posted by Safal on 08 Apr 2014, 03:13.
Last edited by MacFauz on 08 Apr 2014, 03:35, edited 1 time in total.
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Re: Sam and Chris leave City A for City B simultaneously at 6 A.  [#permalink]

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08 Apr 2014, 04:58
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Safal wrote:
Sam and Chris leave City A for City B simultaneously at 6 A.M in the morning driving in two cars at speeds of 60 mph and 80 mph respectively. As soon as Chris reaches City B, he returns back to City A along the same route and meets Sam on the way back. If the distance between the two cities is 210 miles, how far from City A did Sam and Chris meet?
A.60 miles
B.120 miles
C.30 miles
D.180 miles
E.150 miles

You can use ratios to solve it easily.

Let's say, they meet x miles away from city A. Sam has covered a distance of x miles and Chris has covered a distance of 210 + (210 - x) = (420 - x) miles.

The speeds of Sam and Chris are in the ratio 60:80 i.e. 3:4. So in the same time, they will cover distance which will be in the ratio 3:4 too.

x/(420 - x) = 3/4
7x = 1260
x = 180 miles

For more on ratios used in TSD, check: http://www.veritasprep.com/blog/2011/03 ... os-in-tsd/
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Re: Sam and Chris leave City A for City B simultaneously at 6 A.  [#permalink]

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08 Apr 2014, 03:37
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Time taken by Chris to reach City B = 210/80 = More than 2.5 hours

In 2.5 hours, Sam travels 60*2.5 = 150 miles

So distance at which they meet should be greater than 150 miles.

Only D satisfies.

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Re: Sam and Chris leave City A for City B simultaneously at 6 A.  [#permalink]

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08 Apr 2014, 12:35
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Equate the time taken (which will be equal for both).
=> x/60 = [(210) + (210-x)]/80
=> x = 180

Option (D).
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Re: Sam and Chris leave City A for City B simultaneously at 6 A.  [#permalink]

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08 Apr 2014, 20:25
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Look at the diagram, lets say they meet at point P (Which is distance x from start)

Both will take same time to reach point P

Setting up the equation

$$\frac{x}{60} = \frac{210}{80} + \frac{210-x}{80}$$

$$\frac{x}{3} = \frac{420-x}{4}$$

x = 180

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Re: Sam and Chris leave City A for City B simultaneously at 6 A.  [#permalink]

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11 Apr 2014, 20:48
When Sam and Chris meet, they have collectively travelled 210*2 = 420 miles
=> Distance from city A that they meet = 60 * (420/140) = 180 miles

Option (D).
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Re: Sam and Chris leave City A for City B simultaneously at 6 A.  [#permalink]

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26 Feb 2017, 12:50
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Re: Sam and Chris leave City A for City B simultaneously at 6 A.  [#permalink]

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22 Aug 2018, 15:32
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Re: Sam and Chris leave City A for City B simultaneously at 6 A.   [#permalink] 22 Aug 2018, 15:32
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