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City B is 8 miles east of City A. City C is 6 miles north of city B. C

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City B is 8 miles east of City A. City C is 6 miles north of city B. C [#permalink]

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New post 16 Oct 2017, 10:12
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City B is 8 miles east of City A. City C is 6 miles north of city B. City D is 16 miles east of city C, and city E is 12 miles north of city D. What is the distance from city A to city E?

(A) 10 miles
(B) 20 miles
(C) 24 miles
(D) 30 miles
(E) 42 miles
[Reveal] Spoiler: OA

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Kudos [?]: 135256 [0], given: 12679

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Re: City B is 8 miles east of City A. City C is 6 miles north of city B. C [#permalink]

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New post 16 Oct 2017, 12:02
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Bunuel wrote:
City B is 8 miles east of City A. City C is 6 miles north of city B. City D is 16 miles east of city C, and city E is 12 miles north of city D. What is the distance from city A to city E?

(A) 10 miles
(B) 20 miles
(C) 24 miles
(D) 30 miles
(E) 42 miles


there are ONLY two sides movement and we have to find the distance between FIRST and LAST point..

FINALLY we have a RIGHT angled triangle with one sides corresponding to EAST movement and second side corresponding to NORTH movement. The HYPOTENUSE is the distance we are looking for.

so side X = 8(east)+16(east)=24
side Y = 6(N)+12(N)=18

if you look at the sides they are 18:24 0r 6*3:6*4 that is in ratio 3:4 so hypotenuse will be in multiple of 5 as 3:4:5 right angle triangle
so distance = 6*5=30

If you do not this you can find hyp by \(\sqrt{18^2+24^2}=\sqrt{900}=30\)
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Kudos [?]: 6088 [1], given: 121

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Kudos [?]: 396 [1], given: 640

City B is 8 miles east of City A. City C is 6 miles north of city B. C [#permalink]

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New post 16 Oct 2017, 16:46
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Bunuel wrote:
City B is 8 miles east of City A. City C is 6 miles north of city B. City D is 16 miles east of city C, and city E is 12 miles north of city D. What is the distance from city A to city E?

(A) 10 miles
(B) 20 miles
(C) 24 miles
(D) 30 miles
(E) 42 miles

Am I missing something here?

I just figured out the lengths of the hypotenuse of each right triangle and added them.

Both are right triangles whose legs head first east (to the right) and then north (up).

Their hypotenuses have identical slopes of 3/4.

They are both 3x-4x-5x right triangles.

Right ∆ ABC has legs, in miles, of distance 6 and 8. It follows the 3x: 4x: 5x ratio. Its hypotenuse, to keep with the ratio, is 6: 8: 10 miles*

Right ∆ CDE has legs, in miles, of 12 and 16. That's 3(4) and 4(4). So hypotenuse is 5(4) = 20 miles*

Total distance from A to E: 10 + 20 = 30 miles

Answer D

Is this method sound? I can't see why not...

*Or use Pythagorean theorem for each

∆ ABC:
\(6^2 + 8^2 = h^2\)
\(36 + 64 = h^2\)
\(h^2 = 100\)
\(h = 10\)

∆ CDE:
\(12^2 + 16^2 = h^2\)
\(144 + 256 = h^2\)
\(h^2 = 400\)
\(h = 20\)

Kudos [?]: 396 [1], given: 640

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Re: City B is 8 miles east of City A. City C is 6 miles north of city B. C [#permalink]

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New post 23 Oct 2017, 07:33
I sketched it out on paper and if anyone knows how to upload those fancy graphs I see other people doing that would be appreciated. Again, here I am working on actual problem solving mechanics to avoid silly mistakes, so any advice on that would be appreciated.


As I sketched it out, I realized this will create a right triangle. The eastern directions will serve as the base and the northern directions will serve as the height since they are parallel.

B=8+16=24
H=6+12=18.

-Now, squaring those two is sort of hairy, so I checked if this is a 3-4-5 triplet and yes 18/24=3/4 with the common divisor of 6.
-So the hypotenuse which is the length of A to E is 30, answer D.
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I'd love to hear any feedback or ways to improve my problem solving. I make a lot of silly mistakes. If you've had luck improving on stupid mistakes, I'd love to hear how you did it.

Also, I appreciate any kudos.

Kudos [?]: 8 [0], given: 18

Re: City B is 8 miles east of City A. City C is 6 miles north of city B. C   [#permalink] 23 Oct 2017, 07:33
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