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

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Math Expert
Joined: 02 Sep 2009
Posts: 46301
Dan and Karen, who live 10 miles apart, meet at a cafe that is [#permalink]

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24 Aug 2017, 22:06
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Difficulty:

25% (medium)

Question Stats:

85% (01:48) correct 15% (01:30) wrong based on 49 sessions

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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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Dan and Karen, who live 10 miles apart, meet at a cafe that is [#permalink]

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24 Aug 2017, 22:17
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]

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24 Aug 2017, 22:18
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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Math Expert
Joined: 02 Aug 2009
Posts: 5952
Re: Dan and Karen, who live 10 miles apart, meet at a cafe that is [#permalink]

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24 Aug 2017, 23:02
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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Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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Re: Dan and Karen, who live 10 miles apart, meet at a cafe that is   [#permalink] 24 Aug 2017, 23:02
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