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Dan and Karen, who live 10 miles apart, meet at a cafe that is [#permalink]
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Since the cafe is directly east to Karen's house and directly north to Dan's house,
they form the two legs of a right angled triangle. From the question stem,
If we assume that the distance between Karen's house and the cafe is x,
then, distance between Dan's house and the cafe is x-2

Also, the distance between their houses is the hypotenuse of the triangle(which is 10)

In a right triangle, \(Hyp^2 = Side1^2 + Side2^2\)
Substituting the values, \(10^2 = x^2 + (x-2)^2\)
6,8,10 is a Pythagorean triplet which satisfies this equation

Therefore, the distance from Karen's house to the cafe is 8 miles(Option C)
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Dan and Karen, who live 10 miles apart, meet at a cafe that is [#permalink]
Bunuel wrote:
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


From the figure it is evident that
\(10^2 = x^2 +(x-2)^2\) solving this we will get

\(x^2 - 4x + 4 = 100\) or \(x^2 - 2x - 48 = 0\)

Hence \(x = 8\)

option C
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Re: Dan and Karen, who live 10 miles apart, meet at a cafe that is [#permalink]
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Bunuel wrote:
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


Hi..

ofcourse one method is the algebraic as explained above..

the other is to realize that MORE than often, you may be tested for triplet is 3:4:5.. or may be 1:\(\sqrt{3}\):2, a 30-60-90 triangle or 1:1:\(\sqrt{2}\), a 45-45-90 triangle
and so when you see the HYPOTENUSE here as 10, a multiple of 5, try out this triplet ( may save some time)


so 3:4:5 has 5 as hypotenuse and with 10 as hypotenuse it becomes 3*2:4*2:5*2 = 6:8:10..
so Karens distance is TWO more and 6:8 perfectly fit in..
ans 8
C
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Re: Dan and Karen, who live 10 miles apart meet at a cafe that [#permalink]
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Hey guys the simplest approach is to plug in the answer choices and you immediately see that the correct answer has to be C.
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Re: Dan and Karen, who live 10 miles apart meet at a cafe that [#permalink]
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Hi,

Could someone explain with an algebraic equation as to how the distance between Dan's house and the cafe is x-2?

Thanks
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Re: Dan and Karen, who live 10 miles apart meet at a cafe that [#permalink]
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boomtangboy wrote:
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


We see that the three locations (Dan’s house, Karen’s house and the cafe) will form a 6-8-10 right triangle. That is, Dan’s and Karen’s houses are 10 miles apart (which is given) and the cafe is 6 miles from Dan’s house and 8 miles from Karen’s house (notice that this makes the cafe 2 miles closer to Dan’s house than to Karen’s house). Therefore, the cafe is 8 miles from Karen’s house.

Answer: C
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Re: Dan and Karen, who live 10 miles apart meet at a cafe that [#permalink]
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Re: Dan and Karen, who live 10 miles apart meet at a cafe that [#permalink]
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