Distance from City A to City C

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Distance from City A to City C

by Resurgent » Tue Jan 05, 2010 9:03 am
City B is 5 miles east of city A. City C is 10 miles southeast of city B.
Which of the following is the closest to the distance from city A to city C?

a) 11 miles
b) 12 miles
c) 13 miles
d) 14 miles
e) 15 miles

OA not available

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by ace_gre » Tue Jan 05, 2010 10:49 am
Hi, Here is my approach..

Since the position of city C is just given as south east, its distance from A can vary from 15( 10 miles east of B - in straight line) and (125)^1/2south of city B (10 miles below B).

Since shortest distance is asked, eliminate 15 (because this is the longest), and 11(because shortest can only be (125)^1/2.
With no additional information provided, I would go for the next shortest distance in the list - 12 miles...

IMO B

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by gmatv09 » Wed Jan 06, 2010 1:25 pm
IMO B

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by sg1978 » Wed Jan 06, 2010 10:10 pm
Maybe I am slow to pick this up...could anyone please elaborate why the minimum distance is 125^1/2.
Am I assuming correct that angle ABC is 135 degrees and the minimum should be hyp AC formed by the extended
right angled triangle. Kindly correct my thinking.

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by gmatv09 » Wed Jan 06, 2010 10:50 pm
if C is directly below B (i.e south of B), then ABC will be a right angled triangle and AC will be the hypotenuse.
Hence the min. distance will be sqrt(125)
However, since C is south east of B ... ABC is not rt angled triangle... and AC shld be > sqrt(125) which is 12

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by sg1978 » Thu Jan 07, 2010 6:24 am
Thanks for the reply. But C is given southeast of B, not south of B,
hence not directly below but at an angle of 135.

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by sg1978 » Thu Jan 07, 2010 6:27 am
Sorry, I got the logic..ignore my earlier post

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by sg1978 » Thu Jan 07, 2010 6:40 am
Sorry but still confused. If C was exactly
below B, how can we assume BC is 10?
Since BC is 10 at an angle.

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by Shiram » Wed Aug 25, 2010 9:50 am
Hello, Guys! I'm also confused with this question. The book gives answer D as a correct one, while you found that it's B.

Please, can any one help to solve it or find what is wrong there?

Thanks!

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by Mo_Star » Wed Aug 25, 2010 10:15 am
Wow, this one is hard. I've noticed with any direction questions (i.e. city is north, south, etc) it is best to draw a coordinate system.

Also, with the isosceles right triangle, it would be easier to just use the rules of 45-45-90 triangles and know that 10 = x sq rt 2, solve for x to get 5 sq root 2.

I found this:

https://books.google.com/books?id=JpeDFS ... 3F&f=false
Monika

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by Shiram » Wed Aug 25, 2010 10:26 am
Thanks! It's exactly the same explanation as it's in my book! But may be there are other approaches?

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by Mo_Star » Wed Aug 25, 2010 10:42 am
Haha, sorry, I've never seen that question so I didn't have the book solution.

I don't know any other way to solve it...but when I originally looked at it, I definitely knocked out E (15 miles) because for the rules of a triangle, we know that (10-5) < x < (10 +5) so 5 < x < 15 (x being the length of the other side of the triangle).

I also knocked out A with the same reasoning as others used in this post in that if it was a right triangle, the third side would be the sq root of 125 (which is a little bigger than 11 since sq root of 121 = 11). However, the side of the triangle is longer than it would be for a right triangle so it should be greater than 11.

That still left me with B, C, or D.

Hopefully someone else has other ideas on other ways to solve besides the book solution!
Monika