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by shankar.ashwin » Thu Sep 15, 2011 4:34 am
3 people A,B and C start travelling from a same point. A travels eastwards, B travels westwards and C travels in an unknown direction. All of them stop after they cover 50kms. The distance between B & C is 60 kms. What is the distance between A & C.

A) 50
B) 60
C) 70
D) 80
E) Cannot be determined
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by cans » Thu Sep 15, 2011 4:41 am
let same point = 0,0. B-> (-5,0) and A(5,0) (1 unit means 10 km)
C(x,y). x^2 + y^2 = 25
(x+5)^2 + y^2 = 36 -> x^2 + y^2 + 25 + 10x =36 -> x=-1.4
A&C: (x-5)^2 + y^2 -> x^2 + y^2 + 25 - 10x = 50 + 14 = 64..
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by gmatboost » Thu Sep 15, 2011 6:54 am
Cans was very close to getting the right answer.

Let me go through the steps without using the 1 unit = 10 km, because I think it's clearer to use the actual numbers.

As cans said, let the starting point be (0,0).
B is at (-50, 0). A is at (50, 0). C is at (x, y).

Since C is also 50 away from the starting point, x^2 + y^2 = 2500.

(BC)^2 = (x-50)^2 + (y-0)^2 = x^2 - 100x + 2500 + y^2
We know that BC = 60, so
3600 = y^2 + x^2 - 100x + 2500
We know that x^2 + y^2 = 2500, so
3600 = 2500 - 100x + 2500
3600 = 5000 - 100x
-1400 = -100x
x = 14

(AC)^2 = (x+50)^2 + (y-0)^2 = x^2 + 100x + 2500 + y^2
Again, x^2 + y^2 = 2500, so
(AC)^2 = 2500 + 100x + 2500 = 5000 + 100x
Since x = 14
(AC)^2 = 6400
AC = 80

Note that we never needed to solve for y, so we didn't, but we could have used x^2 + y^2 = 2500 and x = 14 to find that y = 48.

Note that if you draw the picture, the distance BC is part of a 3-4-5 right triangle: 36-48-60.
And, the distance from AC is part of a 3-4-5 right triangle as well: 48-64-80.
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by shankar.ashwin » Thu Sep 15, 2011 8:06 am
Once you have figured out that the points are in a semi-circle, we know angle within a semi-circle is 90.
So angle C is 90 and ACB is a right triangle.

So,

100^2 = AC^2 + 60^2. (Phyt theorem)

AC=80.
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by gmatboost » Thu Sep 15, 2011 8:10 am
That's a great approach, didn't cross my mind at all.
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by GMATGuruNY » Thu Sep 15, 2011 12:08 pm
shankar.ashwin wrote:3 people A,B and C start travelling from a same point. A travels eastwards, B travels westwards and C travels in an unknown direction. All of them stop after they cover 50kms. The distance between B & C is 60 kms. What is the distance between A & C.

A) 50
B) 60
C) 70
D) 80
E) Cannot be determined
Don't calculate. DRAW.

Image

Extend CE by 50 kms to form the following quadrilateral:

Image

Since the diagonals of the quadrilateral are equal, the quadrilateral is a rectangle.
Thus, ∆ABC is a right triangle with a side of 60 and a hypotenuse of 100.
Thus, ∆ABC is a multiple of a 6-8-10 triangle in which the sides are 60, 80, and 100.
Thus, AC = 80.

The correct answer is D.
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by saketk » Fri Sep 16, 2011 9:59 am
EDIT: -- did not read above posts. I agree with Mitch.. (the diagrams are of great help!)


we can use "locus of a point".. and solve it.
they all are travelling the same distance. So will form a definite boundary or in other words a CIRCLE.
angle formed at the circumference of the circle with diameter as the base = 90

the answer will be sqrt(100^2-60^2) = 10*2*4 = 80
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