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Dan and Karen, who live 10 miles apart meet at a cafe that

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Dan and Karen, who live 10 miles apart meet at a cafe that [#permalink] New post 20 Feb 2006, 16:30
Dan and Karen, who live 10 miles apart meet at a cafe that is directly north of Dan’s house and directly east of Karen’s house. If the cafe is 2 miles closer to Dan’s house than to Karen’s house, how many miles is the cafe from Karen’s house?
(A) 6
(B) 7
(C) 8
(D) 9
(E) 10
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Re: Problem Solving [#permalink] New post 20 Feb 2006, 16:36
[quote="Avis"]Dan and Karen, who live 10 miles apart meet at a cafe that is directly north of Dan’s house and directly east of Karen’s house. If the cafe is 2 miles closer to Dan’s house than to Karen’s house, how many miles is the cafe from Karen’s house?
(A) 6
(B) 7
(C) 8
(D) 9
(E) 10[/quote

A)
use triangle rectangle 3-4-5
so equivalent 4-6-10

stupid mistake see correction below :stupid2

Last edited by conocieur on 20 Feb 2006, 19:04, edited 1 time in total.
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 [#permalink] New post 20 Feb 2006, 17:46
it should be 8 ... its like a right angle and say the distance of karen from cafe is x so the distance of dan house is x -2


by pythagoras theorem x ^ 2 + (x-2) ^2 = 100

so from above we get x ^2 -2x -48 = 0

==> (x -8)(x+6) =0

since distance cannot be negative so x =8 .
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 [#permalink] New post 20 Feb 2006, 19:18
x
K---C
\ |
\ | x-2
\ |
D

Using pythagoras theroem,

x^2 + (x-2)^2 = 100
x^2 + x^2 - 4x + 4 = 100
2x^2 -4x + 4 = 100
x^2 -2x + 2 = 50
x^2 - 2x -48 = 0
(x+6)(x-8) =0

x = -6 (not valid) or 8 (valid)

Ans C
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 [#permalink] New post 21 Feb 2006, 05:05
ywilfred wrote:
x
K---C
\ |
\ | x-2
\ |
D

Using pythagoras theroem,

x^2 + (x-2)^2 = 100
x^2 + x^2 - 4x + 4 = 100
2x^2 -4x + 4 = 100
x^2 -2x + 2 = 50
x^2 - 2x -48 = 0
(x+6)(x-8) =0

x = -6 (not valid) or 8 (valid)

Ans C


bingo solved and got 8 using the same method... x^2+(x-2)^2 = 100 simplifying = x = 8 or -6 thus 8 :)
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  [#permalink] 21 Feb 2006, 05:05
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